3.394 \(\int \frac{(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt{\tan (c+d x)}} \, dx\)

Optimal. Leaf size=380 \[ -\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt{\tan (c+d x)}}{5 d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b^2 (9 a B+5 A b) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d} \]

[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]]])/(Sq
rt[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))*L
og[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*b*(15*a*A*b + 14*a^2*B - 5*b^2*B)*Sqrt[T
an[c + d*x]])/(5*d) + (2*b^2*(5*A*b + 9*a*B)*Tan[c + d*x]^(3/2))/(15*d) + (2*b*B*Sqrt[Tan[c + d*x]]*(a + b*Tan
[c + d*x])^2)/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.666732, antiderivative size = 380, 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 = {3607, 3637, 3630, 3534, 1168, 1162, 617, 204, 1165, 628} \[ -\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (3 a^2 b (A-B)+a^3 (A+B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}+\frac{2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt{\tan (c+d x)}}{5 d}-\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (-3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{2 b^2 (9 a B+5 A b) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Sqrt[Tan[c + d*x]],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]]])/(Sq
rt[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))*L
og[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + (2*b*(15*a*A*b + 14*a^2*B - 5*b^2*B)*Sqrt[T
an[c + d*x]])/(5*d) + (2*b^2*(5*A*b + 9*a*B)*Tan[c + d*x]^(3/2))/(15*d) + (2*b*B*Sqrt[Tan[c + d*x]]*(a + b*Tan
[c + d*x])^2)/(5*d)

Rule 3607

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*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*
f*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[a^2*A*d*(m +
n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m
- 1) - b*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&
!(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rule 3637

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*Tan[e + f*x]*(c + d*Tan[e + f*x])
^(n + 1))/(d*f*(n + 2)), x] - Dist[1/(d*(n + 2)), Int[(c + d*Tan[e + f*x])^n*Simp[b*c*C - a*A*d*(n + 2) - (A*b
+ a*B - b*C)*d*(n + 2)*Tan[e + f*x] - (a*C*d*(n + 2) - b*(c*C - B*d*(n + 2)))*Tan[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, C, n}, 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))}{\sqrt{\tan (c+d x)}} \, dx &=\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}+\frac{2}{5} \int \frac{(a+b \tan (c+d x)) \left (\frac{1}{2} a (5 a A-b B)+\frac{5}{2} \left (2 a A b+a^2 B-b^2 B\right ) \tan (c+d x)+\frac{1}{2} b (5 A b+9 a B) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{2 b^2 (5 A b+9 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}-\frac{4}{15} \int \frac{-\frac{3}{4} a^2 (5 a A-b B)-\frac{15}{4} \left (3 a^2 A b-A b^3+a^3 B-3 a b^2 B\right ) \tan (c+d x)-\frac{3}{4} b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt{\tan (c+d x)}}{5 d}+\frac{2 b^2 (5 A b+9 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}-\frac{4}{15} \int \frac{-\frac{15}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac{15}{4} \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 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt{\tan (c+d x)}}{5 d}+\frac{2 b^2 (5 A b+9 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}-\frac{8 \operatorname{Subst}\left (\int \frac{-\frac{15}{4} \left (a^3 A-3 a A b^2-3 a^2 b B+b^3 B\right )-\frac{15}{4} \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 )}{15 d}\\ &=\frac{2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt{\tan (c+d x)}}{5 d}+\frac{2 b^2 (5 A b+9 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 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+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 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt{\tan (c+d x)}}{5 d}+\frac{2 b^2 (5 A b+9 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 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 (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 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt{\tan (c+d x)}}{5 d}+\frac{2 b^2 (5 A b+9 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 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 ) \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 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt{\tan (c+d x)}}{5 d}+\frac{2 b^2 (5 A b+9 a B) \tan ^{\frac{3}{2}}(c+d x)}{15 d}+\frac{2 b B \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\\ \end{align*}

Mathematica [C]  time = 1.45073, size = 153, normalized size = 0.4 \[ \frac{2 b \sqrt{\tan (c+d x)} \left (15 \left (3 a^2 B+3 a A b-b^2 B\right )+5 b (3 a B+A b) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )-15 \sqrt [4]{-1} (a-i b)^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )-15 \sqrt [4]{-1} (a+i b)^3 (A+i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Sqrt[Tan[c + d*x]],x]

[Out]

(-15*(-1)^(1/4)*(a - I*b)^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - 15*(-1)^(1/4)*(a + I*b)^3*(A + I
*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 2*b*Sqrt[Tan[c + d*x]]*(15*(3*a*A*b + 3*a^2*B - b^2*B) + 5*b*(A*b
+ 3*a*B)*Tan[c + d*x] + 3*b^2*B*Tan[c + d*x]^2))/(15*d)

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Maple [B]  time = 0.022, size = 1007, normalized size = 2.7 \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)^(1/2),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*tan(d*x+c)^(3/2)*b^3-2/d*B*b^3*tan(d*x+c)^(1/2)+2/5/d*B*b^3*tan(d*x+c)^(5/2)+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+6/d*A*a*b^2*tan(d*x+c)^(1/2)+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+2/d*B*t
an(d*x+c)^(3/2)*a*b^2-1/4/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)+t
an(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*B*a^2*b*tan(d*x+c)^(1/2)

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Maxima [A]  time = 1.63724, size = 441, normalized size = 1.16 \begin{align*} \frac{24 \, B b^{3} \tan \left (d x + c\right )^{\frac{5}{2}} + 30 \, \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 ) + 30 \, \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 ) + 15 \, \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 ) - 15 \, \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 ) + 40 \,{\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{\frac{3}{2}} + 120 \,{\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \sqrt{\tan \left (d x + c\right )}}{60 \, 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)^(1/2),x, algorithm="maxima")

[Out]

1/60*(24*B*b^3*tan(d*x + c)^(5/2) + 30*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)))) + 30*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)))) + 15*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) - 15*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
) + 40*(3*B*a*b^2 + A*b^3)*tan(d*x + c)^(3/2) + 120*(3*B*a^2*b + 3*A*a*b^2 - B*b^3)*sqrt(tan(d*x + c)))/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)^(1/2),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}}{\sqrt{\tan{\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)**(1/2),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3/sqrt(tan(c + d*x)), x)

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Giac [A]  time = 2.32477, size = 690, normalized size = 1.82 \begin{align*} \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 b^{3} d^{4} \tan \left (d x + c\right )^{\frac{5}{2}} + 15 \, B a b^{2} d^{4} \tan \left (d x + c\right )^{\frac{3}{2}} + 5 \, A b^{3} d^{4} \tan \left (d x + c\right )^{\frac{3}{2}} + 45 \, B a^{2} b d^{4} \sqrt{\tan \left (d x + c\right )} + 45 \, A a b^{2} d^{4} \sqrt{\tan \left (d x + c\right )} - 15 \, B b^{3} d^{4} \sqrt{\tan \left (d x + c\right )}\right )}}{15 \, d^{5}} \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)^(1/2),x, algorithm="giac")

[Out]

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(-sqr
t(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1)/d + 2/15*(3*B*b^3*d^4*tan(d*x + c)^(5/2) + 15*B*a*b^2*d^4*tan(d*x
+ c)^(3/2) + 5*A*b^3*d^4*tan(d*x + c)^(3/2) + 45*B*a^2*b*d^4*sqrt(tan(d*x + c)) + 45*A*a*b^2*d^4*sqrt(tan(d*x
+ c)) - 15*B*b^3*d^4*sqrt(tan(d*x + c)))/d^5