\(\int (a x^2+b x^3)^{3/2} \, dx\) [242]

Optimal result

Integrand size = 15, antiderivative size = 108 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=-\frac {32 a^3 \left (a x^2+b x^3\right )^{5/2}}{1155 b^4 x^5}+\frac {16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2} \]

[Out]

-32/1155*a^3*(b*x^3+a*x^2)^(5/2)/b^4/x^5+16/231*a^2*(b*x^3+a*x^2)^(5/2)/b^3/x^4-4/33*a*(b*x^3+a*x^2)^(5/2)/b^2
/x^3+2/11*(b*x^3+a*x^2)^(5/2)/b/x^2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2027, 2041, 2039} \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=-\frac {32 a^3 \left (a x^2+b x^3\right )^{5/2}}{1155 b^4 x^5}+\frac {16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2} \]

[In]

Int[(a*x^2 + b*x^3)^(3/2),x]

[Out]

(-32*a^3*(a*x^2 + b*x^3)^(5/2))/(1155*b^4*x^5) + (16*a^2*(a*x^2 + b*x^3)^(5/2))/(231*b^3*x^4) - (4*a*(a*x^2 +
b*x^3)^(5/2))/(33*b^2*x^3) + (2*(a*x^2 + b*x^3)^(5/2))/(11*b*x^2)

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}}{11 b x^2}-\frac {(6 a) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx}{11 b} \\ & = -\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}+\frac {\left (8 a^2\right ) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx}{33 b^2} \\ & = \frac {16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2}-\frac {\left (16 a^3\right ) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{231 b^3} \\ & = -\frac {32 a^3 \left (a x^2+b x^3\right )^{5/2}}{1155 b^4 x^5}+\frac {16 a^2 \left (a x^2+b x^3\right )^{5/2}}{231 b^3 x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{33 b^2 x^3}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{11 b x^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.54 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 x (a+b x)^3 \left (-16 a^3+40 a^2 b x-70 a b^2 x^2+105 b^3 x^3\right )}{1155 b^4 \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[(a*x^2 + b*x^3)^(3/2),x]

[Out]

(2*x*(a + b*x)^3*(-16*a^3 + 40*a^2*b*x - 70*a*b^2*x^2 + 105*b^3*x^3))/(1155*b^4*Sqrt[x^2*(a + b*x)])

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.12

method	result	size
pseudoelliptic	\(\frac {2 \left (b x +a \right )^{\frac {5}{2}}}{5 b}\)	\(13\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-105 b^{3} x^{3}+70 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{1155 b^{4} x^{3}}\)	\(57\)
default	\(-\frac {2 \left (b x +a \right ) \left (-105 b^{3} x^{3}+70 a \,b^{2} x^{2}-40 a^{2} b x +16 a^{3}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{1155 b^{4} x^{3}}\)	\(57\)
risch	\(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-105 b^{5} x^{5}-140 a \,b^{4} x^{4}-5 a^{2} b^{3} x^{3}+6 a^{3} b^{2} x^{2}-8 a^{4} b x +16 a^{5}\right )}{1155 x \,b^{4}}\)	\(72\)
trager	\(-\frac {2 \left (-105 b^{5} x^{5}-140 a \,b^{4} x^{4}-5 a^{2} b^{3} x^{3}+6 a^{3} b^{2} x^{2}-8 a^{4} b x +16 a^{5}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{1155 b^{4} x}\)	\(74\)

[In]

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

[Out]

2/5*(b*x+a)^(5/2)/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.68 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x^{3} + a x^{2}}}{1155 \, b^{4} x} \]

[In]

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

[Out]

2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^4*b*x - 16*a^5)*sqrt(b*x^3 + a*x^2)/
(b^4*x)

Sympy [F]

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

[In]

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

[Out]

Integral((a*x**2 + b*x**3)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.59 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {2 \, {\left (105 \, b^{5} x^{5} + 140 \, a b^{4} x^{4} + 5 \, a^{2} b^{3} x^{3} - 6 \, a^{3} b^{2} x^{2} + 8 \, a^{4} b x - 16 \, a^{5}\right )} \sqrt {b x + a}}{1155 \, b^{4}} \]

[In]

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

[Out]

2/1155*(105*b^5*x^5 + 140*a*b^4*x^4 + 5*a^2*b^3*x^3 - 6*a^3*b^2*x^2 + 8*a^4*b*x - 16*a^5)*sqrt(b*x + a)/b^4

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 210 vs. \(2 (92) = 184\).

Time = 0.29 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.94 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=\frac {32 \, a^{\frac {11}{2}} \mathrm {sgn}\left (x\right )}{1155 \, b^{4}} + \frac {2 \, {\left (\frac {99 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a^{2} \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {22 \, {\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} a \mathrm {sgn}\left (x\right )}{b^{3}} + \frac {5 \, {\left (63 \, {\left (b x + a\right )}^{\frac {11}{2}} - 385 \, {\left (b x + a\right )}^{\frac {9}{2}} a + 990 \, {\left (b x + a\right )}^{\frac {7}{2}} a^{2} - 1386 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{3} + 1155 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{4} - 693 \, \sqrt {b x + a} a^{5}\right )} \mathrm {sgn}\left (x\right )}{b^{3}}\right )}}{3465 \, b} \]

[In]

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

[Out]

32/1155*a^(11/2)*sgn(x)/b^4 + 2/3465*(99*(5*(b*x + a)^(7/2) - 21*(b*x + a)^(5/2)*a + 35*(b*x + a)^(3/2)*a^2 -
35*sqrt(b*x + a)*a^3)*a^2*sgn(x)/b^3 + 22*(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*sgn(x)/b^3 + 5*(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)*
sgn(x)/b^3)/b

Mupad [B] (verification not implemented)

Time = 9.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.54 \[ \int \left (a x^2+b x^3\right )^{3/2} \, dx=-\frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2\,\left (16\,a^3-40\,a^2\,b\,x+70\,a\,b^2\,x^2-105\,b^3\,x^3\right )}{1155\,b^4\,x} \]

[In]

int((a*x^2 + b*x^3)^(3/2),x)

[Out]

-(2*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^2*(16*a^3 - 105*b^3*x^3 + 70*a*b^2*x^2 - 40*a^2*b*x))/(1155*b^4*x)