\(\int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx\) [259]

Optimal result

Integrand size = 19, antiderivative size = 115 \[ \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx=-\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{7/2}} \]

[Out]

5/8*b^3*arctanh(x*a^(1/2)/(b*x^3+a*x^2)^(1/2))/a^(7/2)-1/3*(b*x^3+a*x^2)^(1/2)/a/x^4+5/12*b*(b*x^3+a*x^2)^(1/2
)/a^2/x^3-5/8*b^2*(b*x^3+a*x^2)^(1/2)/a^3/x^2

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2050, 2033, 212} \[ \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx=\frac {5 b^3 \text {arctanh}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{7/2}}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {\sqrt {a x^2+b x^3}}{3 a x^4} \]

[In]

Int[1/(x^3*Sqrt[a*x^2 + b*x^3]),x]

[Out]

-1/3*Sqrt[a*x^2 + b*x^3]/(a*x^4) + (5*b*Sqrt[a*x^2 + b*x^3])/(12*a^2*x^3) - (5*b^2*Sqrt[a*x^2 + b*x^3])/(8*a^3
*x^2) + (5*b^3*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^3]])/(8*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 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 {\sqrt {a x^2+b x^3}}{3 a x^4}-\frac {(5 b) \int \frac {1}{x^2 \sqrt {a x^2+b x^3}} \, dx}{6 a} \\ & = -\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}+\frac {\left (5 b^2\right ) \int \frac {1}{x \sqrt {a x^2+b x^3}} \, dx}{8 a^2} \\ & = -\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}-\frac {\left (5 b^3\right ) \int \frac {1}{\sqrt {a x^2+b x^3}} \, dx}{16 a^3} \\ & = -\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {\left (5 b^3\right ) \text {Subst}\left (\int \frac {1}{1-a x^2} \, dx,x,\frac {x}{\sqrt {a x^2+b x^3}}\right )}{8 a^3} \\ & = -\frac {\sqrt {a x^2+b x^3}}{3 a x^4}+\frac {5 b \sqrt {a x^2+b x^3}}{12 a^2 x^3}-\frac {5 b^2 \sqrt {a x^2+b x^3}}{8 a^3 x^2}+\frac {5 b^3 \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^3}}\right )}{8 a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx=\frac {-\sqrt {a} \left (8 a^3-2 a^2 b x+5 a b^2 x^2+15 b^3 x^3\right )+15 b^3 x^3 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{24 a^{7/2} x^2 \sqrt {x^2 (a+b x)}} \]

[In]

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

[Out]

(-(Sqrt[a]*(8*a^3 - 2*a^2*b*x + 5*a*b^2*x^2 + 15*b^3*x^3)) + 15*b^3*x^3*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sq
rt[a]])/(24*a^(7/2)*x^2*Sqrt[x^2*(a + b*x)])

Maple [A] (verified)

Time = 2.03 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.49

method	result	size
pseudoelliptic	\(\frac {-3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) b^{2} x^{2}-2 \sqrt {b x +a}\, a^{\frac {3}{2}}+3 b x \sqrt {b x +a}\, \sqrt {a}}{4 x^{2} a^{\frac {5}{2}}}\)	\(56\)
risch	\(-\frac {\left (b x +a \right ) \left (15 b^{2} x^{2}-10 a b x +8 a^{2}\right )}{24 a^{3} x^{2} \sqrt {x^{2} \left (b x +a \right )}}+\frac {5 b^{3} \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {b x +a}\, x}{8 a^{\frac {7}{2}} \sqrt {x^{2} \left (b x +a \right )}}\)	\(84\)
default	\(-\frac {\sqrt {b x +a}\, \left (15 a^{\frac {3}{2}} b^{2} x^{2} \sqrt {b x +a}-15 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) a \,b^{3} x^{3}-10 a^{\frac {5}{2}} b x \sqrt {b x +a}+8 \sqrt {b x +a}\, a^{\frac {7}{2}}\right )}{24 x^{2} \sqrt {b \,x^{3}+a \,x^{2}}\, a^{\frac {9}{2}}}\)	\(95\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.52 \[ \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx=\left [\frac {15 \, \sqrt {a} b^{3} x^{4} \log \left (\frac {b x^{2} + 2 \, a x + 2 \, \sqrt {b x^{3} + a x^{2}} \sqrt {a}}{x^{2}}\right ) - 2 \, {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{48 \, a^{4} x^{4}}, -\frac {15 \, \sqrt {-a} b^{3} x^{4} \arctan \left (\frac {\sqrt {b x^{3} + a x^{2}} \sqrt {-a}}{a x}\right ) + {\left (15 \, a b^{2} x^{2} - 10 \, a^{2} b x + 8 \, a^{3}\right )} \sqrt {b x^{3} + a x^{2}}}{24 \, a^{4} x^{4}}\right ] \]

[In]

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

[Out]

[1/48*(15*sqrt(a)*b^3*x^4*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 - 10*a^2*
b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*x^4), -1/24*(15*sqrt(-a)*b^3*x^4*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(a
*x)) + (15*a*b^2*x^2 - 10*a^2*b*x + 8*a^3)*sqrt(b*x^3 + a*x^2))/(a^4*x^4)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x^{2}} x^{3}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx=-\frac {\frac {15 \, b^{4} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {15 \, {\left (b x + a\right )}^{\frac {5}{2}} b^{4} - 40 \, {\left (b x + a\right )}^{\frac {3}{2}} a b^{4} + 33 \, \sqrt {b x + a} a^{2} b^{4}}{a^{3} b^{3} x^{3}}}{24 \, b \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-1/24*(15*b^4*arctan(sqrt(b*x + a)/sqrt(-a))/(sqrt(-a)*a^3) + (15*(b*x + a)^(5/2)*b^4 - 40*(b*x + a)^(3/2)*a*b
^4 + 33*sqrt(b*x + a)*a^2*b^4)/(a^3*b^3*x^3))/(b*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^3 \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x^3\,\sqrt {b\,x^3+a\,x^2}} \,d x \]

[In]

int(1/(x^3*(a*x^2 + b*x^3)^(1/2)),x)

[Out]

int(1/(x^3*(a*x^2 + b*x^3)^(1/2)), x)