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

Optimal. Leaf size=161 \[ -\frac {512 a^5 \left (a x^2+b x^3\right )^{5/2}}{45045 b^6 x^5}+\frac {256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac {64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b} \]

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Rubi [A]  time = 0.23, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2016, 2002, 2014} \begin {gather*} -\frac {512 a^5 \left (a x^2+b x^3\right )^{5/2}}{45045 b^6 x^5}+\frac {256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac {64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(a*x^2 + b*x^3)^(3/2),x]

[Out]

(2*(a*x^2 + b*x^3)^(5/2))/(15*b) - (512*a^5*(a*x^2 + b*x^3)^(5/2))/(45045*b^6*x^5) + (256*a^4*(a*x^2 + b*x^3)^
(5/2))/(9009*b^5*x^4) - (64*a^3*(a*x^2 + b*x^3)^(5/2))/(1287*b^4*x^3) + (32*a^2*(a*x^2 + b*x^3)^(5/2))/(429*b^
3*x^2) - (4*a*(a*x^2 + b*x^3)^(5/2))/(39*b^2*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^2 \left (a x^2+b x^3\right )^{3/2} \, dx &=\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac {(2 a) \int x \left (a x^2+b x^3\right )^{3/2} \, dx}{3 b}\\ &=\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac {\left (16 a^2\right ) \int \left (a x^2+b x^3\right )^{3/2} \, dx}{39 b^2}\\ &=\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}-\frac {\left (32 a^3\right ) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx}{143 b^3}\\ &=\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac {64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac {\left (128 a^4\right ) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx}{1287 b^4}\\ &=\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b}+\frac {256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac {64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}-\frac {\left (256 a^5\right ) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{9009 b^5}\\ &=\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac {512 a^5 \left (a x^2+b x^3\right )^{5/2}}{45045 b^6 x^5}+\frac {256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac {64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 80, normalized size = 0.50 \begin {gather*} \frac {2 x (a+b x)^3 \left (-256 a^5+640 a^4 b x-1120 a^3 b^2 x^2+1680 a^2 b^3 x^3-2310 a b^4 x^4+3003 b^5 x^5\right )}{45045 b^6 \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(a + b*x)^3*(-256*a^5 + 640*a^4*b*x - 1120*a^3*b^2*x^2 + 1680*a^2*b^3*x^3 - 2310*a*b^4*x^4 + 3003*b^5*x^5
))/(45045*b^6*Sqrt[x^2*(a + b*x)])

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IntegrateAlgebraic [A]  time = 4.70, size = 87, normalized size = 0.54 \begin {gather*} \frac {2 (a+b x) \left (x^2 (a+b x)\right )^{3/2} \left (-9009 a^5+32175 a^4 (a+b x)-50050 a^3 (a+b x)^2+40950 a^2 (a+b x)^3-17325 a (a+b x)^4+3003 (a+b x)^5\right )}{45045 b^6 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x)*(x^2*(a + b*x))^(3/2)*(-9009*a^5 + 32175*a^4*(a + b*x) - 50050*a^3*(a + b*x)^2 + 40950*a^2*(a + b
*x)^3 - 17325*a*(a + b*x)^4 + 3003*(a + b*x)^5))/(45045*b^6*x^3)

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fricas [A]  time = 0.40, size = 95, normalized size = 0.59 \begin {gather*} \frac {2 \, {\left (3003 \, b^{7} x^{7} + 3696 \, a b^{6} x^{6} + 63 \, a^{2} b^{5} x^{5} - 70 \, a^{3} b^{4} x^{4} + 80 \, a^{4} b^{3} x^{3} - 96 \, a^{5} b^{2} x^{2} + 128 \, a^{6} b x - 256 \, a^{7}\right )} \sqrt {b x^{3} + a x^{2}}}{45045 \, b^{6} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*(3003*b^7*x^7 + 3696*a*b^6*x^6 + 63*a^2*b^5*x^5 - 70*a^3*b^4*x^4 + 80*a^4*b^3*x^3 - 96*a^5*b^2*x^2 + 1
28*a^6*b*x - 256*a^7)*sqrt(b*x^3 + a*x^2)/(b^6*x)

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giac [B]  time = 0.18, size = 282, normalized size = 1.75 \begin {gather*} \frac {512 \, a^{\frac {15}{2}} \mathrm {sgn}\relax (x)}{45045 \, b^{6}} + \frac {2 \, {\left (\frac {65 \, {\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^{2} \mathrm {sgn}\relax (x)}{b^{5}} + \frac {30 \, {\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 )} a \mathrm {sgn}\relax (x)}{b^{5}} + \frac {7 \, {\left (429 \, {\left (b x + a\right )}^{\frac {15}{2}} - 3465 \, {\left (b x + a\right )}^{\frac {13}{2}} a + 12285 \, {\left (b x + a\right )}^{\frac {11}{2}} a^{2} - 25025 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{3} + 32175 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{4} - 27027 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} + 15015 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{6} - 6435 \, \sqrt {b x + a} a^{7}\right )} \mathrm {sgn}\relax (x)}{b^{5}}\right )}}{45045 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

512/45045*a^(15/2)*sgn(x)/b^6 + 2/45045*(65*(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^2*sgn(x)/b^5 + 30*(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)*a*sgn(x)/b^5 + 7*(429*(b*x + a)^(15/2) - 346
5*(b*x + a)^(13/2)*a + 12285*(b*x + a)^(11/2)*a^2 - 25025*(b*x + a)^(9/2)*a^3 + 32175*(b*x + a)^(7/2)*a^4 - 27
027*(b*x + a)^(5/2)*a^5 + 15015*(b*x + a)^(3/2)*a^6 - 6435*sqrt(b*x + a)*a^7)*sgn(x)/b^5)/b

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maple [A]  time = 0.05, size = 79, normalized size = 0.49 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (-3003 x^{5} b^{5}+2310 a \,b^{4} x^{4}-1680 a^{2} b^{3} x^{3}+1120 a^{3} b^{2} x^{2}-640 a^{4} b x +256 a^{5}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{45045 b^{6} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45045*(b*x+a)*(-3003*b^5*x^5+2310*a*b^4*x^4-1680*a^2*b^3*x^3+1120*a^3*b^2*x^2-640*a^4*b*x+256*a^5)*(b*x^3+a
*x^2)^(3/2)/b^6/x^3

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maxima [A]  time = 1.52, size = 86, normalized size = 0.53 \begin {gather*} \frac {2 \, {\left (3003 \, b^{7} x^{7} + 3696 \, a b^{6} x^{6} + 63 \, a^{2} b^{5} x^{5} - 70 \, a^{3} b^{4} x^{4} + 80 \, a^{4} b^{3} x^{3} - 96 \, a^{5} b^{2} x^{2} + 128 \, a^{6} b x - 256 \, a^{7}\right )} \sqrt {b x + a}}{45045 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*(3003*b^7*x^7 + 3696*a*b^6*x^6 + 63*a^2*b^5*x^5 - 70*a^3*b^4*x^4 + 80*a^4*b^3*x^3 - 96*a^5*b^2*x^2 + 1
28*a^6*b*x - 256*a^7)*sqrt(b*x + a)/b^6

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mupad [B]  time = 5.24, size = 80, normalized size = 0.50 \begin {gather*} -\frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2\,\left (256\,a^5-640\,a^4\,b\,x+1120\,a^3\,b^2\,x^2-1680\,a^2\,b^3\,x^3+2310\,a\,b^4\,x^4-3003\,b^5\,x^5\right )}{45045\,b^6\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a*x^2 + b*x^3)^(3/2),x)

[Out]

-(2*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^2*(256*a^5 - 3003*b^5*x^5 + 2310*a*b^4*x^4 + 1120*a^3*b^2*x^2 - 1680*a^2*b
^3*x^3 - 640*a^4*b*x))/(45045*b^6*x)

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

[Out]

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

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