3.1.19 \(\int \frac {1}{x^3 (a x+b x^3)} \, dx\) [19]

Optimal. Leaf size=43 \[ -\frac {1}{3 a x^3}+\frac {b}{a^2 x}+\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1598, 331, 211} \begin {gather*} \frac {b^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2}}+\frac {b}{a^2 x}-\frac {1}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a*x + b*x^3)),x]

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1598

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 43, normalized size = 1.00 \begin {gather*} -\frac {1}{3 a x^3}+\frac {b}{a^2 x}+\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a*x + b*x^3)),x]

[Out]

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

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Maple [A]
time = 0.37, size = 39, normalized size = 0.91

method	result	size
default	\(\frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}-\frac {1}{3 a \,x^{3}}+\frac {b}{a^{2} x}\)	\(39\)
risch	\(\frac {\frac {b \,x^{2}}{a^{2}}-\frac {1}{3 a}}{x^{3}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a^{5} \textit {\_Z}^{2}+b^{3}\right )}{\sum }\textit {\_R} \ln \left (\left (3 a^{5} \textit {\_R}^{2}+2 b^{3}\right ) x -a^{3} b \textit {\_R} \right )\right )}{2}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 40, normalized size = 0.93 \begin {gather*} \frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, b x^{2} - a}{3 \, a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.29, size = 106, normalized size = 2.47 \begin {gather*} \left [\frac {3 \, b x^{3} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 6 \, b x^{2} - 2 \, a}{6 \, a^{2} x^{3}}, \frac {3 \, b x^{3} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 3 \, b x^{2} - a}{3 \, a^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (37) = 74\).
time = 0.10, size = 87, normalized size = 2.02 \begin {gather*} - \frac {\sqrt {- \frac {b^{3}}{a^{5}}} \log {\left (- \frac {a^{3} \sqrt {- \frac {b^{3}}{a^{5}}}}{b^{2}} + x \right )}}{2} + \frac {\sqrt {- \frac {b^{3}}{a^{5}}} \log {\left (\frac {a^{3} \sqrt {- \frac {b^{3}}{a^{5}}}}{b^{2}} + x \right )}}{2} + \frac {- a + 3 b x^{2}}{3 a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-b**3/a**5)*log(-a**3*sqrt(-b**3/a**5)/b**2 + x)/2 + sqrt(-b**3/a**5)*log(a**3*sqrt(-b**3/a**5)/b**2 + x
)/2 + (-a + 3*b*x**2)/(3*a**2*x**3)

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Giac [A]
time = 0.92, size = 40, normalized size = 0.93 \begin {gather*} \frac {b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {3 \, b x^{2} - a}{3 \, a^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.94, size = 37, normalized size = 0.86 \begin {gather*} \frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{a^{5/2}}-\frac {\frac {1}{3\,a}-\frac {b\,x^2}{a^2}}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a*x + b*x^3)),x)

[Out]

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

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