3.1621 \(\int \frac{1}{(a+b x)^{11/3} (c+d x)^{4/3}} \, dx\)

Optimal. Leaf size=136 \[ -\frac{243 d^3 \sqrt [3]{a+b x}}{40 \sqrt [3]{c+d x} (b c-a d)^4}-\frac{81 d^2}{40 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)^3}+\frac{27 d}{40 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)^2}-\frac{3}{8 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)} \]

[Out]

-3/(8*(b*c - a*d)*(a + b*x)^(8/3)*(c + d*x)^(1/3)) + (27*d)/(40*(b*c - a*d)^2*(a + b*x)^(5/3)*(c + d*x)^(1/3))
- (81*d^2)/(40*(b*c - a*d)^3*(a + b*x)^(2/3)*(c + d*x)^(1/3)) - (243*d^3*(a + b*x)^(1/3))/(40*(b*c - a*d)^4*(
c + d*x)^(1/3))

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Rubi [A]  time = 0.0296401, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {45, 37} \[ -\frac{243 d^3 \sqrt [3]{a+b x}}{40 \sqrt [3]{c+d x} (b c-a d)^4}-\frac{81 d^2}{40 (a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)^3}+\frac{27 d}{40 (a+b x)^{5/3} \sqrt [3]{c+d x} (b c-a d)^2}-\frac{3}{8 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(11/3)*(c + d*x)^(4/3)),x]

[Out]

-3/(8*(b*c - a*d)*(a + b*x)^(8/3)*(c + d*x)^(1/3)) + (27*d)/(40*(b*c - a*d)^2*(a + b*x)^(5/3)*(c + d*x)^(1/3))
- (81*d^2)/(40*(b*c - a*d)^3*(a + b*x)^(2/3)*(c + d*x)^(1/3)) - (243*d^3*(a + b*x)^(1/3))/(40*(b*c - a*d)^4*(
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)^{11/3} (c+d x)^{4/3}} \, dx &=-\frac{3}{8 (b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}-\frac{(9 d) \int \frac{1}{(a+b x)^{8/3} (c+d x)^{4/3}} \, dx}{8 (b c-a d)}\\ &=-\frac{3}{8 (b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}+\frac{27 d}{40 (b c-a d)^2 (a+b x)^{5/3} \sqrt [3]{c+d x}}+\frac{\left (27 d^2\right ) \int \frac{1}{(a+b x)^{5/3} (c+d x)^{4/3}} \, dx}{20 (b c-a d)^2}\\ &=-\frac{3}{8 (b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}+\frac{27 d}{40 (b c-a d)^2 (a+b x)^{5/3} \sqrt [3]{c+d x}}-\frac{81 d^2}{40 (b c-a d)^3 (a+b x)^{2/3} \sqrt [3]{c+d x}}-\frac{\left (81 d^3\right ) \int \frac{1}{(a+b x)^{2/3} (c+d x)^{4/3}} \, dx}{40 (b c-a d)^3}\\ &=-\frac{3}{8 (b c-a d) (a+b x)^{8/3} \sqrt [3]{c+d x}}+\frac{27 d}{40 (b c-a d)^2 (a+b x)^{5/3} \sqrt [3]{c+d x}}-\frac{81 d^2}{40 (b c-a d)^3 (a+b x)^{2/3} \sqrt [3]{c+d x}}-\frac{243 d^3 \sqrt [3]{a+b x}}{40 (b c-a d)^4 \sqrt [3]{c+d x}}\\ \end{align*}

Mathematica [A]  time = 0.0469954, size = 116, normalized size = 0.85 \[ -\frac{3 \left (60 a^2 b d^2 (c+3 d x)+40 a^3 d^3+24 a b^2 d \left (-c^2+3 c d x+9 d^2 x^2\right )+b^3 \left (-9 c^2 d x+5 c^3+27 c d^2 x^2+81 d^3 x^3\right )\right )}{40 (a+b x)^{8/3} \sqrt [3]{c+d x} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(11/3)*(c + d*x)^(4/3)),x]

[Out]

(-3*(40*a^3*d^3 + 60*a^2*b*d^2*(c + 3*d*x) + 24*a*b^2*d*(-c^2 + 3*c*d*x + 9*d^2*x^2) + b^3*(5*c^3 - 9*c^2*d*x
+ 27*c*d^2*x^2 + 81*d^3*x^3)))/(40*(b*c - a*d)^4*(a + b*x)^(8/3)*(c + d*x)^(1/3))

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Maple [A]  time = 0.008, size = 171, normalized size = 1.3 \begin{align*} -{\frac{243\,{b}^{3}{d}^{3}{x}^{3}+648\,a{b}^{2}{d}^{3}{x}^{2}+81\,{b}^{3}c{d}^{2}{x}^{2}+540\,{a}^{2}b{d}^{3}x+216\,a{b}^{2}c{d}^{2}x-27\,{b}^{3}{c}^{2}dx+120\,{a}^{3}{d}^{3}+180\,{a}^{2}cb{d}^{2}-72\,a{b}^{2}{c}^{2}d+15\,{b}^{3}{c}^{3}}{40\,{d}^{4}{a}^{4}-160\,b{d}^{3}c{a}^{3}+240\,{b}^{2}{d}^{2}{c}^{2}{a}^{2}-160\,{b}^{3}d{c}^{3}a+40\,{b}^{4}{c}^{4}} \left ( bx+a \right ) ^{-{\frac{8}{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)^(11/3)/(d*x+c)^(4/3),x)

[Out]

-3/40*(81*b^3*d^3*x^3+216*a*b^2*d^3*x^2+27*b^3*c*d^2*x^2+180*a^2*b*d^3*x+72*a*b^2*c*d^2*x-9*b^3*c^2*d*x+40*a^3
*d^3+60*a^2*b*c*d^2-24*a*b^2*c^2*d+5*b^3*c^3)/(b*x+a)^(8/3)/(d*x+c)^(1/3)/(a^4*d^4-4*a^3*b*c*d^3+6*a^2*b^2*c^2
*d^2-4*a*b^3*c^3*d+b^4*c^4)

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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{4}{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)^(4/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(4/3)), x)

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Fricas [B]  time = 2.95757, size = 927, normalized size = 6.82 \begin{align*} -\frac{3 \,{\left (81 \, b^{3} d^{3} x^{3} + 5 \, b^{3} c^{3} - 24 \, a b^{2} c^{2} d + 60 \, a^{2} b c d^{2} + 40 \, a^{3} d^{3} + 27 \,{\left (b^{3} c d^{2} + 8 \, a b^{2} d^{3}\right )} x^{2} - 9 \,{\left (b^{3} c^{2} d - 8 \, a b^{2} c d^{2} - 20 \, a^{2} b d^{3}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}{40 \,{\left (a^{3} b^{4} c^{5} - 4 \, a^{4} b^{3} c^{4} d + 6 \, a^{5} b^{2} c^{3} d^{2} - 4 \, a^{6} b c^{2} d^{3} + a^{7} c d^{4} +{\left (b^{7} c^{4} d - 4 \, a b^{6} c^{3} d^{2} + 6 \, a^{2} b^{5} c^{2} d^{3} - 4 \, a^{3} b^{4} c d^{4} + a^{4} b^{3} d^{5}\right )} x^{4} +{\left (b^{7} c^{5} - a b^{6} c^{4} d - 6 \, a^{2} b^{5} c^{3} d^{2} + 14 \, a^{3} b^{4} c^{2} d^{3} - 11 \, a^{4} b^{3} c d^{4} + 3 \, a^{5} b^{2} d^{5}\right )} x^{3} + 3 \,{\left (a b^{6} c^{5} - 3 \, a^{2} b^{5} c^{4} d + 2 \, a^{3} b^{4} c^{3} d^{2} + 2 \, a^{4} b^{3} c^{2} d^{3} - 3 \, a^{5} b^{2} c d^{4} + a^{6} b d^{5}\right )} x^{2} +{\left (3 \, a^{2} b^{5} c^{5} - 11 \, a^{3} b^{4} c^{4} d + 14 \, a^{4} b^{3} c^{3} d^{2} - 6 \, a^{5} b^{2} c^{2} d^{3} - a^{6} b c d^{4} + a^{7} d^{5}\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)^(4/3),x, algorithm="fricas")

[Out]

-3/40*(81*b^3*d^3*x^3 + 5*b^3*c^3 - 24*a*b^2*c^2*d + 60*a^2*b*c*d^2 + 40*a^3*d^3 + 27*(b^3*c*d^2 + 8*a*b^2*d^3
)*x^2 - 9*(b^3*c^2*d - 8*a*b^2*c*d^2 - 20*a^2*b*d^3)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)/(a^3*b^4*c^5 - 4*a^4*b
^3*c^4*d + 6*a^5*b^2*c^3*d^2 - 4*a^6*b*c^2*d^3 + a^7*c*d^4 + (b^7*c^4*d - 4*a*b^6*c^3*d^2 + 6*a^2*b^5*c^2*d^3
- 4*a^3*b^4*c*d^4 + a^4*b^3*d^5)*x^4 + (b^7*c^5 - a*b^6*c^4*d - 6*a^2*b^5*c^3*d^2 + 14*a^3*b^4*c^2*d^3 - 11*a^
4*b^3*c*d^4 + 3*a^5*b^2*d^5)*x^3 + 3*(a*b^6*c^5 - 3*a^2*b^5*c^4*d + 2*a^3*b^4*c^3*d^2 + 2*a^4*b^3*c^2*d^3 - 3*
a^5*b^2*c*d^4 + a^6*b*d^5)*x^2 + (3*a^2*b^5*c^5 - 11*a^3*b^4*c^4*d + 14*a^4*b^3*c^3*d^2 - 6*a^5*b^2*c^2*d^3 -
a^6*b*c*d^4 + a^7*d^5)*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)**(4/3),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{4}{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)^(4/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(11/3)*(d*x + c)^(4/3)), x)