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

Optimal. Leaf size=156 \[ \frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.51, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3596, 3600, 3480, 206, 3599, 63, 208} \[ \frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {2 A \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {A+i B}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {3 A+i B}{2 a d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3596

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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] &&  !GtQ[n,
0]

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

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Mathematica [A]  time = 4.37, size = 192, normalized size = 1.23 \[ \frac {-\frac {12 i (A-i B) e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{3/2}}+18 A \tan (c+d x)+\frac {48 i \sqrt {2} A e^{3 i (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )}{\left (1+e^{2 i (c+d x)}\right )^{3/2}}-22 i A+6 i B \tan (c+d x)+10 B}{12 a d (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-22*I)*A + 10*B - ((12*I)*(A - I*B)*E^((3*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])/(1 + E^((2*I)*(c + d*x)))^
(3/2) + ((48*I)*Sqrt[2]*A*E^((3*I)*(c + d*x))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]]
)/(1 + E^((2*I)*(c + d*x)))^(3/2) + 18*A*Tan[c + d*x] + (6*I)*B*Tan[c + d*x])/(12*a*d*(-I + Tan[c + d*x])*Sqrt
[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.54, size = 621, normalized size = 3.98 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (4 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-4 i \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {A^{2} - 2 i \, A B - B^{2}}{a^{3} d^{2}}} + {\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 6 \, a^{2} d \sqrt {\frac {A^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \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} {\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} 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 {A^{2}}{a^{3} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) + 6 \, a^{2} d \sqrt {\frac {A^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \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} {\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} 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 {A^{2}}{a^{3} d^{2}}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) + \sqrt {2} {\left (2 \, {\left (5 \, A + 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (11 \, A + 5 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sqrt(1/2)*a^2*d*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log((sqrt(2)*sqrt(1/2)*(4*I*
a^2*d*e^(2*I*d*x + 2*I*c) + 4*I*a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2))
+ (4*I*A + 4*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 3*sqrt(1/2)*a^2*d*sqrt((A^2 - 2*I*A*B - B^2)
/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log((sqrt(2)*sqrt(1/2)*(-4*I*a^2*d*e^(2*I*d*x + 2*I*c) - 4*I*a^2*d)*sqrt(a/(e^
(2*I*d*x + 2*I*c) + 1))*sqrt((A^2 - 2*I*A*B - B^2)/(a^3*d^2)) + (4*I*A + 4*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I
*c)/(I*A + B)) - 6*a^2*d*sqrt(A^2/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(16*(3*A*a^2*e^(2*I*d*x + 2*I*c) + A*a^2 +
2*sqrt(2)*(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(A^2/(a^3
*d^2)))*e^(-2*I*d*x - 2*I*c)/A) + 6*a^2*d*sqrt(A^2/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(16*(3*A*a^2*e^(2*I*d*x +
2*I*c) + A*a^2 - 2*sqrt(2)*(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1))*sqrt(A^2/(a^3*d^2)))*e^(-2*I*d*x - 2*I*c)/A) + sqrt(2)*(2*(5*A + 2*I*B)*e^(4*I*d*x + 4*I*c) + (11*A + 5*I*
B)*e^(2*I*d*x + 2*I*c) + A + I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-3*I*d*x - 3*I*c)/(a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 3.40, size = 1026, normalized size = 6.58 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(12*I*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))*sin(d*x+c)+3*I*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*
x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)-12*I*A*(-2*cos(d*x+c)/(1+c
os(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)-3
6*I*A*cos(d*x+c)*sin(d*x+c)-12*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))-16*I*A*cos(d*x+c)^3*sin(d*x+c)-3*A*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
2^(1/2))+16*A*cos(d*x+c)^4+3*I*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/s
in(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*2^(1/2)+3*B*2^(1/2)*sin(d*x+c)*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*2^(1/2))+16*B*cos(d*x+c)^3*sin(d*x+c)+3*I*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-
sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)*2^(1/2)+16*I*B*cos(d*x+c)^4-12
*A*cos(d*x+c)*(-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))-12*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)-3*A*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+4*I*B*cos(d*x+c)^2-12*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arc
tan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+28*A*cos(d*x+c)^2+12*B*cos(d*x+c)*sin(d*x+c))/a^2

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maxima [A]  time = 1.72, size = 161, normalized size = 1.03 \[ -\frac {\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} - \frac {24 \, A \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {4 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (3 \, A + i \, B\right )} + 2 \, {\left (A + i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan
(d*x + c) + a)))/a^(3/2) - 24*A*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(
a)))/a^(3/2) - 4*(3*(I*a*tan(d*x + c) + a)*(3*A + I*B) + 2*(A + I*B)*a)/((I*a*tan(d*x + c) + a)^(3/2)*a))/d

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mupad [B]  time = 6.70, size = 563, normalized size = 3.61 \[ \frac {\frac {A+B\,1{}\mathrm {i}}{3\,d}+\frac {\left (3\,A+B\,1{}\mathrm {i}\right )\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2\,a\,d}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {2\,A\,\mathrm {atanh}\left (\frac {31\,A^3\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^3}\,\left (\frac {31\,A^3\,d}{a}+\frac {A\,B^2\,d}{a}+\frac {A^2\,B\,d\,2{}\mathrm {i}}{a}\right )}+\frac {A\,B^2\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^3}\,\left (\frac {31\,A^3\,d}{a}+\frac {A\,B^2\,d}{a}+\frac {A^2\,B\,d\,2{}\mathrm {i}}{a}\right )}+\frac {A^2\,B\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{\sqrt {a^3}\,\left (\frac {31\,A^3\,d}{a}+\frac {A\,B^2\,d}{a}+\frac {A^2\,B\,d\,2{}\mathrm {i}}{a}\right )}\right )}{d\,\sqrt {a^3}}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,A^3\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,31{}\mathrm {i}}{16\,\left (\frac {31\,d\,A^3\,a^2}{8}-\frac {29{}\mathrm {i}\,d\,A^2\,B\,a^2}{8}+\frac {3\,d\,A\,B^2\,a^2}{8}-\frac {1{}\mathrm {i}\,d\,B^3\,a^2}{8}\right )}+\frac {\sqrt {2}\,B^3\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{16\,\left (\frac {31\,d\,A^3\,a^2}{8}-\frac {29{}\mathrm {i}\,d\,A^2\,B\,a^2}{8}+\frac {3\,d\,A\,B^2\,a^2}{8}-\frac {1{}\mathrm {i}\,d\,B^3\,a^2}{8}\right )}+\frac {\sqrt {2}\,A\,B^2\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,3{}\mathrm {i}}{16\,\left (\frac {31\,d\,A^3\,a^2}{8}-\frac {29{}\mathrm {i}\,d\,A^2\,B\,a^2}{8}+\frac {3\,d\,A\,B^2\,a^2}{8}-\frac {1{}\mathrm {i}\,d\,B^3\,a^2}{8}\right )}+\frac {29\,\sqrt {2}\,A^2\,B\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{16\,\left (\frac {31\,d\,A^3\,a^2}{8}-\frac {29{}\mathrm {i}\,d\,A^2\,B\,a^2}{8}+\frac {3\,d\,A\,B^2\,a^2}{8}-\frac {1{}\mathrm {i}\,d\,B^3\,a^2}{8}\right )}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {-a^3}}{4\,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]

((A + B*1i)/(3*d) + ((3*A + B*1i)*(a + a*tan(c + d*x)*1i))/(2*a*d))/(a + a*tan(c + d*x)*1i)^(3/2) - (2*A*atanh
((31*A^3*d*(a + a*tan(c + d*x)*1i)^(1/2))/((a^3)^(1/2)*((31*A^3*d)/a + (A*B^2*d)/a + (A^2*B*d*2i)/a)) + (A*B^2
*d*(a + a*tan(c + d*x)*1i)^(1/2))/((a^3)^(1/2)*((31*A^3*d)/a + (A*B^2*d)/a + (A^2*B*d*2i)/a)) + (A^2*B*d*(a +
a*tan(c + d*x)*1i)^(1/2)*2i)/((a^3)^(1/2)*((31*A^3*d)/a + (A*B^2*d)/a + (A^2*B*d*2i)/a))))/(d*(a^3)^(1/2)) + (
2^(1/2)*atanh((2^(1/2)*A^3*d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*31i)/(16*((31*A^3*a^2*d)/8 - (B^3*a^2*
d*1i)/8 + (3*A*B^2*a^2*d)/8 - (A^2*B*a^2*d*29i)/8)) + (2^(1/2)*B^3*d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2
))/(16*((31*A^3*a^2*d)/8 - (B^3*a^2*d*1i)/8 + (3*A*B^2*a^2*d)/8 - (A^2*B*a^2*d*29i)/8)) + (2^(1/2)*A*B^2*d*(-a
^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*3i)/(16*((31*A^3*a^2*d)/8 - (B^3*a^2*d*1i)/8 + (3*A*B^2*a^2*d)/8 - (A^
2*B*a^2*d*29i)/8)) + (29*2^(1/2)*A^2*B*d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(16*((31*A^3*a^2*d)/8 - (
B^3*a^2*d*1i)/8 + (3*A*B^2*a^2*d)/8 - (A^2*B*a^2*d*29i)/8)))*(A*1i + B)*(-a^3)^(1/2))/(4*a^3*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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