\(\int \frac {(a x^2+b x^3)^{3/2}}{x} \, dx\) [243]

Optimal result

Integrand size = 19, antiderivative size = 80 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {16 a^2 \left (a x^2+b x^3\right )^{5/2}}{315 b^3 x^5}-\frac {8 a \left (a x^2+b x^3\right )^{5/2}}{63 b^2 x^4}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3} \]

[Out]

16/315*a^2*(b*x^3+a*x^2)^(5/2)/b^3/x^5-8/63*a*(b*x^3+a*x^2)^(5/2)/b^2/x^4+2/9*(b*x^3+a*x^2)^(5/2)/b/x^3

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2041, 2039} \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {16 a^2 \left (a x^2+b x^3\right )^{5/2}}{315 b^3 x^5}-\frac {8 a \left (a x^2+b x^3\right )^{5/2}}{63 b^2 x^4}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3} \]

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x,x]

[Out]

(16*a^2*(a*x^2 + b*x^3)^(5/2))/(315*b^3*x^5) - (8*a*(a*x^2 + b*x^3)^(5/2))/(63*b^2*x^4) + (2*(a*x^2 + b*x^3)^(
5/2))/(9*b*x^3)

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}}{9 b x^3}-\frac {(4 a) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx}{9 b} \\ & = -\frac {8 a \left (a x^2+b x^3\right )^{5/2}}{63 b^2 x^4}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3}+\frac {\left (8 a^2\right ) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{63 b^2} \\ & = \frac {16 a^2 \left (a x^2+b x^3\right )^{5/2}}{315 b^3 x^5}-\frac {8 a \left (a x^2+b x^3\right )^{5/2}}{63 b^2 x^4}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{9 b x^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 x (a+b x)^3 \left (8 a^2-20 a b x+35 b^2 x^2\right )}{315 b^3 \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[(a*x^2 + b*x^3)^(3/2)/x,x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 3 vs. order 2.

Time = 1.82 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.44

method	result	size
pseudoelliptic	\(-2 a^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )+\frac {2 \sqrt {b x +a}\, \left (b x +4 a \right )}{3}\)	\(35\)
gosper	\(\frac {2 \left (b x +a \right ) \left (35 b^{2} x^{2}-20 a b x +8 a^{2}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{315 b^{3} x^{3}}\)	\(46\)
default	\(\frac {2 \left (b x +a \right ) \left (35 b^{2} x^{2}-20 a b x +8 a^{2}\right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{315 b^{3} x^{3}}\)	\(46\)
risch	\(\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (35 b^{4} x^{4}+50 a \,b^{3} x^{3}+3 a^{2} b^{2} x^{2}-4 a^{3} b x +8 a^{4}\right )}{315 x \,b^{3}}\)	\(61\)
trager	\(\frac {2 \left (35 b^{4} x^{4}+50 a \,b^{3} x^{3}+3 a^{2} b^{2} x^{2}-4 a^{3} b x +8 a^{4}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{315 b^{3} x}\)	\(63\)

[In]

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

[Out]

-2*a^(3/2)*arctanh((b*x+a)^(1/2)/a^(1/2))+2/3*(b*x+a)^(1/2)*(b*x+4*a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} + 50 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x^{3} + a x^{2}}}{315 \, b^{3} x} \]

[In]

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

[Out]

2/315*(35*b^4*x^4 + 50*a*b^3*x^3 + 3*a^2*b^2*x^2 - 4*a^3*b*x + 8*a^4)*sqrt(b*x^3 + a*x^2)/(b^3*x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.66 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2 \, {\left (35 \, b^{4} x^{4} + 50 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} - 4 \, a^{3} b x + 8 \, a^{4}\right )} \sqrt {b x + a}}{315 \, b^{3}} \]

[In]

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

[Out]

2/315*(35*b^4*x^4 + 50*a*b^3*x^3 + 3*a^2*b^2*x^2 - 4*a^3*b*x + 8*a^4)*sqrt(b*x + a)/b^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (68) = 136\).

Time = 0.29 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.16 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx=-\frac {16 \, a^{\frac {9}{2}} \mathrm {sgn}\left (x\right )}{315 \, b^{3}} + \frac {2 \, {\left (\frac {21 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a^{2} \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {18 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} a \mathrm {sgn}\left (x\right )}{b^{2}} + \frac {{\left (35 \, {\left (b x + a\right )}^{\frac {9}{2}} - 180 \, {\left (b x + a\right )}^{\frac {7}{2}} a + 378 \, {\left (b x + a\right )}^{\frac {5}{2}} a^{2} - 420 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{3} + 315 \, \sqrt {b x + a} a^{4}\right )} \mathrm {sgn}\left (x\right )}{b^{2}}\right )}}{315 \, b} \]

[In]

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

[Out]

-16/315*a^(9/2)*sgn(x)/b^3 + 2/315*(21*(3*(b*x + a)^(5/2) - 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a^2*s
gn(x)/b^2 + 18*(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*sg
n(x)/b^2 + (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 + 3
15*sqrt(b*x + a)*a^4)*sgn(x)/b^2)/b

Mupad [B] (verification not implemented)

Time = 9.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.59 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x} \, dx=\frac {2\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2\,\left (8\,a^2-20\,a\,b\,x+35\,b^2\,x^2\right )}{315\,b^3\,x} \]

[In]

int((a*x^2 + b*x^3)^(3/2)/x,x)

[Out]

(2*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^2*(8*a^2 + 35*b^2*x^2 - 20*a*b*x))/(315*b^3*x)