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

Optimal. Leaf size=125 \[ -\frac {a^{3/2} (2 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a A \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3593, 3600, 3480, 206, 3599, 63, 208} \[ -\frac {a^{3/2} (2 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {2} a^{3/2} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a A \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 3.32, size = 201, normalized size = 1.61 \[ -\frac {a e^{-\frac {1}{2} i (4 c+5 d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sec (c+d x) \left (\cos \left (\frac {d x}{2}\right )+i \sin \left (\frac {d x}{2}\right )\right ) \left ((-4 B-4 i A) \sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt {2} (2 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )+A \sqrt {1+e^{2 i (c+d x)}} \csc (c+d x)\right )}{2 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(1 + E^((2*I)*(c + d*x)))^(3/2)*(((-4*I)*A - 4
*B)*ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*((3*I)*A + 2*B)*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c
+ d*x))]] + A*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c + d*x])*Sec[c + d*x]*(Cos[(d*x)/2] + I*Sin[(d*x)/2]))/(Sqrt[
2]*d*E^((I/2)*(4*c + 5*d*x)))

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fricas [B]  time = 0.59, size = 686, normalized size = 5.49 \[ -\frac {\sqrt {-\frac {{\left (9 \, A^{2} - 12 i \, A B - 4 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left ({\left (144 i \, A + 96 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (48 i \, A + 32 \, B\right )} a^{2} + 32 \, \sqrt {2} \sqrt {-\frac {{\left (9 \, A^{2} - 12 i \, A B - 4 \, B^{2}\right )} 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 )}}{3 i \, A + 2 \, B}\right ) - \sqrt {-\frac {{\left (9 \, A^{2} - 12 i \, A B - 4 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left ({\left (144 i \, A + 96 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (48 i \, A + 32 \, B\right )} a^{2} - 32 \, \sqrt {2} \sqrt {-\frac {{\left (9 \, A^{2} - 12 i \, A B - 4 \, B^{2}\right )} 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 )}}{3 i \, A + 2 \, B}\right ) - \sqrt {-\frac {{\left (32 \, A^{2} - 64 i \, A B - 32 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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 (d e^{\left (2 i \, d x + 2 i \, c\right )} + 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 {2} {\left (-4 i \, A a e^{\left (3 i \, d x + 3 i \, c\right )} - 4 i \, A a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, {\left (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(cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(sqrt(-(9*A^2 - 12*I*A*B - 4*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(((144*I*A + 96*B)*a^2*e^(2*I*d
*x + 2*I*c) + (48*I*A + 32*B)*a^2 + 32*sqrt(2)*sqrt(-(9*A^2 - 12*I*A*B - 4*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)))*e^(-2*I*d*x - 2*I*c)/(3*I*A + 2*B)) - sqrt(-(9*A^2 -
12*I*A*B - 4*B^2)*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log(((144*I*A + 96*B)*a^2*e^(2*I*d*x + 2*I*c) + (48*I*
A + 32*B)*a^2 - 32*sqrt(2)*sqrt(-(9*A^2 - 12*I*A*B - 4*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)))*e^(-2*I*d*x - 2*I*c)/(3*I*A + 2*B)) - sqrt(-(32*A^2 - 64*I*A*B - 32*B^2)
*a^3/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)*(d*e^(2*I*d*x + 2*I*c) + 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*e^(2*I*d*x + 2*I*c) - 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)*(d*e^(2*I*d*x + 2*I*c) + 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(2)*(-4*I*A*a*e^(3*I*d*x + 3*I*c) - 4*I*A*a*e^(I
*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cot \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^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)^2, x)

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maple [B]  time = 3.58, size = 1117, normalized size = 8.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*sin(d*x+c)*(4*I*B*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+3*I*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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))*cos(d*x+c)^2+4*A*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^2*2^(1/2)-4*I*B*2^(1/2
)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^2+4
*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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))*cos(d*x+c)^2*2^(1/2)-4*I*A*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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*I*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(1/2))+3*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2))*cos(d*x+c)^2+2*I*A*cos(d*x+c)*sin(d*x+c)+2*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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))*cos(d*x+c)^2-4*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+4*I*A*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*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))*cos(d*x+c)^2-3*I*A
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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))-4*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/c
os(d*x+c)*2^(1/2))*2^(1/2)-2*I*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2))*cos(d*x+c)^2-3*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+2*
A*cos(d*x+c)^2-2*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*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*A*cos(d*x+c))/(1+cos(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/(-1+cos(d*x+c))*a

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maxima [A]  time = 0.91, size = 145, normalized size = 1.16 \[ -\frac {i \, {\left (2 \, \sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - {\left (3 \, A - 2 i \, B\right )} \sqrt {a} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - \frac {2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} A}{\tan \left (d x + c\right )}\right )} a}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*(2*sqrt(2)*(A - I*B)*sqrt(a)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqr
t(I*a*tan(d*x + c) + a))) - (3*A - 2*I*B)*sqrt(a)*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)*A/tan(d*x + c))*a/d

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mupad [B]  time = 8.08, size = 2338, normalized size = 18.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2*atanh((6*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((3*B^2*a^3)/(2*d^2) - (17*A^2*a^3)/(8*d^2) - ((A^4*a^18)/d^4 +
(16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)/(8*a^6) + (A*B*a
^3*7i)/(2*d^2))^(1/2)*((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3
*B*a^18*8i)/d^4)^(1/2))/(A^3*a^11*d*10i + 32*B^3*a^11*d + A*B^2*a^11*d*72i - 32*A^2*B*a^11*d + A*a^2*d^3*((A^4
*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)*2i)
+ (2*A^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((3*B^2*a^3)/(2*d^2) - (17*A^2*a^3)/(8*d^2) - ((A^4*a^18)/d^4 +
(16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)/(8*a^6) + (A*B*a
^3*7i)/(2*d^2))^(1/2))/(A^3*a^8*d*10i + 32*B^3*a^8*d + A*B^2*a^8*d*72i - 32*A^2*B*a^8*d + (A*d^3*((A^4*a^18)/d
^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)*2i)/a) + (8*
B^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((3*B^2*a^3)/(2*d^2) - (17*A^2*a^3)/(8*d^2) - ((A^4*a^18)/d^4 + (16*
B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)/(8*a^6) + (A*B*a^3*7i
)/(2*d^2))^(1/2))/(A^3*a^8*d*10i + 32*B^3*a^8*d + A*B^2*a^8*d*72i - 32*A^2*B*a^8*d + (A*d^3*((A^4*a^18)/d^4 +
(16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)*2i)/a) + (A*B*a^6
*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((3*B^2*a^3)/(2*d^2) - (17*A^2*a^3)/(8*d^2) - ((A^4*a^18)/d^4 + (16*B^4*a^1
8)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)/(8*a^6) + (A*B*a^3*7i)/(2*d^
2))^(1/2)*8i)/(A^3*a^8*d*10i + 32*B^3*a^8*d + A*B^2*a^8*d*72i - 32*A^2*B*a^8*d + (A*d^3*((A^4*a^18)/d^4 + (16*
B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)*2i)/a))*((3*B^2*a^3)/
(2*d^2) - (17*A^2*a^3)/(8*d^2) - ((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)
/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)/(8*a^6) + (A*B*a^3*7i)/(2*d^2))^(1/2) - 2*atanh((2*A^2*a^6*d^2*(a + a*tan(c
+ d*x)*1i)^(1/2)*(((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a
^18*8i)/d^4)^(1/2)/(8*a^6) - (17*A^2*a^3)/(8*d^2) + (3*B^2*a^3)/(2*d^2) + (A*B*a^3*7i)/(2*d^2))^(1/2))/(A^3*a^
8*d*10i + 32*B^3*a^8*d + A*B^2*a^8*d*72i - 32*A^2*B*a^8*d - (A*d^3*((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^
2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)*2i)/a) - (6*d^4*(a + a*tan(c + d*x)*1i)^(1
/2)*(((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^
(1/2)/(8*a^6) - (17*A^2*a^3)/(8*d^2) + (3*B^2*a^3)/(2*d^2) + (A*B*a^3*7i)/(2*d^2))^(1/2)*((A^4*a^18)/d^4 + (16
*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2))/(A^3*a^11*d*10i + 3
2*B^3*a^11*d + A*B^2*a^11*d*72i - 32*A^2*B*a^11*d - A*a^2*d^3*((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2
*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)*2i) + (8*B^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1
/2)*(((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^
(1/2)/(8*a^6) - (17*A^2*a^3)/(8*d^2) + (3*B^2*a^3)/(2*d^2) + (A*B*a^3*7i)/(2*d^2))^(1/2))/(A^3*a^8*d*10i + 32*
B^3*a^8*d + A*B^2*a^8*d*72i - 32*A^2*B*a^8*d - (A*d^3*((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d
^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)*2i)/a) + (A*B*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((
A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)/(
8*a^6) - (17*A^2*a^3)/(8*d^2) + (3*B^2*a^3)/(2*d^2) + (A*B*a^3*7i)/(2*d^2))^(1/2)*8i)/(A^3*a^8*d*10i + 32*B^3*
a^8*d + A*B^2*a^8*d*72i - 32*A^2*B*a^8*d - (A*d^3*((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a^18)/d^4 +
(A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)*2i)/a))*(((A^4*a^18)/d^4 + (16*B^4*a^18)/d^4 - (8*A^2*B^2*a
^18)/d^4 + (A*B^3*a^18*32i)/d^4 + (A^3*B*a^18*8i)/d^4)^(1/2)/(8*a^6) - (17*A^2*a^3)/(8*d^2) + (3*B^2*a^3)/(2*d
^2) + (A*B*a^3*7i)/(2*d^2))^(1/2) - (A*a*cot(c + d*x)*(a + a*tan(c + d*x)*1i)^(1/2))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2*(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)**2, x)

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