\(\int \frac {1}{(a x^2+b x^3)^{3/2}} \, dx\) [266]

Optimal result

Integrand size = 15, antiderivative size = 110 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2}{a x \sqrt {a x^2+b x^3}}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}+\frac {15 b \sqrt {a x^2+b x^3}}{4 a^3 x^2}-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{7/2}} \]

[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] (verified)

Time = 0.06 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2031, 2050, 2033, 212} \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {15 b^2 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{4 a^{7/2}}+\frac {15 b \sqrt {a x^2+b x^3}}{4 a^3 x^2}-\frac {5 \sqrt {a x^2+b x^3}}{2 a^2 x^3}+\frac {2}{a x \sqrt {a x^2+b x^3}} \]

[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 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 2031

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 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 2050

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*} \text {integral}& = \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 ) \text {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 [A] (verified)

Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-2 a^2+5 a b x+15 b^2 x^2\right )-15 b^2 x^2 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{7/2} x \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[(a*x^2 + b*x^3)^(-3/2),x]

[Out]

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

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.12

method	result	size
pseudoelliptic	\(-\frac {2}{b \sqrt {b x +a}}\)	\(13\)
default	\(-\frac {x \left (b x +a \right ) \left (15 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-5 a^{\frac {3}{2}} b x -15 \sqrt {a}\, b^{2} x^{2}+2 a^{\frac {5}{2}}\right )}{4 \left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {7}{2}}}\)	\(76\)
risch	\(-\frac {\left (b x +a \right ) \left (-7 b x +2 a \right )}{4 a^{3} x \sqrt {x^{2} \left (b x +a \right )}}+\frac {b^{2} \left (\frac {16}{\sqrt {b x +a}}-\frac {30 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{8 a^{3} \sqrt {x^{2} \left (b x +a \right )}}\)	\(88\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.99 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x - 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) + 2 \, {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{8 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}, \frac {15 \, {\left (b^{3} x^{4} + a b^{2} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (15 \, a b^{2} x^{2} + 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{4 \, {\left (a^{4} b x^{4} + a^{5} x^{3}\right )}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int \frac {1}{\left (a x^{2} + b x^{3}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b x^{3} + a x^{2}\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {15 \, b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{3} \mathrm {sgn}\left (x\right )} + \frac {2 \, b^{2}}{\sqrt {b x + a} a^{3} \mathrm {sgn}\left (x\right )} + \frac {7 \, {\left (b x + a\right )}^{\frac {3}{2}} b^{2} - 9 \, \sqrt {b x + a} a b^{2}}{4 \, a^{3} b^{2} x^{2} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

15/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3*sgn(x)) + 2*b^2/(sqrt(b*x + a)*a^3*sgn(x)) + 1/4*(7*(b*x
+ a)^(3/2)*b^2 - 9*sqrt(b*x + a)*a*b^2)/(a^3*b^2*x^2*sgn(x))

Mupad [B] (verification not implemented)

Time = 9.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.38 \[ \int \frac {1}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {2\,x\,{\left (\frac {a}{b\,x}+1\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {3}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {a}{b\,x}\right )}{7\,{\left (b\,x^3+a\,x^2\right )}^{3/2}} \]

[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))