\(\int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [1064]

Optimal result

Integrand size = 33, antiderivative size = 67 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (1,-\frac {5}{2}+m,-\frac {3}{2},\frac {1}{2} (1-i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m}{5 f (c-i c \tan (e+f x))^{5/2}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(158\) vs. \(2(67)=134\).

Time = 12.43 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.36 \[ \int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i 2^{-\frac {7}{2}+m} \left (e^{i f x}\right )^m \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \left (1+e^{2 i (e+f x)}\right )^4 \operatorname {Hypergeometric2F1}\left (1,\frac {7}{2},1+m,-e^{2 i (e+f x)}\right ) \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m}{c^3 f m} \] Input:

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

Output:

((-I)*2^(-7/2 + m)*(E^(I*f*x))^m*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*(E^(I*( 
e + f*x))/(1 + E^((2*I)*(e + f*x))))^m*(1 + E^((2*I)*(e + f*x)))^4*Hyperge 
ometric2F1[1, 7/2, 1 + m, -E^((2*I)*(e + f*x))]*(a + I*a*Tan[e + f*x])^m)/ 
(c^3*f*m*Sec[e + f*x]^m*(Cos[f*x] + I*Sin[f*x])^m)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {3042, 4006, 80, 79}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 79

\(\displaystyle -\frac {i 2^m (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1-m,-\frac {3}{2},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (c-i c \tan (e+f x))^{5/2}}\)

Input:

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

Output:

((-1/5*I)*2^m*Hypergeometric2F1[-5/2, 1 - m, -3/2, (1 - I*Tan[e + f*x])/2] 
*(a + I*a*Tan[e + f*x])^m)/(f*(1 + I*Tan[e + f*x])^m*(c - I*c*Tan[e + f*x] 
)^(5/2))
 

Defintions of rubi rules used

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 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 [F]

\[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(1/8*sqrt(2)*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m 
*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(e^(6*I*f*x + 6*I*e) + 3*e^(4*I*f*x + 4 
*I*e) + 3*e^(2*I*f*x + 2*I*e) + 1)/c^3, x)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [F(-2)]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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