3.1204 \(\int \frac{1}{(a-i a x)^{11/4} (a+i a x)^{7/4}} \, dx\)

Optimal. Leaf size=114 \[ \frac{10 \left (x^2+1\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \tan ^{-1}(x),2\right )}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}+\frac{10 x}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}-\frac{2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{3/4}} \]

[Out]

((-2*I)/7)/(a^2*(a - I*a*x)^(7/4)*(a + I*a*x)^(3/4)) + (10*x)/(21*a^3*(a - I*a*x)^(3/4)*(a + I*a*x)^(3/4)) + (
10*(1 + x^2)^(3/4)*EllipticF[ArcTan[x]/2, 2])/(21*a^3*(a - I*a*x)^(3/4)*(a + I*a*x)^(3/4))

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Rubi [A]  time = 0.0246717, antiderivative size = 114, 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, 42, 199, 233, 231} \[ \frac{10 \left (x^2+1\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}+\frac{10 x}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}-\frac{2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)/7)/(a^2*(a - I*a*x)^(7/4)*(a + I*a*x)^(3/4)) + (10*x)/(21*a^3*(a - I*a*x)^(3/4)*(a + I*a*x)^(3/4)) + (
10*(1 + x^2)^(3/4)*EllipticF[ArcTan[x]/2, 2])/(21*a^3*(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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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)^{11/4} (a+i a x)^{7/4}} \, dx &=-\frac{2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{3/4}}+\frac{5 \int \frac{1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx}{7 a}\\ &=-\frac{2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{3/4}}+\frac{\left (5 \left (a^2+a^2 x^2\right )^{3/4}\right ) \int \frac{1}{\left (a^2+a^2 x^2\right )^{7/4}} \, dx}{7 a (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=-\frac{2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{3/4}}+\frac{10 x}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}+\frac{\left (5 \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}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=-\frac{2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{3/4}}+\frac{10 x}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}+\frac{\left (5 \left (1+x^2\right )^{3/4}\right ) \int \frac{1}{\left (1+x^2\right )^{3/4}} \, dx}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=-\frac{2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{3/4}}+\frac{10 x}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}+\frac{10 \left (1+x^2\right )^{3/4} F\left (\left .\frac{1}{2} \tan ^{-1}(x)\right |2\right )}{21 a^3 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0239229, size = 70, normalized size = 0.61 \[ -\frac{i \sqrt [4]{2} (1+i x)^{3/4} \, _2F_1\left (-\frac{7}{4},\frac{7}{4};-\frac{3}{4};\frac{1}{2}-\frac{i x}{2}\right )}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a-iax \right ) ^{-{\frac{11}{4}}} \left ( a+iax \right ) ^{-{\frac{7}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-I*a*x)^(11/4)/(a+I*a*x)^(7/4),x)

[Out]

int(1/(a-I*a*x)^(11/4)/(a+I*a*x)^(7/4),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(11/4)/(a+I*a*x)^(7/4),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (21 \, a^{5} x^{3} + 21 i \, a^{5} x^{2} + 21 \, a^{5} x + 21 i \, a^{5}\right )}{\rm integral}\left (\frac{5 \,{\left (i \, a x + a\right )}^{\frac{1}{4}}{\left (-i \, a x + a\right )}^{\frac{1}{4}}}{21 \,{\left (a^{5} x^{2} + a^{5}\right )}}, x\right ) + 2 \,{\left (i \, a x + a\right )}^{\frac{1}{4}}{\left (-i \, a x + a\right )}^{\frac{1}{4}}{\left (5 \, x^{2} + 5 i \, x + 3\right )}}{21 \, a^{5} x^{3} + 21 i \, a^{5} x^{2} + 21 \, a^{5} x + 21 i \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(11/4)/(a+I*a*x)^(7/4),x, algorithm="fricas")

[Out]

((21*a^5*x^3 + 21*I*a^5*x^2 + 21*a^5*x + 21*I*a^5)*integral(5/21*(I*a*x + a)^(1/4)*(-I*a*x + a)^(1/4)/(a^5*x^2
+ a^5), x) + 2*(I*a*x + a)^(1/4)*(-I*a*x + a)^(1/4)*(5*x^2 + 5*I*x + 3))/(21*a^5*x^3 + 21*I*a^5*x^2 + 21*a^5*
x + 21*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)**(11/4)/(a+I*a*x)**(7/4),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)^(11/4)/(a+I*a*x)^(7/4),x, algorithm="giac")

[Out]

Exception raised: TypeError