3.1218 \(\int \frac{(a-i a x)^{3/4}}{(a+i a x)^{9/4}} \, dx\)

Optimal. Leaf size=115 \[ -\frac{6 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{5 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{6 i}{5 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac{4 i (a-i a x)^{3/4}}{5 a (a+i a x)^{5/4}} \]

[Out]

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

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Rubi [A]  time = 0.0252828, antiderivative size = 115, 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, 48, 42, 197, 196} \[ -\frac{6 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{5 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{6 i}{5 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac{4 i (a-i a x)^{3/4}}{5 a (a+i a x)^{5/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a - I*a*x)^(3/4)/(a + I*a*x)^(9/4),x]

[Out]

(((4*I)/5)*(a - I*a*x)^(3/4))/(a*(a + I*a*x)^(5/4)) - ((6*I)/5)/(a*(a - I*a*x)^(1/4)*(a + I*a*x)^(1/4)) - (6*(
1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2])/(5*a*(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 48

Int[1/(((a_) + (b_.)*(x_))^(5/4)*((c_) + (d_.)*(x_))^(1/4)), x_Symbol] :> Simp[-2/(b*(a + b*x)^(1/4)*(c + d*x)
^(1/4)), x] + Dist[c, Int[1/((a + b*x)^(5/4)*(c + d*x)^(5/4)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a
*d, 0] && NegQ[a^2*b^2]

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

Mathematica [C]  time = 0.0227565, size = 70, normalized size = 0.61 \[ \frac{i \sqrt [4]{1+i x} (a-i a x)^{7/4} \, _2F_1\left (\frac{7}{4},\frac{9}{4};\frac{11}{4};\frac{1}{2}-\frac{i x}{2}\right )}{7 \sqrt [4]{2} a^3 \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - I*a*x)^(3/4)/(a + I*a*x)^(9/4),x]

[Out]

((I/7)*(1 + I*x)^(1/4)*(a - I*a*x)^(7/4)*Hypergeometric2F1[7/4, 9/4, 11/4, 1/2 - (I/2)*x])/(2^(1/4)*a^3*(a + I
*a*x)^(1/4))

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Maple [C]  time = 0.034, size = 107, normalized size = 0.9 \begin{align*} -{\frac{6\,{x}^{2}+2+4\,ix}{ \left ( 5\,x-5\,i \right ) a}{\frac{1}{\sqrt [4]{-a \left ( -1+ix \right ) }}}{\frac{1}{\sqrt [4]{a \left ( 1+ix \right ) }}}}+{\frac{3\,x}{5\,a}{\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((a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x)

[Out]

-2/5*(3*x^2+1+2*I*x)/(x-I)/a/(-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*(-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{9}{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)^(9/4),x, algorithm="maxima")

[Out]

integrate((-I*a*x + a)^(3/4)/(I*a*x + a)^(9/4), x)

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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 (5 i \, x + 3\right )} -{\left (5 \, a^{3} x^{3} - 10 i \, a^{3} x^{2} - 5 \, a^{3} 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}}}{5 \,{\left (a^{3} x^{4} + a^{3} x^{2}\right )}}, x\right )}{5 \, a^{3} x^{3} - 10 i \, a^{3} x^{2} - 5 \, a^{3} 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)^(9/4),x, algorithm="fricas")

[Out]

-(2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)*(5*I*x + 3) - (5*a^3*x^3 - 10*I*a^3*x^2 - 5*a^3*x)*integral(6/5*(I*a*
x + a)^(3/4)*(-I*a*x + a)^(3/4)/(a^3*x^4 + a^3*x^2), x))/(5*a^3*x^3 - 10*I*a^3*x^2 - 5*a^3*x)

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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((a-I*a*x)**(3/4)/(a+I*a*x)**(9/4),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((a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x, algorithm="giac")

[Out]

Exception raised: TypeError