3.1207 \(\int \frac{(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx\)

Optimal. Leaf size=102 \[ \frac{6 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{6 x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac{4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}} \]

[Out]

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

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Rubi [A]  time = 0.0199544, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {47, 42, 229, 227, 196} \[ \frac{6 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{6 x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac{4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 47

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] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && 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 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 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(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]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && 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{(a-i a x)^{3/4}}{(a+i a x)^{5/4}} \, dx &=\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]{a+i a x}} \, dx\\ &=\frac{4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-\frac{\left (3 \sqrt [4]{a^2+a^2 x^2}\right ) \int \frac{1}{\sqrt [4]{a^2+a^2 x^2}} \, dx}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=\frac{4 i (a-i a x)^{3/4}}{a \sqrt [4]{a+i a x}}-\frac{\left (3 \sqrt [4]{1+x^2}\right ) \int \frac{1}{\sqrt [4]{1+x^2}} \, dx}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac{6 x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac{4 i (a-i a x)^{3/4}}{a \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}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac{6 x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac{4 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{1}{2} \tan ^{-1}(x)\right |2\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ \end{align*}

Mathematica [C]  time = 0.0220003, size = 70, normalized size = 0.69 \[ \frac{i 2^{3/4} \sqrt [4]{1+i x} (a-i a x)^{7/4} \, _2F_1\left (\frac{5}{4},\frac{7}{4};\frac{11}{4};\frac{1}{2}-\frac{i x}{2}\right )}{7 a^2 \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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))

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Maple [C]  time = 0.042, size = 88, normalized size = 0.9 \begin{align*} 4\,{\frac{x+i}{\sqrt [4]{-a \left ( -1+ix \right ) }\sqrt [4]{a \left ( 1+ix \right ) }}}-3\,{\frac{x{\mbox{$_2$F$_1$}(1/4,1/2;\,3/2;\,-{x}^{2})}\sqrt [4]{-{a}^{2} \left ( -1+ix \right ) \left ( 1+ix \right ) }}{\sqrt [4]{{a}^{2}}\sqrt [4]{-a \left ( -1+ix \right ) }\sqrt [4]{a \left ( 1+ix \right ) }}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*(x+I)/(-a*(-1+I*x))^(1/4)/(a*(1+I*x))^(1/4)-3/(a^2)^(1/4)*x*hypergeom([1/4,1/2],[3/2],-x^2)*(-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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{{\left (i \, a x + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}{\left (2 \, x - 6 i\right )} -{\left (a^{2} x^{2} - i \, a^{2} x\right )}{\rm integral}\left (-\frac{6 \,{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{a^{2} x^{4} + a^{2} x^{2}}, x\right )}{a^{2} x^{2} - i \, a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-a*(I*x - 1))**(3/4)/(a*(I*x + 1))**(5/4), x)

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

[Out]

Exception raised: TypeError