3.1620 \(\int \frac{1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx\)

Optimal. Leaf size=101 \[ \frac{27 d^2 \sqrt [3]{a+b x}}{5 \sqrt [3]{c+d x} (b c-a d)^3}+\frac{9 d}{5 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)^2}-\frac{3}{5 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)} \]

[Out]

-3/(5*(b*c - a*d)*(a + b*x)^(5/3)*(c + d*x)^(1/3)) + (9*d)/(5*(b*c - a*d)^2*(a + b*x)^(2/3)*(c + d*x)^(1/3)) +
(27*d^2*(a + b*x)^(1/3))/(5*(b*c - a*d)^3*(c + d*x)^(1/3))

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Rubi [A]  time = 0.0180775, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {45, 37} \[ \frac{27 d^2 \sqrt [3]{a+b x}}{5 \sqrt [3]{c+d x} (b c-a d)^3}+\frac{9 d}{5 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)^2}-\frac{3}{5 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(8/3)*(c + d*x)^(4/3)),x]

[Out]

-3/(5*(b*c - a*d)*(a + b*x)^(5/3)*(c + d*x)^(1/3)) + (9*d)/(5*(b*c - a*d)^2*(a + b*x)^(2/3)*(c + d*x)^(1/3)) +
(27*d^2*(a + b*x)^(1/3))/(5*(b*c - a*d)^3*(c + d*x)^(1/3))

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+b x)^{8/3} (c+d x)^{4/3}} \, dx &=-\frac{3}{5 (b c-a d) (a+b x)^{5/3} \sqrt [3]{c+d x}}-\frac{(6 d) \int \frac{1}{(a+b x)^{5/3} (c+d x)^{4/3}} \, dx}{5 (b c-a d)}\\ &=-\frac{3}{5 (b c-a d) (a+b x)^{5/3} \sqrt [3]{c+d x}}+\frac{9 d}{5 (b c-a d)^2 (a+b x)^{2/3} \sqrt [3]{c+d x}}+\frac{\left (9 d^2\right ) \int \frac{1}{(a+b x)^{2/3} (c+d x)^{4/3}} \, dx}{5 (b c-a d)^2}\\ &=-\frac{3}{5 (b c-a d) (a+b x)^{5/3} \sqrt [3]{c+d x}}+\frac{9 d}{5 (b c-a d)^2 (a+b x)^{2/3} \sqrt [3]{c+d x}}+\frac{27 d^2 \sqrt [3]{a+b x}}{5 (b c-a d)^3 \sqrt [3]{c+d x}}\\ \end{align*}

Mathematica [A]  time = 0.0292202, size = 75, normalized size = 0.74 \[ \frac{3 \left (5 a^2 d^2+5 a b d (c+3 d x)+b^2 \left (-c^2+3 c d x+9 d^2 x^2\right )\right )}{5 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(8/3)*(c + d*x)^(4/3)),x]

[Out]

(3*(5*a^2*d^2 + 5*a*b*d*(c + 3*d*x) + b^2*(-c^2 + 3*c*d*x + 9*d^2*x^2)))/(5*(b*c - a*d)^3*(a + b*x)^(5/3)*(c +
d*x)^(1/3))

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Maple [A]  time = 0.004, size = 105, normalized size = 1. \begin{align*} -{\frac{27\,{b}^{2}{d}^{2}{x}^{2}+45\,ab{d}^{2}x+9\,{b}^{2}cdx+15\,{a}^{2}{d}^{2}+15\,abcd-3\,{b}^{2}{c}^{2}}{5\,{a}^{3}{d}^{3}-15\,{a}^{2}cb{d}^{2}+15\,a{b}^{2}{c}^{2}d-5\,{b}^{3}{c}^{3}} \left ( bx+a \right ) ^{-{\frac{5}{3}}}{\frac{1}{\sqrt [3]{dx+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x)

[Out]

-3/5*(9*b^2*d^2*x^2+15*a*b*d^2*x+3*b^2*c*d*x+5*a^2*d^2+5*a*b*c*d-b^2*c^2)/(b*x+a)^(5/3)/(d*x+c)^(1/3)/(a^3*d^3
-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{8}{3}}{\left (d x + c\right )}^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(8/3)*(d*x + c)^(4/3)), x)

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Fricas [B]  time = 2.37148, size = 549, normalized size = 5.44 \begin{align*} \frac{3 \,{\left (9 \, b^{2} d^{2} x^{2} - b^{2} c^{2} + 5 \, a b c d + 5 \, a^{2} d^{2} + 3 \,{\left (b^{2} c d + 5 \, a b d^{2}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}{5 \,{\left (a^{2} b^{3} c^{4} - 3 \, a^{3} b^{2} c^{3} d + 3 \, a^{4} b c^{2} d^{2} - a^{5} c d^{3} +{\left (b^{5} c^{3} d - 3 \, a b^{4} c^{2} d^{2} + 3 \, a^{2} b^{3} c d^{3} - a^{3} b^{2} d^{4}\right )} x^{3} +{\left (b^{5} c^{4} - a b^{4} c^{3} d - 3 \, a^{2} b^{3} c^{2} d^{2} + 5 \, a^{3} b^{2} c d^{3} - 2 \, a^{4} b d^{4}\right )} x^{2} +{\left (2 \, a b^{4} c^{4} - 5 \, a^{2} b^{3} c^{3} d + 3 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b c d^{3} - a^{5} d^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

3/5*(9*b^2*d^2*x^2 - b^2*c^2 + 5*a*b*c*d + 5*a^2*d^2 + 3*(b^2*c*d + 5*a*b*d^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2
/3)/(a^2*b^3*c^4 - 3*a^3*b^2*c^3*d + 3*a^4*b*c^2*d^2 - a^5*c*d^3 + (b^5*c^3*d - 3*a*b^4*c^2*d^2 + 3*a^2*b^3*c*
d^3 - a^3*b^2*d^4)*x^3 + (b^5*c^4 - a*b^4*c^3*d - 3*a^2*b^3*c^2*d^2 + 5*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^2 + (2*
a*b^4*c^4 - 5*a^2*b^3*c^3*d + 3*a^3*b^2*c^2*d^2 + a^4*b*c*d^3 - a^5*d^4)*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b x\right )^{\frac{8}{3}} \left (c + d x\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**(8/3)/(d*x+c)**(4/3),x)

[Out]

Integral(1/((a + b*x)**(8/3)*(c + d*x)**(4/3)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{8}{3}}{\left (d x + c\right )}^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(8/3)/(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(8/3)*(d*x + c)^(4/3)), x)