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

Optimal. Leaf size=148 \[ \frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(7 A+i B) \sqrt {\tan (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

(1/4-1/4*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(3/2)/d+1/6*(7*A+I*B)*t
an(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)+1/3*(A+I*B)*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.38, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3596, 12, 3544, 205} \[ \frac {\left (\frac {1}{4}-\frac {i}{4}\right ) (A-i B) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(7 A+i B) \sqrt {\tan (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/4 - I/4)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) +
((A + I*B)*Sqrt[Tan[c + d*x]])/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((7*A + I*B)*Sqrt[Tan[c + d*x]])/(6*a*d*S
qrt[a + I*a*Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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]

Rubi steps

\begin {align*} \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{2} a (5 A-i B)-a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(7 A+i B) \sqrt {\tan (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {3 a^2 (A-i B) \sqrt {a+i a \tan (c+d x)}}{4 \sqrt {\tan (c+d x)}} \, dx}{3 a^4}\\ &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(7 A+i B) \sqrt {\tan (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac {(A+i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(7 A+i B) \sqrt {\tan (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {(i A+B) \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 )}{2 d}\\ &=-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) (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 )}{a^{3/2} d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(7 A+i B) \sqrt {\tan (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 5.69, size = 230, normalized size = 1.55 \[ \frac {e^{-i (c+d x)} \sqrt {\tan (c+d x)} \sec ^{\frac {3}{2}}(c+d x) \left (\sqrt {-1+e^{2 i (c+d x)}} \left (-i A \left (1+8 e^{2 i (c+d x)}\right )+2 B e^{2 i (c+d x)}+B\right )-3 i (A-i B) e^{3 i (c+d x)} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right )}{6 \sqrt {2} a d \sqrt {-1+e^{2 i (c+d x)}} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[-1 + E^((2*I)*(c + d*x))]*(B + 2*B*E^((2*I)*(c + d*x)) - I*A*(1 + 8*E^((2*I)*(c + d*x)))) - (3*I)*(A -
I*B)*E^((3*I)*(c + d*x))*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sec[c + d*x]^(3/2)*Sqrt[Tan[
c + d*x]])/(6*Sqrt[2]*a*d*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c
+ d*x)))]*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.80, size = 441, normalized size = 2.98 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {2 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {-2 i \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + \sqrt {2} {\left (2 \, {\left (4 \, A + i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (3 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/12*(3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log((2*I*sqrt(1/2)*a^2*d*
sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*s
qrt(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)))/(4*I*A + 4*B))
- 3*sqrt(1/2)*a^2*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log((-2*I*sqrt(1/2)*a^2*d*sqr
t((-I*A^2 - 2*A*B + I*B^2)/(a^3*d^2))*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) + s
qrt(2)*(2*(4*A + I*B)*e^(4*I*d*x + 4*I*c) + 3*(3*A + I*B)*e^(2*I*d*x + 2*I*c) + A + I*B)*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^(-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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sqrt {\tan \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.35, size = 868, normalized size = 5.86 \[ \frac {\left (\sqrt {\tan }\left (d x +c \right )\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (3 i A \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a -9 i B \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a +3 B \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{3}\left (d x +c \right )\right ) a -9 i A \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right ) a +28 i A \left (\tan ^{2}\left (d x +c \right )\right ) \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+9 A \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \left (\tan ^{2}\left (d x +c \right )\right ) a +3 i B \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +16 i B \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-9 B \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \tan \left (d x +c \right ) a -4 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{2}\left (d x +c \right )\right )-36 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-3 A \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +64 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+12 B \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{24 d \,a^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(3*I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(
1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a-9*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)
^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^2*a+3*B*2^(1/2)*ln
(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c
)^3*a-9*I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(ta
n(d*x+c)+I))*tan(d*x+c)*a+28*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+9*A*2^(1/2)*l
n(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+
c)^2*a+3*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(t
an(d*x+c)+I))*a+16*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)-9*B*2^(1/2)*ln(-(-2*2^(1/
2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)*a-4*B*(a*
tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-36*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*
a)^(1/2)-3*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(t
an(d*x+c)+I))*a+64*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+12*B*(a*tan(d*x+c)*(1+I*tan
(d*x+c)))^(1/2)*(-I*a)^(1/2))/a^2/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a)^(1/2)/(-tan(d*x+c)+I)^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/tan(d*x+c)**(1/2)/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

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

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