3.1201 \(\int \frac {\sqrt [4]{a-i a x}}{(a+i a x)^{7/4}} \, dx\)

Optimal. Leaf size=79 \[ \frac {4 i \sqrt [4]{a-i a x}}{3 a (a+i a x)^{3/4}}-\frac {2 \left (x^2+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \tan ^{-1}(x),2\right )}{3 (a-i a x)^{3/4} (a+i a x)^{3/4}} \]

[Out]

4/3*I*(a-I*a*x)^(1/4)/a/(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))*Elli
pticF(sin(1/2*arctan(x)),2^(1/2))/(a-I*a*x)^(3/4)/(a+I*a*x)^(3/4)

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Rubi [A]  time = 0.01, antiderivative size = 79, 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 = {47, 42, 233, 231} \[ \frac {4 i \sqrt [4]{a-i a x}}{3 a (a+i a x)^{3/4}}-\frac {2 \left (x^2+1\right )^{3/4} F\left (\left .\frac {1}{2} \tan ^{-1}(x)\right |2\right )}{3 (a-i a x)^{3/4} (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((4*I)/3)*(a - I*a*x)^(1/4))/(a*(a + I*a*x)^(3/4)) - (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 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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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

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Mathematica [C]  time = 0.02, size = 70, normalized size = 0.89 \[ \frac {i \sqrt [4]{2} (1+i x)^{3/4} (a-i a x)^{5/4} \, _2F_1\left (\frac {5}{4},\frac {7}{4};\frac {9}{4};\frac {1}{2}-\frac {i x}{2}\right )}{5 a^2 (a+i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/5)*2^(1/4)*(1 + I*x)^(3/4)*(a - I*a*x)^(5/4)*Hypergeometric2F1[5/4, 7/4, 9/4, 1/2 - (I/2)*x])/(a^2*(a + I*
a*x)^(3/4))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ \frac {3 \, {\left (a^{2} x - i \, a^{2}\right )} {\rm integral}\left (-\frac {{\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{3 \, {\left (a^{2} x^{2} + a^{2}\right )}}, x\right ) + 4 \, {\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{3 \, {\left (a^{2} x - i \, a^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (-i a x +a \right )^{\frac {1}{4}}}{\left (i a x +a \right )^{\frac {7}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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