\(\int \frac {x^3}{(a x+b x^2)^{5/2}} \, dx\) [63]

Optimal result

Integrand size = 17, antiderivative size = 23 \[ \int \frac {x^3}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 x^3}{3 a \left (a x+b x^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {664} \[ \int \frac {x^3}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 x^3}{3 a \left (a x+b x^2\right )^{3/2}} \]

[In]

Int[x^3/(a*x + b*x^2)^(5/2),x]

[Out]

(2*x^3)/(3*a*(a*x + b*x^2)^(3/2))

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^m*((a +
b*x + c*x^2)^(p + 1)/((p + 1)*(2*c*d - b*e))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 x^3}{3 a \left (a x+b x^2\right )^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {x^3}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 x^3}{3 a (x (a+b x))^{3/2}} \]

[In]

Integrate[x^3/(a*x + b*x^2)^(5/2),x]

[Out]

(2*x^3)/(3*a*(x*(a + b*x))^(3/2))

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
gosper	\(\frac {2 x^{4} \left (b x +a \right )}{3 a \left (b \,x^{2}+a x \right )^{\frac {5}{2}}}\)	\(25\)
trager	\(\frac {2 x \sqrt {b \,x^{2}+a x}}{3 \left (b x +a \right )^{2} a}\)	\(25\)
pseudoelliptic	\(\frac {2 x^{2}}{3 a \left (b x +a \right ) \sqrt {x \left (b x +a \right )}}\)	\(25\)
default	\(-\frac {x^{2}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {a \left (-\frac {x}{2 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\right )}{4 b}\right )}{2 b}\)	\(120\)

[In]

int(x^3/(b*x^2+a*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.52 \[ \int \frac {x^3}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \, \sqrt {b x^{2} + a x} x}{3 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )}} \]

[In]

integrate(x^3/(b*x^2+a*x)^(5/2),x, algorithm="fricas")

[Out]

2/3*sqrt(b*x^2 + a*x)*x/(a*b^2*x^2 + 2*a^2*b*x + a^3)

Sympy [F]

\[ \int \frac {x^3}{\left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {x^{3}}{\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(x**3/(b*x**2+a*x)**(5/2),x)

[Out]

Integral(x**3/(x*(a + b*x))**(5/2), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (19) = 38\).

Time = 0.20 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.22 \[ \int \frac {x^3}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {x^{2}}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} - \frac {a x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} b^{2}} + \frac {2 \, x}{3 \, \sqrt {b x^{2} + a x} a b} + \frac {1}{3 \, \sqrt {b x^{2} + a x} b^{2}} \]

[In]

integrate(x^3/(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (19) = 38\).

Time = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 3.70 \[ \int \frac {x^3}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} b + 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a \sqrt {b} + a^{2}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a\right )}^{3} b^{\frac {3}{2}}} \]

[In]

integrate(x^3/(b*x^2+a*x)^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {x^3}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {2\,x\,\sqrt {b\,x^2+a\,x}}{3\,a\,{\left (a+b\,x\right )}^2} \]

[In]

int(x^3/(a*x + b*x^2)^(5/2),x)

[Out]

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