3.1592 \(\int \frac{1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx\)

Optimal. Leaf size=101 \[ -\frac{27 d^2 (c+d x)^{2/3}}{40 (a+b x)^{2/3} (b c-a d)^3}+\frac{9 d (c+d x)^{2/3}}{20 (a+b x)^{5/3} (b c-a d)^2}-\frac{3 (c+d x)^{2/3}}{8 (a+b x)^{8/3} (b c-a d)} \]

[Out]

(-3*(c + d*x)^(2/3))/(8*(b*c - a*d)*(a + b*x)^(8/3)) + (9*d*(c + d*x)^(2/3))/(20*(b*c - a*d)^2*(a + b*x)^(5/3)
) - (27*d^2*(c + d*x)^(2/3))/(40*(b*c - a*d)^3*(a + b*x)^(2/3))

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Rubi [A]  time = 0.0176758, 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} \[ -\frac{27 d^2 (c+d x)^{2/3}}{40 (a+b x)^{2/3} (b c-a d)^3}+\frac{9 d (c+d x)^{2/3}}{20 (a+b x)^{5/3} (b c-a d)^2}-\frac{3 (c+d x)^{2/3}}{8 (a+b x)^{8/3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(11/3)*(c + d*x)^(1/3)),x]

[Out]

(-3*(c + d*x)^(2/3))/(8*(b*c - a*d)*(a + b*x)^(8/3)) + (9*d*(c + d*x)^(2/3))/(20*(b*c - a*d)^2*(a + b*x)^(5/3)
) - (27*d^2*(c + d*x)^(2/3))/(40*(b*c - a*d)^3*(a + b*x)^(2/3))

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)^{11/3} \sqrt [3]{c+d x}} \, dx &=-\frac{3 (c+d x)^{2/3}}{8 (b c-a d) (a+b x)^{8/3}}-\frac{(3 d) \int \frac{1}{(a+b x)^{8/3} \sqrt [3]{c+d x}} \, dx}{4 (b c-a d)}\\ &=-\frac{3 (c+d x)^{2/3}}{8 (b c-a d) (a+b x)^{8/3}}+\frac{9 d (c+d x)^{2/3}}{20 (b c-a d)^2 (a+b x)^{5/3}}+\frac{\left (9 d^2\right ) \int \frac{1}{(a+b x)^{5/3} \sqrt [3]{c+d x}} \, dx}{20 (b c-a d)^2}\\ &=-\frac{3 (c+d x)^{2/3}}{8 (b c-a d) (a+b x)^{8/3}}+\frac{9 d (c+d x)^{2/3}}{20 (b c-a d)^2 (a+b x)^{5/3}}-\frac{27 d^2 (c+d x)^{2/3}}{40 (b c-a d)^3 (a+b x)^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0339313, size = 77, normalized size = 0.76 \[ -\frac{3 (c+d x)^{2/3} \left (20 a^2 d^2+8 a b d (3 d x-2 c)+b^2 \left (5 c^2-6 c d x+9 d^2 x^2\right )\right )}{40 (a+b x)^{8/3} (b c-a d)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(11/3)*(c + d*x)^(1/3)),x]

[Out]

(-3*(c + d*x)^(2/3)*(20*a^2*d^2 + 8*a*b*d*(-2*c + 3*d*x) + b^2*(5*c^2 - 6*c*d*x + 9*d^2*x^2)))/(40*(b*c - a*d)
^3*(a + b*x)^(8/3))

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Maple [A]  time = 0.005, size = 105, normalized size = 1. \begin{align*}{\frac{27\,{b}^{2}{d}^{2}{x}^{2}+72\,ab{d}^{2}x-18\,{b}^{2}cdx+60\,{a}^{2}{d}^{2}-48\,abcd+15\,{b}^{2}{c}^{2}}{40\,{a}^{3}{d}^{3}-120\,{a}^{2}cb{d}^{2}+120\,a{b}^{2}{c}^{2}d-40\,{b}^{3}{c}^{3}} \left ( dx+c \right ) ^{{\frac{2}{3}}} \left ( bx+a \right ) ^{-{\frac{8}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(11/3)/(d*x+c)^(1/3),x)

[Out]

3/40*(d*x+c)^(2/3)*(9*b^2*d^2*x^2+24*a*b*d^2*x-6*b^2*c*d*x+20*a^2*d^2-16*a*b*c*d+5*b^2*c^2)/(b*x+a)^(8/3)/(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{11}{3}}{\left (d x + c\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(1/3)), x)

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Fricas [B]  time = 2.09935, size = 514, normalized size = 5.09 \begin{align*} -\frac{3 \,{\left (9 \, b^{2} d^{2} x^{2} + 5 \, b^{2} c^{2} - 16 \, a b c d + 20 \, a^{2} d^{2} - 6 \,{\left (b^{2} c d - 4 \, a b d^{2}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}{40 \,{\left (a^{3} b^{3} c^{3} - 3 \, a^{4} b^{2} c^{2} d + 3 \, a^{5} b c d^{2} - a^{6} d^{3} +{\left (b^{6} c^{3} - 3 \, a b^{5} c^{2} d + 3 \, a^{2} b^{4} c d^{2} - a^{3} b^{3} d^{3}\right )} x^{3} + 3 \,{\left (a b^{5} c^{3} - 3 \, a^{2} b^{4} c^{2} d + 3 \, a^{3} b^{3} c d^{2} - a^{4} b^{2} d^{3}\right )} x^{2} + 3 \,{\left (a^{2} b^{4} c^{3} - 3 \, a^{3} b^{3} c^{2} d + 3 \, a^{4} b^{2} c d^{2} - a^{5} b d^{3}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(1/3),x, algorithm="fricas")

[Out]

-3/40*(9*b^2*d^2*x^2 + 5*b^2*c^2 - 16*a*b*c*d + 20*a^2*d^2 - 6*(b^2*c*d - 4*a*b*d^2)*x)*(b*x + a)^(1/3)*(d*x +
c)^(2/3)/(a^3*b^3*c^3 - 3*a^4*b^2*c^2*d + 3*a^5*b*c*d^2 - a^6*d^3 + (b^6*c^3 - 3*a*b^5*c^2*d + 3*a^2*b^4*c*d^
2 - a^3*b^3*d^3)*x^3 + 3*(a*b^5*c^3 - 3*a^2*b^4*c^2*d + 3*a^3*b^3*c*d^2 - a^4*b^2*d^3)*x^2 + 3*(a^2*b^4*c^3 -
3*a^3*b^3*c^2*d + 3*a^4*b^2*c*d^2 - a^5*b*d^3)*x)

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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/(b*x+a)**(11/3)/(d*x+c)**(1/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x + a\right )}^{\frac{11}{3}}{\left (d x + c\right )}^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(11/3)/(d*x+c)^(1/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(1/3)), x)