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

Optimal. Leaf size=75 \[ \frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{5/2}}-\frac {3 \sqrt {a x^2+b x^3}}{a^2 x^2}+\frac {2}{a \sqrt {a x^2+b x^3}} \]

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Rubi [A]  time = 0.09, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2023, 2025, 2008, 206} \begin {gather*} -\frac {3 \sqrt {a x^2+b x^3}}{a^2 x^2}+\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{5/2}}+\frac {2}{a \sqrt {a x^2+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a*x^2 + b*x^3)^(3/2),x]

[Out]

2/(a*Sqrt[a*x^2 + b*x^3]) - (3*Sqrt[a*x^2 + b*x^3])/(a^2*x^2) + (3*b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])
/a^(5/2)

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]

Rule 2025

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, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx &=\frac {2}{a \sqrt {a x^2+b x^3}}+\frac {3 \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{a}\\ &=\frac {2}{a \sqrt {a x^2+b x^3}}-\frac {3 \sqrt {a x^2+b x^3}}{a^2 x^2}-\frac {(3 b) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{2 a^2}\\ &=\frac {2}{a \sqrt {a x^2+b x^3}}-\frac {3 \sqrt {a x^2+b x^3}}{a^2 x^2}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{a^2}\\ &=\frac {2}{a \sqrt {a x^2+b x^3}}-\frac {3 \sqrt {a x^2+b x^3}}{a^2 x^2}+\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.48 \begin {gather*} -\frac {2 b x \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {b x}{a}+1\right )}{a^2 \sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a*x^2 + b*x^3)^(3/2),x]

[Out]

(-2*b*x*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (b*x)/a])/(a^2*Sqrt[x^2*(a + b*x)])

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IntegrateAlgebraic [A]  time = 3.79, size = 76, normalized size = 1.01 \begin {gather*} \frac {x \sqrt {a+b x} \left (\frac {3 b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {2 a-3 (a+b x)}{a^2 x \sqrt {a+b x}}\right )}{\sqrt {x^2 (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[a + b*x]*((2*a - 3*(a + b*x))/(a^2*x*Sqrt[a + b*x]) + (3*b*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(5/2)))/S
qrt[x^2*(a + b*x)]

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fricas [A]  time = 0.41, size = 189, normalized size = 2.52 \begin {gather*} \left [\frac {3 \, {\left (b^{2} x^{3} + a b x^{2}\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}} {\left (3 \, a b x + a^{2}\right )}}{2 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac {3 \, {\left (b^{2} x^{3} + a b x^{2}\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}} {\left (3 \, a b x + a^{2}\right )}}{a^{3} b x^{3} + a^{4} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,0,0]%%%}+%%%{-2,[0,1,1]%%%},0,%%%{1,[0,2,2]%%%}] at parameters values [62.4600259969,-13,46]-4*a^2/4/a^4*sqrt
(a*(1/x)^2+b/x)+2*(-b^2/a^2/(-a*(sqrt(a*(1/x)^2+b/x)-sqrt(a)/x)+sqrt(a)*b)-3*b/4/a^2/sqrt(a)*ln(abs(2*sqrt(a)*
(sqrt(a*(1/x)^2+b/x)-sqrt(a)/x)-b)))

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maple [A]  time = 0.06, size = 62, normalized size = 0.83 \begin {gather*} \frac {\left (b x +a \right ) \left (3 \sqrt {b x +a}\, b x \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-3 \sqrt {a}\, b x -a^{\frac {3}{2}}\right ) x^{2}}{\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/(b*x^3+a*x^2)^(3/2),x)

[Out]

x^2*(b*x+a)*(3*(b*x+a)^(1/2)*arctanh((b*x+a)^(1/2)/a^(1/2))*x*b-3*x*b*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}{{\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/(b*x^3+a*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x/(b*x^3 + a*x^2)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x}{{\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/(a*x^2 + b*x^3)^(3/2),x)

[Out]

int(x/(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}{\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/(b*x**3+a*x**2)**(3/2),x)

[Out]

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

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