3.131 \(\int \frac{(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=136 \[ \frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 (3 B+7 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{\tan (c+d x)}}-\frac{16 a^3 A \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

[Out]

(8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (16*a^3*A*Sqrt[Tan[c + d*x]])/(3*d) - (
2*a*A*(a + I*a*Tan[c + d*x])^2)/(3*d*Tan[c + d*x]^(3/2)) - (2*((7*I)*A + 3*B)*(a^3 + I*a^3*Tan[c + d*x]))/(3*d
*Sqrt[Tan[c + d*x]])

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Rubi [A]  time = 0.357701, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {3593, 3592, 3533, 205} \[ \frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{2 (3 B+7 i A) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{\tan (c+d x)}}-\frac{16 a^3 A \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

(8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (16*a^3*A*Sqrt[Tan[c + d*x]])/(3*d) - (
2*a*A*(a + I*a*Tan[c + d*x])^2)/(3*d*Tan[c + d*x]^(3/2)) - (2*((7*I)*A + 3*B)*(a^3 + I*a^3*Tan[c + d*x]))/(3*d
*Sqrt[Tan[c + d*x]])

Rule 3593

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(a^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3592

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(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*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] &&  !LeQ[m, -1]

Rule 3533

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(2*c^2)/f, S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{(a+i a \tan (c+d x))^2 \left (\frac{1}{2} a (7 i A+3 B)+\frac{1}{2} a (A+3 i B) \tan (c+d x)\right )}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{\tan (c+d x)}}+\frac{4}{3} \int \frac{(a+i a \tan (c+d x)) \left (-a^2 (5 A-3 i B)+2 i a^2 A \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{16 a^3 A \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{\tan (c+d x)}}+\frac{4}{3} \int \frac{-3 a^3 (A-i B)-3 a^3 (i A+B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{16 a^3 A \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{\tan (c+d x)}}+\frac{\left (24 a^6 (A-i B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-3 a^3 (A-i B)+3 a^3 (i A+B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{8 \sqrt [4]{-1} a^3 (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{16 a^3 A \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (7 i A+3 B) \left (a^3+i a^3 \tan (c+d x)\right )}{3 d \sqrt{\tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 6.69537, size = 266, normalized size = 1.96 \[ \frac{\cos ^4(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \left (-\frac{8 e^{-3 i c} (A-i B) \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}}-\frac{1}{3} (\cos (3 c)-i \sin (3 c)) \sqrt{\tan (c+d x)} \csc ^2(c+d x) (3 (B+3 i A) \sin (2 (c+d x))+(A-3 i B) \cos (2 (c+d x))+A+3 i B)\right )}{d (\cos (d x)+i \sin (d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

(Cos[c + d*x]^4*((-8*(A - I*B)*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*ArcTanh[Sqrt[
(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/(E^((3*I)*c)*Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((
2*I)*(c + d*x)))]) - (Csc[c + d*x]^2*(Cos[3*c] - I*Sin[3*c])*(A + (3*I)*B + (A - (3*I)*B)*Cos[2*(c + d*x)] + 3
*((3*I)*A + B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/3)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/(d*(Cos
[d*x] + I*Sin[d*x])^3*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.017, size = 522, normalized size = 3.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x)

[Out]

2*I/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)-I/d*a^3*A*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)))*2^(1/2)-2/d*a^3/tan(d*x+c)^(1/2)*B-2*I/d*
a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2*I/d*a^3*B*tan(d*x+c)^(1/2)-6*I/d*a^3/tan(d*x+c)^(1/2)*A-2/d
*a^3*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2/d*a^3*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/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)))-2*I/d*a^3*A*ar
ctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+2*I/d*a^3*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+I/d*a^3*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)))-1/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)-2/d*a^3*B*arctan(1+2^(1/2)*ta
n(d*x+c)^(1/2))*2^(1/2)-2/d*a^3*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)

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Maxima [A]  time = 2.09837, size = 255, normalized size = 1.88 \begin{align*} -\frac{6 i \, B a^{3} \sqrt{\tan \left (d x + c\right )} - 3 \,{\left (\sqrt{2}{\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\right ) \, B\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + \sqrt{2}{\left (-\left (2 i + 2\right ) \, A + \left (2 i - 2\right ) \, B\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3} - \frac{2 \,{\left (3 \,{\left (-3 i \, A - B\right )} a^{3} \tan \left (d x + c\right ) - A a^{3}\right )}}{\tan \left (d x + c\right )^{\frac{3}{2}}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

-1/3*(6*I*B*a^3*sqrt(tan(d*x + c)) - 3*(sqrt(2)*(-(2*I + 2)*A + (2*I - 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*s
qrt(tan(d*x + c)))) + sqrt(2)*(-(2*I + 2)*A + (2*I - 2)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))
)) + sqrt(2)*((I - 1)*A + (I + 1)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*((I - 1)*A +
(I + 1)*B)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*a^3 - 2*(3*(-3*I*A - B)*a^3*tan(d*x + c) - A*
a^3)/tan(d*x + c)^(3/2))/d

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Fricas [B]  time = 1.8594, size = 1185, normalized size = 8.71 \begin{align*} -\frac{3 \, \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 3 \, \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 16 \,{\left ({\left (5 \, A - 3 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (A + 3 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \, A a^{3}\right )} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(3*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*
log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*
I*c) + I*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*
a^3)) - 3*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
*log((8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x +
2*I*c) - I*d)*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B
)*a^3)) - 16*((5*A - 3*I*B)*a^3*e^(4*I*d*x + 4*I*c) + (A + 3*I*B)*a^3*e^(2*I*d*x + 2*I*c) - 4*A*a^3)*sqrt((-I*
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int \frac{A}{\tan ^{\frac{5}{2}}{\left (c + d x \right )}}\, dx + \int - \frac{3 A}{\sqrt{\tan{\left (c + d x \right )}}}\, dx + \int \frac{B}{\tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int - 3 B \sqrt{\tan{\left (c + d x \right )}}\, dx + \int \frac{3 i A}{\tan ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int - i A \sqrt{\tan{\left (c + d x \right )}}\, dx + \int \frac{3 i B}{\sqrt{\tan{\left (c + d x \right )}}}\, dx + \int - i B \tan ^{\frac{3}{2}}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)

[Out]

a**3*(Integral(A/tan(c + d*x)**(5/2), x) + Integral(-3*A/sqrt(tan(c + d*x)), x) + Integral(B/tan(c + d*x)**(3/
2), x) + Integral(-3*B*sqrt(tan(c + d*x)), x) + Integral(3*I*A/tan(c + d*x)**(3/2), x) + Integral(-I*A*sqrt(ta
n(c + d*x)), x) + Integral(3*I*B/sqrt(tan(c + d*x)), x) + Integral(-I*B*tan(c + d*x)**(3/2), x))

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Giac [A]  time = 1.36722, size = 131, normalized size = 0.96 \begin{align*} -\frac{2 i \, B a^{3} \sqrt{\tan \left (d x + c\right )}}{d} - \frac{\left (4 i - 4\right ) \, \sqrt{2}{\left (-i \, A a^{3} - B a^{3}\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{d} + \frac{-18 i \, A a^{3} \tan \left (d x + c\right ) - 6 \, B a^{3} \tan \left (d x + c\right ) - 2 \, A a^{3}}{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+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-2*I*B*a^3*sqrt(tan(d*x + c))/d - (4*I - 4)*sqrt(2)*(-I*A*a^3 - B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(
d*x + c)))/d + 1/3*(-18*I*A*a^3*tan(d*x + c) - 6*B*a^3*tan(d*x + c) - 2*A*a^3)/(d*tan(d*x + c)^(3/2))