\(\int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx\) [743]

Optimal result

Integrand size = 41, antiderivative size = 60 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 a (i A+B) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a B (c-i c \tan (e+f x))^{3/2}}{3 c f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 60, 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} \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 a (B+i A) \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 a B (c-i c \tan (e+f x))^{3/2}}{3 c f} \]

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

Mathematica [A] (verified)

Time = 1.18 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.75 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 a (3 i A+2 B+i B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{3 f} \]

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-i \sqrt {c -i c \tan \left (f x +e \right )}\, B c +\sqrt {c -i c \tan \left (f x +e \right )}\, c A \right )}{f c}\)	\(66\)
default	\(\frac {2 i a \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-i \sqrt {c -i c \tan \left (f x +e \right )}\, B c +\sqrt {c -i c \tan \left (f x +e \right )}\, c A \right )}{f c}\)	\(66\)
parts	\(\frac {a \left (i A +B \right ) \left (2 \sqrt {c -i c \tan \left (f x +e \right )}-\sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}+\frac {i A a \sqrt {c}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{f}-\frac {2 a B \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-\frac {c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{2}\right )}{f c}\)	\(156\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.10 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {2 \, \sqrt {2} {\left (3 \, {\left (-i \, A - B\right )} a e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (-3 i \, A - B\right )} a\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.80 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {2 i \, {\left (i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B a + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A - i \, B\right )} a c\right )}}{3 \, c f} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))*(A+B*tan(f*x+e)),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 = 0.57 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.70 \[ \int (a+i a \tan (e+f x)) (A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} \, dx=\frac {a\,\sqrt {-\frac {c\,\left (-2\,{\cos \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (e+f\,x\right )}^2}}\,\left (A\,3{}\mathrm {i}+2\,B+A\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )\,3{}\mathrm {i}+2\,B\,\left (2\,{\cos \left (e+f\,x\right )}^2-1\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{3\,f\,{\cos \left (e+f\,x\right )}^2} \]

[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*(-(c*(sin(2*e + 2*f*x)*1i - 2*cos(e + f*x)^2))/(2*cos(e + f*x)^2))^(1/2)*(A*3i + 2*B + A*(2*cos(e + f*x)^2
- 1)*3i + 2*B*(2*cos(e + f*x)^2 - 1) + B*sin(2*e + 2*f*x)*1i))/(3*f*cos(e + f*x)^2)