3.1183 \(\int \frac{1}{(a-i a x)^{19/4} \sqrt [4]{a+i a x}} \, dx\)

Optimal. Leaf size=133 \[ -\frac{32 i (a+i a x)^{3/4}}{1155 a^5 (a-i a x)^{3/4}}-\frac{16 i (a+i a x)^{3/4}}{385 a^4 (a-i a x)^{7/4}}-\frac{4 i (a+i a x)^{3/4}}{55 a^3 (a-i a x)^{11/4}}-\frac{2 i (a+i a x)^{3/4}}{15 a^2 (a-i a x)^{15/4}} \]

[Out]

(((-2*I)/15)*(a + I*a*x)^(3/4))/(a^2*(a - I*a*x)^(15/4)) - (((4*I)/55)*(a + I*a*x)^(3/4))/(a^3*(a - I*a*x)^(11
/4)) - (((16*I)/385)*(a + I*a*x)^(3/4))/(a^4*(a - I*a*x)^(7/4)) - (((32*I)/1155)*(a + I*a*x)^(3/4))/(a^5*(a -
I*a*x)^(3/4))

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Rubi [A]  time = 0.0283329, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.08, Rules used = {45, 37} \[ -\frac{32 i (a+i a x)^{3/4}}{1155 a^5 (a-i a x)^{3/4}}-\frac{16 i (a+i a x)^{3/4}}{385 a^4 (a-i a x)^{7/4}}-\frac{4 i (a+i a x)^{3/4}}{55 a^3 (a-i a x)^{11/4}}-\frac{2 i (a+i a x)^{3/4}}{15 a^2 (a-i a x)^{15/4}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a - I*a*x)^(19/4)*(a + I*a*x)^(1/4)),x]

[Out]

(((-2*I)/15)*(a + I*a*x)^(3/4))/(a^2*(a - I*a*x)^(15/4)) - (((4*I)/55)*(a + I*a*x)^(3/4))/(a^3*(a - I*a*x)^(11
/4)) - (((16*I)/385)*(a + I*a*x)^(3/4))/(a^4*(a - I*a*x)^(7/4)) - (((32*I)/1155)*(a + I*a*x)^(3/4))/(a^5*(a -
I*a*x)^(3/4))

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])

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]

Rubi steps

\begin{align*} \int \frac{1}{(a-i a x)^{19/4} \sqrt [4]{a+i a x}} \, dx &=-\frac{2 i (a+i a x)^{3/4}}{15 a^2 (a-i a x)^{15/4}}+\frac{2 \int \frac{1}{(a-i a x)^{15/4} \sqrt [4]{a+i a x}} \, dx}{5 a}\\ &=-\frac{2 i (a+i a x)^{3/4}}{15 a^2 (a-i a x)^{15/4}}-\frac{4 i (a+i a x)^{3/4}}{55 a^3 (a-i a x)^{11/4}}+\frac{8 \int \frac{1}{(a-i a x)^{11/4} \sqrt [4]{a+i a x}} \, dx}{55 a^2}\\ &=-\frac{2 i (a+i a x)^{3/4}}{15 a^2 (a-i a x)^{15/4}}-\frac{4 i (a+i a x)^{3/4}}{55 a^3 (a-i a x)^{11/4}}-\frac{16 i (a+i a x)^{3/4}}{385 a^4 (a-i a x)^{7/4}}+\frac{16 \int \frac{1}{(a-i a x)^{7/4} \sqrt [4]{a+i a x}} \, dx}{385 a^3}\\ &=-\frac{2 i (a+i a x)^{3/4}}{15 a^2 (a-i a x)^{15/4}}-\frac{4 i (a+i a x)^{3/4}}{55 a^3 (a-i a x)^{11/4}}-\frac{16 i (a+i a x)^{3/4}}{385 a^4 (a-i a x)^{7/4}}-\frac{32 i (a+i a x)^{3/4}}{1155 a^5 (a-i a x)^{3/4}}\\ \end{align*}

Mathematica [A]  time = 0.0284822, size = 57, normalized size = 0.43 \[ \frac{2 \left (-16 i x^3+72 x^2+138 i x-159\right ) (a+i a x)^{3/4}}{1155 a^5 (x+i)^3 (a-i a x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a - I*a*x)^(19/4)*(a + I*a*x)^(1/4)),x]

[Out]

(2*(a + I*a*x)^(3/4)*(-159 + (138*I)*x + 72*x^2 - (16*I)*x^3))/(1155*a^5*(I + x)^3*(a - I*a*x)^(3/4))

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Maple [A]  time = 0.039, size = 55, normalized size = 0.4 \begin{align*}{\frac{112\,i{x}^{3}+32\,{x}^{4}-42\,ix-318-132\,{x}^{2}}{1155\,{a}^{4} \left ( x+i \right ) ^{3}} \left ( -a \left ( -1+ix \right ) \right ) ^{-{\frac{3}{4}}}{\frac{1}{\sqrt [4]{a \left ( 1+ix \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a-I*a*x)^(19/4)/(a+I*a*x)^(1/4),x)

[Out]

2/1155/a^4/(-a*(-1+I*x))^(3/4)/(a*(1+I*x))^(1/4)*(56*I*x^3+16*x^4-21*I*x-159-66*x^2)/(x+I)^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (i \, a x + a\right )}^{\frac{1}{4}}{\left (-i \, a x + a\right )}^{\frac{19}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(19/4)/(a+I*a*x)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((I*a*x + a)^(1/4)*(-I*a*x + a)^(19/4)), x)

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Fricas [A]  time = 1.52049, size = 200, normalized size = 1.5 \begin{align*} \frac{{\left (32 \, x^{3} + 144 i \, x^{2} - 276 \, x - 318 i\right )}{\left (i \, a x + a\right )}^{\frac{3}{4}}{\left (-i \, a x + a\right )}^{\frac{1}{4}}}{1155 \, a^{6} x^{4} + 4620 i \, a^{6} x^{3} - 6930 \, a^{6} x^{2} - 4620 i \, a^{6} x + 1155 \, a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(19/4)/(a+I*a*x)^(1/4),x, algorithm="fricas")

[Out]

(32*x^3 + 144*I*x^2 - 276*x - 318*I)*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)/(1155*a^6*x^4 + 4620*I*a^6*x^3 - 693
0*a^6*x^2 - 4620*I*a^6*x + 1155*a^6)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)**(19/4)/(a+I*a*x)**(1/4),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a-I*a*x)^(19/4)/(a+I*a*x)^(1/4),x, algorithm="giac")

[Out]

Exception raised: TypeError