3.1210 \(\int \frac{1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx\)

Optimal. Leaf size=82 \[ \frac{6 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{5 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}} \]

[Out]

((-2*I)/5)/(a^2*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4)) + (6*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2])/(5*a^3*(a
- I*a*x)^(1/4)*(a + I*a*x)^(1/4))

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Rubi [A]  time = 0.0160064, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {51, 42, 197, 196} \[ \frac{6 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{5 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)/5)/(a^2*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4)) + (6*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2])/(5*a^3*(a
- I*a*x)^(1/4)*(a + I*a*x)^(1/4))

Rule 51

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 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 197

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]

Rule 196

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

Rubi steps

\begin{align*} \int \frac{1}{(a-i a x)^{9/4} (a+i a x)^{5/4}} \, dx &=-\frac{2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac{3 \int \frac{1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx}{5 a}\\ &=-\frac{2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac{\left (3 \sqrt [4]{a^2+a^2 x^2}\right ) \int \frac{1}{\left (a^2+a^2 x^2\right )^{5/4}} \, dx}{5 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac{2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac{\left (3 \sqrt [4]{1+x^2}\right ) \int \frac{1}{\left (1+x^2\right )^{5/4}} \, dx}{5 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac{2 i}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}+\frac{6 \sqrt [4]{1+x^2} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{5 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ \end{align*}

Mathematica [C]  time = 0.024226, size = 70, normalized size = 0.85 \[ -\frac{i 2^{3/4} \sqrt [4]{1+i x} \, _2F_1\left (-\frac{5}{4},\frac{5}{4};-\frac{1}{4};\frac{1}{2}-\frac{i x}{2}\right )}{5 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.059, size = 107, normalized size = 1.3 \begin{align*}{\frac{6\,ix+6\,{x}^{2}+2}{ \left ( 5\,x+5\,i \right ){a}^{3}}{\frac{1}{\sqrt [4]{-a \left ( -1+ix \right ) }}}{\frac{1}{\sqrt [4]{a \left ( 1+ix \right ) }}}}-{\frac{3\,x}{5\,{a}^{3}}{\mbox{$_2$F$_1$}({\frac{1}{4}},{\frac{1}{2}};\,{\frac{3}{2}};\,-{x}^{2})}\sqrt [4]{-{a}^{2} \left ( -1+ix \right ) \left ( 1+ix \right ) }{\frac{1}{\sqrt [4]{{a}^{2}}}}{\frac{1}{\sqrt [4]{-a \left ( -1+ix \right ) }}}{\frac{1}{\sqrt [4]{a \left ( 1+ix \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-I*a*x)^(9/4)/(a+I*a*x)^(5/4),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(9/4)/(a+I*a*x)^(5/4),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \,{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}{\left (3 \, x^{2} + 3 i \, x + 1\right )} +{\left (5 \, a^{5} x^{3} + 5 i \, a^{5} x^{2} + 5 \, a^{5} x + 5 i \, a^{5}\right )}{\rm integral}\left (-\frac{3 \,{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{5 \,{\left (a^{5} x^{2} + a^{5}\right )}}, x\right )}{5 \, a^{5} x^{3} + 5 i \, a^{5} x^{2} + 5 \, a^{5} x + 5 i \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(9/4)/(a+I*a*x)^(5/4),x, algorithm="fricas")

[Out]

(2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)*(3*x^2 + 3*I*x + 1) + (5*a^5*x^3 + 5*I*a^5*x^2 + 5*a^5*x + 5*I*a^5)*in
tegral(-3/5*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)/(a^5*x^2 + a^5), x))/(5*a^5*x^3 + 5*I*a^5*x^2 + 5*a^5*x + 5*I
*a^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)**(9/4)/(a+I*a*x)**(5/4),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(9/4)/(a+I*a*x)^(5/4),x, algorithm="giac")

[Out]

Exception raised: TypeError