3.1172 \(\int \frac{(a-i a x)^{3/4}}{\sqrt [4]{a+i a x}} \, dx\)

Optimal. Leaf size=106 \[ -\frac{2 a \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac{2 a x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{2 i (a-i a x)^{3/4} (a+i a x)^{3/4}}{3 a} \]

[Out]

(2*a*x)/((a - I*a*x)^(1/4)*(a + I*a*x)^(1/4)) - (((2*I)/3)*(a - I*a*x)^(3/4)*(a + I*a*x)^(3/4))/a - (2*a*(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 = 0.0223588, antiderivative size = 106, 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 = {50, 42, 229, 227, 196} \[ -\frac{2 a \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac{2 a x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{2 i (a-i a x)^{3/4} (a+i a x)^{3/4}}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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}}{\sqrt [4]{a+i a x}} \, dx &=-\frac{2 i (a-i a x)^{3/4} (a+i a x)^{3/4}}{3 a}+a \int \frac{1}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \, dx\\ &=-\frac{2 i (a-i a x)^{3/4} (a+i a x)^{3/4}}{3 a}+\frac{\left (a \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{2 i (a-i a x)^{3/4} (a+i a x)^{3/4}}{3 a}+\frac{\left (a \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{2 a x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{2 i (a-i a x)^{3/4} (a+i a x)^{3/4}}{3 a}-\frac{\left (a \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{2 a x}{\sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{2 i (a-i a x)^{3/4} (a+i a x)^{3/4}}{3 a}-\frac{2 a \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 = 0.0227682, size = 70, normalized size = 0.66 \[ \frac{2 i 2^{3/4} \sqrt [4]{1+i x} (a-i a x)^{7/4} \, _2F_1\left (\frac{1}{4},\frac{7}{4};\frac{11}{4};\frac{1}{2}-\frac{i x}{2}\right )}{7 a \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.039, size = 94, normalized size = 0.9 \begin{align*}{-{\frac{2\,i}{3}} \left ( x+i \right ) \left ( x-i \right ) a{\frac{1}{\sqrt [4]{-a \left ( -1+ix \right ) }}}{\frac{1}{\sqrt [4]{a \left ( 1+ix \right ) }}}}+{ax{\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)^(1/4),x)

[Out]

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

[Out]

integrate((-I*a*x + a)^(3/4)/(I*a*x + a)^(1/4), x)

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

[Out]

1/3*(3*a*x*integral(2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)/(a*x^4 + a*x^2), x) - 2*(I*a*x + a)^(3/4)*(-I*a*x +
a)^(3/4)*(I*x - 3))/(a*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}}}{\sqrt [4]{a \left (i x + 1\right )}}\, 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)**(1/4),x)

[Out]

Integral((-a*(I*x - 1))**(3/4)/(a*(I*x + 1))**(1/4), 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)^(1/4),x, algorithm="giac")

[Out]

Exception raised: TypeError