3.1176 \(\int \frac{1}{(a-i a x)^{13/4} \sqrt [4]{a+i a x}} \, dx\)

Optimal. Leaf size=115 \[ \frac{2 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{15 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{2 i (a+i a x)^{3/4}}{9 a^2 (a-i a x)^{9/4}}-\frac{4 i}{15 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}} \]

[Out]

((-4*I)/15)/(a^2*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4)) - (((2*I)/9)*(a + I*a*x)^(3/4))/(a^2*(a - I*a*x)^(9/4))
+ (2*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2])/(15*a^3*(a - I*a*x)^(1/4)*(a + I*a*x)^(1/4))

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Rubi [A]  time = 0.02639, 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 = {51, 46, 42, 197, 196} \[ \frac{2 \sqrt [4]{x^2+1} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{15 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac{2 i (a+i a x)^{3/4}}{9 a^2 (a-i a x)^{9/4}}-\frac{4 i}{15 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*I)/15)/(a^2*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4)) - (((2*I)/9)*(a + I*a*x)^(3/4))/(a^2*(a - I*a*x)^(9/4))
+ (2*(1 + x^2)^(1/4)*EllipticE[ArcTan[x]/2, 2])/(15*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 46

Int[1/(((a_) + (b_.)*(x_))^(9/4)*((c_) + (d_.)*(x_))^(1/4)), x_Symbol] :> Simp[-4/(5*b*(a + b*x)^(5/4)*(c + d*
x)^(1/4)), x] - Dist[d/(5*b), 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{1}{(a-i a x)^{13/4} \sqrt [4]{a+i a x}} \, dx &=-\frac{2 i (a+i a x)^{3/4}}{9 a^2 (a-i a x)^{9/4}}+\frac{\int \frac{1}{(a-i a x)^{9/4} \sqrt [4]{a+i a x}} \, dx}{3 a}\\ &=-\frac{4 i}{15 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac{2 i (a+i a x)^{3/4}}{9 a^2 (a-i a x)^{9/4}}+\frac{\int \frac{1}{(a-i a x)^{5/4} (a+i a x)^{5/4}} \, dx}{15 a}\\ &=-\frac{4 i}{15 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac{2 i (a+i a x)^{3/4}}{9 a^2 (a-i a x)^{9/4}}+\frac{\sqrt [4]{a^2+a^2 x^2} \int \frac{1}{\left (a^2+a^2 x^2\right )^{5/4}} \, dx}{15 a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac{4 i}{15 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac{2 i (a+i a x)^{3/4}}{9 a^2 (a-i a x)^{9/4}}+\frac{\sqrt [4]{1+x^2} \int \frac{1}{\left (1+x^2\right )^{5/4}} \, dx}{15 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ &=-\frac{4 i}{15 a^2 (a-i a x)^{5/4} \sqrt [4]{a+i a x}}-\frac{2 i (a+i a x)^{3/4}}{9 a^2 (a-i a x)^{9/4}}+\frac{2 \sqrt [4]{1+x^2} E\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{15 a^3 \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}\\ \end{align*}

Mathematica [C]  time = 0.0277512, size = 70, normalized size = 0.61 \[ -\frac{2 i 2^{3/4} \sqrt [4]{1+i x} \, _2F_1\left (-\frac{9}{4},\frac{1}{4};-\frac{5}{4};\frac{1}{2}-\frac{i x}{2}\right )}{9 a (a-i a x)^{9/4} \sqrt [4]{a+i a x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.059, size = 113, normalized size = 1. \begin{align*}{\frac{12\,i{x}^{2}+6\,{x}^{3}-4\,x+22\,i}{45\, \left ( x+i \right ) ^{2}{a}^{3}}{\frac{1}{\sqrt [4]{-a \left ( -1+ix \right ) }}}{\frac{1}{\sqrt [4]{a \left ( 1+ix \right ) }}}}-{\frac{x}{15\,{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)^(13/4)/(a+I*a*x)^(1/4),x)

[Out]

2/45*(6*I*x^2+3*x^3-2*x+11*I)/(x+I)^2/a^3/(-a*(-1+I*x))^(1/4)/(a*(1+I*x))^(1/4)-1/15/(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a x + a\right )}^{\frac{1}{4}}{\left (-i \, a x + a\right )}^{\frac{13}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((I*a*x + a)^(1/4)*(-I*a*x + a)^(13/4)), 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 (3 \, x^{2} + 9 i \, x - 11\right )} +{\left (45 \, a^{5} x^{3} + 135 i \, a^{5} x^{2} - 135 \, a^{5} x - 45 i \, a^{5}\right )}{\rm integral}\left (-\frac{{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{3}{4}}}{15 \,{\left (a^{5} x^{2} + a^{5}\right )}}, x\right )}{45 \, a^{5} x^{3} + 135 i \, a^{5} x^{2} - 135 \, a^{5} x - 45 i \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)*(3*x^2 + 9*I*x - 11) + (45*a^5*x^3 + 135*I*a^5*x^2 - 135*a^5*x - 45*I*
a^5)*integral(-1/15*(I*a*x + a)^(3/4)*(-I*a*x + a)^(3/4)/(a^5*x^2 + a^5), x))/(45*a^5*x^3 + 135*I*a^5*x^2 - 13
5*a^5*x - 45*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)**(13/4)/(a+I*a*x)**(1/4),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)^(13/4)/(a+I*a*x)^(1/4),x, algorithm="giac")

[Out]

Exception raised: TypeError