3.13.4 \(\int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{7/4}} \, dx\) [1204]

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

[Out]

-2/7*I/a^2/(a-I*a*x)^(7/4)/(a+I*a*x)^(3/4)+10/21*x/a^3/(a-I*a*x)^(3/4)/(a+I*a*x)^(3/4)+10/21*(x^2+1)^(3/4)*(co
s(1/2*arctan(x))^2)^(1/2)/cos(1/2*arctan(x))*EllipticF(sin(1/2*arctan(x)),2^(1/2))/a^3/(a-I*a*x)^(3/4)/(a+I*a*
x)^(3/4)

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {53, 42, 205, 239, 237} \begin {gather*} \frac {10 \left (x^2+1\right )^{3/4} F\left (\left .\frac {\text {ArcTan}(x)}{2}\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}} \end {gather*}

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 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 53

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 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 239

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)^{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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.03, size = 70, normalized size = 0.61 \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/7*I)*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))

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (-i a x +a \right )^{\frac {11}{4}} \left (i a x +a \right )^{\frac {7}{4}}}\, dx\]

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)

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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 >> ECL says: THROW: The catch RAT-ERR is undefined.

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

1/21*(21*(a^5*x^3 + I*a^5*x^2 + a^5*x + 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))/(a^5*x^3 + I*a^5*x^2 + a^5*x + I*a^5)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (x - i\right )\right )^{\frac {7}{4}} \left (- i a \left (x + i\right )\right )^{\frac {11}{4}}}\, dx \end {gather*}

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]

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:ext_reduce Error: Bad Argument TypeDone

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (a-a\,x\,1{}\mathrm {i}\right )}^{11/4}\,{\left (a+a\,x\,1{}\mathrm {i}\right )}^{7/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________