\(\int \frac {(a x^2+b x^3)^{3/2}}{x^2} \, dx\) [244]

Optimal result

Integrand size = 19, antiderivative size = 52 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4} \]

[Out]

-4/35*a*(b*x^3+a*x^2)^(5/2)/b^2/x^5+2/7*(b*x^3+a*x^2)^(5/2)/b/x^4

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2041, 2039} \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4}-\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5} \]

[In]

Int[(a*x^2 + b*x^3)^(3/2)/x^2,x]

[Out]

(-4*a*(a*x^2 + b*x^3)^(5/2))/(35*b^2*x^5) + (2*(a*x^2 + b*x^3)^(5/2))/(7*b*x^4)

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}}{7 b x^4}-\frac {(2 a) \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^3} \, dx}{7 b} \\ & = -\frac {4 a \left (a x^2+b x^3\right )^{5/2}}{35 b^2 x^5}+\frac {2 \left (a x^2+b x^3\right )^{5/2}}{7 b x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.60 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 \left (x^2 (a+b x)\right )^{5/2} (-2 a+5 b x)}{35 b^2 x^5} \]

[In]

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

[Out]

(2*(x^2*(a + b*x))^(5/2)*(-2*a + 5*b*x))/(35*b^2*x^5)

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.67

method	result	size
gosper	\(-\frac {2 \left (b x +a \right ) \left (-5 b x +2 a \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{3}}\)	\(35\)
default	\(-\frac {2 \left (b x +a \right ) \left (-5 b x +2 a \right ) \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}}}{35 b^{2} x^{3}}\)	\(35\)
risch	\(-\frac {2 \sqrt {x^{2} \left (b x +a \right )}\, \left (-5 b^{3} x^{3}-8 a \,b^{2} x^{2}-a^{2} b x +2 a^{3}\right )}{35 x \,b^{2}}\)	\(50\)
trager	\(-\frac {2 \left (-5 b^{3} x^{3}-8 a \,b^{2} x^{2}-a^{2} b x +2 a^{3}\right ) \sqrt {b \,x^{3}+a \,x^{2}}}{35 b^{2} x}\)	\(52\)
pseudoelliptic	\(\frac {2 b x \sqrt {b x +a}\, \sqrt {a}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a b x -\sqrt {b x +a}\, a^{\frac {3}{2}}}{x \sqrt {a}}\)	\(52\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 \, {\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{35 \, b^{2} x} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.79 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {2 \, {\left (5 \, b^{3} x^{3} + 8 \, a b^{2} x^{2} + a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x + a}}{35 \, b^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (44) = 88\).

Time = 0.29 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.62 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=\frac {4 \, a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )}{35 \, b^{2}} + \frac {2 \, {\left (\frac {35 \, {\left ({\left (b x + a\right )}^{\frac {3}{2}} - 3 \, \sqrt {b x + a} a\right )} a^{2} \mathrm {sgn}\left (x\right )}{b} + \frac {14 \, {\left (3 \, {\left (b x + a\right )}^{\frac {5}{2}} - 10 \, {\left (b x + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {b x + a} a^{2}\right )} a \mathrm {sgn}\left (x\right )}{b} + \frac {3 \, {\left (5 \, {\left (b x + a\right )}^{\frac {7}{2}} - 21 \, {\left (b x + a\right )}^{\frac {5}{2}} a + 35 \, {\left (b x + a\right )}^{\frac {3}{2}} a^{2} - 35 \, \sqrt {b x + a} a^{3}\right )} \mathrm {sgn}\left (x\right )}{b}\right )}}{105 \, b} \]

[In]

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

[Out]

4/35*a^(7/2)*sgn(x)/b^2 + 2/105*(35*((b*x + a)^(3/2) - 3*sqrt(b*x + a)*a)*a^2*sgn(x)/b + 14*(3*(b*x + a)^(5/2)
- 10*(b*x + a)^(3/2)*a + 15*sqrt(b*x + a)*a^2)*a*sgn(x)/b + 3*(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)*sgn(x)/b)/b

Mupad [B] (verification not implemented)

Time = 8.92 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.69 \[ \int \frac {\left (a x^2+b x^3\right )^{3/2}}{x^2} \, dx=-\frac {2\,\left (2\,a-5\,b\,x\right )\,\sqrt {b\,x^3+a\,x^2}\,{\left (a+b\,x\right )}^2}{35\,b^2\,x} \]

[In]

int((a*x^2 + b*x^3)^(3/2)/x^2,x)

[Out]

-(2*(2*a - 5*b*x)*(a*x^2 + b*x^3)^(1/2)*(a + b*x)^2)/(35*b^2*x)