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

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

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Rubi [A]  time = 0.06, 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 = {2023, 2008, 206} \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 206

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

Rule 2008

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 2023

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 \operatorname {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*}

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Mathematica [C]  time = 0.03, size = 35, normalized size = 0.67 \begin {gather*} \frac {2 x \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b x}{a}+1\right )}{a \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*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*x)/a])/(a*Sqrt[x^2*(a + b*x)])

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IntegrateAlgebraic [A]  time = 2.41, size = 63, normalized size = 1.21 \begin {gather*} \frac {2 \sqrt {a x^2+b x^3}}{a x (a+b x)}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, 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)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Unable to divide, perhaps due to rounding error%%%{%%%{1,[1]%%%},[2,2]%%%}+%%%{%%{[-2,0]:[1,0,%%%{
-1,[1]%%%}]%%},[1,3]%%%}+%%%{1,[0,4]%%%} / %%%{%%%{1,[2]%%%},[2,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0,%%%{-1,[
1]%%%}]%%},[1,1]%%%}+%%%{%%%{1,[1]%%%},[0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.05, size = 54, normalized size = 1.04 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (\sqrt {b x +a}\, a \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-a^{\frac {3}{2}}\right ) x^{3}}{\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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

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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)

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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)

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