\(\int \frac {x}{(a x^2+b x^3)^{3/2}} \, dx\) [265]

Optimal result

Integrand size = 17, antiderivative size = 75 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {2}{a \sqrt {a x^2+b x^3}}-\frac {3 \sqrt {a x^2+b x^3}}{a^2 x^2}+\frac {3 b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2048, 2050, 2033, 212} \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {3 b \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{a^{5/2}}-\frac {3 \sqrt {a x^2+b x^3}}{a^2 x^2}+\frac {2}{a \sqrt {a x^2+b x^3}} \]

[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 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 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 2048

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 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 \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) \text {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*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.83 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\frac {-\sqrt {a} (a+3 b x)+3 b x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{5/2} \sqrt {x^2 (a+b x)}} \]

[In]

Integrate[x/(a*x^2 + b*x^3)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 1.82 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.27

method	result	size
pseudoelliptic	\(\frac {2 b x +4 a}{b^{2} \sqrt {b x +a}}\)	\(20\)
default	\(\frac {x^{2} \left (b x +a \right ) \left (3 \sqrt {b x +a}\, \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b x -3 \sqrt {a}\, b x -a^{\frac {3}{2}}\right )}{\left (b \,x^{3}+a \,x^{2}\right )^{\frac {3}{2}} a^{\frac {5}{2}}}\)	\(62\)
risch	\(-\frac {b x +a}{a^{2} \sqrt {x^{2} \left (b x +a \right )}}-\frac {b \left (\frac {4}{\sqrt {b x +a}}-\frac {6 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{2 a^{2} \sqrt {x^{2} \left (b x +a \right )}}\)	\(75\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 189, normalized size of antiderivative = 2.52 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {a} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, \sqrt {b x^{3} + a x^{2}} {\left (3 \, a b x + a^{2}\right )}}{2 \, {\left (a^{3} b x^{3} + a^{4} x^{2}\right )}}, -\frac {3 \, {\left (b^{2} x^{3} + a b x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + \sqrt {b x^{3} + a x^{2}} {\left (3 \, a b x + a^{2}\right )}}{a^{3} b x^{3} + a^{4} x^{2}}\right ] \]

[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)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\left (a x^2+b x^3\right )^{3/2}} \, dx=-\frac {3 \, b \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (x\right )} - \frac {3 \, {\left (b x + a\right )} b - 2 \, a b}{{\left ({\left (b x + a\right )}^{\frac {3}{2}} - \sqrt {b x + a} a\right )} a^{2} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-3*b*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^2*sgn(x)) - (3*(b*x + a)*b - 2*a*b)/(((b*x + a)^(3/2) - sqrt(b
*x + a)*a)*a^2*sgn(x))

Mupad [F(-1)]

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

[In]

int(x/(a*x^2 + b*x^3)^(3/2),x)

[Out]

int(x/(a*x^2 + b*x^3)^(3/2), x)