3.1189 \(\int \frac{(a-i a x)^{5/4}}{(a+i a x)^{3/4}} \, dx\)

Optimal. Leaf size=112 \[ \frac{10 a^2 \left (x^2+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}(x),2\right )}{3 (a-i a x)^{3/4} (a+i a x)^{3/4}}-\frac{10}{3} i \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}-\frac{2 i (a-i a x)^{5/4} \sqrt [4]{a+i a x}}{3 a} \]

[Out]

((-10*I)/3)*(a - I*a*x)^(1/4)*(a + I*a*x)^(1/4) - (((2*I)/3)*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4))/a + (10*a^2*
(1 + x^2)^(3/4)*EllipticF[ArcTan[x]/2, 2])/(3*(a - I*a*x)^(3/4)*(a + I*a*x)^(3/4))

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Rubi [A]  time = 0.0241203, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {50, 42, 233, 231} \[ \frac{10 a^2 \left (x^2+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{3 (a-i a x)^{3/4} (a+i a x)^{3/4}}-\frac{10}{3} i \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}-\frac{2 i (a-i a x)^{5/4} \sqrt [4]{a+i a x}}{3 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-10*I)/3)*(a - I*a*x)^(1/4)*(a + I*a*x)^(1/4) - (((2*I)/3)*(a - I*a*x)^(5/4)*(a + I*a*x)^(1/4))/a + (10*a^2*
(1 + x^2)^(3/4)*EllipticF[ArcTan[x]/2, 2])/(3*(a - I*a*x)^(3/4)*(a + I*a*x)^(3/4))

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 233

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + (b*x^2
)/a)^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*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)^{5/4}}{(a+i a x)^{3/4}} \, dx &=-\frac{2 i (a-i a x)^{5/4} \sqrt [4]{a+i a x}}{3 a}+\frac{1}{3} (5 a) \int \frac{\sqrt [4]{a-i a x}}{(a+i a x)^{3/4}} \, dx\\ &=-\frac{10}{3} i \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}-\frac{2 i (a-i a x)^{5/4} \sqrt [4]{a+i a x}}{3 a}+\frac{1}{3} \left (5 a^2\right ) \int \frac{1}{(a-i a x)^{3/4} (a+i a x)^{3/4}} \, dx\\ &=-\frac{10}{3} i \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}-\frac{2 i (a-i a x)^{5/4} \sqrt [4]{a+i a x}}{3 a}+\frac{\left (5 a^2 \left (a^2+a^2 x^2\right )^{3/4}\right ) \int \frac{1}{\left (a^2+a^2 x^2\right )^{3/4}} \, dx}{3 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=-\frac{10}{3} i \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}-\frac{2 i (a-i a x)^{5/4} \sqrt [4]{a+i a x}}{3 a}+\frac{\left (5 a^2 \left (1+x^2\right )^{3/4}\right ) \int \frac{1}{\left (1+x^2\right )^{3/4}} \, dx}{3 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=-\frac{10}{3} i \sqrt [4]{a-i a x} \sqrt [4]{a+i a x}-\frac{2 i (a-i a x)^{5/4} \sqrt [4]{a+i a x}}{3 a}+\frac{10 a^2 \left (1+x^2\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{3 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0261663, size = 70, normalized size = 0.62 \[ \frac{2 i \sqrt [4]{2} (1+i x)^{3/4} (a-i a x)^{9/4} \, _2F_1\left (\frac{3}{4},\frac{9}{4};\frac{13}{4};\frac{1}{2}-\frac{i x}{2}\right )}{9 a (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)/9)*2^(1/4)*(1 + I*x)^(3/4)*(a - I*a*x)^(9/4)*Hypergeometric2F1[3/4, 9/4, 13/4, 1/2 - (I/2)*x])/(a*(a +
I*a*x)^(3/4))

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Maple [F]  time = 0.054, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a-iax \right ) ^{{\frac{5}{4}}} \left ( a+iax \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{3} \,{\left (i \, a x + a\right )}^{\frac{1}{4}}{\left (-i \, a x + a\right )}^{\frac{1}{4}}{\left (2 \, x + 12 i\right )} +{\rm integral}\left (\frac{5 \,{\left (i \, a x + a\right )}^{\frac{1}{4}}{\left (-i \, a x + a\right )}^{\frac{1}{4}}}{3 \,{\left (x^{2} + 1\right )}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(I*a*x + a)^(1/4)*(-I*a*x + a)^(1/4)*(2*x + 12*I) + integral(5/3*(I*a*x + a)^(1/4)*(-I*a*x + a)^(1/4)/(x^
2 + 1), 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{5}{4}}}{\left (a \left (i x + 1\right )\right )^{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((-a*(I*x - 1))**(5/4)/(a*(I*x + 1))**(3/4), 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)^(5/4)/(a+I*a*x)^(3/4),x, algorithm="giac")

[Out]

Exception raised: TypeError