3.3.64 \(\int \frac {x^2}{(a x^2+b x^3)^{3/2}} \, dx\) [264]

Optimal. Leaf size=52 \[ \frac {2 x}{a \sqrt {a x^2+b x^3}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2048, 2033, 212} \begin {gather*} \frac {2 x}{a \sqrt {a x^2+b x^3}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a*x^2 + b*x^3)^(3/2),x]

[Out]

(2*x)/(a*Sqrt[a*x^2 + b*x^3]) - (2*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/a^(3/2)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2033

Int[1/Sqrt[(a_.)*(x_)^2 + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[2/(2 - n), Subst[Int[1/(1 - a*x^2), x], x, x/Sq
rt[a*x^2 + b*x^n]], x] /; FreeQ[{a, b, n}, x] && NeQ[n, 2]

Rule 2048

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] + Dist[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))
, Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n]
&& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a x^2+b x^3\right )^{3/2}} \, dx &=\frac {2 x}{a \sqrt {a x^2+b x^3}}+\frac {\int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{a}\\ &=\frac {2 x}{a \sqrt {a x^2+b x^3}}-\frac {2 \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{a}\\ &=\frac {2 x}{a \sqrt {a x^2+b x^3}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.03, size = 54, normalized size = 1.04 \begin {gather*} \frac {2 x \left (\sqrt {a}-\sqrt {a+b x} \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{a^{3/2} \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(Sqrt[a] - Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(a^(3/2)*Sqrt[x^2*(a + b*x)])

________________________________________________________________________________________

Maple [A]
time = 0.37, size = 54, normalized size = 1.04

method	result	size
default	\(-\frac {2 x^{3} \left (b x +a \right ) \left (\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \sqrt {b x +a}-a^{\frac {3}{2}}\right )}{\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {5}{2}}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 1.47, size = 156, normalized size = 3.00 \begin {gather*} \left [\frac {{\left (b x^{2} + a x\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, \sqrt {b x^{3} + a x^{2}} a}{a^{2} b x^{2} + a^{3} x}, \frac {2 \, {\left ({\left (b x^{2} + a x\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + \sqrt {b x^{3} + a x^{2}} a\right )}}{a^{2} b x^{2} + a^{3} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((b*x^2 + a*x)*sqrt(a)*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*sqrt(b*x^3 + a*x^2)*a)/(a
^2*b*x^2 + a^3*x), 2*((b*x^2 + a*x)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) + sqrt(b*x^3 + a*x^2)*
a)/(a^2*b*x^2 + a^3*x)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (x^{2} \left (a + b x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

Giac [A]
time = 1.31, size = 77, normalized size = 1.48 \begin {gather*} -\frac {2 \, {\left (\sqrt {a} \arctan \left (\frac {\sqrt {a}}{\sqrt {-a}}\right ) + \sqrt {-a}\right )} \mathrm {sgn}\left (x\right )}{\sqrt {-a} a^{\frac {3}{2}}} + \frac {2 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (x\right )} + \frac {2}{\sqrt {b x + a} a \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{{\left (b\,x^3+a\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x^2 + b*x^3)^(3/2),x)

[Out]

int(x^2/(a*x^2 + b*x^3)^(3/2), x)

________________________________________________________________________________________