3.12.6 \(\int \frac {1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx\)

Optimal. Leaf size=100 \[ -\frac {16 i (a+i a x)^{3/4}}{231 a^4 (a-i a x)^{3/4}}-\frac {8 i (a+i a x)^{3/4}}{77 a^3 (a-i a x)^{7/4}}-\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}} \]

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Rubi [A]  time = 0.02, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {45, 37} \begin {gather*} -\frac {16 i (a+i a x)^{3/4}}{231 a^4 (a-i a x)^{3/4}}-\frac {8 i (a+i a x)^{3/4}}{77 a^3 (a-i a x)^{7/4}}-\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a - I*a*x)^(15/4)*(a + I*a*x)^(1/4)),x]

[Out]

(((-2*I)/11)*(a + I*a*x)^(3/4))/(a^2*(a - I*a*x)^(11/4)) - (((8*I)/77)*(a + I*a*x)^(3/4))/(a^3*(a - I*a*x)^(7/
4)) - (((16*I)/231)*(a + I*a*x)^(3/4))/(a^4*(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 45

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)^{15/4} \sqrt [4]{a+i a x}} \, dx &=-\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}}+\frac {4 \int \frac {1}{(a-i a x)^{11/4} \sqrt [4]{a+i a x}} \, dx}{11 a}\\ &=-\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}}-\frac {8 i (a+i a x)^{3/4}}{77 a^3 (a-i a x)^{7/4}}+\frac {8 \int \frac {1}{(a-i a x)^{7/4} \sqrt [4]{a+i a x}} \, dx}{77 a^2}\\ &=-\frac {2 i (a+i a x)^{3/4}}{11 a^2 (a-i a x)^{11/4}}-\frac {8 i (a+i a x)^{3/4}}{77 a^3 (a-i a x)^{7/4}}-\frac {16 i (a+i a x)^{3/4}}{231 a^4 (a-i a x)^{3/4}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 52, normalized size = 0.52 \begin {gather*} \frac {2 \left (-8 i x^2+28 x+41 i\right ) (a+i a x)^{3/4}}{231 a^4 (x+i)^2 (a-i a x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a - I*a*x)^(15/4)*(a + I*a*x)^(1/4)),x]

[Out]

(2*(a + I*a*x)^(3/4)*(41*I + 28*x - (8*I)*x^2))/(231*a^4*(I + x)^2*(a - I*a*x)^(3/4))

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IntegrateAlgebraic [A]  time = 0.13, size = 77, normalized size = 0.77 \begin {gather*} -\frac {i (a+i a x)^{11/4} \left (\frac {77 (a-i a x)^2}{(a+i a x)^2}+\frac {66 (a-i a x)}{a+i a x}+21\right )}{462 a^4 (a-i a x)^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((a - I*a*x)^(15/4)*(a + I*a*x)^(1/4)),x]

[Out]

((-1/462*I)*(a + I*a*x)^(11/4)*(21 + (77*(a - I*a*x)^2)/(a + I*a*x)^2 + (66*(a - I*a*x))/(a + I*a*x)))/(a^4*(a
- I*a*x)^(11/4))

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fricas [A]  time = 1.42, size = 58, normalized size = 0.58 \begin {gather*} \frac {2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} {\left (8 \, x^{2} + 28 i \, x - 41\right )}}{231 \, a^{5} x^{3} + 693 i \, a^{5} x^{2} - 693 \, a^{5} x - 231 i \, a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(15/4)/(a+I*a*x)^(1/4),x, algorithm="fricas")

[Out]

2*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)*(8*x^2 + 28*I*x - 41)/(231*a^5*x^3 + 693*I*a^5*x^2 - 693*a^5*x - 231*I*
a^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {15}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(15/4)/(a+I*a*x)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((I*a*x + a)^(1/4)*(-I*a*x + a)^(15/4)), x)

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maple [A]  time = 0.05, size = 50, normalized size = 0.50 \begin {gather*} \frac {\frac {16}{231} x^{3}+\frac {40}{231} i x^{2}-\frac {26}{231} x +\frac {82}{231} i}{\left (-\left (i x -1\right ) a \right )^{\frac {3}{4}} \left (\left (i x +1\right ) a \right )^{\frac {1}{4}} \left (x +i\right )^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-I*a*x+a)^(15/4)/(I*a*x+a)^(1/4),x)

[Out]

2/231/a^3/(-(I*x-1)*a)^(3/4)/((I*x+1)*a)^(1/4)*(20*I*x^2+8*x^3-13*x+41*I)/(x+I)^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (i \, a x + a\right )}^{\frac {1}{4}} {\left (-i \, a x + a\right )}^{\frac {15}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(15/4)/(a+I*a*x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(1/4)*(-I*a*x + a)^(15/4)), x)

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mupad [B]  time = 0.75, size = 51, normalized size = 0.51 \begin {gather*} \frac {{\left (x-\mathrm {i}\right )}^4\,{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}\,\left (8\,x^2+x\,28{}\mathrm {i}-41\right )\,2{}\mathrm {i}}{231\,a^4\,{\left (x^2+1\right )}^3\,{\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)^(15/4)*(a + a*x*1i)^(1/4)),x)

[Out]

((x - 1i)^4*(-a*(x*1i - 1))^(1/4)*(x*28i + 8*x^2 - 41)*2i)/(231*a^4*(x^2 + 1)^3*(a*(x*1i + 1))^(1/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)**(15/4)/(a+I*a*x)**(1/4),x)

[Out]

Timed out

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