3.1192 \(\int \frac{1}{(a-i a x)^{7/4} (a+i a x)^{3/4}} \, dx\)

Optimal. Leaf size=82 \[ \frac{2 \left (x^2+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}(x),2\right )}{3 a (a-i a x)^{3/4} (a+i a x)^{3/4}}-\frac{2 i \sqrt [4]{a+i a x}}{3 a^2 (a-i a x)^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.0147231, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {51, 42, 233, 231} \[ \frac{2 \left (x^2+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{3 a (a-i a x)^{3/4} (a+i a x)^{3/4}}-\frac{2 i \sqrt [4]{a+i a x}}{3 a^2 (a-i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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{1}{(a-i a x)^{7/4} (a+i a x)^{3/4}} \, dx &=-\frac{2 i \sqrt [4]{a+i a x}}{3 a^2 (a-i a x)^{3/4}}+\frac{\int \frac{1}{(a-i a x)^{3/4} (a+i a x)^{3/4}} \, dx}{3 a}\\ &=-\frac{2 i \sqrt [4]{a+i a x}}{3 a^2 (a-i a x)^{3/4}}+\frac{\left (a^2+a^2 x^2\right )^{3/4} \int \frac{1}{\left (a^2+a^2 x^2\right )^{3/4}} \, dx}{3 a (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=-\frac{2 i \sqrt [4]{a+i a x}}{3 a^2 (a-i a x)^{3/4}}+\frac{\left (1+x^2\right )^{3/4} \int \frac{1}{\left (1+x^2\right )^{3/4}} \, dx}{3 a (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=-\frac{2 i \sqrt [4]{a+i a x}}{3 a^2 (a-i a x)^{3/4}}+\frac{2 \left (1+x^2\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{3 a (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0206751, size = 70, normalized size = 0.85 \[ -\frac{2 i \sqrt [4]{2} (1+i x)^{3/4} \, _2F_1\left (-\frac{3}{4},\frac{3}{4};\frac{1}{4};\frac{1}{2}-\frac{i x}{2}\right )}{3 a (a-i a x)^{3/4} (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a-iax \right ) ^{-{\frac{7}{4}}} \left ( a+iax \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(3/4),x)

[Out]

int(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{7}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(3/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(3/4)*(-I*a*x + a)^(7/4)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(3/4),x, algorithm="fricas")

[Out]

1/3*(3*(a^3*x + I*a^3)*integral(1/3*(I*a*x + a)^(1/4)*(-I*a*x + a)^(1/4)/(a^3*x^2 + a^3), x) + 2*(I*a*x + a)^(
1/4)*(-I*a*x + a)^(1/4))/(a^3*x + I*a^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a \left (i x + 1\right )\right )^{\frac{3}{4}} \left (- a \left (i x - 1\right )\right )^{\frac{7}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)**(7/4)/(a+I*a*x)**(3/4),x)

[Out]

Integral(1/((a*(I*x + 1))**(3/4)*(-a*(I*x - 1))**(7/4)), 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(1/(a-I*a*x)^(7/4)/(a+I*a*x)^(3/4),x, algorithm="giac")

[Out]

Exception raised: TypeError