3.312 \(\int \frac {x^2}{\sqrt {a x^3+b x^4}} \, dx\)

Optimal. Leaf size=56 \[ \frac {\sqrt {a x^3+b x^4}}{b x}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 56, 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 = {2024, 2029, 206} \[ \frac {\sqrt {a x^3+b x^4}}{b x}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^2/Sqrt[a*x^3 + b*x^4],x]

[Out]

Sqrt[a*x^3 + b*x^4]/(b*x) - (a*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a*x^3 + b*x^4]])/b^(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 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2029

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

Rubi steps

\begin {align*} \int \frac {x^2}{\sqrt {a x^3+b x^4}} \, dx &=\frac {\sqrt {a x^3+b x^4}}{b x}-\frac {a \int \frac {x}{\sqrt {a x^3+b x^4}} \, dx}{2 b}\\ &=\frac {\sqrt {a x^3+b x^4}}{b x}-\frac {a \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a x^3+b x^4}}\right )}{b}\\ &=\frac {\sqrt {a x^3+b x^4}}{b x}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x^3+b x^4}}\right )}{b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 75, normalized size = 1.34 \[ \frac {\sqrt {b} x^2 (a+b x)-a^{3/2} x^{3/2} \sqrt {\frac {b x}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{b^{3/2} \sqrt {x^3 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/Sqrt[a*x^3 + b*x^4],x]

[Out]

(Sqrt[b]*x^2*(a + b*x) - a^(3/2)*x^(3/2)*Sqrt[1 + (b*x)/a]*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(b^(3/2)*Sqrt[x
^3*(a + b*x)])

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fricas [A]  time = 0.41, size = 122, normalized size = 2.18 \[ \left [\frac {a \sqrt {b} x \log \left (\frac {2 \, b x^{2} + a x - 2 \, \sqrt {b x^{4} + a x^{3}} \sqrt {b}}{x}\right ) + 2 \, \sqrt {b x^{4} + a x^{3}} b}{2 \, b^{2} x}, \frac {a \sqrt {-b} x \arctan \left (\frac {\sqrt {b x^{4} + a x^{3}} \sqrt {-b}}{b x^{2}}\right ) + \sqrt {b x^{4} + a x^{3}} b}{b^{2} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^4+a*x^3)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(a*sqrt(b)*x*log((2*b*x^2 + a*x - 2*sqrt(b*x^4 + a*x^3)*sqrt(b))/x) + 2*sqrt(b*x^4 + a*x^3)*b)/(b^2*x), (
a*sqrt(-b)*x*arctan(sqrt(b*x^4 + a*x^3)*sqrt(-b)/(b*x^2)) + sqrt(b*x^4 + a*x^3)*b)/(b^2*x)]

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giac [A]  time = 0.23, size = 48, normalized size = 0.86 \[ \frac {\frac {a \sqrt {b + \frac {a}{x}} x}{b} + \frac {a^{2} \arctan \left (\frac {\sqrt {b + \frac {a}{x}}}{\sqrt {-b}}\right )}{\sqrt {-b} b}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^4+a*x^3)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 78, normalized size = 1.39 \[ \frac {\sqrt {\left (b x +a \right ) x}\, \left (-a b \ln \left (\frac {2 b x +a +2 \sqrt {b \,x^{2}+a x}\, \sqrt {b}}{2 \sqrt {b}}\right )+2 \sqrt {b \,x^{2}+a x}\, b^{\frac {3}{2}}\right ) x}{2 \sqrt {b \,x^{4}+a \,x^{3}}\, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b*x^4+a*x^3)^(1/2),x)

[Out]

1/2*x*((b*x+a)*x)^(1/2)*(2*(b*x^2+a*x)^(1/2)*b^(3/2)-a*ln(1/2*(2*b*x+a+2*(b*x^2+a*x)^(1/2)*b^(1/2))/b^(1/2))*b
)/(b*x^4+a*x^3)^(1/2)/b^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {b x^{4} + a x^{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^4+a*x^3)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^2/sqrt(b*x^4 + a*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^2}{\sqrt {b\,x^4+a\,x^3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a*x^3 + b*x^4)^(1/2),x)

[Out]

int(x^2/(a*x^3 + b*x^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{2}}{\sqrt {x^{3} \left (a + b x\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b*x**4+a*x**3)**(1/2),x)

[Out]

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

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