3.13.11 \(\int \frac {1}{(a-i a x)^{13/4} (a+i a x)^{5/4}} \, dx\) [1211]

Optimal. Leaf size=115 \[ -\frac {2 i}{9 a^2 (a-i a x)^{9/4} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {2 \sqrt [4]{1+x^2} E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{3 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[Out]

-2/9*I/a^2/(a-I*a*x)^(9/4)/(a+I*a*x)^(1/4)-2/9*I/a^3/(a-I*a*x)^(5/4)/(a+I*a*x)^(1/4)+2/3*(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^4/(a-I*a*x)^(1/4)/(a+I*a*x)^(
1/4)

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {53, 42, 203, 202} \begin {gather*} \frac {2 \sqrt [4]{x^2+1} E\left (\left .\frac {\text {ArcTan}(x)}{2}\right |2\right )}{3 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^2 (a-i a x)^{9/4} \sqrt [4]{a+i a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a - I*a*x)^(13/4)*(a + I*a*x)^(5/4)),x]

[Out]

((-2*I)/9)/(a^2*(a - I*a*x)^(9/4)*(a + I*a*x)^(1/4)) - ((2*I)/9)/(a^3*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4)) + (
2*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2])/(3*a^4*(a - I*a*x)^(1/4)*(a + I*a*x)^(1/4))

Rule 42

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[(a + b*x)^FracPart[m]*((c + d*x)^Frac
Part[m]/(a*c + b*d*x^2)^FracPart[m]), Int[(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 53

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] - Dist[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] && NeQ[b*c - a*d, 0] && 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 202

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 203

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(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]

Rubi steps

\begin {align*} \int \frac {1}{(a-i a x)^{13/4} (a+i a x)^{5/4}} \, dx &=-\frac {2 i}{9 a^2 (a-i a x)^{9/4} \sqrt [4]{a+i a x}}+\frac {5 \int \frac {1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx}{9 a}\\ &=-\frac {2 i}{9 a^2 (a-i a x)^{9/4} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {\int \frac {1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx}{3 a^2}\\ &=-\frac {2 i}{9 a^2 (a-i a x)^{9/4} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {\sqrt [4]{a^2+a^2 x^2} \int \frac {1}{\left (a^2+a^2 x^2\right )^{5/4}} \, dx}{3 a^2 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac {2 i}{9 a^2 (a-i a x)^{9/4} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {\sqrt [4]{1+x^2} \int \frac {1}{\left (1+x^2\right )^{5/4}} \, dx}{3 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac {2 i}{9 a^2 (a-i a x)^{9/4} \sqrt [4]{a+i a x}}-\frac {2 i}{9 a^3 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac {2 \sqrt [4]{1+x^2} E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{3 a^4 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.03, size = 70, normalized size = 0.61 \begin {gather*} -\frac {i 2^{3/4} \sqrt [4]{1+i x} \, _2F_1\left (-\frac {9}{4},\frac {5}{4};-\frac {5}{4};\frac {1}{2}-\frac {i x}{2}\right )}{9 a^2 (a-i a x)^{9/4} \sqrt [4]{a+i a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a - I*a*x)^(13/4)*(a + I*a*x)^(5/4)),x]

[Out]

((-1/9*I)*2^(3/4)*(1 + I*x)^(1/4)*Hypergeometric2F1[-9/4, 5/4, -5/4, 1/2 - (I/2)*x])/(a^2*(a - I*a*x)^(9/4)*(a
+ I*a*x)^(1/4))

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.18, size = 113, normalized size = 0.98

method	result	size
risch	\(\frac {\frac {2}{3} x^{3}+\frac {4}{3} i x^{2}-\frac {4}{9} x +\frac {4}{9} i}{\left (x +i\right )^{2} a^{4} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{3 \left (a^{2}\right )^{\frac {1}{4}} a^{4} \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-I*a*x)^(13/4)/(a+I*a*x)^(5/4),x,method=_RETURNVERBOSE)

[Out]

2/9*(6*I*x^2+3*x^3-2*x+2*I)/(x+I)^2/a^4/(-a*(-1+I*x))^(1/4)/(a*(1+I*x))^(1/4)-1/3/(a^2)^(1/4)*x*hypergeom([1/4
,1/2],[3/2],-x^2)/a^4*(-a^2*(-1+I*x)*(1+I*x))^(1/4)/(-a*(-1+I*x))^(1/4)/(a*(1+I*x))^(1/4)

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(13/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(13/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

1/9*(2*(3*x^3 + 6*I*x^2 - 2*x + 2*I)*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4) + 9*(a^6*x^4 + 2*I*a^6*x^3 + 2*I*a^6
*x - a^6)*integral(-1/3*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)/(a^6*x^2 + a^6), x))/(a^6*x^4 + 2*I*a^6*x^3 + 2*I
*a^6*x - a^6)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)**(13/4)/(a+I*a*x)**(5/4),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3656 deep

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(13/4)/(a+I*a*x)^(5/4),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:ext_reduce Error: Bad Argument TypeDone

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{13/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a - a*x*1i)^(13/4)*(a + a*x*1i)^(5/4)),x)

[Out]

int(1/((a - a*x*1i)^(13/4)*(a + a*x*1i)^(5/4)), x)

________________________________________________________________________________________