3.1607 \(\int \frac{1}{(a+b x)^{13/3} (c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=136 \[ \frac{243 d^3 \sqrt [3]{c+d x}}{140 \sqrt [3]{a+b x} (b c-a d)^4}-\frac{81 d^2 \sqrt [3]{c+d x}}{140 (a+b x)^{4/3} (b c-a d)^3}+\frac{27 d \sqrt [3]{c+d x}}{70 (a+b x)^{7/3} (b c-a d)^2}-\frac{3 \sqrt [3]{c+d x}}{10 (a+b x)^{10/3} (b c-a d)} \]

[Out]

(-3*(c + d*x)^(1/3))/(10*(b*c - a*d)*(a + b*x)^(10/3)) + (27*d*(c + d*x)^(1/3))/(70*(b*c - a*d)^2*(a + b*x)^(7
/3)) - (81*d^2*(c + d*x)^(1/3))/(140*(b*c - a*d)^3*(a + b*x)^(4/3)) + (243*d^3*(c + d*x)^(1/3))/(140*(b*c - a*
d)^4*(a + b*x)^(1/3))

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Rubi [A]  time = 0.0295051, 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} \[ \frac{243 d^3 \sqrt [3]{c+d x}}{140 \sqrt [3]{a+b x} (b c-a d)^4}-\frac{81 d^2 \sqrt [3]{c+d x}}{140 (a+b x)^{4/3} (b c-a d)^3}+\frac{27 d \sqrt [3]{c+d x}}{70 (a+b x)^{7/3} (b c-a d)^2}-\frac{3 \sqrt [3]{c+d x}}{10 (a+b x)^{10/3} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^(13/3)*(c + d*x)^(2/3)),x]

[Out]

(-3*(c + d*x)^(1/3))/(10*(b*c - a*d)*(a + b*x)^(10/3)) + (27*d*(c + d*x)^(1/3))/(70*(b*c - a*d)^2*(a + b*x)^(7
/3)) - (81*d^2*(c + d*x)^(1/3))/(140*(b*c - a*d)^3*(a + b*x)^(4/3)) + (243*d^3*(c + d*x)^(1/3))/(140*(b*c - a*
d)^4*(a + b*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)^{13/3} (c+d x)^{2/3}} \, dx &=-\frac{3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}-\frac{(9 d) \int \frac{1}{(a+b x)^{10/3} (c+d x)^{2/3}} \, dx}{10 (b c-a d)}\\ &=-\frac{3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}+\frac{27 d \sqrt [3]{c+d x}}{70 (b c-a d)^2 (a+b x)^{7/3}}+\frac{\left (27 d^2\right ) \int \frac{1}{(a+b x)^{7/3} (c+d x)^{2/3}} \, dx}{35 (b c-a d)^2}\\ &=-\frac{3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}+\frac{27 d \sqrt [3]{c+d x}}{70 (b c-a d)^2 (a+b x)^{7/3}}-\frac{81 d^2 \sqrt [3]{c+d x}}{140 (b c-a d)^3 (a+b x)^{4/3}}-\frac{\left (81 d^3\right ) \int \frac{1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx}{140 (b c-a d)^3}\\ &=-\frac{3 \sqrt [3]{c+d x}}{10 (b c-a d) (a+b x)^{10/3}}+\frac{27 d \sqrt [3]{c+d x}}{70 (b c-a d)^2 (a+b x)^{7/3}}-\frac{81 d^2 \sqrt [3]{c+d x}}{140 (b c-a d)^3 (a+b x)^{4/3}}+\frac{243 d^3 \sqrt [3]{c+d x}}{140 (b c-a d)^4 \sqrt [3]{a+b x}}\\ \end{align*}

Mathematica [A]  time = 0.0484734, size = 116, normalized size = 0.85 \[ \frac{3 \sqrt [3]{c+d x} \left (-105 a^2 b d^2 (c-3 d x)+140 a^3 d^3+30 a b^2 d \left (2 c^2-3 c d x+9 d^2 x^2\right )+b^3 \left (18 c^2 d x-14 c^3-27 c d^2 x^2+81 d^3 x^3\right )\right )}{140 (a+b x)^{10/3} (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^(13/3)*(c + d*x)^(2/3)),x]

[Out]

(3*(c + d*x)^(1/3)*(140*a^3*d^3 - 105*a^2*b*d^2*(c - 3*d*x) + 30*a*b^2*d*(2*c^2 - 3*c*d*x + 9*d^2*x^2) + b^3*(
-14*c^3 + 18*c^2*d*x - 27*c*d^2*x^2 + 81*d^3*x^3)))/(140*(b*c - a*d)^4*(a + b*x)^(10/3))

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Maple [A]  time = 0.005, size = 171, normalized size = 1.3 \begin{align*}{\frac{243\,{b}^{3}{d}^{3}{x}^{3}+810\,a{b}^{2}{d}^{3}{x}^{2}-81\,{b}^{3}c{d}^{2}{x}^{2}+945\,{a}^{2}b{d}^{3}x-270\,a{b}^{2}c{d}^{2}x+54\,{b}^{3}{c}^{2}dx+420\,{a}^{3}{d}^{3}-315\,{a}^{2}cb{d}^{2}+180\,a{b}^{2}{c}^{2}d-42\,{b}^{3}{c}^{3}}{140\,{d}^{4}{a}^{4}-560\,b{d}^{3}c{a}^{3}+840\,{b}^{2}{d}^{2}{c}^{2}{a}^{2}-560\,{b}^{3}d{c}^{3}a+140\,{b}^{4}{c}^{4}}\sqrt [3]{dx+c} \left ( bx+a \right ) ^{-{\frac{10}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^(13/3)/(d*x+c)^(2/3),x)

[Out]

3/140*(d*x+c)^(1/3)*(81*b^3*d^3*x^3+270*a*b^2*d^3*x^2-27*b^3*c*d^2*x^2+315*a^2*b*d^3*x-90*a*b^2*c*d^2*x+18*b^3
*c^2*d*x+140*a^3*d^3-105*a^2*b*c*d^2+60*a*b^2*c^2*d-14*b^3*c^3)/(b*x+a)^(10/3)/(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{13}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(13/3)/(d*x+c)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(13/3)*(d*x + c)^(2/3)), x)

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Fricas [B]  time = 2.23929, size = 861, normalized size = 6.33 \begin{align*} \frac{3 \,{\left (81 \, b^{3} d^{3} x^{3} - 14 \, b^{3} c^{3} + 60 \, a b^{2} c^{2} d - 105 \, a^{2} b c d^{2} + 140 \, a^{3} d^{3} - 27 \,{\left (b^{3} c d^{2} - 10 \, a b^{2} d^{3}\right )} x^{2} + 9 \,{\left (2 \, b^{3} c^{2} d - 10 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x\right )}{\left (b x + a\right )}^{\frac{2}{3}}{\left (d x + c\right )}^{\frac{1}{3}}}{140 \,{\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)^(13/3)/(d*x+c)^(2/3),x, algorithm="fricas")

[Out]

3/140*(81*b^3*d^3*x^3 - 14*b^3*c^3 + 60*a*b^2*c^2*d - 105*a^2*b*c*d^2 + 140*a^3*d^3 - 27*(b^3*c*d^2 - 10*a*b^2
*d^3)*x^2 + 9*(2*b^3*c^2*d - 10*a*b^2*c*d^2 + 35*a^2*b*d^3)*x)*(b*x + a)^(2/3)*(d*x + c)^(1/3)/(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)**(13/3)/(d*x+c)**(2/3),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{13}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^(13/3)/(d*x+c)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(13/3)*(d*x + c)^(2/3)), x)