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

Optimal. Leaf size=134 \[ -\frac{8 \sqrt [4]{-1} a^3 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 (-B+3 i A) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{3 d}-\frac{16 a^3 B \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{d \sqrt{\tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.354862, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3593, 3594, 3592, 3533, 205} \[ -\frac{8 \sqrt [4]{-1} a^3 (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 (-B+3 i A) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{3 d}-\frac{16 a^3 B \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3594

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 - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, 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{3}{2}}(c+d x)} \, dx &=-\frac{2 a A (a+i a \tan (c+d x))^2}{d \sqrt{\tan (c+d x)}}+2 \int \frac{(a+i a \tan (c+d x))^2 \left (\frac{1}{2} a (5 i A+B)+\frac{1}{2} a (3 A+i B) \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{2 a A (a+i a \tan (c+d x))^2}{d \sqrt{\tan (c+d x)}}+\frac{2 (3 i A-B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{3 d}+\frac{4}{3} \int \frac{(a+i a \tan (c+d x)) \left (a^2 (3 i A+B)+2 i a^2 B \tan (c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{16 a^3 B \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{d \sqrt{\tan (c+d x)}}+\frac{2 (3 i A-B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{3 d}+\frac{4}{3} \int \frac{3 a^3 (i A+B)-3 a^3 (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{16 a^3 B \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{d \sqrt{\tan (c+d x)}}+\frac{2 (3 i A-B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{3 d}+\frac{\left (24 a^6 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{3 a^3 (i A+B)+3 a^3 (A-i B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{8 \sqrt [4]{-1} a^3 (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}-\frac{16 a^3 B \sqrt{\tan (c+d x)}}{3 d}-\frac{2 a A (a+i a \tan (c+d x))^2}{d \sqrt{\tan (c+d x)}}+\frac{2 (3 i A-B) \sqrt{\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{3 d}\\ \end{align*}

Mathematica [A]  time = 6.5167, size = 151, normalized size = 1.13 \[ -\frac{a^3 \sqrt{i \tan (c+d x)} \sqrt{\tan (c+d x)} \csc ^2(c+d x) \left (\sqrt{i \tan (c+d x)} (3 (A-3 i B) \sin (2 (c+d x))+(-B-3 i A) \cos (2 (c+d x))-3 i A+B)-12 (A-i B) \sin (2 (c+d x)) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Csc[c + d*x]^2*(-12*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Sin[2*
(c + d*x)] + ((-3*I)*A + B + ((-3*I)*A - B)*Cos[2*(c + d*x)] + 3*(A - (3*I)*B)*Sin[2*(c + d*x)])*Sqrt[I*Tan[c
+ d*x]])*Sqrt[I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/(3*d)

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Maple [B]  time = 0.017, size = 521, normalized size = 3.9 \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)^(3/2),x)

[Out]

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

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Maxima [A]  time = 2.5345, size = 254, normalized size = 1.9 \begin{align*} -\frac{2 i \, B a^{3} \tan \left (d x + c\right )^{\frac{3}{2}} + 6 \,{\left (i \, A + 3 \, B\right )} 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{6 \, A a^{3}}{\sqrt{\tan \left (d x + c\right )}}}{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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.81921, size = 1069, normalized size = 7.98 \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 )} - 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 )} - 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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 ) +{\left ({\left (-48 i \, A - 80 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-48 i \, A + 16 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 64 \, B 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 )} - 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)^(3/2),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) - 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)*(d*e^(2*I*d*x + 2*I*c) + 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) - 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)*(d*e^(2*I*d*x + 2*I*c) + 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)) + ((-48*I*A - 80*B)*a^3*e^(4*I*d*x + 4*I*c) + (-48*I*A + 1
6*B)*a^3*e^(2*I*d*x + 2*I*c) + 64*B*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) - d)

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

[Out]

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

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Giac [A]  time = 1.39826, size = 149, normalized size = 1.11 \begin{align*} -\frac{2 \, A a^{3}}{d \sqrt{\tan \left (d x + c\right )}} + \frac{\left (i + 1\right ) \, \sqrt{2}{\left (16 i \, A a^{3} + 16 \, 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 )}{4 \, d} - \frac{2 i \, B a^{3} d^{2} \tan \left (d x + c\right )^{\frac{3}{2}} + 6 i \, A a^{3} d^{2} \sqrt{\tan \left (d x + c\right )} + 18 \, B a^{3} d^{2} \sqrt{\tan \left (d x + c\right )}}{3 \, d^{3}} \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)^(3/2),x, algorithm="giac")

[Out]

-2*A*a^3/(d*sqrt(tan(d*x + c))) + (1/4*I + 1/4)*sqrt(2)*(16*I*A*a^3 + 16*B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*
sqrt(tan(d*x + c)))/d - 1/3*(2*I*B*a^3*d^2*tan(d*x + c)^(3/2) + 6*I*A*a^3*d^2*sqrt(tan(d*x + c)) + 18*B*a^3*d^
2*sqrt(tan(d*x + c)))/d^3