\(\int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [792]

Optimal result

Integrand size = 45, antiderivative size = 102 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac {(i A-2 B) \sqrt {a+i a \tan (e+f x)}}{3 c f \sqrt {c-i c \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 3.00 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {a+i a \tan (e+f x)} (2 A+i B+(-i A+2 B) \tan (e+f x))}{3 c f (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3042, 4071, 87, 48}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {a c \left (\frac {(A+2 i B) \int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 c}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {a c \left (-\frac {i (A+2 i B) \sqrt {a+i a \tan (e+f x)}}{3 a c^2 \sqrt {c-i c \tan (e+f x)}}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\)

Input:

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

Output:

(a*c*(-1/3*((I*A + B)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*(c - I*c*Tan[e + f* 
x])^(3/2)) - ((I/3)*(A + (2*I)*B)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c^2*Sqrt[ 
c - I*c*Tan[e + f*x]])))/f
 

Defintions of rubi rules used

rule 48

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4071

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[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]
 

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.82

method	result	size
risch	\(-\frac {\sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (i A \,{\mathrm e}^{2 i \left (f x +e \right )}+B \,{\mathrm e}^{2 i \left (f x +e \right )}+3 i A -3 B \right )}{6 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(84\)
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i B \tan \left (f x +e \right )^{2}+3 i \tan \left (f x +e \right ) A +A \tan \left (f x +e \right )^{2}-i B -3 B \tan \left (f x +e \right )-2 A \right )}{3 f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{3}}\)	\(100\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i B \tan \left (f x +e \right )^{2}+3 i \tan \left (f x +e \right ) A +A \tan \left (f x +e \right )^{2}-i B -3 B \tan \left (f x +e \right )-2 A \right )}{3 f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{3}}\)	\(100\)
parts	\(\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (3 i \tan \left (f x +e \right )+\tan \left (f x +e \right )^{2}-2\right )}{3 f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{3}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (3 i \tan \left (f x +e \right )+2 \tan \left (f x +e \right )^{2}-1\right )}{3 f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{3}}\)	\(145\)

Input:

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

Output:

-1/6/c*(a*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^(1/2)/(c/(exp(2*I*(f*x+e) 
)+1))^(1/2)*(I*A*exp(2*I*(f*x+e))+B*exp(2*I*(f*x+e))+3*I*A-3*B)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left ({\left (-i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} - 2 \, {\left (2 i \, A - B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} - 3 \, {\left (i \, A - B\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{6 \, c^{2} f} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.42 \[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,3{}\mathrm {i}-3\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+B\,\cos \left (2\,e+2\,f\,x\right )-A\,\sin \left (2\,e+2\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{6\,c\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:

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

Output:

-(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1) 
)^(1/2)*(A*3i - 3*B + A*cos(2*e + 2*f*x)*1i + B*cos(2*e + 2*f*x) - A*sin(2 
*e + 2*f*x) + B*sin(2*e + 2*f*x)*1i))/(6*c*f*((c*(cos(2*e + 2*f*x) - sin(2 
*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))
 

Reduce [F]

\[ \int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, \left (-\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) a +2 \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) b i +\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, b +2 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )+i}d x \right ) \tan \left (f x +e \right )^{2} a f i +2 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )+i}d x \right ) \tan \left (f x +e \right )^{2} b f +2 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )+i}d x \right ) a f i +2 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )+i}d x \right ) b f \right )}{c^{2} f \left (\tan \left (f x +e \right )^{2}+1\right )} \] Input:

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

Output:

(sqrt(c)*sqrt(a)*( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)* 
tan(e + f*x)*a + 2*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*ta 
n(e + f*x)*b*i + sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*b + 
2*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x) 
**3 + tan(e + f*x)**2*i + tan(e + f*x) + i),x)*tan(e + f*x)**2*a*f*i + 2*i 
nt((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**3 
 + tan(e + f*x)**2*i + tan(e + f*x) + i),x)*tan(e + f*x)**2*b*f + 2*int((s 
qrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**3 + ta 
n(e + f*x)**2*i + tan(e + f*x) + i),x)*a*f*i + 2*int((sqrt(tan(e + f*x)*i 
+ 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**3 + tan(e + f*x)**2*i + t 
an(e + f*x) + i),x)*b*f))/(c**2*f*(tan(e + f*x)**2 + 1))