\(\int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [760]

Optimal result

Integrand size = 43, antiderivative size = 140 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {8 a^3 (i A+B)}{f \sqrt {c-i c \tan (e+f x)}}-\frac {8 a^3 (i A+2 B) \sqrt {c-i c \tan (e+f x)}}{c f}+\frac {2 a^3 (i A+5 B) (c-i c \tan (e+f x))^{3/2}}{3 c^2 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{5/2}}{5 c^3 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 140, 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} \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 a^3 (5 B+i A) (c-i c \tan (e+f x))^{3/2}}{3 c^2 f}-\frac {8 a^3 (2 B+i A) \sqrt {c-i c \tan (e+f x)}}{c f}-\frac {8 a^3 (B+i A)}{f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a^3 B (c-i c \tan (e+f x))^{5/2}}{5 c^3 f} \]

[In]

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

[Out]

(-8*a^3*(I*A + B))/(f*Sqrt[c - I*c*Tan[e + f*x]]) - (8*a^3*(I*A + 2*B)*Sqrt[c - I*c*Tan[e + f*x]])/(c*f) + (2*
a^3*(I*A + 5*B)*(c - I*c*Tan[e + f*x])^(3/2))/(3*c^2*f) - (2*a^3*B*(c - I*c*Tan[e + f*x])^(5/2))/(5*c^3*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)^2 (A+B x)}{(c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a c) \text {Subst}\left (\int \left (\frac {4 a^2 (A-i B)}{(c-i c x)^{3/2}}-\frac {4 a^2 (A-2 i B)}{c \sqrt {c-i c x}}+\frac {a^2 (A-5 i B) \sqrt {c-i c x}}{c^2}+\frac {i a^2 B (c-i c x)^{3/2}}{c^3}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {8 a^3 (i A+B)}{f \sqrt {c-i c \tan (e+f x)}}-\frac {8 a^3 (i A+2 B) \sqrt {c-i c \tan (e+f x)}}{c f}+\frac {2 a^3 (i A+5 B) (c-i c \tan (e+f x))^{3/2}}{3 c^2 f}-\frac {2 a^3 B (c-i c \tan (e+f x))^{5/2}}{5 c^3 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.59 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i a^3 \left (115 A-158 i B+(-50 i A-79 B) \tan (e+f x)+(5 A-16 i B) \tan ^2(e+f x)+3 B \tan ^3(e+f x)\right )}{15 f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(((-2*I)/15)*a^3*(115*A - (158*I)*B + ((-50*I)*A - 79*B)*Tan[e + f*x] + (5*A - (16*I)*B)*Tan[e + f*x]^2 + 3*B*
Tan[e + f*x]^3))/(f*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {5 i B c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\frac {A c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+8 i \sqrt {c -i c \tan \left (f x +e \right )}\, B \,c^{2}-4 \sqrt {c -i c \tan \left (f x +e \right )}\, A \,c^{2}-\frac {4 c^{3} \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(135\)
default	\(\frac {2 i a^{3} \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {5 i B c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+\frac {A c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+8 i \sqrt {c -i c \tan \left (f x +e \right )}\, B \,c^{2}-4 \sqrt {c -i c \tan \left (f x +e \right )}\, A \,c^{2}-\frac {4 c^{3} \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f \,c^{3}}\)	\(135\)
parts	\(\frac {2 i a^{3} A c \left (-\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}\right )}{f}+\frac {a^{3} \left (3 i A +B \right ) \left (-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}\right )}{f}+\frac {2 B \,a^{3} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 \sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}-\frac {c^{3}}{2 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}\right )}{f \,c^{3}}-\frac {6 i a^{3} \left (-i B +A \right ) \left (\sqrt {c -i c \tan \left (f x +e \right )}+\frac {c}{2 \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}\right )}{f c}-\frac {2 a^{3} \left (i A +3 B \right ) \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+c \sqrt {c -i c \tan \left (f x +e \right )}+\frac {c^{2}}{2 \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}\right )}{f \,c^{2}}\)	\(423\)

[In]

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

[Out]

2*I/f*a^3/c^3*(1/5*I*B*(c-I*c*tan(f*x+e))^(5/2)-5/3*I*B*c*(c-I*c*tan(f*x+e))^(3/2)+1/3*A*c*(c-I*c*tan(f*x+e))^
(3/2)+8*I*(c-I*c*tan(f*x+e))^(1/2)*B*c^2-4*(c-I*c*tan(f*x+e))^(1/2)*A*c^2-4*c^3*(A-I*B)/(c-I*c*tan(f*x+e))^(1/
2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.91 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {4 \, \sqrt {2} {\left (15 \, {\left (i \, A + B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 15 \, {\left (5 i \, A + 7 \, B\right )} a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 20 \, {\left (5 i \, A + 7 \, B\right )} a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 8 \, {\left (5 i \, A + 7 \, B\right )} a^{3}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{15 \, {\left (c f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \]

[In]

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

[Out]

-4/15*sqrt(2)*(15*(I*A + B)*a^3*e^(6*I*f*x + 6*I*e) + 15*(5*I*A + 7*B)*a^3*e^(4*I*f*x + 4*I*e) + 20*(5*I*A + 7
*B)*a^3*e^(2*I*f*x + 2*I*e) + 8*(5*I*A + 7*B)*a^3)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c*f*e^(4*I*f*x + 4*I*e)
+ 2*c*f*e^(2*I*f*x + 2*I*e) + c*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 i \, {\left (\frac {60 \, {\left (A - i \, B\right )} a^{3} c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}} - \frac {3 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} B a^{3} + 5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (A - 5 i \, B\right )} a^{3} c - 60 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - 2 i \, B\right )} a^{3} c^{2}}{c^{2}}\right )}}{15 \, c f} \]

[In]

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

[Out]

-2/15*I*(60*(A - I*B)*a^3*c/sqrt(-I*c*tan(f*x + e) + c) - (3*I*(-I*c*tan(f*x + e) + c)^(5/2)*B*a^3 + 5*(-I*c*t
an(f*x + e) + c)^(3/2)*(A - 5*I*B)*a^3*c - 60*sqrt(-I*c*tan(f*x + e) + c)*(A - 2*I*B)*a^3*c^2)/c^2)/(c*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.89 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.51 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f}+\frac {a^3\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f}\right )-\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f}+\frac {a^3\,\left (A-B\,3{}\mathrm {i}\right )\,4{}\mathrm {i}}{c\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}-\frac {\left (\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{5\,c\,f}-\frac {a^3\,\left (A+B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{5\,c\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\left (\frac {16\,B\,a^3}{3\,c\,f}+\frac {a^3\,\left (A-B\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{3\,c\,f}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1} \]

[In]

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

[Out]

(((a^3*(A - B*1i)*4i)/(3*c*f) + (16*B*a^3)/(3*c*f))*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2
i) + 1))^(1/2))/(exp(e*2i + f*x*2i) + 1) - ((a^3*(A - B*1i)*4i)/(c*f) + (a^3*(A - B*3i)*4i)/(c*f))*(c + (c*(ex
p(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2) - (((a^3*(A - B*1i)*4i)/(5*c*f) - (a^3*(A + B*1i
)*4i)/(5*c*f))*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2))/(exp(e*2i + f*x*2i) +
1)^2 - (c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2)*((a^3*(A - B*1i)*4i)/(c*f) +
(a^3*exp(e*2i + f*x*2i)*(A - B*1i)*4i)/(c*f))