\(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx\) [782]

Optimal result

Integrand size = 43, antiderivative size = 211 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {(i A+3 B) c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}+\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}-\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[Out]

-1/64*(I*A+3*B)*c^(3/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/f*2^(1/2)+1/8*(I*A+3*B)*c*(c
-I*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e))^2-1/32*(I*A+3*B)*c*(c-I*c*tan(f*x+e))^(1/2)/a^3/f/(1+I*tan(f*x+e
))+1/6*(I*A-B)*(c-I*c*tan(f*x+e))^(3/2)/a^3/f/(1+I*tan(f*x+e))^3

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3669, 79, 43, 44, 65, 214} \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=-\frac {c^{3/2} (3 B+i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}-\frac {c (3 B+i A) \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {c (3 B+i A) \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

-1/32*((I*A + 3*B)*c^(3/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(Sqrt[2]*a^3*f) + ((I*A + 3*
B)*c*Sqrt[c - I*c*Tan[e + f*x]])/(8*a^3*f*(1 + I*Tan[e + f*x])^2) - ((I*A + 3*B)*c*Sqrt[c - I*c*Tan[e + f*x]])
/(32*a^3*f*(1 + I*Tan[e + f*x])) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(3/2))/(6*a^3*f*(1 + I*Tan[e + f*x])^3)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

Rule 214

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

Rule 3669

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*(c/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {(A+B x) \sqrt {c-i c x}}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac {((A-3 i B) c) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = \frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left ((A-3 i B) c^2\right ) \text {Subst}\left (\int \frac {1}{(a+i a x)^2 \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{16 a f} \\ & = \frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}-\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {\left ((A-3 i B) c^2\right ) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{64 a^2 f} \\ & = \frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}-\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac {((i A+3 B) c) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{32 a^2 f} \\ & = -\frac {(i A+3 B) c^{3/2} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{32 \sqrt {2} a^3 f}+\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{8 a^3 f (1+i \tan (e+f x))^2}-\frac {(i A+3 B) c \sqrt {c-i c \tan (e+f x)}}{32 a^3 f (1+i \tan (e+f x))}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{6 a^3 f (1+i \tan (e+f x))^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.44 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {c \sec ^3(e+f x) \left (3 \sqrt {2} (A-3 i B) \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right ) (\cos (3 (e+f x))+i \sin (3 (e+f x)))-2 \cos (e+f x) (2 (7 A-5 i B)+(11 A-i B) \cos (2 (e+f x))+(-5 i A+17 B) \sin (2 (e+f x))) \sqrt {c-i c \tan (e+f x)}\right )}{192 a^3 f (-i+\tan (e+f x))^3} \]

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(c*Sec[e + f*x]^3*(3*Sqrt[2]*(A - (3*I)*B)*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*(Cos[
3*(e + f*x)] + I*Sin[3*(e + f*x)]) - 2*Cos[e + f*x]*(2*(7*A - (5*I)*B) + (11*A - I*B)*Cos[2*(e + f*x)] + ((-5*
I)*A + 17*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]]))/(192*a^3*f*(-I + Tan[e + f*x])^3)

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(\frac {2 i c^{3} \left (\frac {-\frac {\left (-3 i B +A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{64 c}+8 \left (\frac {i B}{96}+\frac {A}{96}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\frac {c \left (-3 i B +A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{16}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}-\frac {\left (-3 i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{128 c^{\frac {3}{2}}}\right )}{f \,a^{3}}\)	\(139\)
default	\(\frac {2 i c^{3} \left (\frac {-\frac {\left (-3 i B +A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{64 c}+8 \left (\frac {i B}{96}+\frac {A}{96}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+\frac {c \left (-3 i B +A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{16}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}-\frac {\left (-3 i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{128 c^{\frac {3}{2}}}\right )}{f \,a^{3}}\)	\(139\)

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

2*I/f/a^3*c^3*(8*(-1/512/c*(A-3*I*B)*(c-I*c*tan(f*x+e))^(5/2)+(1/96*I*B+1/96*A)*(c-I*c*tan(f*x+e))^(3/2)+1/128
*c*(A-3*I*B)*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^3-1/128/c^(3/2)*(A-3*I*B)*2^(1/2)*arctanh(1/2*(c-I*c
*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 395 vs. \(2 (164) = 328\).

Time = 0.26 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.87 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c^{3}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left ({\left (-i \, A - 3 \, B\right )} c^{2} + \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c^{3}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{3} f}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c^{3}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {{\left ({\left (-i \, A - 3 \, B\right )} c^{2} - \sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 6 i \, A B - 9 \, B^{2}\right )} c^{3}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{16 \, a^{3} f}\right ) - \sqrt {2} {\left (3 \, {\left (-i \, A - 3 \, B\right )} c e^{\left (6 i \, f x + 6 i \, e\right )} - {\left (17 i \, A + 19 \, B\right )} c e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-11 i \, A - B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (-i \, A + B\right )} c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{192 \, a^{3} f} \]

[In]

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

[Out]

1/192*(3*sqrt(1/2)*a^3*f*sqrt(-(A^2 - 6*I*A*B - 9*B^2)*c^3/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16*((-I*A - 3*
B)*c^2 + sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(A^2 -
6*I*A*B - 9*B^2)*c^3/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f)) - 3*sqrt(1/2)*a^3*f*sqrt(-(A^2 - 6*I*A*B - 9*B^2)*c
^3/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/16*((-I*A - 3*B)*c^2 - sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) +
a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(A^2 - 6*I*A*B - 9*B^2)*c^3/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f
)) - sqrt(2)*(3*(-I*A - 3*B)*c*e^(6*I*f*x + 6*I*e) - (17*I*A + 19*B)*c*e^(4*I*f*x + 4*I*e) + 2*(-11*I*A - B)*c
*e^(2*I*f*x + 2*I*e) + 8*(-I*A + B)*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)

Sympy [F]

\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \left (\int \frac {A c \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx + \int \frac {B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\, dx + \int \left (- \frac {i A c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\right )\, dx + \int \left (- \frac {i B c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}}{\tan ^{3}{\left (e + f x \right )} - 3 i \tan ^{2}{\left (e + f x \right )} - 3 \tan {\left (e + f x \right )} + i}\right )\, dx\right )}{a^{3}} \]

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**3,x)

[Out]

I*(Integral(A*c*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x) + I), x) +
Integral(B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 - 3*tan(e + f*x)
+ I), x) + Integral(-I*A*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**3 - 3*I*tan(e + f*x)**2 -
3*tan(e + f*x) + I), x) + Integral(-I*B*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2/(tan(e + f*x)**3 - 3*I*t
an(e + f*x)**2 - 3*tan(e + f*x) + I), x))/a**3

Maxima [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (A - 3 i \, B\right )} c^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (A - 3 i \, B\right )} c^{3} - 16 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A + i \, B\right )} c^{4} - 12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - 3 i \, B\right )} c^{5}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{3} c^{2} - 8 \, a^{3} c^{3}}\right )}}{384 \, c f} \]

[In]

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

[Out]

1/384*I*(3*sqrt(2)*(A - 3*I*B)*c^(5/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) +
sqrt(-I*c*tan(f*x + e) + c)))/a^3 + 4*(3*(-I*c*tan(f*x + e) + c)^(5/2)*(A - 3*I*B)*c^3 - 16*(-I*c*tan(f*x + e
) + c)^(3/2)*(A + I*B)*c^4 - 12*sqrt(-I*c*tan(f*x + e) + c)*(A - 3*I*B)*c^5)/((-I*c*tan(f*x + e) + c)^3*a^3 -
6*(-I*c*tan(f*x + e) + c)^2*a^3*c + 12*(-I*c*tan(f*x + e) + c)*a^3*c^2 - 8*a^3*c^3))/(c*f)

Giac [F]

\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}} \,d x } \]

[In]

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

[Out]

integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(3/2)/(I*a*tan(f*x + e) + a)^3, x)

Mupad [B] (verification not implemented)

Time = 9.10 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.71 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^3} \, dx=\frac {\frac {A\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8\,a^3\,f}+\frac {A\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,1{}\mathrm {i}}{6\,a^3\,f}-\frac {A\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}}{32\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+8\,c^3}-\frac {-\frac {3\,B\,c^4\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{8}+\frac {B\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{6}+\frac {3\,B\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{32}}{8\,a^3\,c^3\,f-a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3+6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-12\,a^3\,c^2\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\sqrt {2}\,A\,{\left (-c\right )}^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,1{}\mathrm {i}}{64\,a^3\,f}-\frac {3\,\sqrt {2}\,B\,c^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{64\,a^3\,f} \]

[In]

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

[Out]

((A*c^4*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(8*a^3*f) + (A*c^3*(c - c*tan(e + f*x)*1i)^(3/2)*1i)/(6*a^3*f) - (A*
c^2*(c - c*tan(e + f*x)*1i)^(5/2)*1i)/(32*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^2 - 12*c^2*(c - c*tan(e + f*x)*
1i) - (c - c*tan(e + f*x)*1i)^3 + 8*c^3) - ((B*c^3*(c - c*tan(e + f*x)*1i)^(3/2))/6 - (3*B*c^4*(c - c*tan(e +
f*x)*1i)^(1/2))/8 + (3*B*c^2*(c - c*tan(e + f*x)*1i)^(5/2))/32)/(8*a^3*c^3*f - a^3*f*(c - c*tan(e + f*x)*1i)^3
+ 6*a^3*c*f*(c - c*tan(e + f*x)*1i)^2 - 12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)) + (2^(1/2)*A*(-c)^(3/2)*atan((2
^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*1i)/(64*a^3*f) - (3*2^(1/2)*B*c^(3/2)*atanh((2^(1/2)*(c
- c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(64*a^3*f)