3.266 \(\int \frac {1}{(a x^2+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{7/2}}+\frac {15 b \sqrt {a x^2+b x^3}}{4 a^3 x^2}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}+\frac {2}{a x \sqrt {a x^2+b x^3}} \]

[Out]

-15/4*b^2*arctanh(x*a^(1/2)/(b*x^3+a*x^2)^(1/2))/a^(7/2)+2/a/x/(b*x^3+a*x^2)^(1/2)-5/2*(b*x^3+a*x^2)^(1/2)/a^2
/x^3+15/4*b*(b*x^3+a*x^2)^(1/2)/a^3/x^2

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2006, 2025, 2008, 206} \[ -\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{7/2}}+\frac {15 b \sqrt {a x^2+b x^3}}{4 a^3 x^2}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}+\frac {2}{a x \sqrt {a x^2+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[(a*x^2 + b*x^3)^(-3/2),x]

[Out]

2/(a*x*Sqrt[a*x^2 + b*x^3]) - (5*Sqrt[a*x^2 + b*x^3])/(2*a^2*x^3) + (15*b*Sqrt[a*x^2 + b*x^3])/(4*a^3*x^2) - (
15*b^2*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(4*a^(7/2))

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 2006

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

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 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 {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx &=\frac {2}{a x \sqrt {a x^2+b x^3}}+\frac {5 \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx}{a}\\ &=\frac {2}{a x \sqrt {a x^2+b x^3}}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}-\frac {(15 b) \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{4 a^2}\\ &=\frac {2}{a x \sqrt {a x^2+b x^3}}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}+\frac {15 b \sqrt {a x^2+b x^3}}{4 a^3 x^2}+\frac {\left (15 b^2\right ) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{8 a^3}\\ &=\frac {2}{a x \sqrt {a x^2+b x^3}}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}+\frac {15 b \sqrt {a x^2+b x^3}}{4 a^3 x^2}-\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{4 a^3}\\ &=\frac {2}{a x \sqrt {a x^2+b x^3}}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}+\frac {15 b \sqrt {a x^2+b x^3}}{4 a^3 x^2}-\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 38, normalized size = 0.35 \[ \frac {2 b^2 x \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {b x}{a}+1\right )}{a^3 \sqrt {x^2 (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x^2 + b*x^3)^(-3/2),x]

[Out]

(2*b^2*x*Hypergeometric2F1[-1/2, 3, 1/2, 1 + (b*x)/a])/(a^3*Sqrt[x^2*(a + b*x)])

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 219, normalized size = 1.99 \[ \left [\frac {15 \, {\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{8 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}, \frac {15 \, {\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{4 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(15*(b^3*x^4 + a*b^2*x^3)*sqrt(a)*log((b*x^2 + 2*a*x - 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) + 2*(15*a*b^2*
x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*b*x^4 + a^5*x^3), 1/4*(15*(b^3*x^4 + a*b^2*x^3)*sqrt(-a)*ar
ctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a*x)) + (15*a*b^2*x^2 + 5*a^2*b*x - 2*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*b*x^4
+ a^5*x^3)]

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

________________________________________________________________________________________

maple [A]  time = 0.06, size = 76, normalized size = 0.69 \[ -\frac {\left (b x +a \right ) \left (15 \sqrt {b x +a}\, b^{2} x^{2} \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )-15 \sqrt {a}\, b^{2} x^{2}-5 a^{\frac {3}{2}} b x +2 a^{\frac {5}{2}}\right ) x}{4 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x*(b*x+a)*(15*arctanh((b*x+a)^(1/2)/a^(1/2))*(b*x+a)^(1/2)*x^2*b^2-5*a^(3/2)*x*b-15*x^2*b^2*a^(1/2)+2*a^(
5/2))/(b*x^3+a*x^2)^(3/2)/a^(7/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x^2)^(-3/2), x)

________________________________________________________________________________________

mupad [B]  time = 5.43, size = 42, normalized size = 0.38 \[ -\frac {2\,x\,{\left (\frac {a}{b\,x}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {a}{b\,x}\right )}{7\,{\left (b\,x^3+a\,x^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x^2 + b*x^3)^(3/2),x)

[Out]

-(2*x*(a/(b*x) + 1)^(3/2)*hypergeom([3/2, 7/2], 9/2, -a/(b*x)))/(7*(a*x^2 + b*x^3)^(3/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a x^{2} + b x^{3}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________