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 \[ \frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}-\frac{2 a^2 (3 a B+7 A b)}{3 d \sqrt{\tan (c+d x)}}+\frac{2 b^2 (a A+3 b B) \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+b \tan (c+d x))^2}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

[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))

________________________________________________________________________________________

Rubi [A]  time = 0.611542, antiderivative size = 372, normalized size of antiderivative = 1., 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 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 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]

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 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 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 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 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 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 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 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]

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*}

Mathematica [C]  time = 1.22226, size = 165, normalized size = 0.44 \[ -\frac{2 \left (\left (a^3 A-3 a^2 b B-3 a A b^2+b^3 B\right ) \text{Hypergeometric2F1}\left (-\frac{3}{4},1,\frac{1}{4},-\tan ^2(c+d x)\right )+3 \left (3 a^2 A b+a^3 B-3 a b^2 B-A b^3\right ) \tan (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{4},1,\frac{3}{4},-\tan ^2(c+d x)\right )+b \left (3 a^2 B+3 b (3 a B+A b) \tan (c+d x)+3 a A b-3 b^2 B \tan ^2(c+d x)-b^2 B\right )\right )}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

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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Maple [B]  time = 0.028, size = 971, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

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))*b^3-1/2/d*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*b^
3-3/2/d*A*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+3/2/d*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(
1/2))*a^2*b-1/2/d*a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/2/d*a^3*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1
/2))*2^(1/2)-1/4/d*a^3*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)))-1/4/d*a^3*B*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)))*2^(1/2)
-1/2/d*a^3*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/2/d*a^3*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2
)-2/3/d*a^3*A/tan(d*x+c)^(3/2)-2/d*a^3/tan(d*x+c)^(1/2)*B+2/d*B*b^3*tan(d*x+c)^(1/2)-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*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)*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*b^2-3/2/d*A*2^(1/2)*arctan(-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*b^2+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-1/4/d*B*2^(1/2)*ln((1+2^(1/2)*t
an(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*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

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Maxima [A]  time = 1.67636, size = 419, normalized size = 1.13 \begin{align*} \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} \end{align*}

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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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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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Giac [A]  time = 1.67064, size = 622, normalized size = 1.67 \begin{align*} \frac{2 \, B b^{3} \sqrt{\tan \left (d x + c\right )}}{d} - \frac{{\left (\sqrt{2} A a^{3} + \sqrt{2} B a^{3} + 3 \, \sqrt{2} A a^{2} b - 3 \, \sqrt{2} B a^{2} b - 3 \, \sqrt{2} A a b^{2} - 3 \, \sqrt{2} B a b^{2} - \sqrt{2} A b^{3} + \sqrt{2} B b^{3}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} - \frac{{\left (\sqrt{2} A a^{3} + \sqrt{2} B a^{3} + 3 \, \sqrt{2} A a^{2} b - 3 \, \sqrt{2} B a^{2} b - 3 \, \sqrt{2} A a b^{2} - 3 \, \sqrt{2} B a b^{2} - \sqrt{2} A b^{3} + \sqrt{2} B b^{3}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right )}{2 \, d} - \frac{{\left (\sqrt{2} A a^{3} - \sqrt{2} B a^{3} - 3 \, \sqrt{2} A a^{2} b - 3 \, \sqrt{2} B a^{2} b - 3 \, \sqrt{2} A a b^{2} + 3 \, \sqrt{2} B a b^{2} + \sqrt{2} A b^{3} + \sqrt{2} B b^{3}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} + \frac{{\left (\sqrt{2} A a^{3} - \sqrt{2} B a^{3} - 3 \, \sqrt{2} A a^{2} b - 3 \, \sqrt{2} B a^{2} b - 3 \, \sqrt{2} A a b^{2} + 3 \, \sqrt{2} B a b^{2} + \sqrt{2} A b^{3} + \sqrt{2} B b^{3}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{4 \, d} - \frac{2 \,{\left (3 \, B a^{3} \tan \left (d x + c\right ) + 9 \, A a^{2} b \tan \left (d x + c\right ) + A a^{3}\right )}}{3 \, d \tan \left (d x + c\right )^{\frac{3}{2}}} \end{align*}

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]

2*B*b^3*sqrt(tan(d*x + c))/d - 1/2*(sqrt(2)*A*a^3 + sqrt(2)*B*a^3 + 3*sqrt(2)*A*a^2*b - 3*sqrt(2)*B*a^2*b - 3*
sqrt(2)*A*a*b^2 - 3*sqrt(2)*B*a*b^2 - sqrt(2)*A*b^3 + sqrt(2)*B*b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(
d*x + c))))/d - 1/2*(sqrt(2)*A*a^3 + sqrt(2)*B*a^3 + 3*sqrt(2)*A*a^2*b - 3*sqrt(2)*B*a^2*b - 3*sqrt(2)*A*a*b^2
- 3*sqrt(2)*B*a*b^2 - sqrt(2)*A*b^3 + sqrt(2)*B*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))))/d
- 1/4*(sqrt(2)*A*a^3 - sqrt(2)*B*a^3 - 3*sqrt(2)*A*a^2*b - 3*sqrt(2)*B*a^2*b - 3*sqrt(2)*A*a*b^2 + 3*sqrt(2)*B
*a*b^2 + sqrt(2)*A*b^3 + sqrt(2)*B*b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/d + 1/4*(sqrt(2)*A*
a^3 - sqrt(2)*B*a^3 - 3*sqrt(2)*A*a^2*b - 3*sqrt(2)*B*a^2*b - 3*sqrt(2)*A*a*b^2 + 3*sqrt(2)*B*a*b^2 + sqrt(2)*
A*b^3 + sqrt(2)*B*b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/d - 2/3*(3*B*a^3*tan(d*x + c) + 9*A
*a^2*b*tan(d*x + c) + A*a^3)/(d*tan(d*x + c)^(3/2))