3.12.74 \(\int \frac {1}{(a-i a x)^{5/4} \sqrt [4]{a+i a x}} \, dx\) [1174]

Optimal. Leaf size=78 \[ -\frac {2 i}{a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}+\frac {2 \sqrt [4]{1+x^2} E\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \]

[Out]

-2*I/a/(a-I*a*x)^(1/4)/(a+I*a*x)^(1/4)+2*(x^2+1)^(1/4)*(cos(1/2*arctan(x))^2)^(1/2)/cos(1/2*arctan(x))*Ellipti
cE(sin(1/2*arctan(x)),2^(1/2))/a/(a-I*a*x)^(1/4)/(a+I*a*x)^(1/4)

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Rubi [A]
time = 0.01, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {50, 42, 203, 202} \begin {gather*} \frac {2 \sqrt [4]{x^2+1} E\left (\left .\frac {\text {ArcTan}(x)}{2}\right |2\right )}{a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}}-\frac {2 i}{a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 50

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 202

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

Rule 203

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]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.03, size = 68, normalized size = 0.87 \begin {gather*} -\frac {2 i 2^{3/4} \sqrt [4]{1+i x} \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};\frac {1}{2}-\frac {i x}{2}\right )}{a \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 5 vs. order 4.
time = 0.17, size = 94, normalized size = 1.21

method	result	size
risch	\(\frac {2 x -2 i}{a \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}-\frac {x \hypergeom \left (\left [\frac {1}{4}, \frac {1}{2}\right ], \left [\frac {3}{2}\right ], -x^{2}\right ) \left (-a^{2} \left (i x -1\right ) \left (i x +1\right )\right )^{\frac {1}{4}}}{\left (a^{2}\right )^{\frac {1}{4}} a \left (-a \left (i x -1\right )\right )^{\frac {1}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}}}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-I*a*x)^(5/4)/(a+I*a*x)^(1/4),x,method=_RETURNVERBOSE)

[Out]

2*(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{i a \left (x - i\right )} \left (- i a \left (x + i\right )\right )^{\frac {5}{4}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:ext_reduce Error: Bad Argument TypeDone

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{5/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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