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

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

[Out]

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

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Rubi [A]  time = 0.226374, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3581, 3591, 3529, 3533, 208} \[ \frac{2 a (B+i A)}{3 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{2 a (A-i B)}{d \sqrt{\cot (c+d x)}}-\frac{2 \sqrt [4]{-1} a (B+i A) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i a B}{5 d \cot ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3591

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

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[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 208

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

Rubi steps

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

Mathematica [A]  time = 3.50123, size = 133, normalized size = 1.27 \[ \frac{a \left (\sec ^2(c+d x) (5 (B+i A) \sin (2 (c+d x))+3 (5 A-6 i B) \cos (2 (c+d x))+3 (5 A-4 i B))-\frac{30 (A-i B) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{i \tan (c+d x)}}\right )}{15 d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(Sec[c + d*x]^2*(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)]) - (
30*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]])/Sqrt[I*Tan[c + d*x]]))/(15*d
*Sqrt[Cot[c + d*x]])

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Maple [C]  time = 0.459, size = 971, normalized size = 9.3 \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))*(A+B*tan(d*x+c))/cot(d*x+c)^(3/2),x)

[Out]

-1/15*a/d*2^(1/2)*(cos(d*x+c)-1)*(-5*I*A*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+5*I*A*2^(1/2)*cos(d*x+c)*sin(d*x+c)+1
5*I*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c
))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/
2))*sin(d*x+c)+15*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+
c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^
(1/2))*sin(d*x+c)-15*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/
2+1/2*I,1/2*2^(1/2))*sin(d*x+c)+15*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^
(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*
x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)+3*I*2^(1/2)*B-18*I*B*2^(1/2)*cos(d*x+c)^2-15*I*B*((cos(d*x+c)-1)
/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*
cos(d*x+c)^2*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*sin(d*x+c)-15*A*cos(d*x+c)^3
*2^(1/2)-3*I*B*2^(1/2)*cos(d*x+c)-5*B*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+15*A*2^(1/2)*cos(d*x+c)^2+18*I*B*2^(1/2)
*cos(d*x+c)^3+5*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)+15*I*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*
x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-
sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c))*(cos(d*x+c)+1)^2/cos(d*x+c)/sin(d*x+c)^5/(cos
(d*x+c)/sin(d*x+c))^(3/2)

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Maxima [B]  time = 1.56457, size = 261, normalized size = 2.49 \begin{align*} -\frac{8 \,{\left (-3 i \, B a - \frac{5 \,{\left (i \, A + B\right )} a}{\tan \left (d x + c\right )} - \frac{{\left (15 \, A - 15 i \, B\right )} a}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac{5}{2}} + 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} + \frac{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} - \frac{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 (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\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((a+I*a*tan(d*x+c))*(A+B*tan(d*x+c))/cot(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.65072, size = 1310, normalized size = 12.48 \begin{align*} -\frac{15 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 (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}}} \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 (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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 (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}}} \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 ) -{\left ({\left (-160 i \, A - 184 \, B\right )} a e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (-80 i \, A - 8 \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (160 i \, A + 88 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (80 i \, A + 104 \, 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 (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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))*(A+B*tan(d*x+c))/cot(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

-1/60*(15*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*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) + (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)*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^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*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) - (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)*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)) - ((-160*I*A - 184*B)*a*e^(6*I*d*x + 6*I*c) + (-80*I*A - 8*B)*a*e^(4*I*d*x + 4*I*c) + (16
0*I*A + 88*B)*a*e^(2*I*d*x + 2*I*c) + (80*I*A + 104*B)*a)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c
) - 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*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 \left (\int \frac{A}{\cot ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{B \tan{\left (c + d x \right )}}{\cot ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{i A \tan{\left (c + d x \right )}}{\cot ^{\frac{3}{2}}{\left (c + d x \right )}}\, dx + \int \frac{i B \tan ^{2}{\left (c + d x \right )}}{\cot ^{\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))*(A+B*tan(d*x+c))/cot(d*x+c)**(3/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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