3.241 \(\int x (a x^2+b x^3)^{3/2} \, dx\)

Optimal. Leaf size=136 \[ \frac{256 a^4 \left (a x^2+b x^3\right )^{5/2}}{15015 b^5 x^5}-\frac{128 a^3 \left (a x^2+b x^3\right )^{5/2}}{3003 b^4 x^4}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^3}-\frac{16 a \left (a x^2+b x^3\right )^{5/2}}{143 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{13 b x} \]

[Out]

(256*a^4*(a*x^2 + b*x^3)^(5/2))/(15015*b^5*x^5) - (128*a^3*(a*x^2 + b*x^3)^(5/2))/(3003*b^4*x^4) + (32*a^2*(a*
x^2 + b*x^3)^(5/2))/(429*b^3*x^3) - (16*a*(a*x^2 + b*x^3)^(5/2))/(143*b^2*x^2) + (2*(a*x^2 + b*x^3)^(5/2))/(13
*b*x)

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Rubi [A]  time = 0.171624, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2016, 2002, 2014} \[ \frac{256 a^4 \left (a x^2+b x^3\right )^{5/2}}{15015 b^5 x^5}-\frac{128 a^3 \left (a x^2+b x^3\right )^{5/2}}{3003 b^4 x^4}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^3}-\frac{16 a \left (a x^2+b x^3\right )^{5/2}}{143 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{13 b x} \]

Antiderivative was successfully verified.

[In]

Int[x*(a*x^2 + b*x^3)^(3/2),x]

[Out]

(256*a^4*(a*x^2 + b*x^3)^(5/2))/(15015*b^5*x^5) - (128*a^3*(a*x^2 + b*x^3)^(5/2))/(3003*b^4*x^4) + (32*a^2*(a*
x^2 + b*x^3)^(5/2))/(429*b^3*x^3) - (16*a*(a*x^2 + b*x^3)^(5/2))/(143*b^2*x^2) + (2*(a*x^2 + b*x^3)^(5/2))/(13
*b*x)

Rule 2016

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(j - 1)*(c*x)^(m - j +
1)*(a*x^j + b*x^n)^(p + 1))/(a*(m + j*p + 1)), x] - Dist[(b*(m + n*p + n - j + 1))/(a*c^(n - j)*(m + j*p + 1)
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2014

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(c^(j - 1)*(c*x)^(m - j
+ 1)*(a*x^j + b*x^n)^(p + 1))/(a*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin{align*} \int x \left (a x^2+b x^3\right )^{3/2} \, dx &=\frac{2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac{(8 a) \int \left (a x^2+b x^3\right )^{3/2} \, dx}{13 b}\\ &=-\frac{16 a \left (a x^2+b x^3\right )^{5/2}}{143 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}+\frac{\left (48 a^2\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x} \, dx}{143 b^2}\\ &=\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^3}-\frac{16 a \left (a x^2+b x^3\right )^{5/2}}{143 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}-\frac{\left (64 a^3\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx}{429 b^3}\\ &=-\frac{128 a^3 \left (a x^2+b x^3\right )^{5/2}}{3003 b^4 x^4}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^3}-\frac{16 a \left (a x^2+b x^3\right )^{5/2}}{143 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}+\frac{\left (128 a^4\right ) \int \frac{\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{3003 b^4}\\ &=\frac{256 a^4 \left (a x^2+b x^3\right )^{5/2}}{15015 b^5 x^5}-\frac{128 a^3 \left (a x^2+b x^3\right )^{5/2}}{3003 b^4 x^4}+\frac{32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^3}-\frac{16 a \left (a x^2+b x^3\right )^{5/2}}{143 b^2 x^2}+\frac{2 \left (a x^2+b x^3\right )^{5/2}}{13 b x}\\ \end{align*}

Mathematica [A]  time = 0.0332815, size = 69, normalized size = 0.51 \[ \frac{2 x (a+b x)^3 \left (560 a^2 b^2 x^2-320 a^3 b x+128 a^4-840 a b^3 x^3+1155 b^4 x^4\right )}{15015 b^5 \sqrt{x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a*x^2 + b*x^3)^(3/2),x]

[Out]

(2*x*(a + b*x)^3*(128*a^4 - 320*a^3*b*x + 560*a^2*b^2*x^2 - 840*a*b^3*x^3 + 1155*b^4*x^4))/(15015*b^5*Sqrt[x^2
*(a + b*x)])

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Maple [A]  time = 0.004, size = 68, normalized size = 0.5 \begin{align*}{\frac{ \left ( 2\,bx+2\,a \right ) \left ( 1155\,{x}^{4}{b}^{4}-840\,a{b}^{3}{x}^{3}+560\,{a}^{2}{x}^{2}{b}^{2}-320\,x{a}^{3}b+128\,{a}^{4} \right ) }{15015\,{b}^{5}{x}^{3}} \left ( b{x}^{3}+a{x}^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^3+a*x^2)^(3/2),x)

[Out]

2/15015*(b*x+a)*(1155*b^4*x^4-840*a*b^3*x^3+560*a^2*b^2*x^2-320*a^3*b*x+128*a^4)*(b*x^3+a*x^2)^(3/2)/b^5/x^3

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Maxima [A]  time = 1.05686, size = 101, normalized size = 0.74 \begin{align*} \frac{2 \,{\left (1155 \, b^{6} x^{6} + 1470 \, a b^{5} x^{5} + 35 \, a^{2} b^{4} x^{4} - 40 \, a^{3} b^{3} x^{3} + 48 \, a^{4} b^{2} x^{2} - 64 \, a^{5} b x + 128 \, a^{6}\right )} \sqrt{b x + a}}{15015 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015*(1155*b^6*x^6 + 1470*a*b^5*x^5 + 35*a^2*b^4*x^4 - 40*a^3*b^3*x^3 + 48*a^4*b^2*x^2 - 64*a^5*b*x + 128*a
^6)*sqrt(b*x + a)/b^5

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Fricas [A]  time = 0.808423, size = 193, normalized size = 1.42 \begin{align*} \frac{2 \,{\left (1155 \, b^{6} x^{6} + 1470 \, a b^{5} x^{5} + 35 \, a^{2} b^{4} x^{4} - 40 \, a^{3} b^{3} x^{3} + 48 \, a^{4} b^{2} x^{2} - 64 \, a^{5} b x + 128 \, a^{6}\right )} \sqrt{b x^{3} + a x^{2}}}{15015 \, b^{5} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^3+a*x^2)^(3/2),x, algorithm="fricas")

[Out]

2/15015*(1155*b^6*x^6 + 1470*a*b^5*x^5 + 35*a^2*b^4*x^4 - 40*a^3*b^3*x^3 + 48*a^4*b^2*x^2 - 64*a^5*b*x + 128*a
^6)*sqrt(b*x^3 + a*x^2)/(b^5*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (x^{2} \left (a + b x\right )\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**3+a*x**2)**(3/2),x)

[Out]

Integral(x*(x**2*(a + b*x))**(3/2), x)

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Giac [A]  time = 1.17089, size = 211, normalized size = 1.55 \begin{align*} -\frac{256 \, a^{\frac{13}{2}} \mathrm{sgn}\left (x\right )}{15015 \, b^{5}} + \frac{2 \,{\left (\frac{13 \,{\left (315 \,{\left (b x + a\right )}^{\frac{11}{2}} - 1540 \,{\left (b x + a\right )}^{\frac{9}{2}} a + 2970 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{2} - 2772 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{3} + 1155 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{4}\right )} a \mathrm{sgn}\left (x\right )}{b^{4}} + \frac{5 \,{\left (693 \,{\left (b x + a\right )}^{\frac{13}{2}} - 4095 \,{\left (b x + a\right )}^{\frac{11}{2}} a + 10010 \,{\left (b x + a\right )}^{\frac{9}{2}} a^{2} - 12870 \,{\left (b x + a\right )}^{\frac{7}{2}} a^{3} + 9009 \,{\left (b x + a\right )}^{\frac{5}{2}} a^{4} - 3003 \,{\left (b x + a\right )}^{\frac{3}{2}} a^{5}\right )} \mathrm{sgn}\left (x\right )}{b^{4}}\right )}}{45045 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-256/15015*a^(13/2)*sgn(x)/b^5 + 2/45045*(13*(315*(b*x + a)^(11/2) - 1540*(b*x + a)^(9/2)*a + 2970*(b*x + a)^(
7/2)*a^2 - 2772*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4)*a*sgn(x)/b^4 + 5*(693*(b*x + a)^(13/2) - 4095*
(b*x + a)^(11/2)*a + 10010*(b*x + a)^(9/2)*a^2 - 12870*(b*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 3003*(
b*x + a)^(3/2)*a^5)*sgn(x)/b^4)/b