\(\int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [744]

Optimal result

Integrand size = 41, antiderivative size = 58 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 a (i A+B)}{f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a B \sqrt {c-i c \tan (e+f x)}}{c f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.049, Rules used = {3669, 45} \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {2 a (B+i A)}{f \sqrt {c-i c \tan (e+f x)}}-\frac {2 a B \sqrt {c-i c \tan (e+f x)}}{c f} \]

[In]

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

[Out]

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

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.72 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i a (-A+2 i B+B \tan (e+f x))}{f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {2 i a \left (i \sqrt {c -i c \tan \left (f x +e \right )}\, B -\frac {c \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\)	\(53\)
default	\(\frac {2 i a \left (i \sqrt {c -i c \tan \left (f x +e \right )}\, B -\frac {c \left (-i B +A \right )}{\sqrt {c -i c \tan \left (f x +e \right )}}\right )}{f c}\)	\(53\)
parts	\(\frac {2 i A a c \left (-\frac {1}{2 c \sqrt {c -i c \tan \left (f x +e \right )}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 c^{\frac {3}{2}}}\right )}{f}+\frac {a \left (i A +B \right ) \left (-\frac {1}{\sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2 \sqrt {c}}\right )}{f}-\frac {2 a B \left (\sqrt {c -i c \tan \left (f x +e \right )}+\frac {c}{2 \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4}\right )}{f c}\)	\(194\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {2} {\left ({\left (-i \, A - B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-i \, A - 3 \, B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{c f} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i \, {\left (i \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} B a - \frac {{\left (A - i \, B\right )} a c}{\sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}}{c f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

Mupad [B] (verification not implemented)

Time = 8.43 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.83 \[ \int \frac {(a+i a \tan (e+f x)) (A+B \tan (e+f x))}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {a\,\sqrt {\frac {2\,c}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left (A\,1{}\mathrm {i}+3\,B+A\,\left (\frac {{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}}{2}\right )\,1{}\mathrm {i}-A\,\left (\frac {{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )+B\,\left (\frac {{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}}{2}\right )+B\,\left (\frac {{\mathrm {e}}^{-e\,2{}\mathrm {i}-f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}\right )}{c\,f} \]

[In]

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

[Out]

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