\(\int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx\) [246]

Optimal result

Integrand size = 25, antiderivative size = 103 \[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\frac {6 i}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}+\frac {6 \sqrt [4]{1+x^2} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \] Output:

6*I/(a-I*a*x)^(1/4)/(a+I*a*x)^(1/4)-2*I*(a-I*a*x)^(3/4)/a/(a+I*a*x)^(1/4)+ 
6*(x^2+1)^(1/4)*EllipticE(sin(1/2*arctan(x)),2^(1/2))/(a-I*a*x)^(1/4)/(a+I 
*a*x)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.68 \[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\frac {i 2^{3/4} \sqrt [4]{1+i x} (a-i a x)^{7/4} \operatorname {Hypergeometric2F1}\left (\frac {5}{4},\frac {7}{4},\frac {11}{4},\frac {1}{2}-\frac {i x}{2}\right )}{7 a^2 \sqrt [4]{a+i a x}} \] Input:

Integrate[(a - I*a*x)^(3/4)/(a + I*a*x)^(5/4),x]
 

Output:

((I/7)*2^(3/4)*(1 + I*x)^(1/4)*(a - I*a*x)^(7/4)*Hypergeometric2F1[5/4, 7/ 
4, 11/4, 1/2 - (I/2)*x])/(a^2*(a + I*a*x)^(1/4))
 

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {57, 46, 227, 225, 212}

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.

\(\displaystyle \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-3 \int \frac {1}{\sqrt [4]{a-i a x} \sqrt [4]{i x a+a}}dx\)

\(\Big \downarrow \) 46

\(\displaystyle \frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-\frac {3 \sqrt [4]{a^2 x^2+a^2} \int \frac {1}{\sqrt [4]{x^2 a^2+a^2}}dx}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-\frac {3 \sqrt [4]{x^2+1} \int \frac {1}{\sqrt [4]{x^2+1}}dx}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-\frac {3 \sqrt [4]{x^2+1} \left (\frac {2 x}{\sqrt [4]{x^2+1}}-\int \frac {1}{\left (x^2+1\right )^{5/4}}dx\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-\frac {3 \sqrt [4]{x^2+1} \left (\frac {2 x}{\sqrt [4]{x^2+1}}-2 E\left (\left .\frac {\arctan (x)}{2}\right |2\right )\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\)

Input:

Int[(a - I*a*x)^(3/4)/(a + I*a*x)^(5/4),x]
 

Output:

((4*I)*(a - I*a*x)^(3/4))/(a*(a + I*a*x)^(1/4)) - (3*(1 + x^2)^(1/4)*((2*x 
)/(1 + x^2)^(1/4) - 2*EllipticE[ArcTan[x]/2, 2]))/((a - I*a*x)^(1/4)*(a + 
I*a*x)^(1/4))
 

Defintions of rubi rules used

rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(a 
 + b*x)^FracPart[m]*((c + d*x)^FracPart[m]/(a*c + b*d*x^2)^FracPart[m])   I 
nt[(a*c + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 
 0] &&  !IntegerQ[2*m]
 

rule 57

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

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 225

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( 
a + b*x^2)^(1/4)   Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4.

Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.85

method	result	size
risch	\(\frac {4 x +4 i}{\left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {3 x \operatorname {hypergeom}\left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{\left (a^{2}\right )^{\frac {1}{4}} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(88\)

Input:

int((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)
 

Output:

4*(x+I)/(-a*(I*x-1))^(1/4)/(a*(I*x+1))^(1/4)-3/(a^2)^(1/4)*x*hypergeom([1/ 
4,1/2],[3/2],-x^2)*(-a^2*(I*x-1)*(I*x+1))^(1/4)/(-a*(I*x-1))^(1/4)/(a*(I*x 
+1))^(1/4)
 

Fricas [F]

\[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {3}{4}}}{{\left (i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")
 

Output:

-(2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)*(x - 3*I) - (a^2*x^2 - I*a^2*x)*i 
ntegral(-6*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)/(a^2*x^4 + a^2*x^2), x))/( 
a^2*x^2 - I*a^2*x)
 

Sympy [F]

\[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int \frac {\left (- i a \left (x + i\right )\right )^{\frac {3}{4}}}{\left (i a \left (x - i\right )\right )^{\frac {5}{4}}}\, dx \] Input:

integrate((a-I*a*x)**(3/4)/(a+I*a*x)**(5/4),x)
 

Output:

Integral((-I*a*(x + I))**(3/4)/(I*a*(x - I))**(5/4), x)
 

Maxima [F]

\[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {3}{4}}}{{\left (i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")
 

Output:

integrate((-I*a*x + a)^(3/4)/(I*a*x + a)^(5/4), x)
 

Giac [F]

\[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int { \frac {{\left (-i \, a x + a\right )}^{\frac {3}{4}}}{{\left (i \, a x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x, algorithm="giac")
 

Output:

integrate((-I*a*x + a)^(3/4)/(I*a*x + a)^(5/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx=\int \frac {{\left (a-a\,x\,1{}\mathrm {i}\right )}^{3/4}}{{\left (a+a\,x\,1{}\mathrm {i}\right )}^{5/4}} \,d x \] Input:

int((a - a*x*1i)^(3/4)/(a + a*x*1i)^(5/4),x)
 

Output:

int((a - a*x*1i)^(3/4)/(a + a*x*1i)^(5/4), x)
 

Reduce [F]

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

int((a-I*a*x)^(3/4)/(a+I*a*x)^(5/4),x)
 

Output:

(sqrt(a)*int(( - i*x + 1)**(3/4)/((i*x + 1)**(1/4)*i*x + (i*x + 1)**(1/4)) 
,x))/a