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

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

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Rubi [A]  time = 0.17, antiderivative size = 136, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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 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])

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])

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*}

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Mathematica [A]  time = 0.04, size = 69, normalized size = 0.51 \begin {gather*} \frac {2 x (a+b x)^3 \left (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\right )}{15015 b^5 \sqrt {x^2 (a+b x)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 4.56, size = 75, normalized size = 0.55 \begin {gather*} \frac {2 (a+b x) \left (x^2 (a+b x)\right )^{3/2} \left (3003 a^4-8580 a^3 (a+b x)+10010 a^2 (a+b x)^2-5460 a (a+b x)^3+1155 (a+b x)^4\right )}{15015 b^5 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x*(a*x^2 + b*x^3)^(3/2),x]

[Out]

(2*(a + b*x)*(x^2*(a + b*x))^(3/2)*(3003*a^4 - 8580*a^3*(a + b*x) + 10010*a^2*(a + b*x)^2 - 5460*a*(a + b*x)^3
+ 1155*(a + b*x)^4))/(15015*b^5*x^3)

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fricas [A]  time = 0.39, size = 84, normalized size = 0.62 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.19, size = 246, normalized size = 1.81 \begin {gather*} -\frac {256 \, a^{\frac {13}{2}} \mathrm {sgn}\relax (x)}{15015 \, b^{5}} + \frac {2 \, {\left (\frac {143 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a^{2} \mathrm {sgn}\relax (x)}{b^{4}} + \frac {130 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a \mathrm {sgn}\relax (x)}{b^{4}} + \frac {15 \, {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )} \mathrm {sgn}\relax (x)}{b^{4}}\right )}}{45045 \, b} \end {gather*}

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*(143*(35*(b*x + a)^(9/2) - 180*(b*x + a)^(7/2)*a + 378*(b*x + a)^(5/2
)*a^2 - 420*(b*x + a)^(3/2)*a^3 + 315*sqrt(b*x + a)*a^4)*a^2*sgn(x)/b^4 + 130*(63*(b*x + a)^(11/2) - 385*(b*x
+ a)^(9/2)*a + 990*(b*x + a)^(7/2)*a^2 - 1386*(b*x + a)^(5/2)*a^3 + 1155*(b*x + a)^(3/2)*a^4 - 693*sqrt(b*x +
a)*a^5)*a*sgn(x)/b^4 + 15*(231*(b*x + a)^(13/2) - 1638*(b*x + a)^(11/2)*a + 5005*(b*x + a)^(9/2)*a^2 - 8580*(b
*x + a)^(7/2)*a^3 + 9009*(b*x + a)^(5/2)*a^4 - 6006*(b*x + a)^(3/2)*a^5 + 3003*sqrt(b*x + a)*a^6)*sgn(x)/b^4)/
b

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maple [A]  time = 0.05, size = 68, normalized size = 0.50 \begin {gather*} \frac {2 \left (b x +a \right ) \left (1155 x^{4} b^{4}-840 a \,b^{3} x^{3}+560 a^{2} x^{2} b^{2}-320 a^{3} x b +128 a^{4}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{15015 b^{5} x^{3}} \end {gather*}

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 = 1.55, size = 75, normalized size = 0.55 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 5.24, size = 69, normalized size = 0.51 \begin {gather*} \frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2\,\left (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\right )}{15015\,b^5\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a*x^2 + b*x^3)^(3/2),x)

[Out]

(2*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^2*(128*a^4 + 1155*b^4*x^4 - 840*a*b^3*x^3 + 560*a^2*b^2*x^2 - 320*a^3*b*x))
/(15015*b^5*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}\, dx \end {gather*}

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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