\(\int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [753]

Optimal result

Integrand size = 43, antiderivative size = 101 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {4 a^2 (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^2 (i A+3 B)}{c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a^2 B \sqrt {c-i c \tan (e+f x)}}{c^2 f} \]

[Out]

2*a^2*(I*A+3*B)/c/f/(c-I*c*tan(f*x+e))^(1/2)+2*a^2*B*(c-I*c*tan(f*x+e))^(1/2)/c^2/f-4/3*a^2*(I*A+B)/f/(c-I*c*t
an(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {3669, 78} \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 a^2 (3 B+i A)}{c f \sqrt {c-i c \tan (e+f x)}}-\frac {4 a^2 (B+i A)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^2 B \sqrt {c-i c \tan (e+f x)}}{c^2 f} \]

[In]

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

[Out]

(-4*a^2*(I*A + B))/(3*f*(c - I*c*Tan[e + f*x])^(3/2)) + (2*a^2*(I*A + 3*B))/(c*f*Sqrt[c - I*c*Tan[e + f*x]]) +
(2*a^2*B*Sqrt[c - I*c*Tan[e + f*x]])/(c^2*f)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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+i a x) (A+B x)}{(c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {2 a (A-i B)}{(c-i c x)^{5/2}}-\frac {a (A-3 i B)}{c (c-i c x)^{3/2}}-\frac {i a B}{c^2 \sqrt {c-i c x}}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {4 a^2 (i A+B)}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {2 a^2 (i A+3 B)}{c f \sqrt {c-i c \tan (e+f x)}}+\frac {2 a^2 B \sqrt {c-i c \tan (e+f x)}}{c^2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.54 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {2 a^2 \left (-A+10 i B+3 (i A+5 B) \tan (e+f x)-3 i B \tan ^2(e+f x)\right )}{3 c f (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(2*a^2*(-A + (10*I)*B + 3*(I*A + 5*B)*Tan[e + f*x] - (3*I)*B*Tan[e + f*x]^2))/(3*c*f*(I + Tan[e + f*x])*Sqrt[c
- I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (-i \sqrt {c -i c \tan \left (f x +e \right )}\, B +\frac {c \left (-3 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {2 c^{2} \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,c^{2}}\)	\(79\)
default	\(\frac {2 i a^{2} \left (-i \sqrt {c -i c \tan \left (f x +e \right )}\, B +\frac {c \left (-3 i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {2 c^{2} \left (-i B +A \right )}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,c^{2}}\)	\(79\)
parts	\(\frac {2 i A \,a^{2} c \left (-\frac {1}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{6 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 c^{\frac {5}{2}}}\right )}{f}+\frac {a^{2} \left (2 i A +B \right ) \left (-\frac {1}{3 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}\right )}{f}-\frac {2 B \,a^{2} \left (-\sqrt {c -i c \tan \left (f x +e \right )}-\frac {5 c}{4 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c^{2}}{6 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8}\right )}{f \,c^{2}}-\frac {2 i a^{2} \left (-2 i B +A \right ) \left (-\frac {3}{4 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c}{6 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )}{f c}\)	\(342\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.80 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {2} {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (i \, A + 7 \, B\right )} a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 \, {\left (-i \, A - 7 \, B\right )} a^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, c^{2} f} \]

[In]

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

[Out]

1/3*sqrt(2)*((-I*A - B)*a^2*e^(4*I*f*x + 4*I*e) + (I*A + 7*B)*a^2*e^(2*I*f*x + 2*I*e) - 2*(-I*A - 7*B)*a^2)*sq
rt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^2*f)

Sympy [F]

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

[In]

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

[Out]

-a**2*(Integral(-A/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + Integ
ral(A*tan(e + f*x)**2/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + In
tegral(-B*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x) + I
ntegral(B*tan(e + f*x)**3/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c)), x)
+ Integral(-2*I*A*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*x) + c))
, x) + Integral(-2*I*B*tan(e + f*x)**2/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x) + c*sqrt(-I*c*tan(e + f*
x) + c)), x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {2 i \, {\left (\frac {3 i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B a^{2}}{c} - \frac {3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (A - 3 i \, B\right )} a^{2} - 2 \, {\left (A - i \, B\right )} a^{2} c}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\right )}}{3 \, c f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

Mupad [B] (verification not implemented)

Time = 9.01 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.56 \[ \int \frac {(a+i a \tan (e+f x))^2 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {a^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,2{}\mathrm {i}+14\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+7\,B\,\cos \left (2\,e+2\,f\,x\right )-B\,\cos \left (4\,e+4\,f\,x\right )-A\,\sin \left (2\,e+2\,f\,x\right )+A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,7{}\mathrm {i}-B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{3\,c^2\,f} \]

[In]

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

[Out]

(a^2*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*2i + 14*B + A*cos(2*e
+ 2*f*x)*1i - A*cos(4*e + 4*f*x)*1i + 7*B*cos(2*e + 2*f*x) - B*cos(4*e + 4*f*x) - A*sin(2*e + 2*f*x) + A*sin(4
*e + 4*f*x) + B*sin(2*e + 2*f*x)*7i - B*sin(4*e + 4*f*x)*1i))/(3*c^2*f)