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

Optimal. Leaf size=113 \[ \frac {2 \sqrt {2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 i a B \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.37, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3594, 3600, 3480, 206, 3599, 63, 208} \[ \frac {2 \sqrt {2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 i a B \sqrt {a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

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 3599

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

Rule 3600

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

Rubi steps

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

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Mathematica [A]  time = 2.26, size = 157, normalized size = 1.39 \[ \frac {\sqrt {2} a e^{-i (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (\sqrt {2} (A-i B) \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-A \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )+i \sqrt {2} B e^{i (c+d x)}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 518, normalized size = 4.58 \[ \frac {8 i \, \sqrt {2} B a \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 2 \, \sqrt {\frac {A^{2} a^{3}}{d^{2}}} d \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} + 2 \, \sqrt {2} \sqrt {\frac {A^{2} a^{3}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) + 2 \, \sqrt {\frac {A^{2} a^{3}}{d^{2}}} d \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} - 2 \, \sqrt {2} \sqrt {\frac {A^{2} a^{3}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) + \sqrt {\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} d \log \left (\frac {{\left ({\left (8 i \, A + 8 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) - \sqrt {\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} d \log \left (\frac {{\left ({\left (8 i \, A + 8 \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*I*sqrt(2)*B*a*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 2*sqrt(A^2*a^3/d^2)*d*log(16*(3*A*a^2
*e^(2*I*d*x + 2*I*c) + A*a^2 + 2*sqrt(2)*sqrt(A^2*a^3/d^2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a/
(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/A) + 2*sqrt(A^2*a^3/d^2)*d*log(16*(3*A*a^2*e^(2*I*d*x + 2*I*c
) + A*a^2 - 2*sqrt(2)*sqrt(A^2*a^3/d^2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c
) + 1)))*e^(-2*I*d*x - 2*I*c)/A) + sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*d*log(((8*I*A + 8*B)*a^2*e^(I*d*
x + I*c) + sqrt(2)*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt(a/(e^(2*I*d
*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) - sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*d*log(((8*
I*A + 8*B)*a^2*e^(I*d*x + I*c) + sqrt(2)*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^3/d^2)*(-I*d*e^(2*I*d*x + 2*I*c)
- I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)))/d

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 3.46, size = 467, normalized size = 4.13 \[ -\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (2 i A \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+2 i B \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )-2 A \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+i A \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+2 B \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )-A \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-2 i B \cos \left (d x +c \right )+2 B \sin \left (d x +c \right )+2 i B \right ) a}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(2*I*A*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(
1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*sin(d*x+c)+2*I*B*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*sin(d*x+c)-2*A*arctanh(1/2*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/
2)*sin(d*x+c)+I*A*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+
c)+2*B*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*s
in(d*x+c)-A*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*sin(d*x+c)-2*I*B*cos(d*x+c)+2*B*sin(d*x+c)+2*I*B)/(I*sin(d*x+c)+cos(d*x+c)-1)*a

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maxima [A]  time = 0.95, size = 130, normalized size = 1.15 \[ -\frac {\sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - A a^{\frac {3}{2}} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - 2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(2)*(A - I*B)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*ta
n(d*x + c) + a))) - A*a^(3/2)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)
)) - 2*I*sqrt(I*a*tan(d*x + c) + a)*B*a)/d

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mupad [B]  time = 6.60, size = 553, normalized size = 4.89 \[ \frac {B\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{d}-\frac {2\,A\,\mathrm {atanh}\left (-\frac {32\,A^3\,a^6\,d\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{-32\,d\,A^3\,a^8+128{}\mathrm {i}\,d\,A^2\,B\,a^8+64\,d\,A\,B^2\,a^8}+\frac {64\,A\,B^2\,a^6\,d\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{-32\,d\,A^3\,a^8+128{}\mathrm {i}\,d\,A^2\,B\,a^8+64\,d\,A\,B^2\,a^8}+\frac {A^2\,B\,a^6\,d\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,128{}\mathrm {i}}{-32\,d\,A^3\,a^8+128{}\mathrm {i}\,d\,A^2\,B\,a^8+64\,d\,A\,B^2\,a^8}\right )\,\sqrt {a^3}}{d}+\frac {2\,\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,A^3\,a^6\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,16{}\mathrm {i}}{32\,d\,A^3\,a^8-160{}\mathrm {i}\,d\,A^2\,B\,a^8-192\,d\,A\,B^2\,a^8+64{}\mathrm {i}\,d\,B^3\,a^8}-\frac {32\,\sqrt {2}\,B^3\,a^6\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{32\,d\,A^3\,a^8-160{}\mathrm {i}\,d\,A^2\,B\,a^8-192\,d\,A\,B^2\,a^8+64{}\mathrm {i}\,d\,B^3\,a^8}-\frac {\sqrt {2}\,A\,B^2\,a^6\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,96{}\mathrm {i}}{32\,d\,A^3\,a^8-160{}\mathrm {i}\,d\,A^2\,B\,a^8-192\,d\,A\,B^2\,a^8+64{}\mathrm {i}\,d\,B^3\,a^8}+\frac {80\,\sqrt {2}\,A^2\,B\,a^6\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{32\,d\,A^3\,a^8-160{}\mathrm {i}\,d\,A^2\,B\,a^8-192\,d\,A\,B^2\,a^8+64{}\mathrm {i}\,d\,B^3\,a^8}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {-a^3}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

(B*a*(a + a*tan(c + d*x)*1i)^(1/2)*2i)/d - (2*A*atanh((64*A*B^2*a^6*d*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2
))/(64*A*B^2*a^8*d - 32*A^3*a^8*d + A^2*B*a^8*d*128i) - (32*A^3*a^6*d*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2
))/(64*A*B^2*a^8*d - 32*A^3*a^8*d + A^2*B*a^8*d*128i) + (A^2*B*a^6*d*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)
*128i)/(64*A*B^2*a^8*d - 32*A^3*a^8*d + A^2*B*a^8*d*128i))*(a^3)^(1/2))/d + (2*2^(1/2)*atanh((2^(1/2)*A^3*a^6*
d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*16i)/(32*A^3*a^8*d + B^3*a^8*d*64i - 192*A*B^2*a^8*d - A^2*B*a^8*
d*160i) - (32*2^(1/2)*B^3*a^6*d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(32*A^3*a^8*d + B^3*a^8*d*64i - 19
2*A*B^2*a^8*d - A^2*B*a^8*d*160i) - (2^(1/2)*A*B^2*a^6*d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*96i)/(32*A
^3*a^8*d + B^3*a^8*d*64i - 192*A*B^2*a^8*d - A^2*B*a^8*d*160i) + (80*2^(1/2)*A^2*B*a^6*d*(-a^3)^(1/2)*(a + a*t
an(c + d*x)*1i)^(1/2))/(32*A^3*a^8*d + B^3*a^8*d*64i - 192*A*B^2*a^8*d - A^2*B*a^8*d*160i))*(A*1i + B)*(-a^3)^
(1/2))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (c + d x \right )}\right ) \cot {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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