\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx\) [785]

Optimal result

Integrand size = 43, antiderivative size = 274 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {35 (3 i A+B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 (3 i A+B)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

35/512*(3*I*A+B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a^3/c^(3/2)/f*2^(1/2)-35/256*(3*I*A+B)/
a^3/c/f/(c-I*c*tan(f*x+e))^(1/2)-35/384*(3*I*A+B)/a^3/f/(c-I*c*tan(f*x+e))^(3/2)+1/6*(I*A-B)/a^3/f/(1+I*tan(f*
x+e))^3/(c-I*c*tan(f*x+e))^(3/2)+1/16*(3*I*A+B)/a^3/f/(1+I*tan(f*x+e))^2/(c-I*c*tan(f*x+e))^(3/2)+7/64*(3*I*A+
B)/a^3/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {3669, 79, 44, 53, 65, 214} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {35 (B+3 i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}+\frac {-B+i A}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}-\frac {35 (B+3 i A)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {35 (B+3 i A)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {7 (B+3 i A)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {B+3 i A}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}} \]

[In]

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

[Out]

(35*((3*I)*A + B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(256*Sqrt[2]*a^3*c^(3/2)*f) - (35*((3
*I)*A + B))/(384*a^3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(6*a^3*f*(1 + I*Tan[e + f*x])^3*(c - I*c*Tan[
e + f*x])^(3/2)) + ((3*I)*A + B)/(16*a^3*f*(1 + I*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(3/2)) + (7*((3*I)*A
+ B))/(64*a^3*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) - (35*((3*I)*A + B))/(256*a^3*c*f*Sqrt[c -
I*c*Tan[e + f*x]])

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 53

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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}{(a+i a x)^4 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {((3 A-i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x)^3 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = \frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {(7 (3 A-i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x)^2 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{32 a f} \\ & = \frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(35 (3 A-i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{128 a^2 f} \\ & = -\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(35 (3 A-i B)) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{256 a^2 f} \\ & = -\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 (3 i A+B)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(35 (3 A-i B)) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{512 a^2 c f} \\ & = -\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 (3 i A+B)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {(35 (3 i A+B)) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{256 a^2 c^2 f} \\ & = \frac {35 (3 i A+B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {35 (3 i A+B)}{384 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i A+B}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {7 (3 i A+B)}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {35 (3 i A+B)}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 6.14 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.70 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sec ^4(e+f x) \left (105 (3 i A+B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+2 \cos (e+f x) (3 (-55 i A+3 B) \cos (e+f x)+8 (i A+3 B) \cos (3 (e+f x))+(3 A-i B) (27 \sin (e+f x)-8 \sin (3 (e+f x))))\right )}{768 a^3 c f (-i+\tan (e+f x))^3 (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(Sec[e + f*x]^4*(105*((3*I)*A + B)*Hypergeometric2F1[-1/2, 1, 1/2, (-1/2*I)*(I + Tan[e + f*x])]*(Cos[2*(e + f*
x)] + I*Sin[2*(e + f*x)]) + 2*Cos[e + f*x]*(3*((-55*I)*A + 3*B)*Cos[e + f*x] + 8*(I*A + 3*B)*Cos[3*(e + f*x)]
+ (3*A - I*B)*(27*Sin[e + f*x] - 8*Sin[3*(e + f*x)]))))/(768*a^3*c*f*(-I + Tan[e + f*x])^3*(I + Tan[e + f*x])*
Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {2 i c^{3} \left (\frac {\frac {8 \left (-\frac {3 i B}{256}+\frac {41 A}{256}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {1}{48} i B c -\frac {35}{48} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (\frac {3}{64} i B \,c^{2}+\frac {55}{64} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {35 \left (\frac {3 A}{8}-\frac {i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}}{16 c^{4}}-\frac {-i B +2 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{3}}\)	\(205\)
default	\(\frac {2 i c^{3} \left (\frac {\frac {8 \left (-\frac {3 i B}{256}+\frac {41 A}{256}\right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}+8 \left (\frac {1}{48} i B c -\frac {35}{48} c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}+8 \left (\frac {3}{64} i B \,c^{2}+\frac {55}{64} c^{2} A \right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {35 \left (\frac {3 A}{8}-\frac {i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}}{16 c^{4}}-\frac {-i B +2 A}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{48 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{3}}\)	\(205\)

[In]

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

[Out]

2*I/f/a^3*c^3*(1/16/c^4*(8*((-3/256*I*B+41/256*A)*(c-I*c*tan(f*x+e))^(5/2)+(1/48*I*B*c-35/48*c*A)*(c-I*c*tan(f
*x+e))^(3/2)+(3/64*I*B*c^2+55/64*c^2*A)*(c-I*c*tan(f*x+e))^(1/2))/(c+I*c*tan(f*x+e))^3+35/8*(3/8*A-1/8*I*B)*2^
(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-1/16/c^4*(2*A-I*B)/(c-I*c*tan(f*x+e))^(1/
2)-1/48/c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(3/2))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.60 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (105 \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{6} c^{3} f^{2}}} + 3 i \, A + B\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) - 105 \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {35 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {9 \, A^{2} - 6 i \, A B - B^{2}}{a^{6} c^{3} f^{2}}} - 3 i \, A - B\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) - \sqrt {2} {\left (16 \, {\left (i \, A + B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + 32 \, {\left (7 i \, A + 4 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - {\left (-43 i \, A - 121 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 5 \, {\left (-43 i \, A + 7 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-29 i \, A + 17 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, A + 8 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} c^{2} f} \]

[In]

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

[Out]

1/1536*(105*sqrt(1/2)*a^3*c^2*f*sqrt(-(9*A^2 - 6*I*A*B - B^2)/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log(35/128*(s
qrt(2)*sqrt(1/2)*(a^3*c*f*e^(2*I*f*x + 2*I*e) + a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 6*I*
A*B - B^2)/(a^6*c^3*f^2)) + 3*I*A + B)*e^(-I*f*x - I*e)/(a^3*c*f)) - 105*sqrt(1/2)*a^3*c^2*f*sqrt(-(9*A^2 - 6*
I*A*B - B^2)/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I*e)*log(-35/128*(sqrt(2)*sqrt(1/2)*(a^3*c*f*e^(2*I*f*x + 2*I*e) +
a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 6*I*A*B - B^2)/(a^6*c^3*f^2)) - 3*I*A - B)*e^(-I*f*x
- I*e)/(a^3*c*f)) - sqrt(2)*(16*(I*A + B)*e^(10*I*f*x + 10*I*e) + 32*(7*I*A + 4*B)*e^(8*I*f*x + 8*I*e) - (-43
*I*A - 121*B)*e^(6*I*f*x + 6*I*e) + 5*(-43*I*A + 7*B)*e^(4*I*f*x + 4*I*e) + 2*(-29*I*A + 17*B)*e^(2*I*f*x + 2*
I*e) - 8*I*A + 8*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e)/(a^3*c^2*f)

Sympy [F]

\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {i \left (\int \frac {A}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx + \int \frac {B \tan {\left (e + f x \right )}}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx\right )}{a^{3}} \]

[In]

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

[Out]

I*(Integral(A/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4 - 2*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)
**3 - 2*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*c*sqrt(-I*c*tan(e + f*x) + c)), x) + Integral(B*tan(e +
f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4 - 2*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 2*
c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + I*c*sqrt(-I*c*tan(e + f*x) + c)), x))/a**3

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.96 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} {\left (3 \, A - i \, B\right )} - 560 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} {\left (3 \, A - i \, B\right )} c + 924 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (3 \, A - i \, B\right )} c^{2} - 384 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (3 \, A - i \, B\right )} c^{3} - 256 \, {\left (A - i \, B\right )} c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{2} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{3}} + \frac {105 \, \sqrt {2} {\left (3 \, A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} \sqrt {c}}\right )}}{3072 \, c f} \]

[In]

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

[Out]

-1/3072*I*(4*(105*(-I*c*tan(f*x + e) + c)^4*(3*A - I*B) - 560*(-I*c*tan(f*x + e) + c)^3*(3*A - I*B)*c + 924*(-
I*c*tan(f*x + e) + c)^2*(3*A - I*B)*c^2 - 384*(-I*c*tan(f*x + e) + c)*(3*A - I*B)*c^3 - 256*(A - I*B)*c^4)/((-
I*c*tan(f*x + e) + c)^(9/2)*a^3 - 6*(-I*c*tan(f*x + e) + c)^(7/2)*a^3*c + 12*(-I*c*tan(f*x + e) + c)^(5/2)*a^3
*c^2 - 8*(-I*c*tan(f*x + e) + c)^(3/2)*a^3*c^3) + 105*sqrt(2)*(3*A - I*B)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*ta
n(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan(f*x + e) + c)))/(a^3*sqrt(c)))/(c*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.49 (sec) , antiderivative size = 443, normalized size of antiderivative = 1.62 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{16\,a^3\,f}+\frac {A\,c^3\,1{}\mathrm {i}}{3\,a^3\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,105{}\mathrm {i}}{256\,a^3\,c\,f}-\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,231{}\mathrm {i}}{64\,a^3\,f}+\frac {A\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\frac {B\,c^3}{3}+\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3}{48}-\frac {77\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{64}+\frac {B\,c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{2}-\frac {35\,B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4}{256\,c}}{a^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}-6\,a^3\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-8\,a^3\,c^3\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}+12\,a^3\,c^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,105{}\mathrm {i}}{512\,a^3\,{\left (-c\right )}^{3/2}\,f}+\frac {35\,\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{512\,a^3\,c^{3/2}\,f} \]

[In]

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

[Out]

((B*c^3)/3 + (35*B*(c - c*tan(e + f*x)*1i)^3)/48 - (77*B*c*(c - c*tan(e + f*x)*1i)^2)/64 + (B*c^2*(c - c*tan(e
+ f*x)*1i))/2 - (35*B*(c - c*tan(e + f*x)*1i)^4)/(256*c))/(a^3*f*(c - c*tan(e + f*x)*1i)^(9/2) - 6*a^3*c*f*(c
- c*tan(e + f*x)*1i)^(7/2) - 8*a^3*c^3*f*(c - c*tan(e + f*x)*1i)^(3/2) + 12*a^3*c^2*f*(c - c*tan(e + f*x)*1i)
^(5/2)) - ((A*(c - c*tan(e + f*x)*1i)^3*35i)/(16*a^3*f) + (A*c^3*1i)/(3*a^3*f) - (A*(c - c*tan(e + f*x)*1i)^4*
105i)/(256*a^3*c*f) - (A*c*(c - c*tan(e + f*x)*1i)^2*231i)/(64*a^3*f) + (A*c^2*(c - c*tan(e + f*x)*1i)*3i)/(2*
a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^(7/2) - (c - c*tan(e + f*x)*1i)^(9/2) + 8*c^3*(c - c*tan(e + f*x)*1i)^(3/
2) - 12*c^2*(c - c*tan(e + f*x)*1i)^(5/2)) + (2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(
1/2)))*105i)/(512*a^3*(-c)^(3/2)*f) + (35*2^(1/2)*B*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2)))
)/(512*a^3*c^(3/2)*f)