\(\int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx\) [1022]

Optimal result

Integrand size = 35, antiderivative size = 140 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {i}{5 f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {i}{5 a f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 \tan (e+f x)}{5 a^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Output:

1/5*I/f/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2)+1/5*I/a/f/(a+I*a 
*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/2)+2/5*tan(f*x+e)/a^2/f/(a+I*a*ta 
n(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.64 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {-2 i-\tan (e+f x)-4 i \tan ^2(e+f x)+2 \tan ^3(e+f x)}{5 a^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]
 

Output:

(-2*I - Tan[e + f*x] - (4*I)*Tan[e + f*x]^2 + 2*Tan[e + f*x]^3)/(5*a^2*f*( 
-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]] 
)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4006, 55, 55, 41}

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.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}dx\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a c \int \frac {1}{(i \tan (e+f x) a+a)^{7/2} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {3 \int \frac {1}{(i \tan (e+f x) a+a)^{5/2} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{5 a}+\frac {i}{5 a c (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {3 \left (\frac {2 \int \frac {1}{(i \tan (e+f x) a+a)^{3/2} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}\right )}{5 a}+\frac {i}{5 a c (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}\right )}{f}\)

\(\Big \downarrow \) 41

\(\displaystyle \frac {a c \left (\frac {3 \left (\frac {2 \tan (e+f x)}{3 a^2 c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}\right )}{5 a}+\frac {i}{5 a c (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}\right )}{f}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]]),x]
 

Output:

(a*c*((I/5)/(a*c*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c - I*c*Tan[e + f*x]]) 
+ (3*((I/3)/(a*c*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]]) 
+ (2*Tan[e + f*x])/(3*a^2*c*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e 
+ f*x]])))/(5*a)))/f
 

Defintions of rubi rules used

rule 41

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> S 
imp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && Eq 
Q[b*c + a*d, 0]
 

rule 55

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] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

rule 3042

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

rule 4006

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (4 i \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{5}+6 i \tan \left (f x +e \right )^{2}-\tan \left (f x +e \right )^{3}+2 i+\tan \left (f x +e \right )\right )}{5 f \,a^{3} c \left (-\tan \left (f x +e \right )+i\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(118\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (4 i \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{5}+6 i \tan \left (f x +e \right )^{2}-\tan \left (f x +e \right )^{3}+2 i+\tan \left (f x +e \right )\right )}{5 f \,a^{3} c \left (-\tan \left (f x +e \right )+i\right )^{4} \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(118\)

Input:

int(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVE 
RBOSE)
 

Output:

-1/5/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^3/c*(4*I*t 
an(f*x+e)^4-2*tan(f*x+e)^5+6*I*tan(f*x+e)^2-tan(f*x+e)^3+2*I+tan(f*x+e))/( 
-tan(f*x+e)+I)^4/(I+tan(f*x+e))^2
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-5 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 16 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 10 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 16 i \, e^{\left (5 i \, f x + 5 i \, e\right )} + 20 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 6 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{40 \, a^{3} c f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm 
="fricas")
 

Output:

1/40*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*( 
-5*I*e^(8*I*f*x + 8*I*e) - 16*I*e^(7*I*f*x + 7*I*e) + 10*I*e^(6*I*f*x + 6* 
I*e) - 16*I*e^(5*I*f*x + 5*I*e) + 20*I*e^(4*I*f*x + 4*I*e) + 6*I*e^(2*I*f* 
x + 2*I*e) + I)*e^(-5*I*f*x - 5*I*e)/(a^3*c*f)
 

Sympy [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \] Input:

integrate(1/(a+I*a*tan(f*x+e))**(5/2)/(c-I*c*tan(f*x+e))**(1/2),x)
 

Output:

Integral(1/((I*a*(tan(e + f*x) - I))**(5/2)*sqrt(-I*c*(tan(e + f*x) + I))) 
, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm 
="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} \sqrt {-i \, c \tan \left (f x + e\right ) + c}} \,d x } \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm 
="giac")
 

Output:

integrate(1/((I*a*tan(f*x + e) + a)^(5/2)*sqrt(-I*c*tan(f*x + e) + c)), x)
 

Mupad [B] (verification not implemented)

Time = 2.95 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.13 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,5{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}+15\,\sin \left (2\,e+2\,f\,x\right )+5\,\sin \left (4\,e+4\,f\,x\right )+\sin \left (6\,e+6\,f\,x\right )-5{}\mathrm {i}\right )}{40\,a^3\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:

int(1/((a + a*tan(e + f*x)*1i)^(5/2)*(c - c*tan(e + f*x)*1i)^(1/2)),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)*(cos(2*e + 2*f*x)*15i + cos(4*e + 4*f*x)*5i + cos(6*e + 6*f*x)*1i + 
 15*sin(2*e + 2*f*x) + 5*sin(4*e + 4*f*x) + sin(6*e + 6*f*x) - 5i))/(40*a^ 
3*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]

\[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {\int \frac {1}{\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}d x}{\sqrt {c}\, \sqrt {a}\, a^{2}} \] Input:

int(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(1/2),x)
 

Output:

( - int(1/(sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f* 
x)**2 - 2*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x 
)*i - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)),x))/(sqrt(c)*s 
qrt(a)*a**2)