3.506 \(\int \frac{(a+i a \tan (c+d x)) (A+B \tan (c+d x))}{\sqrt{\cot (c+d x)}} \, dx\)

Optimal. Leaf size=80 \[ \frac{2 a (B+i A)}{d \sqrt{\cot (c+d x)}}+\frac{2 \sqrt [4]{-1} a (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i a B}{3 d \cot ^{\frac{3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.187618, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{2 a (B+i A)}{d \sqrt{\cot (c+d x)}}+\frac{2 \sqrt [4]{-1} a (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i a B}{3 d \cot ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 2.64355, size = 96, normalized size = 1.2 \[ \frac{2 a \left (3 (B+i A) \cot (c+d x)+\frac{3 (A-i B) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{(i \tan (c+d x))^{3/2}}+i B\right )}{3 d \cot ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.501, size = 889, normalized size = 11.1 \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)^(1/2),x)

[Out]

1/3*a/d*2^(1/2)*(cos(d*x+c)+1)^2*(cos(d*x+c)-1)*(3*I*A*cos(d*x+c)*sin(d*x+c)*((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)*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))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-3*I*A*cos(d*x+c)*sin(d*x+c)*((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)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/s
in(d*x+c))^(1/2),1/2*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)-3*I*B*cos(d*x+c)*sin(d*x+c)*(-(cos(d*x+c)-1-si
n(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)-1)/sin(d*x+c))^(1/2)*Ell
ipticPi((-(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))+3*A*cos(d*x+c)*sin(d*x+c)*(-(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)-1)/sin(d*x+c)
)^(1/2)*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))+3*B*cos(d*x+c)*sin(d*x
+c)*(-(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)-1)
/sin(d*x+c))^(1/2)*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))-3*B*cos(d*x
+c)*sin(d*x+c)*((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)*Elli
pticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)+3*I*A*cos(d
*x+c)^2*2^(1/2)+I*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)-3*I*A*cos(d*x+c)*2^(1/2)+3*B*cos(d*x+c)^2*2^(1/2)-I*B*sin(d*
x+c)*2^(1/2)-3*B*cos(d*x+c)*2^(1/2))/cos(d*x+c)/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)

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Maxima [B]  time = 1.55281, size = 239, normalized size = 2.99 \begin{align*} \frac{8 \,{\left (i \, B a - \frac{3 \,{\left (-i \, A - B\right )} a}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac{3}{2}} + 3 \,{\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}{12 \, 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)^(1/2),x, algorithm="maxima")

[Out]

1/12*(8*(I*B*a - 3*(-I*A - B)*a/tan(d*x + c))*tan(d*x + c)^(3/2) + 3*(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 = 1.47274, size = 1135, normalized size = 14.19 \begin{align*} -\frac{3 \,{\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 ) - 3 \,{\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 ) -{\left (8 \,{\left (3 \, A - 4 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, B a e^{\left (2 i \, d x + 2 i \, c\right )} - 8 \,{\left (3 \, A - 2 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}}}{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))*(A+B*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

-1/12*(3*(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)) - 3*(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*(3*A - 4*I*B)*a*e^(4*I*
d*x + 4*I*c) + 16*I*B*a*e^(2*I*d*x + 2*I*c) - 8*(3*A - 2*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]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \frac{A}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int \frac{B \tan{\left (c + d x \right )}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int \frac{i A \tan{\left (c + d x \right )}}{\sqrt{\cot{\left (c + d x \right )}}}\, dx + \int \frac{i B \tan ^{2}{\left (c + d x \right )}}{\sqrt{\cot{\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)**(1/2),x)

[Out]

a*(Integral(A/sqrt(cot(c + d*x)), x) + Integral(B*tan(c + d*x)/sqrt(cot(c + d*x)), x) + Integral(I*A*tan(c + d
*x)/sqrt(cot(c + d*x)), x) + Integral(I*B*tan(c + d*x)**2/sqrt(cot(c + d*x)), 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 )}}{\sqrt{\cot \left (d x + c\right )}}\,{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)^(1/2),x, algorithm="giac")

[Out]

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