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)*cot(d*x+c)^(1/2))/d+2/3*I*a*B/d/cot(d*x+c)^(3/2)+2*a*(I*A+B)/d/cot(d
*x+c)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 2.82, size = 96, normalized size = 1.20 \[ \frac {2 a \left (3 (B+i A) \cot (c+d x)+\frac {3 (A-i B) \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{(i \tan (c+d x))^{3/2}}+i B\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)} \]

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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fricas [B]  time = 0.50, size = 428, normalized size = 5.35 \[ -\frac {3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (-\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (-\frac {{\left (2 \, {\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - {\left (8 \, {\left (3 \, A - 4 i \, B\right )} a e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, B a e^{\left (2 i \, d x + 2 i \, c\right )} - 8 \, {\left (3 \, A - 2 i \, B\right )} a\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{12 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}}{\sqrt {\cot \left (d x + c\right )}}\,{d x} \]

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)

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maple [C]  time = 1.66, size = 889, normalized size = 11.11 \[ \text {result too large to display} \]

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*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(3*I*A*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+
1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*
x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)-3*I*A*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*
x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/
sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)-3*I*B*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(-sin(d*
x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)+3*A*(-(-sin(d*x+c)-1+cos(d*x
+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*Ellipt
icPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)+3*B*(-(-sin(d
*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c
)-3*B*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(d*x+c)*sin(d*x+c)*((-1+cos(d*x
+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^
(1/2)+3*I*A*2^(1/2)*cos(d*x+c)^2+I*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)-3*I*A*cos(d*x+c)*2^(1/2)-I*B*sin(d*x+c)*2^(
1/2)+3*B*2^(1/2)*cos(d*x+c)^2-3*B*2^(1/2)*cos(d*x+c))/cos(d*x+c)/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)*2^
(1/2)

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maxima [B]  time = 1.00, size = 176, normalized size = 2.20 \[ \frac {8 \, {\left (i \, B a + \frac {{\left (3 i \, A + 3 \, B\right )} a}{\tan \left (d x + c\right )}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + 3 \, {\left (2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a}{12 \, d} \]

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 + 3*B)*a/tan(d*x + c))*tan(d*x + c)^(3/2) + 3*(2*sqrt(2)*((I - 1)*A + (I + 1)*B)*arcta
n(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i))/cot(c + d*x)^(1/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i))/cot(c + d*x)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ i a \left (\int \left (- \frac {i A}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {A \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \frac {B \tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {i B \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx\right ) \]

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]

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

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