\(\int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx\) [1031]

Optimal result

Integrand size = 35, antiderivative size = 147 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {i}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 \tan (e+f x)}{15 a f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 \tan (e+f x)}{15 a^2 c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.91 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {3-12 i \tan (e+f x)+12 \tan ^2(e+f x)-8 i \tan ^3(e+f x)+8 \tan ^4(e+f x)}{15 a^2 c f (-i+\tan (e+f x))^2 (i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)),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^2*c*f*(-I + Tan[e + f*x])^2*(I + Tan[e + f*x])*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 = 161, normalized size of antiderivative = 1.10, 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, 42, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 42

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

\(\Big \downarrow \) 41

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

Input:

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

Output:

(a*c*((I/5)/(a*c*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2) 
) + (4*(Tan[e + f*x]/(3*a*c*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + 
f*x])^(3/2)) + (2*Tan[e + f*x])/(3*a^2*c^2*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 42

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(- 
x)*(a + b*x)^(m + 1)*((c + d*x)^(m + 1)/(2*a*c*(m + 1))), x] + Simp[(2*m + 
3)/(2*a*c*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /; Fre 
eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 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 = 0.34 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88

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

Input:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-5 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 65 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 48 i \, e^{\left (7 i \, f x + 7 i \, e\right )} + 30 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 48 i \, e^{\left (5 i \, f x + 5 i \, e\right )} + 110 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 23 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{240 \, a^{3} c^{2} f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(3/2),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))* 
(-5*I*e^(10*I*f*x + 10*I*e) - 65*I*e^(8*I*f*x + 8*I*e) - 48*I*e^(7*I*f*x + 
 7*I*e) + 30*I*e^(6*I*f*x + 6*I*e) - 48*I*e^(5*I*f*x + 5*I*e) + 110*I*e^(4 
*I*f*x + 4*I*e) + 23*I*e^(2*I*f*x + 2*I*e) + 3*I)*e^(-5*I*f*x - 5*I*e)/(a^ 
3*c^2*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(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]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 2.96 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (\cos \left (2\,e+2\,f\,x\right )\,85{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,20{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+95\,\sin \left (2\,e+2\,f\,x\right )+20\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )-60{}\mathrm {i}\right )}{240\,a^3\,c\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:

int(1/((a + a*tan(e + f*x)*1i)^(5/2)*(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)*(cos(2*e + 2*f*x)*85i + cos(4*e + 4*f*x)*20i + cos(6*e + 6*f*x)*3i 
+ 95*sin(2*e + 2*f*x) + 20*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x) - 60i))/( 
240*a^3*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]

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

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