3.12.81 \(\int \frac {1}{(a-i a x)^{11/4} \sqrt [4]{a+i a x}} \, dx\) [1181]

Optimal. Leaf size=67 \[ -\frac {2 i (a+i a x)^{3/4}}{7 a^2 (a-i a x)^{7/4}}-\frac {4 i (a+i a x)^{3/4}}{21 a^3 (a-i a x)^{3/4}} \]

[Out]

-2/7*I*(a+I*a*x)^(3/4)/a^2/(a-I*a*x)^(7/4)-4/21*I*(a+I*a*x)^(3/4)/a^3/(a-I*a*x)^(3/4)

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Rubi [A]
time = 0.01, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, 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 {4 i (a+i a x)^{3/4}}{21 a^3 (a-i a x)^{3/4}}-\frac {2 i (a+i a x)^{3/4}}{7 a^2 (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)^(1/4)),x]

[Out]

(((-2*I)/7)*(a + I*a*x)^(3/4))/(a^2*(a - I*a*x)^(7/4)) - (((4*I)/21)*(a + I*a*x)^(3/4))/(a^3*(a - I*a*x)^(3/4)
)

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} \sqrt [4]{a+i a x}} \, dx &=-\frac {2 i (a+i a x)^{3/4}}{7 a^2 (a-i a x)^{7/4}}+\frac {2 \int \frac {1}{(a-i a x)^{7/4} \sqrt [4]{a+i a x}} \, dx}{7 a}\\ &=-\frac {2 i (a+i a x)^{3/4}}{7 a^2 (a-i a x)^{7/4}}-\frac {4 i (a+i a x)^{3/4}}{21 a^3 (a-i a x)^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 45, normalized size = 0.67 \begin {gather*} \frac {2 (5-2 i x) (a+i a x)^{3/4}}{21 a^3 (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)^(1/4)),x]

[Out]

(2*(5 - (2*I)*x)*(a + I*a*x)^(3/4))/(21*a^3*(I + x)*(a - I*a*x)^(3/4))

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Maple [A]
time = 0.15, size = 44, normalized size = 0.66

method	result	size
risch	\(\frac {\frac {4}{21} x^{2}+\frac {2}{7} i x +\frac {10}{21}}{a^{2} \left (-a \left (i x -1\right )\right )^{\frac {3}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}} \left (x +i\right )}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 2.39, size = 44, normalized size = 0.66 \begin {gather*} \frac {2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} {\left (2 \, x + 5 i\right )}}{21 \, {\left (a^{4} x^{2} + 2 i \, a^{4} x - a^{4}\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)^(1/4),x, algorithm="fricas")

[Out]

2/21*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)*(2*x + 5*I)/(a^4*x^2 + 2*I*a^4*x - a^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt [4]{i a \left (x - i\right )} \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)**(1/4),x)

[Out]

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

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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)^(1/4),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 = 0.67, size = 46, normalized size = 0.69 \begin {gather*} -\frac {{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}\,\left (2\,x^2+x\,3{}\mathrm {i}+5\right )\,2{}\mathrm {i}}{21\,a^3\,{\left (-1+x\,1{}\mathrm {i}\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)^(1/4)),x)

[Out]

-((-a*(x*1i - 1))^(1/4)*(x*3i + 2*x^2 + 5)*2i)/(21*a^3*(x*1i - 1)^2*(a*(x*1i + 1))^(1/4))

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