\(\int x^2 (a x^2+b x^3)^{3/2} \, dx\) [240]

Optimal result

Integrand size = 19, antiderivative size = 161 \[ \int x^2 \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b}-\frac {512 a^5 \left (a x^2+b x^3\right )^{5/2}}{45045 b^6 x^5}+\frac {256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac {64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x} \]

[Out]

2/15*(b*x^3+a*x^2)^(5/2)/b-512/45045*a^5*(b*x^3+a*x^2)^(5/2)/b^6/x^5+256/9009*a^4*(b*x^3+a*x^2)^(5/2)/b^5/x^4-
64/1287*a^3*(b*x^3+a*x^2)^(5/2)/b^4/x^3+32/429*a^2*(b*x^3+a*x^2)^(5/2)/b^3/x^2-4/39*a*(b*x^3+a*x^2)^(5/2)/b^2/
x

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2041, 2027, 2039} \[ \int x^2 \left (a x^2+b x^3\right )^{3/2} \, dx=-\frac {512 a^5 \left (a x^2+b x^3\right )^{5/2}}{45045 b^6 x^5}+\frac {256 a^4 \left (a x^2+b x^3\right )^{5/2}}{9009 b^5 x^4}-\frac {64 a^3 \left (a x^2+b x^3\right )^{5/2}}{1287 b^4 x^3}+\frac {32 a^2 \left (a x^2+b x^3\right )^{5/2}}{429 b^3 x^2}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{39 b^2 x}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{15 b} \]

[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 2027

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 2039

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] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.50 \[ \int x^2 \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 x (a+b x)^3 \left (-256 a^5+640 a^4 b x-1120 a^3 b^2 x^2+1680 a^2 b^3 x^3-2310 a b^4 x^4+3003 b^5 x^5\right )}{45045 b^6 \sqrt {x^2 (a+b x)}} \]

[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)])

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.20

method	result	size
pseudoelliptic	\(\frac {2 \left (b x +a \right )^{\frac {5}{2}} \left (35 b^{2} x^{2}-20 a b x +8 a^{2}\right )}{315 b^{3}}\)	\(32\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-3003 b^{5} x^{5}+2310 a \,b^{4} x^{4}-1680 a^{2} b^{3} x^{3}+1120 a^{3} b^{2} x^{2}-640 a^{4} b x +256 a^{5}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{45045 b^{6} x^{3}}\)	\(79\)
default	\(-\frac {2 \left (b x +a \right ) \left (-3003 b^{5} x^{5}+2310 a \,b^{4} x^{4}-1680 a^{2} b^{3} x^{3}+1120 a^{3} b^{2} x^{2}-640 a^{4} b x +256 a^{5}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{45045 b^{6} x^{3}}\)	\(79\)
risch	\(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-3003 b^{7} x^{7}-3696 a \,b^{6} x^{6}-63 a^{2} b^{5} x^{5}+70 a^{3} b^{4} x^{4}-80 a^{4} b^{3} x^{3}+96 a^{5} b^{2} x^{2}-128 a^{6} b x +256 a^{7}\right )}{45045 x \,b^{6}}\)	\(94\)
trager	\(-\frac {2 \left (-3003 b^{7} x^{7}-3696 a \,b^{6} x^{6}-63 a^{2} b^{5} x^{5}+70 a^{3} b^{4} x^{4}-80 a^{4} b^{3} x^{3}+96 a^{5} b^{2} x^{2}-128 a^{6} b x +256 a^{7}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{45045 b^{6} x}\)	\(96\)

[In]

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

[Out]

2/315*(b*x+a)^(5/2)*(35*b^2*x^2-20*a*b*x+8*a^2)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.59 \[ \int x^2 \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (3003 \, b^{7} x^{7} + 3696 \, a b^{6} x^{6} + 63 \, a^{2} b^{5} x^{5} - 70 \, a^{3} b^{4} x^{4} + 80 \, a^{4} b^{3} x^{3} - 96 \, a^{5} b^{2} x^{2} + 128 \, a^{6} b x - 256 \, a^{7}\right )} \sqrt {b x^{3} + a x^{2}}}{45045 \, b^{6} x} \]

[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)

Sympy [F]

\[ \int x^2 \left (a x^2+b x^3\right )^{3/2} \, dx=\int x^{2} \left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.53 \[ \int x^2 \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (3003 \, b^{7} x^{7} + 3696 \, a b^{6} x^{6} + 63 \, a^{2} b^{5} x^{5} - 70 \, a^{3} b^{4} x^{4} + 80 \, a^{4} b^{3} x^{3} - 96 \, a^{5} b^{2} x^{2} + 128 \, a^{6} b x - 256 \, a^{7}\right )} \sqrt {b x + a}}{45045 \, b^{6}} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (137) = 274\).

Time = 0.28 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.75 \[ \int x^2 \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {512 \, a^{\frac {15}{2}} \mathrm {sgn}\left (x\right )}{45045 \, b^{6}} + \frac {2 \, {\left (\frac {65 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} a^{2} \mathrm {sgn}\left (x\right )}{b^{5}} + \frac {30 \, {\left (231 \, {\left (b x + a\right )}^{\frac {13}{2}} - 1638 \, {\left (b x + a\right )}^{\frac {11}{2}} a + 5005 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{2} - 8580 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{3} + 9009 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{4} - 6006 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{5} + 3003 \, \sqrt {b x + a} a^{6}\right )} a \mathrm {sgn}\left (x\right )}{b^{5}} + \frac {7 \, {\left (429 \, {\left (b x + a\right )}^{\frac {15}{2}} - 3465 \, {\left (b x + a\right )}^{\frac {13}{2}} a + 12285 \, {\left (b x + a\right )}^{\frac {11}{2}} a^{2} - 25025 \, {\left (b x + a\right )}^{\frac {9}{2}} a^{3} + 32175 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{4} - 27027 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{5} + 15015 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{6} - 6435 \, \sqrt {b x + a} a^{7}\right )} \mathrm {sgn}\left (x\right )}{b^{5}}\right )}}{45045 \, b} \]

[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

Mupad [B] (verification not implemented)

Time = 9.14 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.50 \[ \int x^2 \left (a x^2+b x^3\right )^{3/2} \, dx=-\frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2\,\left (256\,a^5-640\,a^4\,b\,x+1120\,a^3\,b^2\,x^2-1680\,a^2\,b^3\,x^3+2310\,a\,b^4\,x^4-3003\,b^5\,x^5\right )}{45045\,b^6\,x} \]

[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)