3.3.88 \(\int \frac {1}{x^3 \sqrt {a x^2+b x^5}} \, dx\) [288]

Optimal. Leaf size=59 \[ -\frac {\sqrt {a x^2+b x^5}}{3 a x^4}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^5}}\right )}{3 a^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2050, 2033, 212} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+b x^5}}\right )}{3 a^{3/2}}-\frac {\sqrt {a x^2+b x^5}}{3 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a*x^2 + b*x^5]),x]

[Out]

-1/3*Sqrt[a*x^2 + b*x^5]/(a*x^4) + (b*ArcTanh[(Sqrt[a]*x)/Sqrt[a*x^2 + b*x^5]])/(3*a^(3/2))

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

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Mathematica [A]
time = 0.06, size = 76, normalized size = 1.29 \begin {gather*} \frac {-\sqrt {a} \left (a+b x^3\right )+b x^3 \sqrt {a+b x^3} \tanh ^{-1}\left (\frac {\sqrt {a+b x^3}}{\sqrt {a}}\right )}{3 a^{3/2} x^2 \sqrt {x^2 \left (a+b x^3\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a*x^2 + b*x^5]),x]

[Out]

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

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Maple [A]
time = 0.43, size = 66, normalized size = 1.12

method	result	size
default	\(\frac {\sqrt {b \,x^{3}+a}\, \left (b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) a \,x^{3}-\sqrt {b \,x^{3}+a}\, a^{\frac {3}{2}}\right )}{3 x^{2} \sqrt {b \,x^{5}+a \,x^{2}}\, a^{\frac {5}{2}}}\)	\(66\)
risch	\(-\frac {b \,x^{3}+a}{3 a \,x^{2} \sqrt {x^{2} \left (b \,x^{3}+a \right )}}+\frac {b \arctanh \left (\frac {\sqrt {b \,x^{3}+a}}{\sqrt {a}}\right ) \sqrt {b \,x^{3}+a}\, x}{3 a^{\frac {3}{2}} \sqrt {x^{2} \left (b \,x^{3}+a \right )}}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(b*x^5+a*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(b*x^5 + a*x^2)*x^3), x)

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Fricas [A]
time = 1.78, size = 127, normalized size = 2.15 \begin {gather*} \left [\frac {\sqrt {a} b x^{4} \log \left (\frac {b x^{4} + 2 \, a x + 2 \, \sqrt {b x^{5} + a x^{2}} \sqrt {a}}{x^{4}}\right ) - 2 \, \sqrt {b x^{5} + a x^{2}} a}{6 \, a^{2} x^{4}}, -\frac {\sqrt {-a} b x^{4} \arctan \left (\frac {\sqrt {b x^{5} + a x^{2}} \sqrt {-a}}{a x}\right ) + \sqrt {b x^{5} + a x^{2}} a}{3 \, a^{2} x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(sqrt(a)*b*x^4*log((b*x^4 + 2*a*x + 2*sqrt(b*x^5 + a*x^2)*sqrt(a))/x^4) - 2*sqrt(b*x^5 + a*x^2)*a)/(a^2*x
^4), -1/3*(sqrt(-a)*b*x^4*arctan(sqrt(b*x^5 + a*x^2)*sqrt(-a)/(a*x)) + sqrt(b*x^5 + a*x^2)*a)/(a^2*x^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**5+a*x**2)**(1/2),x)

[Out]

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

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Giac [A]
time = 2.05, size = 55, normalized size = 0.93 \begin {gather*} -\frac {\frac {b^{2} \arctan \left (\frac {\sqrt {b x^{3} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \frac {\sqrt {b x^{3} + a} b}{a x^{3}}}{3 \, b \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^5+a*x^2)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x^3\,\sqrt {b\,x^5+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a*x^2 + b*x^5)^(1/2)),x)

[Out]

int(1/(x^3*(a*x^2 + b*x^5)^(1/2)), x)

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