3.1203 \(\int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx\)

Optimal. Leaf size=81 \[ \frac {2 \left (x^2+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}(x),2\right )}{3 a^2 (a-i a x)^{3/4} (a+i a x)^{3/4}}+\frac {2 x}{3 a^2 (a-i a x)^{3/4} (a+i a x)^{3/4}} \]

[Out]

2/3*x/a^2/(a-I*a*x)^(3/4)/(a+I*a*x)^(3/4)+2/3*(x^2+1)^(3/4)*(cos(1/2*arctan(x))^2)^(1/2)/cos(1/2*arctan(x))*El
lipticF(sin(1/2*arctan(x)),2^(1/2))/a^2/(a-I*a*x)^(3/4)/(a+I*a*x)^(3/4)

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Rubi [A]  time = 0.01, antiderivative size = 81, 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 = {42, 199, 233, 231} \[ \frac {2 \left (x^2+1\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{3 a^2 (a-i a x)^{3/4} (a+i a x)^{3/4}}+\frac {2 x}{3 a^2 (a-i a x)^{3/4} (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{7/4}} \, dx &=\frac {\left (a^2+a^2 x^2\right )^{3/4} \int \frac {1}{\left (a^2+a^2 x^2\right )^{7/4}} \, dx}{(a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=\frac {2 x}{3 a^2 (a-i a x)^{3/4} (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^2 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=\frac {2 x}{3 a^2 (a-i a x)^{3/4} (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^2 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ &=\frac {2 x}{3 a^2 (a-i a x)^{3/4} (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^2 (a-i a x)^{3/4} (a+i a x)^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 70, normalized size = 0.86 \[ -\frac {i \sqrt [4]{2} (1+i x)^{3/4} \, _2F_1\left (-\frac {3}{4},\frac {7}{4};\frac {1}{4};\frac {1}{2}-\frac {i x}{2}\right )}{3 a^2 (a-i a x)^{3/4} (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ \frac {3 \, {\left (a^{4} x^{2} + a^{4}\right )} {\rm integral}\left (\frac {{\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{3 \, {\left (a^{4} x^{2} + a^{4}\right )}}, x\right ) + 2 \, {\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} x}{3 \, {\left (a^{4} x^{2} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a x + a\right )}^{\frac {7}{4}} {\left (-i \, a x + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (-i a x +a \right )^{\frac {7}{4}} \left (i a x +a \right )^{\frac {7}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a x + a\right )}^{\frac {7}{4}} {\left (-i \, a x + a\right )}^{\frac {7}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{7/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{7/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 36.70, size = 95, normalized size = 1.17 \[ - \frac {i {G_{6, 6}^{5, 3}\left (\begin {matrix} \frac {7}{8}, \frac {11}{8}, 1 & \frac {1}{2}, \frac {7}{4}, \frac {9}{4} \\\frac {7}{8}, \frac {5}{4}, \frac {11}{8}, \frac {7}{4}, \frac {9}{4} & 0 \end {matrix} \middle | {\frac {e^{- 3 i \pi }}{x^{2}}} \right )} e^{- \frac {i \pi }{4}}}{4 \pi a^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right )} + \frac {i {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, 0, \frac {3}{8}, \frac {1}{2}, \frac {7}{8}, 1 & \\\frac {3}{8}, \frac {7}{8} & - \frac {1}{2}, 0, \frac {5}{4}, 0 \end {matrix} \middle | {\frac {e^{- i \pi }}{x^{2}}} \right )}}{4 \pi a^{\frac {7}{2}} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*meijerg(((7/8, 11/8, 1), (1/2, 7/4, 9/4)), ((7/8, 5/4, 11/8, 7/4, 9/4), (0,)), exp_polar(-3*I*pi)/x**2)*exp
(-I*pi/4)/(4*pi*a**(7/2)*gamma(7/4)) + I*meijerg(((-1/2, 0, 3/8, 1/2, 7/8, 1), ()), ((3/8, 7/8), (-1/2, 0, 5/4
, 0)), exp_polar(-I*pi)/x**2)/(4*pi*a**(7/2)*gamma(7/4))

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