3.13.23 \(\int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx\) [1223]

3.13.23.1 Optimal result
3.13.23.2 Mathematica [C] (verified)
3.13.23.3 Rubi [A] (verified)
3.13.23.4 Maple [C] (verified)
3.13.23.5 Fricas [F]
3.13.23.6 Sympy [F(-1)]
3.13.23.7 Maxima [F(-2)]
3.13.23.8 Giac [F(-2)]
3.13.23.9 Mupad [F(-1)]

3.13.23.1 Optimal result

Integrand size = 25, antiderivative size = 154 \[ \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx=-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}-\frac {2 i}{13 a^3 (a-i a x)^{9/4} (a+i a x)^{5/4}}+\frac {14 x}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x} \left (1+x^2\right )}+\frac {42 \sqrt [4]{1+x^2} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{65 a^6 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

output
-2/13*I/a^2/(a-I*a*x)^(13/4)/(a+I*a*x)^(5/4)-2/13*I/a^3/(a-I*a*x)^(9/4)/(a 
+I*a*x)^(5/4)+14/65*x/a^6/(a-I*a*x)^(1/4)/(a+I*a*x)^(1/4)/(x^2+1)+42/65*(x 
^2+1)^(1/4)*(cos(1/2*arctan(x))^2)^(1/2)/cos(1/2*arctan(x))*EllipticE(sin( 
1/2*arctan(x)),2^(1/2))/a^6/(a-I*a*x)^(1/4)/(a+I*a*x)^(1/4)
 
3.13.23.2 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 = 0.45 \[ \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx=-\frac {i \sqrt [4]{1+i x} \operatorname {Hypergeometric2F1}\left (-\frac {13}{4},\frac {9}{4},-\frac {9}{4},\frac {1}{2}-\frac {i x}{2}\right )}{13 \sqrt [4]{2} a^3 (a-i a x)^{13/4} \sqrt [4]{a+i a x}} \]

input
Integrate[1/((a - I*a*x)^(17/4)*(a + I*a*x)^(9/4)),x]
 
output
((-1/13*I)*(1 + I*x)^(1/4)*Hypergeometric2F1[-13/4, 9/4, -9/4, 1/2 - (I/2) 
*x])/(2^(1/4)*a^3*(a - I*a*x)^(13/4)*(a + I*a*x)^(1/4))
 
3.13.23.3 Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.19, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {61, 61, 46, 215, 213, 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.

\(\displaystyle \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {9 \int \frac {1}{(a-i a x)^{13/4} (i x a+a)^{9/4}}dx}{13 a}-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {9 \left (\frac {7 \int \frac {1}{(a-i a x)^{9/4} (i x a+a)^{9/4}}dx}{9 a}-\frac {2 i}{9 a^2 (a-i a x)^{9/4} (a+i a x)^{5/4}}\right )}{13 a}-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}\)

\(\Big \downarrow \) 46

\(\displaystyle \frac {9 \left (\frac {7 \sqrt [4]{a^2 x^2+a^2} \int \frac {1}{\left (x^2 a^2+a^2\right )^{9/4}}dx}{9 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^2 (a-i a x)^{9/4} (a+i a x)^{5/4}}\right )}{13 a}-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {9 \left (\frac {7 \sqrt [4]{a^2 x^2+a^2} \left (\frac {3 \int \frac {1}{\left (x^2 a^2+a^2\right )^{5/4}}dx}{5 a^2}+\frac {2 x}{5 a^2 \left (a^2 x^2+a^2\right )^{5/4}}\right )}{9 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^2 (a-i a x)^{9/4} (a+i a x)^{5/4}}\right )}{13 a}-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}\)

\(\Big \downarrow \) 213

\(\displaystyle \frac {9 \left (\frac {7 \sqrt [4]{a^2 x^2+a^2} \left (\frac {3 \sqrt [4]{x^2+1} \int \frac {1}{\left (x^2+1\right )^{5/4}}dx}{5 a^4 \sqrt [4]{a^2 x^2+a^2}}+\frac {2 x}{5 a^2 \left (a^2 x^2+a^2\right )^{5/4}}\right )}{9 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^2 (a-i a x)^{9/4} (a+i a x)^{5/4}}\right )}{13 a}-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {9 \left (\frac {7 \sqrt [4]{a^2 x^2+a^2} \left (\frac {2 x}{5 a^2 \left (a^2 x^2+a^2\right )^{5/4}}+\frac {6 \sqrt [4]{x^2+1} E\left (\left .\frac {\arctan (x)}{2}\right |2\right )}{5 a^4 \sqrt [4]{a^2 x^2+a^2}}\right )}{9 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^2 (a-i a x)^{9/4} (a+i a x)^{5/4}}\right )}{13 a}-\frac {2 i}{13 a^2 (a-i a x)^{13/4} (a+i a x)^{5/4}}\)

input
Int[1/((a - I*a*x)^(17/4)*(a + I*a*x)^(9/4)),x]
 
output
((-2*I)/13)/(a^2*(a - I*a*x)^(13/4)*(a + I*a*x)^(5/4)) + (9*(((-2*I)/9)/(a 
^2*(a - I*a*x)^(9/4)*(a + I*a*x)^(5/4)) + (7*(a^2 + a^2*x^2)^(1/4)*((2*x)/ 
(5*a^2*(a^2 + a^2*x^2)^(5/4)) + (6*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 
2])/(5*a^4*(a^2 + a^2*x^2)^(1/4))))/(9*a*(a - I*a*x)^(1/4)*(a + I*a*x)^(1/ 
4))))/(13*a)
 

3.13.23.3.1 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 61
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*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 213
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( 
a*(a + b*x^2)^(1/4))   Int[1/(1 + b*(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, 
 x] && PosQ[a] && PosQ[b/a]
 

rule 215
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 
3.13.23.4 Maple [C] (verified)

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

Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84

method result size
risch \(\frac {\frac {42}{65} x^{5}+\frac {84}{65} i x^{4}+\frac {14}{65} x^{3}+\frac {112}{65} i x^{2}-\frac {46}{65} x +\frac {4}{13} i}{\left (x -i\right ) \left (x +i\right )^{3} a^{6} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {21 x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{65 \left (a^{2}\right )^{\frac {1}{4}} a^{6} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\) \(130\)

input
int(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(9/4),x,method=_RETURNVERBOSE)
 
output
2/65*(42*I*x^4+21*x^5+56*I*x^2-23*x+7*x^3+10*I)/(x-I)/(x+I)^3/a^6/(-a*(I*x 
-1))^(1/4)/(a*(I*x+1))^(1/4)-21/65/(a^2)^(1/4)*x*hypergeom([1/4,1/2],[3/2] 
,-x^2)/a^6*(-a^2*(I*x-1)*(I*x+1))^(1/4)/(-a*(I*x-1))^(1/4)/(a*(I*x+1))^(1/ 
4)
 
3.13.23.5 Fricas [F]

\[ \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {9}{4}} {\left (-i \, a x + a\right )}^{\frac {17}{4}}} \,d x } \]

input
integrate(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(9/4),x, algorithm="fricas")
 
output
1/65*(2*(21*x^5 + 42*I*x^4 + 7*x^3 + 56*I*x^2 - 23*x + 10*I)*(I*a*x + a)^( 
3/4)*(-I*a*x + a)^(3/4) + 65*(a^8*x^6 + 2*I*a^8*x^5 + a^8*x^4 + 4*I*a^8*x^ 
3 - a^8*x^2 + 2*I*a^8*x - a^8)*integral(-21/65*(I*a*x + a)^(3/4)*(-I*a*x + 
 a)^(3/4)/(a^8*x^2 + a^8), x))/(a^8*x^6 + 2*I*a^8*x^5 + a^8*x^4 + 4*I*a^8* 
x^3 - a^8*x^2 + 2*I*a^8*x - a^8)
 
3.13.23.6 Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx=\text {Timed out} \]

input
integrate(1/(a-I*a*x)**(17/4)/(a+I*a*x)**(9/4),x)
 
output
Timed out
 
3.13.23.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx=\text {Exception raised: RuntimeError} \]

input
integrate(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(9/4),x, algorithm="maxima")
 
output
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 
3.13.23.8 Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx=\text {Exception raised: TypeError} \]

input
integrate(1/(a-I*a*x)^(17/4)/(a+I*a*x)^(9/4),x, algorithm="giac")
 
output
Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 
]ext_reduce Error: Bad Argument TypeDone
 
3.13.23.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a-i a x)^{17/4} (a+i a x)^{9/4}} \, dx=\int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{17/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{9/4}} \,d x \]

input
int(1/((a - a*x*1i)^(17/4)*(a + a*x*1i)^(9/4)),x)
 
output
int(1/((a - a*x*1i)^(17/4)*(a + a*x*1i)^(9/4)), x)