\(\int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx\) [1040]

Optimal result

Integrand size = 35, antiderivative size = 234 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {i}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 i}{3 a f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}-\frac {4 i \sqrt {a+i a \tan (e+f x)}}{5 a^2 f (c-i c \tan (e+f x))^{5/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c f (c-i c \tan (e+f x))^{3/2}}-\frac {8 i \sqrt {a+i a \tan (e+f x)}}{15 a^2 c^2 f \sqrt {c-i c \tan (e+f x)}} \] Output:

1/3*I/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2)+4/3*I/a/f/(a+I*a 
*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(5/2)-4/5*I*(a+I*a*tan(f*x+e))^(1/2) 
/a^2/f/(c-I*c*tan(f*x+e))^(5/2)-8/15*I*(a+I*a*tan(f*x+e))^(1/2)/a^2/c/f/(c 
-I*c*tan(f*x+e))^(3/2)-8/15*I*(a+I*a*tan(f*x+e))^(1/2)/a^2/c^2/f/(c-I*c*ta 
n(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {3+12 i \tan (e+f x)+12 \tan ^2(e+f x)+8 i \tan ^3(e+f x)+8 \tan ^4(e+f x)}{15 a c^2 f (-i+\tan (e+f x)) (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]
 

Output:

(3 + (12*I)*Tan[e + f*x] + 12*Tan[e + f*x]^2 + (8*I)*Tan[e + f*x]^3 + 8*Ta 
n[e + f*x]^4)/(15*a*c^2*f*(-I + Tan[e + f*x])*(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 = 0.38 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4006, 55, 55, 55, 55, 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.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a c \int \frac {1}{(i \tan (e+f x) a+a)^{5/2} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {4 \int \frac {1}{(i \tan (e+f x) a+a)^{3/2} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {4 \left (\frac {3 \int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{a}+\frac {i}{a c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}\right )}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {4 \left (\frac {3 \left (\frac {2 \int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{a}+\frac {i}{a c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}\right )}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {4 \left (\frac {3 \left (\frac {2 \left (\frac {\int \frac {1}{\sqrt {i \tan (e+f x) a+a} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{a}+\frac {i}{a c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}\right )}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {a c \left (\frac {4 \left (\frac {3 \left (\frac {2 \left (-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c^2 \sqrt {c-i c \tan (e+f x)}}-\frac {i \sqrt {a+i a \tan (e+f x)}}{3 a c (c-i c \tan (e+f x))^{3/2}}\right )}{5 c}-\frac {i \sqrt {a+i a \tan (e+f x)}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{a}+\frac {i}{a c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}}\right )}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2)),x]
 

Output:

(a*c*((I/3)/(a*c*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(5/2) 
) + (4*(I/(a*c*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2)) + 
(3*(((-1/5*I)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*(c - I*c*Tan[e + f*x])^(5/2 
)) + (2*(((-1/3*I)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c*(c - I*c*Tan[e + f*x]) 
^(3/2)) - ((I/3)*Sqrt[a + I*a*Tan[e + f*x]])/(a*c^2*Sqrt[c - I*c*Tan[e + f 
*x]])))/(5*c)))/a))/(3*a)))/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 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 = 0.35 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.56

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \tan \left (f x +e \right )^{5}+8 \tan \left (f x +e \right )^{6}+20 i \tan \left (f x +e \right )^{3}+20 \tan \left (f x +e \right )^{4}+12 i \tan \left (f x +e \right )+15 \tan \left (f x +e \right )^{2}+3\right )}{15 f \,a^{2} c^{3} \left (-\tan \left (f x +e \right )+i\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(130\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (8 i \tan \left (f x +e \right )^{5}+8 \tan \left (f x +e \right )^{6}+20 i \tan \left (f x +e \right )^{3}+20 \tan \left (f x +e \right )^{4}+12 i \tan \left (f x +e \right )+15 \tan \left (f x +e \right )^{2}+3\right )}{15 f \,a^{2} c^{3} \left (-\tan \left (f x +e \right )+i\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(130\)

Input:

int(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVE 
RBOSE)
 

Output:

-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^2/c^3*(8* 
I*tan(f*x+e)^5+8*tan(f*x+e)^6+20*I*tan(f*x+e)^3+20*tan(f*x+e)^4+12*I*tan(f 
*x+e)+15*tan(f*x+e)^2+3)/(-tan(f*x+e)+I)^3/(I+tan(f*x+e))^4
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-3 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 23 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 110 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 48 i \, e^{\left (5 i \, f x + 5 i \, e\right )} - 30 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 48 i \, e^{\left (3 i \, f x + 3 i \, e\right )} + 65 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{240 \, a^{2} c^{3} f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm 
="fricas")
 

Output:

1/240*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))* 
(-3*I*e^(10*I*f*x + 10*I*e) - 23*I*e^(8*I*f*x + 8*I*e) - 110*I*e^(6*I*f*x 
+ 6*I*e) + 48*I*e^(5*I*f*x + 5*I*e) - 30*I*e^(4*I*f*x + 4*I*e) + 48*I*e^(3 
*I*f*x + 3*I*e) + 65*I*e^(2*I*f*x + 2*I*e) + 5*I)*e^(-3*I*f*x - 3*I*e)/(a^ 
2*c^3*f)
 

Sympy [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(a+I*a*tan(f*x+e))**(3/2)/(c-I*c*tan(f*x+e))**(5/2),x)
 

Output:

Integral(1/((I*a*(tan(e + f*x) - I))**(3/2)*(-I*c*(tan(e + f*x) + I))**(5/ 
2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm 
="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(5/2),x, algorithm 
="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.61 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.60 \[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,20{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+40\,\sin \left (2\,e+2\,f\,x\right )+4\,\sin \left (4\,e+4\,f\,x\right )-45{}\mathrm {i}\right )}{120\,a^2\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:

int(1/((a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(5/2)),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)*20i + cos(4*e + 4*f*x)*1i + 40*sin(2*e + 2*f*x) + 
 4*sin(4*e + 4*f*x) - 45i))/(120*a^2*c^2*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]

\[ \int \frac {1}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c}\, \sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{5} i -\tan \left (f x +e \right )^{4}+2 \tan \left (f x +e \right )^{3} i -2 \tan \left (f x +e \right )^{2}+\tan \left (f x +e \right ) i -1}d x \right )}{a^{2} c^{3}} \] Input:

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

Output:

( - sqrt(c)*sqrt(a)*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 
 1))/(tan(e + f*x)**5*i - tan(e + f*x)**4 + 2*tan(e + f*x)**3*i - 2*tan(e 
+ f*x)**2 + tan(e + f*x)*i - 1),x))/(a**2*c**3)