3.1593 \(\int \frac{1}{(a+b x)^{14/3} \sqrt [3]{c+d x}} \, dx\)

Optimal. Leaf size=136 \[ \frac{243 d^3 (c+d x)^{2/3}}{440 (a+b x)^{2/3} (b c-a d)^4}-\frac{81 d^2 (c+d x)^{2/3}}{220 (a+b x)^{5/3} (b c-a d)^3}+\frac{27 d (c+d x)^{2/3}}{88 (a+b x)^{8/3} (b c-a d)^2}-\frac{3 (c+d x)^{2/3}}{11 (a+b x)^{11/3} (b c-a d)} \]

[Out]

(-3*(c + d*x)^(2/3))/(11*(b*c - a*d)*(a + b*x)^(11/3)) + (27*d*(c + d*x)^(2/3))/(88*(b*c - a*d)^2*(a + b*x)^(8
/3)) - (81*d^2*(c + d*x)^(2/3))/(220*(b*c - a*d)^3*(a + b*x)^(5/3)) + (243*d^3*(c + d*x)^(2/3))/(440*(b*c - a*
d)^4*(a + b*x)^(2/3))

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Rubi [A]  time = 0.0282084, 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} \[ \frac{243 d^3 (c+d x)^{2/3}}{440 (a+b x)^{2/3} (b c-a d)^4}-\frac{81 d^2 (c+d x)^{2/3}}{220 (a+b x)^{5/3} (b c-a d)^3}+\frac{27 d (c+d x)^{2/3}}{88 (a+b x)^{8/3} (b c-a d)^2}-\frac{3 (c+d x)^{2/3}}{11 (a+b x)^{11/3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(14/3)*(c + d*x)^(1/3)),x]

[Out]

(-3*(c + d*x)^(2/3))/(11*(b*c - a*d)*(a + b*x)^(11/3)) + (27*d*(c + d*x)^(2/3))/(88*(b*c - a*d)^2*(a + b*x)^(8
/3)) - (81*d^2*(c + d*x)^(2/3))/(220*(b*c - a*d)^3*(a + b*x)^(5/3)) + (243*d^3*(c + d*x)^(2/3))/(440*(b*c - a*
d)^4*(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)^{14/3} \sqrt [3]{c+d x}} \, dx &=-\frac{3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}-\frac{(9 d) \int \frac{1}{(a+b x)^{11/3} \sqrt [3]{c+d x}} \, dx}{11 (b c-a d)}\\ &=-\frac{3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}+\frac{27 d (c+d x)^{2/3}}{88 (b c-a d)^2 (a+b x)^{8/3}}+\frac{\left (27 d^2\right ) \int \frac{1}{(a+b x)^{8/3} \sqrt [3]{c+d x}} \, dx}{44 (b c-a d)^2}\\ &=-\frac{3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}+\frac{27 d (c+d x)^{2/3}}{88 (b c-a d)^2 (a+b x)^{8/3}}-\frac{81 d^2 (c+d x)^{2/3}}{220 (b c-a d)^3 (a+b x)^{5/3}}-\frac{\left (81 d^3\right ) \int \frac{1}{(a+b x)^{5/3} \sqrt [3]{c+d x}} \, dx}{220 (b c-a d)^3}\\ &=-\frac{3 (c+d x)^{2/3}}{11 (b c-a d) (a+b x)^{11/3}}+\frac{27 d (c+d x)^{2/3}}{88 (b c-a d)^2 (a+b x)^{8/3}}-\frac{81 d^2 (c+d x)^{2/3}}{220 (b c-a d)^3 (a+b x)^{5/3}}+\frac{243 d^3 (c+d x)^{2/3}}{440 (b c-a d)^4 (a+b x)^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0468981, size = 118, normalized size = 0.87 \[ \frac{3 (c+d x)^{2/3} \left (132 a^2 b d^2 (3 d x-2 c)+220 a^3 d^3+33 a b^2 d \left (5 c^2-6 c d x+9 d^2 x^2\right )+b^3 \left (45 c^2 d x-40 c^3-54 c d^2 x^2+81 d^3 x^3\right )\right )}{440 (a+b x)^{11/3} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(14/3)*(c + d*x)^(1/3)),x]

[Out]

(3*(c + d*x)^(2/3)*(220*a^3*d^3 + 132*a^2*b*d^2*(-2*c + 3*d*x) + 33*a*b^2*d*(5*c^2 - 6*c*d*x + 9*d^2*x^2) + b^
3*(-40*c^3 + 45*c^2*d*x - 54*c*d^2*x^2 + 81*d^3*x^3)))/(440*(b*c - a*d)^4*(a + b*x)^(11/3))

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Maple [A]  time = 0.007, size = 171, normalized size = 1.3 \begin{align*}{\frac{243\,{b}^{3}{d}^{3}{x}^{3}+891\,a{b}^{2}{d}^{3}{x}^{2}-162\,{b}^{3}c{d}^{2}{x}^{2}+1188\,{a}^{2}b{d}^{3}x-594\,a{b}^{2}c{d}^{2}x+135\,{b}^{3}{c}^{2}dx+660\,{a}^{3}{d}^{3}-792\,{a}^{2}cb{d}^{2}+495\,a{b}^{2}{c}^{2}d-120\,{b}^{3}{c}^{3}}{440\,{d}^{4}{a}^{4}-1760\,b{d}^{3}c{a}^{3}+2640\,{b}^{2}{d}^{2}{c}^{2}{a}^{2}-1760\,{b}^{3}d{c}^{3}a+440\,{b}^{4}{c}^{4}} \left ( dx+c \right ) ^{{\frac{2}{3}}} \left ( bx+a \right ) ^{-{\frac{11}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(14/3)/(d*x+c)^(1/3),x)

[Out]

3/440*(d*x+c)^(2/3)*(81*b^3*d^3*x^3+297*a*b^2*d^3*x^2-54*b^3*c*d^2*x^2+396*a^2*b*d^3*x-198*a*b^2*c*d^2*x+45*b^
3*c^2*d*x+220*a^3*d^3-264*a^2*b*c*d^2+165*a*b^2*c^2*d-40*b^3*c^3)/(b*x+a)^(11/3)/(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{14}{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)^(14/3)/(d*x+c)^(1/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(14/3)*(d*x + c)^(1/3)), x)

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Fricas [B]  time = 2.747, size = 865, normalized size = 6.36 \begin{align*} \frac{3 \,{\left (81 \, b^{3} d^{3} x^{3} - 40 \, b^{3} c^{3} + 165 \, a b^{2} c^{2} d - 264 \, a^{2} b c d^{2} + 220 \, a^{3} d^{3} - 27 \,{\left (2 \, b^{3} c d^{2} - 11 \, a b^{2} d^{3}\right )} x^{2} + 9 \,{\left (5 \, b^{3} c^{2} d - 22 \, a b^{2} c d^{2} + 44 \, a^{2} b d^{3}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}{440 \,{\left (a^{4} b^{4} c^{4} - 4 \, a^{5} b^{3} c^{3} d + 6 \, a^{6} b^{2} c^{2} d^{2} - 4 \, a^{7} b c d^{3} + a^{8} d^{4} +{\left (b^{8} c^{4} - 4 \, a b^{7} c^{3} d + 6 \, a^{2} b^{6} c^{2} d^{2} - 4 \, a^{3} b^{5} c d^{3} + a^{4} b^{4} d^{4}\right )} x^{4} + 4 \,{\left (a b^{7} c^{4} - 4 \, a^{2} b^{6} c^{3} d + 6 \, a^{3} b^{5} c^{2} d^{2} - 4 \, a^{4} b^{4} c d^{3} + a^{5} b^{3} d^{4}\right )} x^{3} + 6 \,{\left (a^{2} b^{6} c^{4} - 4 \, a^{3} b^{5} c^{3} d + 6 \, a^{4} b^{4} c^{2} d^{2} - 4 \, a^{5} b^{3} c d^{3} + a^{6} b^{2} d^{4}\right )} x^{2} + 4 \,{\left (a^{3} b^{5} c^{4} - 4 \, a^{4} b^{4} c^{3} d + 6 \, a^{5} b^{3} c^{2} d^{2} - 4 \, a^{6} b^{2} c d^{3} + a^{7} b d^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(14/3)/(d*x+c)^(1/3),x, algorithm="fricas")

[Out]

3/440*(81*b^3*d^3*x^3 - 40*b^3*c^3 + 165*a*b^2*c^2*d - 264*a^2*b*c*d^2 + 220*a^3*d^3 - 27*(2*b^3*c*d^2 - 11*a*
b^2*d^3)*x^2 + 9*(5*b^3*c^2*d - 22*a*b^2*c*d^2 + 44*a^2*b*d^3)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3)/(a^4*b^4*c^4
- 4*a^5*b^3*c^3*d + 6*a^6*b^2*c^2*d^2 - 4*a^7*b*c*d^3 + a^8*d^4 + (b^8*c^4 - 4*a*b^7*c^3*d + 6*a^2*b^6*c^2*d^
2 - 4*a^3*b^5*c*d^3 + a^4*b^4*d^4)*x^4 + 4*(a*b^7*c^4 - 4*a^2*b^6*c^3*d + 6*a^3*b^5*c^2*d^2 - 4*a^4*b^4*c*d^3
+ a^5*b^3*d^4)*x^3 + 6*(a^2*b^6*c^4 - 4*a^3*b^5*c^3*d + 6*a^4*b^4*c^2*d^2 - 4*a^5*b^3*c*d^3 + a^6*b^2*d^4)*x^2
+ 4*(a^3*b^5*c^4 - 4*a^4*b^4*c^3*d + 6*a^5*b^3*c^2*d^2 - 4*a^6*b^2*c*d^3 + a^7*b*d^4)*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)**(14/3)/(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{14}{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)^(14/3)/(d*x+c)^(1/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(14/3)*(d*x + c)^(1/3)), x)