\(\int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx\) [750]

Optimal result

Integrand size = 43, antiderivative size = 105 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {4 a^2 (i A+B) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (i A+3 B) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{7/2}}{7 c^2 f} \]

[Out]

4/3*a^2*(I*A+B)*(c-I*c*tan(f*x+e))^(3/2)/f-2/5*a^2*(I*A+3*B)*(c-I*c*tan(f*x+e))^(5/2)/c/f+2/7*a^2*B*(c-I*c*tan
(f*x+e))^(7/2)/c^2/f

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 105, 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} \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {2 a^2 (3 B+i A) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {4 a^2 (B+i A) (c-i c \tan (e+f x))^{3/2}}{3 f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{7/2}}{7 c^2 f} \]

[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)*(c - I*c*Tan[e + f*x])^(3/2))/(3*f) - (2*a^2*(I*A + 3*B)*(c - I*c*Tan[e + f*x])^(5/2))/(5*c*f
) + (2*a^2*B*(c - I*c*Tan[e + f*x])^(7/2))/(7*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 (a+i a x) (A+B x) \sqrt {c-i c x} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (2 a (A-i B) \sqrt {c-i c x}-\frac {a (A-3 i B) (c-i c x)^{3/2}}{c}-\frac {i a B (c-i c x)^{5/2}}{c^2}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {4 a^2 (i A+B) (c-i c \tan (e+f x))^{3/2}}{3 f}-\frac {2 a^2 (i A+3 B) (c-i c \tan (e+f x))^{5/2}}{5 c f}+\frac {2 a^2 B (c-i c \tan (e+f x))^{7/2}}{7 c^2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.24 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {a^2 c^2 \sec ^4(e+f x) (\cos (2 (e+f x))-i \sin (2 (e+f x))) (7 (7 i A+B)+(49 i A+37 B) \cos (2 (e+f x))+(-21 A+33 i B) \sin (2 (e+f x)))}{105 f \sqrt {c-i c \tan (e+f x)}} \]

[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]

(a^2*c^2*Sec[e + f*x]^4*(Cos[2*(e + f*x)] - I*Sin[2*(e + f*x)])*(7*((7*I)*A + B) + ((49*I)*A + 37*B)*Cos[2*(e
+ f*x)] + (-21*A + (33*I)*B)*Sin[2*(e + f*x)]))/(105*f*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(\frac {2 i a^{2} \left (-\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {\left (3 i B c -c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (-i B c +c A \right ) c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f \,c^{2}}\)	\(84\)
default	\(\frac {2 i a^{2} \left (-\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {\left (3 i B c -c A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 \left (-i B c +c A \right ) c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}\right )}{f \,c^{2}}\)	\(84\)
parts	\(\frac {2 i A \,a^{2} c \left (-\sqrt {c -i c \tan \left (f x +e \right )}+\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}+\frac {a^{2} \left (2 i A +B \right ) \left (\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 c \sqrt {c -i c \tan \left (f x +e \right )}-2 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 B \,a^{2} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}} c^{2}}{3}-\sqrt {c -i c \tan \left (f x +e \right )}\, c^{3}+c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f \,c^{2}}-\frac {2 i a^{2} \left (-2 i B +A \right ) \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}-c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f c}\)	\(336\)

[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*(-1/7*I*B*(c-I*c*tan(f*x+e))^(7/2)+1/5*(3*I*B*c-c*A)*(c-I*c*tan(f*x+e))^(5/2)+2/3*(-I*B*c+c*A)*c
*(c-I*c*tan(f*x+e))^(3/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.13 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {8 \, \sqrt {2} {\left (35 \, {\left (-i \, A - B\right )} a^{2} c e^{\left (4 i \, f x + 4 i \, e\right )} + 7 \, {\left (-7 i \, A - B\right )} a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + 2 \, {\left (-7 i \, A - B\right )} a^{2} c\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{105 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[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, algorithm="fricas")

[Out]

-8/105*sqrt(2)*(35*(-I*A - B)*a^2*c*e^(4*I*f*x + 4*I*e) + 7*(-7*I*A - B)*a^2*c*e^(2*I*f*x + 2*I*e) + 2*(-7*I*A
- B)*a^2*c)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f
*x + 2*I*e) + f)

Sympy [F]

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

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.74 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {2 i \, {\left (15 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} B a^{2} + 21 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (A - 3 i \, B\right )} a^{2} c - 70 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a^{2} c^{2}\right )}}{105 \, c^{2} f} \]

[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="maxima")

[Out]

-2/105*I*(15*I*(-I*c*tan(f*x + e) + c)^(7/2)*B*a^2 + 21*(-I*c*tan(f*x + e) + c)^(5/2)*(A - 3*I*B)*a^2*c - 70*(
-I*c*tan(f*x + e) + c)^(3/2)*(A - I*B)*a^2*c^2)/(c^2*f)

Giac [F]

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

[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 = 11.70 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.24 \[ \int (a+i a \tan (e+f x))^2 (A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2} \, dx=\frac {8\,a^2\,c\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (A\,14{}\mathrm {i}+2\,B+A\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,49{}\mathrm {i}+A\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,35{}\mathrm {i}+7\,B\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+35\,B\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\right )}{105\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3} \]

[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]

(8*a^2*c*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2)*(A*14i + 2*B + A*exp(e*2i +
f*x*2i)*49i + A*exp(e*4i + f*x*4i)*35i + 7*B*exp(e*2i + f*x*2i) + 35*B*exp(e*4i + f*x*4i)))/(105*f*(exp(e*2i +
f*x*2i) + 1)^3)