3.13.28 \(\int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{9/4}} \, dx\) [1228]

Optimal. Leaf size=133 \[ -\frac {2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{5/4}}-\frac {4 i}{7 a^3 (a-i a x)^{3/4} (a+i a x)^{5/4}}+\frac {16 i \sqrt [4]{a-i a x}}{35 a^4 (a+i a x)^{5/4}}+\frac {32 i \sqrt [4]{a-i a x}}{35 a^5 \sqrt [4]{a+i a x}} \]

[Out]

-2/7*I/a^2/(a-I*a*x)^(7/4)/(a+I*a*x)^(5/4)-4/7*I/a^3/(a-I*a*x)^(3/4)/(a+I*a*x)^(5/4)+16/35*I*(a-I*a*x)^(1/4)/a
^4/(a+I*a*x)^(5/4)+32/35*I*(a-I*a*x)^(1/4)/a^5/(a+I*a*x)^(1/4)

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Rubi [A]
time = 0.02, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {47, 37} \begin {gather*} \frac {32 i \sqrt [4]{a-i a x}}{35 a^5 \sqrt [4]{a+i a x}}+\frac {16 i \sqrt [4]{a-i a x}}{35 a^4 (a+i a x)^{5/4}}-\frac {4 i}{7 a^3 (a+i a x)^{5/4} (a-i a x)^{3/4}}-\frac {2 i}{7 a^2 (a+i a x)^{5/4} (a-i a x)^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a - I*a*x)^(11/4)*(a + I*a*x)^(9/4)),x]

[Out]

((-2*I)/7)/(a^2*(a - I*a*x)^(7/4)*(a + I*a*x)^(5/4)) - ((4*I)/7)/(a^3*(a - I*a*x)^(3/4)*(a + I*a*x)^(5/4)) + (
((16*I)/35)*(a - I*a*x)^(1/4))/(a^4*(a + I*a*x)^(5/4)) + (((32*I)/35)*(a - I*a*x)^(1/4))/(a^5*(a + I*a*x)^(1/4
))

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
 NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rubi steps

\begin {align*} \int \frac {1}{(a-i a x)^{11/4} (a+i a x)^{9/4}} \, dx &=-\frac {2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{5/4}}+\frac {6 \int \frac {1}{(a-i a x)^{7/4} (a+i a x)^{9/4}} \, dx}{7 a}\\ &=-\frac {2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{5/4}}-\frac {4 i}{7 a^3 (a-i a x)^{3/4} (a+i a x)^{5/4}}+\frac {8 \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx}{7 a^2}\\ &=-\frac {2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{5/4}}-\frac {4 i}{7 a^3 (a-i a x)^{3/4} (a+i a x)^{5/4}}+\frac {16 i \sqrt [4]{a-i a x}}{35 a^4 (a+i a x)^{5/4}}+\frac {16 \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{5/4}} \, dx}{35 a^3}\\ &=-\frac {2 i}{7 a^2 (a-i a x)^{7/4} (a+i a x)^{5/4}}-\frac {4 i}{7 a^3 (a-i a x)^{3/4} (a+i a x)^{5/4}}+\frac {16 i \sqrt [4]{a-i a x}}{35 a^4 (a+i a x)^{5/4}}+\frac {32 i \sqrt [4]{a-i a x}}{35 a^5 \sqrt [4]{a+i a x}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 64, normalized size = 0.48 \begin {gather*} \frac {2 (a+i a x)^{3/4} \left (9-22 i x+8 x^2-16 i x^3\right )}{35 a^5 (-i+x)^2 (i+x) (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)^(9/4)),x]

[Out]

(2*(a + I*a*x)^(3/4)*(9 - (22*I)*x + 8*x^2 - (16*I)*x^3))/(35*a^5*(-I + x)^2*(I + x)*(a - I*a*x)^(3/4))

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Maple [A]
time = 0.16, size = 56, normalized size = 0.42

method result size
risch \(\frac {\frac {32}{35} x^{3}+\frac {16}{35} i x^{2}+\frac {44}{35} x +\frac {18}{35} i}{a^{4} \left (-a \left (i x -1\right )\right )^{\frac {3}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}} \left (x -i\right ) \left (x +i\right )}\) \(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-I*a*x)^(11/4)/(a+I*a*x)^(9/4),x,method=_RETURNVERBOSE)

[Out]

2/35/a^4/(-a*(-1+I*x))^(3/4)/(a*(1+I*x))^(1/4)*(16*x^3+8*I*x^2+22*x+9*I)/(x-I)/(x+I)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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)^(9/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(9/4)*(-I*a*x + a)^(11/4)), x)

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Fricas [A]
time = 0.96, size = 54, normalized size = 0.41 \begin {gather*} \frac {2 \, {\left (16 \, x^{3} + 8 i \, x^{2} + 22 \, x + 9 i\right )} {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}}}{35 \, {\left (a^{6} x^{4} + 2 \, a^{6} x^{2} + a^{6}\right )}} \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)^(9/4),x, algorithm="fricas")

[Out]

2/35*(16*x^3 + 8*I*x^2 + 22*x + 9*I)*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)/(a^6*x^4 + 2*a^6*x^2 + a^6)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \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)**(9/4),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3656 deep

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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)^(9/4),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

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Mupad [B]
time = 0.69, size = 56, normalized size = 0.42 \begin {gather*} \frac {2\,{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}\,\left (x^4\,16{}\mathrm {i}+8\,x^3+x^2\,30{}\mathrm {i}+13\,x+9{}\mathrm {i}\right )}{35\,a^5\,{\left (x^2+1\right )}^2\,{\left (a\,\left (1+x\,1{}\mathrm {i}\right )\right )}^{1/4}} \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)^(9/4)),x)

[Out]

(2*(-a*(x*1i - 1))^(1/4)*(13*x + x^2*30i + 8*x^3 + x^4*16i + 9i))/(35*a^5*(x^2 + 1)^2*(a*(x*1i + 1))^(1/4))

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