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

Optimal. Leaf size=105 \[ \frac{2 a (B+i A) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 a (A-i B) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.165715, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3592, 3528, 3533, 205} \[ \frac{2 a (B+i A) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{2 \sqrt [4]{-1} a (A-i B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 a (A-i B) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{5}{2}}(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3528

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

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

Mathematica [B]  time = 2.66962, size = 266, normalized size = 2.53 \[ \frac{\cos ^2(c+d x) (\cos (d x)-i \sin (d x)) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \left (\frac{2 e^{-i c} (B+i A) \sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}}}+\frac{1}{15} (\cos (c)-i \sin (c)) \sqrt{\tan (c+d x)} \sec ^2(c+d x) (5 (B+i A) \sin (2 (c+d x))+3 (5 A-6 i B) \cos (2 (c+d x))+3 (5 A-4 i B))\right )}{d (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]^2*(Cos[d*x] - I*Sin[d*x])*((2*(I*A + B)*Sqrt[(-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^(I*c)*Sqrt[((-I)*(-1 + E^((2*I)*(c
+ d*x))))/(1 + E^((2*I)*(c + d*x)))]) + (Sec[c + d*x]^2*(Cos[c] - I*Sin[c])*(3*(5*A - (4*I)*B) + 3*(5*A - (6*I
)*B)*Cos[2*(c + d*x)] + 5*(I*A + B)*Sin[2*(c + d*x)])*Sqrt[Tan[c + d*x]])/15)*(a + I*a*Tan[c + d*x])*(A + B*Ta
n[c + d*x]))/(d*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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Maple [B]  time = 0.012, size = 506, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*I/d*a*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)))+1/2*I/
d*a*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+2/3/d*a*B*tan(d*x+c)^(3/2)+2/3*I/d*a*A*tan(d*x+c)^(3/2)+2/d*
a*A*tan(d*x+c)^(1/2)+1/2*I/d*a*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-2*I/d*a*B*tan(d*x+c)^(1/2)-1/2*I/d
*a*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/2/d*a*A*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/4/d*a
*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)))-1/2/d*a*A*arcta
n(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)+2/5*I*a*B*tan(d*x+c)^(5/2)/d-1/4*I/d*a*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)))-1/2*I/d*a*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(
1/2))-1/2/d*a*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)-1/4/d*a*B*2^(1/2)*ln((1-2^(1/2)*tan(d*x+c)^(1/2)+t
an(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))-1/2/d*a*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)

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Maxima [B]  time = 1.76801, size = 254, normalized size = 2.42 \begin{align*} -\frac{-24 i \, B a \tan \left (d x + c\right )^{\frac{5}{2}} + 40 \,{\left (-i \, A - B\right )} a \tan \left (d x + c\right )^{\frac{3}{2}} - 8 \,{\left (15 \, A - 15 i \, B\right )} a \sqrt{\tan \left (d x + c\right )} - 15 \,{\left (2 \, \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt{2}{\left (-\left (i + 1\right ) \, A + \left (i - 1\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}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(-24*I*B*a*tan(d*x + c)^(5/2) + 40*(-I*A - B)*a*tan(d*x + c)^(3/2) - 8*(15*A - 15*I*B)*a*sqrt(tan(d*x +
c)) - 15*(2*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*
(-(I + 1)*A + (I - 1)*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)/d

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Fricas [B]  time = 2.00713, size = 1160, normalized size = 11.05 \begin{align*} -\frac{15 \,{\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 )} \sqrt{\frac{{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac{{\left (2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt{\frac{{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \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 (i \, A + B\right )} a}\right ) - 15 \,{\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 )} \sqrt{\frac{{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac{{\left (2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt{\frac{{\left (-4 i \, A^{2} - 8 \, A B + 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \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 (i \, A + B\right )} a}\right ) - 8 \,{\left ({\left (20 \, A - 23 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \,{\left (5 \, A - 4 i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (10 \, A - 13 i \, B\right )} a\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}}}{60 \,{\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(tan(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/60*(15*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^2)*log
((2*(A - I*B)*a*e^(2*I*d*x + 2*I*c) + (I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^
2)*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)/((I*A + B)*a)) - 15*(d*e
^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^2)*log((2*(A - I*B)*
a*e^(2*I*d*x + 2*I*c) + (-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((-4*I*A^2 - 8*A*B + 4*I*B^2)*a^2/d^2)*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)/((I*A + B)*a)) - 8*((20*A - 23*I*B)*a
*e^(4*I*d*x + 4*I*c) + 6*(5*A - 4*I*B)*a*e^(2*I*d*x + 2*I*c) + (10*A - 13*I*B)*a)*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(-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(tan(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.21819, size = 151, normalized size = 1.44 \begin{align*} \frac{\left (i - 1\right ) \, \sqrt{2}{\left (i \, A a + B a\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{d} - \frac{-6 i \, B a d^{4} \tan \left (d x + c\right )^{\frac{5}{2}} - 10 i \, A a d^{4} \tan \left (d x + c\right )^{\frac{3}{2}} - 10 \, B a d^{4} \tan \left (d x + c\right )^{\frac{3}{2}} - 30 \, A a d^{4} \sqrt{\tan \left (d x + c\right )} + 30 i \, B a d^{4} \sqrt{\tan \left (d x + c\right )}}{15 \, d^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(I - 1)*sqrt(2)*(I*A*a + B*a)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 1/15*(-6*I*B*a*d^4*tan(d*x
+ c)^(5/2) - 10*I*A*a*d^4*tan(d*x + c)^(3/2) - 10*B*a*d^4*tan(d*x + c)^(3/2) - 30*A*a*d^4*sqrt(tan(d*x + c))
+ 30*I*B*a*d^4*sqrt(tan(d*x + c)))/d^5