3.396 \(\int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=372 \[ -\frac {2 a^2 (3 a B+7 A b)}{3 d \sqrt {\tan (c+d x)}}+\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 b^2 (a A+3 b B) \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
[Out]
-1/2*(3*a^2*b*(A-B)-b^3*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*(3*a^
2*b*(A-B)-b^3*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)+1/4*(a^3*(A-B)-3*a*b
^2*(A-B)-3*a^2*b*(A+B)+b^3*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-1/4*(a^3*(A-B)-3*a*b^2*(
A-B)-3*a^2*b*(A+B)+b^3*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-2/3*a^2*(7*A*b+3*B*a)/d/tan(
d*x+c)^(1/2)+2/3*b^2*(A*a+3*B*b)*tan(d*x+c)^(1/2)/d-2/3*a*A*(a+b*tan(d*x+c))^2/d/tan(d*x+c)^(3/2)
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Rubi [A] time = 0.61, antiderivative size = 372, normalized size of antiderivative = 1.00,
number of steps used = 13, number of rules used = 10, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used
= {3605, 3635, 3630, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a^2 (3 a B+7 A b)}{3 d \sqrt {\tan (c+d x)}}+\frac {2 b^2 (a A+3 b B) \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]
[Out]
((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt
[2]*d) - ((3*a^2*b*(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]
]])/(Sqrt[2]*d) + ((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c
+ d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(A + B))*Log[
1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (2*a^2*(7*A*b + 3*a*B))/(3*d*Sqrt[Tan[c + d*x]
]) + (2*b^2*(a*A + 3*b*B)*Sqrt[Tan[c + d*x]])/(3*d) - (2*a*A*(a + b*Tan[c + d*x])^2)/(3*d*Tan[c + d*x]^(3/2))
Rule 204
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 617
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 628
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Rule 1162
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Rule 1165
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Rule 1168
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]
Rule 3534
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]
Rule 3605
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e
+ f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])
Rule 3630
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])
^m*Simp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&& !LeQ[m, -1]
Rule 3635
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e
_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(c^2*C - B*c*d + A*d^2)*
(c + d*Tan[e + f*x])^(n + 1))/(d^2*f*(n + 1)*(c^2 + d^2)), x] + Dist[1/(d*(c^2 + d^2)), Int[(c + d*Tan[e + f*x
])^(n + 1)*Simp[a*d*(A*c - c*C + B*d) + b*(c^2*C - B*c*d + A*d^2) + d*(A*b*c + a*B*c - b*c*C - a*A*d + b*B*d +
a*C*d)*Tan[e + f*x] + b*C*(c^2 + d^2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] &&
NeQ[b*c - a*d, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1]
Rubi steps
\begin {align*} \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {(a+b \tan (c+d x)) \left (\frac {1}{2} a (7 A b+3 a B)-\frac {3}{2} \left (a^2 A-A b^2-2 a b B\right ) \tan (c+d x)+\frac {1}{2} b (a A+3 b B) \tan ^2(c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (7 A b+3 a B)}{3 d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {-\frac {1}{2} a \left (3 a^2 A-10 A b^2-9 a b B\right )-\frac {3}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)+\frac {1}{2} b^2 (a A+3 b B) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 (7 A b+3 a B)}{3 d \sqrt {\tan (c+d x)}}+\frac {2 b^2 (a A+3 b B) \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {-\frac {3}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac {3}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 (7 A b+3 a B)}{3 d \sqrt {\tan (c+d x)}}+\frac {2 b^2 (a A+3 b B) \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \operatorname {Subst}\left (\int \frac {-\frac {3}{2} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac {3}{2} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{3 d}\\ &=-\frac {2 a^2 (7 A b+3 a B)}{3 d \sqrt {\tan (c+d x)}}+\frac {2 b^2 (a A+3 b B) \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}\\ &=-\frac {2 a^2 (7 A b+3 a B)}{3 d \sqrt {\tan (c+d x)}}+\frac {2 b^2 (a A+3 b B) \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 \sqrt {2} d}\\ &=\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2 (7 A b+3 a B)}{3 d \sqrt {\tan (c+d x)}}+\frac {2 b^2 (a A+3 b B) \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}\\ &=\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a^2 (7 A b+3 a B)}{3 d \sqrt {\tan (c+d x)}}+\frac {2 b^2 (a A+3 b B) \sqrt {\tan (c+d x)}}{3 d}-\frac {2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac {3}{2}}(c+d x)}\\ \end {align*}
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Mathematica [C] time = 1.41, size = 165, normalized size = 0.44 \[ -\frac {2 \left (b \left (3 a^2 B+3 b (3 a B+A b) \tan (c+d x)+3 a A b-3 b^2 B \tan ^2(c+d x)-b^2 B\right )+\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\tan ^2(c+d x)\right )+3 \left (a^3 B+3 a^2 A b-3 a b^2 B-A b^3\right ) \tan (c+d x) \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};-\tan ^2(c+d x)\right )\right )}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]
[Out]
(-2*((a^3*A - 3*a*A*b^2 - 3*a^2*b*B + b^3*B)*Hypergeometric2F1[-3/4, 1, 1/4, -Tan[c + d*x]^2] + 3*(3*a^2*A*b -
A*b^3 + a^3*B - 3*a*b^2*B)*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[c + d*x]^2]*Tan[c + d*x] + b*(3*a*A*b + 3*a^2
*B - b^2*B + 3*b*(A*b + 3*a*B)*Tan[c + d*x] - 3*b^2*B*Tan[c + d*x]^2)))/(3*d*Tan[c + d*x]^(3/2))
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
Timed out
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{3}}{\tan \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^3/tan(d*x + c)^(5/2), x)
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maple [B] time = 0.12, size = 971, normalized size = 2.61 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x)
[Out]
-1/2/d*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+3/2/d*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*
b^2-1/2/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^3-1/4/d*A*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan
(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^3-1/2/d*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^3-
2/d*a^3/tan(d*x+c)^(1/2)*B-2/3/d*A*a^3/tan(d*x+c)^(3/2)+2/d*B*tan(d*x+c)^(1/2)*b^3-1/2/d*B*2^(1/2)*arctan(1+2^
(1/2)*tan(d*x+c)^(1/2))*b^3+1/2/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*b^3-6/d*a^2/tan(d*x+c)^(1/2)*A
*b-1/4/d*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^3-1/4
/d*B*2^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*b^3+1/4/d*A*2
^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*b^3+1/2/d*A*2^(1/2)
*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^3-1/2/d*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^3-1/2/d*A*2^(1/2)
*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^3+3/2/d*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2-3/2/d*A*2^(1/
2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b+3/4/d*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(
1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a*b^2+3/2/d*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b+3/2/d*B*2^(
1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2+3/2/d*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*a*b^2-3/4/d
*A*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^2*b+3/4/d*B*2
^(1/2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a^2*b+3/4/d*A*2^(1/
2)*ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*a*b^2+3/2/d*B*2^(1/2)*a
rctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b-3/2/d*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*a^2*b
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maxima [A] time = 0.44, size = 310, normalized size = 0.83 \[ \frac {24 \, B b^{3} \sqrt {\tan \left (d x + c\right )} - 6 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 6 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 3 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 3 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac {8 \, {\left (A a^{3} + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
1/12*(24*B*b^3*sqrt(tan(d*x + c)) - 6*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 - (A - B)*b^3)*
arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) - 6*sqrt(2)*((A + B)*a^3 + 3*(A - B)*a^2*b - 3*(A + B)*a*
b^2 - (A - B)*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - 3*sqrt(2)*((A - B)*a^3 - 3*(A + B)*
a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 3*sqrt(2)*((A - B)
*a^3 - 3*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B)*b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) -
8*(A*a^3 + 3*(B*a^3 + 3*A*a^2*b)*tan(d*x + c))/tan(d*x + c)^(3/2))/d
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mupad [B] time = 10.15, size = 7578, normalized size = 20.37 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3)/tan(c + d*x)^(5/2),x)
[Out]
2*atanh((32*A^2*a^6*d^3*tan(c + d*x)^(1/2)*((5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^
12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4)
- (3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/(16*A*a^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^
12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 16*A^3
*b^9*d^2 - 288*A^3*a^2*b^7*d^2 + 960*A^3*a^4*b^5*d^2 - 736*A^3*a^6*b^3*d^2 - 48*A*a*b^2*(30*A^4*a^2*b^10*d^4 -
A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b
^2*d^4)^(1/2) + 48*A^3*a^8*b*d^2) - (32*A^2*b^6*d^3*tan(c + d*x)^(1/2)*((5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10
*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*
a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/(16*A*a^3*(30*A^4*a^2*b^10
*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*
a^10*b^2*d^4)^(1/2) + 16*A^3*b^9*d^2 - 288*A^3*a^2*b^7*d^2 + 960*A^3*a^4*b^5*d^2 - 736*A^3*a^6*b^3*d^2 - 48*A*
a*b^2*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4
*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a^8*b*d^2) + (480*A^2*a^2*b^4*d^3*tan(c + d*x)^(1/2)*((5*A^
2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^
4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^
(1/2))/(16*A*a^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^
4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 16*A^3*b^9*d^2 - 288*A^3*a^2*b^7*d^2 + 960*A^3*a^4*b^5*
d^2 - 736*A^3*a^6*b^3*d^2 - 48*A*a*b^2*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^
4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a^8*b*d^2) - (480*A^2*a^4*
b^2*d^3*tan(c + d*x)^(1/2)*((5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4
*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*A^2*a*b^5)/
(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/(16*A*a^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4
*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 16*A^3*b^9*d^2 - 288*A
^3*a^2*b^7*d^2 + 960*A^3*a^4*b^5*d^2 - 736*A^3*a^6*b^3*d^2 - 48*A*a*b^2*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 -
A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) +
48*A^3*a^8*b*d^2))*((5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8
*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3*A^2*a*b^5)/(2*d^2)
- (3*A^2*a^5*b)/(2*d^2))^(1/2) - atan((B^2*a^6*d^3*tan(c + d*x)^(1/2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B
^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*
d^4) - (5*B^2*a^3*b^3)/d^2 + (3*B^2*a*b^5)/(2*d^2) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*32i)/(16*B^3*a^9*d^2 + 16*B*
b^3*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a
^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2) - 736*B^3*a^3*b^6*d^2 + 960*B^3*a^5*b^4*d^2 - 288*B^3*a^7*b^2*d^2 + 48
*B^3*a*b^8*d^2 - 48*B*a^2*b*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4
*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)) - (B^2*b^6*d^3*tan(c + d*x)^(1/2)*((30*B^4*a^
2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 3
0*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*B^2*a^3*b^3)/d^2 + (3*B^2*a*b^5)/(2*d^2) + (3*B^2*a^5*b)/(2*d^2))^(1/2)
*32i)/(16*B^3*a^9*d^2 + 16*B*b^3*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 45
2*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2) - 736*B^3*a^3*b^6*d^2 + 960*B^3*a^5*b^4*d
^2 - 288*B^3*a^7*b^2*d^2 + 48*B^3*a*b^8*d^2 - 48*B*a^2*b*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 -
255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)) + (B^2*a^2*b^4*d
^3*tan(c + d*x)^(1/2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*
b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*B^2*a^3*b^3)/d^2 + (3*B^2*a*b^5)/(2*d^
2) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*480i)/(16*B^3*a^9*d^2 + 16*B*b^3*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a
^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2) - 736*B
^3*a^3*b^6*d^2 + 960*B^3*a^5*b^4*d^2 - 288*B^3*a^7*b^2*d^2 + 48*B^3*a*b^8*d^2 - 48*B*a^2*b*(30*B^4*a^2*b^10*d^
4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^1
0*b^2*d^4)^(1/2)) - (B^2*a^4*b^2*d^3*tan(c + d*x)^(1/2)*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 -
255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*B^2*
a^3*b^3)/d^2 + (3*B^2*a*b^5)/(2*d^2) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*480i)/(16*B^3*a^9*d^2 + 16*B*b^3*(30*B^4*a
^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 +
30*B^4*a^10*b^2*d^4)^(1/2) - 736*B^3*a^3*b^6*d^2 + 960*B^3*a^5*b^4*d^2 - 288*B^3*a^7*b^2*d^2 + 48*B^3*a*b^8*d^
2 - 48*B*a^2*b*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4
- 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)))*((30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255
*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (5*B^2*a^3
*b^3)/d^2 + (3*B^2*a*b^5)/(2*d^2) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*2i - atan((B^2*a^6*d^3*tan(c + d*x)^(1/2)*((3
*B^2*a*b^5)/(2*d^2) - (5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b
^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*B^2*a^5*b)/(2*d^2
))^(1/2)*32i)/(16*B^3*a^9*d^2 - 16*B*b^3*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*
d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2) - 736*B^3*a^3*b^6*d^2 + 960*B^3*a
^5*b^4*d^2 - 288*B^3*a^7*b^2*d^2 + 48*B^3*a*b^8*d^2 + 48*B*a^2*b*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^1
2*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)) - (B^2*b
^6*d^3*tan(c + d*x)^(1/2)*((3*B^2*a*b^5)/(2*d^2) - (5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 -
B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(
4*d^4) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*32i)/(16*B^3*a^9*d^2 - 16*B*b^3*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^
4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2) - 73
6*B^3*a^3*b^6*d^2 + 960*B^3*a^5*b^4*d^2 - 288*B^3*a^7*b^2*d^2 + 48*B^3*a*b^8*d^2 + 48*B*a^2*b*(30*B^4*a^2*b^10
*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*
a^10*b^2*d^4)^(1/2)) + (B^2*a^2*b^4*d^3*tan(c + d*x)^(1/2)*((3*B^2*a*b^5)/(2*d^2) - (5*B^2*a^3*b^3)/d^2 - (30*
B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d
^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*480i)/(16*B^3*a^9*d^2 - 16*B*b^3*(30*B^
4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4
+ 30*B^4*a^10*b^2*d^4)^(1/2) - 736*B^3*a^3*b^6*d^2 + 960*B^3*a^5*b^4*d^2 - 288*B^3*a^7*b^2*d^2 + 48*B^3*a*b^8
*d^2 + 48*B*a^2*b*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d
^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)) - (B^2*a^4*b^2*d^3*tan(c + d*x)^(1/2)*((3*B^2*a*b^5)/(2
*d^2) - (5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B
^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*480i)
/(16*B^3*a^9*d^2 - 16*B*b^3*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4
*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2) - 736*B^3*a^3*b^6*d^2 + 960*B^3*a^5*b^4*d^2 -
288*B^3*a^7*b^2*d^2 + 48*B^3*a*b^8*d^2 + 48*B*a^2*b*(30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B
^4*a^4*b^8*d^4 + 452*B^4*a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)))*((3*B^2*a*b^5)/(2*d^
2) - (5*B^2*a^3*b^3)/d^2 - (30*B^4*a^2*b^10*d^4 - B^4*b^12*d^4 - B^4*a^12*d^4 - 255*B^4*a^4*b^8*d^4 + 452*B^4*
a^6*b^6*d^4 - 255*B^4*a^8*b^4*d^4 + 30*B^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (3*B^2*a^5*b)/(2*d^2))^(1/2)*2i + 2*a
tanh((32*A^2*a^6*d^3*tan(c + d*x)^(1/2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*
d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*A^2*a^3*b^3)/d^2 - (
3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/(16*A^3*b^9*d^2 - 16*A*a^3*(30*A^4*a^2*b^10*d^4 - A^4*b^1
2*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^
(1/2) - 288*A^3*a^2*b^7*d^2 + 960*A^3*a^4*b^5*d^2 - 736*A^3*a^6*b^3*d^2 + 48*A*a*b^2*(30*A^4*a^2*b^10*d^4 - A^
4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*
d^4)^(1/2) + 48*A^3*a^8*b*d^2) - (32*A^2*b^6*d^3*tan(c + d*x)^(1/2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4
*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^
4) + (5*A^2*a^3*b^3)/d^2 - (3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/(16*A^3*b^9*d^2 - 16*A*a^3*(3
0*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4
*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) - 288*A^3*a^2*b^7*d^2 + 960*A^3*a^4*b^5*d^2 - 736*A^3*a^6*b^3*d^2 + 48*A*a*b
^2*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^
8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a^8*b*d^2) + (480*A^2*a^2*b^4*d^3*tan(c + d*x)^(1/2)*((30*A^4*
a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 +
30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*A^2*a^3*b^3)/d^2 - (3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/
2))/(16*A^3*b^9*d^2 - 16*A*a^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*
A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) - 288*A^3*a^2*b^7*d^2 + 960*A^3*a^4*b^5*d^2
- 736*A^3*a^6*b^3*d^2 + 48*A*a*b^2*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 +
452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*A^3*a^8*b*d^2) - (480*A^2*a^4*b^2
*d^3*tan(c + d*x)^(1/2)*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^
6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*A^2*a^3*b^3)/d^2 - (3*A^2*a*b^5)/(2*
d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/(16*A^3*b^9*d^2 - 16*A*a^3*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12
*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) - 288*A^3*
a^2*b^7*d^2 + 960*A^3*a^4*b^5*d^2 - 736*A^3*a^6*b^3*d^2 + 48*A*a*b^2*(30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4
*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2) + 48*
A^3*a^8*b*d^2))*((30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^
4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) + (5*A^2*a^3*b^3)/d^2 - (3*A^2*a*b^5)/(2*d^2) - (
3*A^2*a^5*b)/(2*d^2))^(1/2) - ((2*A*a^3)/3 + 6*A*a^2*b*tan(c + d*x))/(d*tan(c + d*x)^(3/2)) - (2*B*a^3)/(d*tan
(c + d*x)^(1/2)) + (2*B*b^3*tan(c + d*x)^(1/2))/d
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{3}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)
[Out]
Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3/tan(c + d*x)**(5/2), x)
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