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

Optimal. Leaf size=186 \[ \frac {(2+2 i) a^{3/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {2 \sqrt [4]{-1} a^{3/2} B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

2*(-1)^(1/4)*a^(3/2)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*t
an(d*x+c)^(1/2)/d+(2+2*I)*a^(3/2)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot
(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d-2*a*A*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.63, antiderivative size = 186, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.237, Rules used = {4241, 3593, 3601, 3544, 205, 3599, 63, 217, 203} \[ \frac {(2+2 i) a^{3/2} (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {2 \sqrt [4]{-1} a^{3/2} B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a A \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-1)^(1/4)*a^(3/2)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c
+ d*x]]*Sqrt[Tan[c + d*x]])/d + ((2 + 2*I)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt
[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - (2*a*A*Sqrt[Cot[c + d*x]]*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 203

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3544

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

Rule 3593

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(a^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 3601

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

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps

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

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Mathematica [A]  time = 5.06, size = 286, normalized size = 1.54 \[ -\frac {a \cos (c+d x) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)} \left (-4 \sqrt {2} (A-i B) \sqrt {-1+e^{2 i (c+d x)}} \log \left (\sqrt {-1+e^{2 i (c+d x)}}+e^{i (c+d x)}\right )+4 \sqrt {2} A e^{i (c+d x)}-i B \sqrt {-1+e^{2 i (c+d x)}} \log \left (-2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )+i B \sqrt {-1+e^{2 i (c+d x)}} \log \left (2 \sqrt {2} e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}}-3 e^{2 i (c+d x)}+1\right )\right )}{\sqrt {2} d \left (1+e^{2 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.56, size = 671, normalized size = 3.61 \[ -\frac {8 \, \sqrt {2} A a \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \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}} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {4 i \, B^{2} a^{3}}{d^{2}}} d \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} + \sqrt {2} \sqrt {-\frac {4 i \, B^{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}} \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 )}}{B}\right ) - \sqrt {-\frac {4 i \, B^{2} a^{3}}{d^{2}}} d \log \left (-\frac {16 \, {\left (3 \, B a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - B a^{2} - \sqrt {2} \sqrt {-\frac {4 i \, B^{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}} \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 )}}{B}\right ) - \sqrt {\frac {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{3}}{d^{2}}} d \log \left (-\frac {{\left (8 \, {\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, 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}} \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 (-i \, d x - i \, c\right )}}{{\left (2 i \, A + 2 \, B\right )} a}\right ) + \sqrt {\frac {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, B^{2}\right )} a^{3}}{d^{2}}} d \log \left (-\frac {{\left (8 \, {\left (A - i \, B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {{\left (32 i \, A^{2} + 64 \, A B - 32 i \, 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}} \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 (-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)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(8*sqrt(2)*A*a*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) -
1))*e^(I*d*x + I*c) + sqrt(-4*I*B^2*a^3/d^2)*d*log(-16*(3*B*a^2*e^(2*I*d*x + 2*I*c) - B*a^2 + sqrt(2)*sqrt(-4*
I*B^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))*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)/B) - sqrt(-4*I*B^2*a^3/d^2)*d*log(-16*(3*B*a^
2*e^(2*I*d*x + 2*I*c) - B*a^2 - sqrt(2)*sqrt(-4*I*B^2*a^3/d^2)*(d*e^(3*I*d*x + 3*I*c) - d*e^(I*d*x + I*c))*sqr
t(a/(e^(2*I*d*x + 2*I*c) + 1))*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)/B) - sqrt((32*I*A^2 + 64*A*B - 32*I*B^2)*a^3/d^2)*d*log(-(8*(A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(2)*sqrt((3
2*I*A^2 + 64*A*B - 32*I*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))*sqrt((
I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-I*d*x - I*c)/((2*I*A + 2*B)*a)) + sqrt((32*I*A^2 +
64*A*B - 32*I*B^2)*a^3/d^2)*d*log(-(8*(A - I*B)*a^2*e^(I*d*x + I*c) - sqrt(2)*sqrt((32*I*A^2 + 64*A*B - 32*I*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))*sqrt((I*e^(2*I*d*x + 2*I*c) +
I)/(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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(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)^(3/2), x)

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maple [B]  time = 4.20, size = 1366, normalized size = 7.34 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*(4*I*B*sin(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/
2)+4*I*B*sin(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+2*
I*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-c
os(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+4*I*A
*sin(d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+4*I*A*sin(
d*x+c)*arctan(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-2*I*B*sin(d*x+c
)*2^(1/2)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-B*sin(d*x+c)*2^(1/2)*l
n(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+B*sin(d*x+c)*2^(1/2)*ln(((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-2*B*sin(d*x+c)*2^(1/2)*arctan(((-1+cos(d*x+c))/s
in(d*x+c))^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+2*I*A*sin(d*x+c)*2^(1/2)+I*B*sin(d*x+c)*2^(1/2)*ln(((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)-1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-I*B*sin(d*x+c)*2^(1/2)*ln(((-1+cos(d*x+c))/
sin(d*x+c))^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+2*I*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln
(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/si
n(d*x+c))^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))-4*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(
2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)-4*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)
*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)-2*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)*((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin
(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))+4*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*
x+c))/sin(d*x+c))^(1/2)+1)+4*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*arctan(2^(1/2)*((-1+cos(d*x+c))/s
in(d*x+c))^(1/2)-1)+2*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*ln(-(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+cos(d*x+c)+
sin(d*x+c)-1))+2*A*2^(1/2)*cos(d*x+c)-2*A*2^(1/2))*sin(d*x+c)*(cos(d*x+c)/sin(d*x+c))^(3/2)*(a*(I*sin(d*x+c)+c
os(d*x+c))/cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)*a*2^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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