3.2.41 \(\int \frac {1}{x^6 (a+b x^2)} \, dx\) [141]

Optimal. Leaf size=58 \[ -\frac {1}{5 a x^5}+\frac {b}{3 a^2 x^3}-\frac {b^2}{a^3 x}-\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2}} \]

[Out]

-1/5/a/x^5+1/3*b/a^2/x^3-b^2/a^3/x-b^(5/2)*arctan(x*b^(1/2)/a^(1/2))/a^(7/2)

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Rubi [A]
time = 0.02, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {331, 211} \begin {gather*} -\frac {b^{5/2} \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2}}-\frac {b^2}{a^3 x}+\frac {b}{3 a^2 x^3}-\frac {1}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^2)),x]

[Out]

-1/5*1/(a*x^5) + b/(3*a^2*x^3) - b^2/(a^3*x) - (b^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(7/2)

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]

Rubi steps

\begin {align*} \int \frac {1}{x^6 \left (a+b x^2\right )} \, dx &=-\frac {1}{5 a x^5}-\frac {b \int \frac {1}{x^4 \left (a+b x^2\right )} \, dx}{a}\\ &=-\frac {1}{5 a x^5}+\frac {b}{3 a^2 x^3}+\frac {b^2 \int \frac {1}{x^2 \left (a+b x^2\right )} \, dx}{a^2}\\ &=-\frac {1}{5 a x^5}+\frac {b}{3 a^2 x^3}-\frac {b^2}{a^3 x}-\frac {b^3 \int \frac {1}{a+b x^2} \, dx}{a^3}\\ &=-\frac {1}{5 a x^5}+\frac {b}{3 a^2 x^3}-\frac {b^2}{a^3 x}-\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 58, normalized size = 1.00 \begin {gather*} -\frac {1}{5 a x^5}+\frac {b}{3 a^2 x^3}-\frac {b^2}{a^3 x}-\frac {b^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a + b*x^2)),x]

[Out]

-1/5*1/(a*x^5) + b/(3*a^2*x^3) - b^2/(a^3*x) - (b^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(7/2)

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Maple [A]
time = 0.05, size = 52, normalized size = 0.90

method	result	size
default	\(-\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{a^{3} \sqrt {a b}}-\frac {1}{5 a \,x^{5}}-\frac {b^{2}}{a^{3} x}+\frac {b}{3 a^{2} x^{3}}\)	\(52\)
risch	\(\frac {-\frac {b^{2} x^{4}}{a^{3}}+\frac {b \,x^{2}}{3 a^{2}}-\frac {1}{5 a}}{x^{5}}+\frac {\sqrt {-a b}\, b^{2} \ln \left (-b x +\sqrt {-a b}\right )}{2 a^{4}}-\frac {\sqrt {-a b}\, b^{2} \ln \left (-b x -\sqrt {-a b}\right )}{2 a^{4}}\)	\(86\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(b*x^2+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.53, size = 52, normalized size = 0.90 \begin {gather*} -\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {15 \, b^{2} x^{4} - 5 \, a b x^{2} + 3 \, a^{2}}{15 \, a^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.18, size = 132, normalized size = 2.28 \begin {gather*} \left [\frac {15 \, b^{2} x^{5} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) - 30 \, b^{2} x^{4} + 10 \, a b x^{2} - 6 \, a^{2}}{30 \, a^{3} x^{5}}, -\frac {15 \, b^{2} x^{5} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 15 \, b^{2} x^{4} - 5 \, a b x^{2} + 3 \, a^{2}}{15 \, a^{3} x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a),x, algorithm="fricas")

[Out]

[1/30*(15*b^2*x^5*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) - 30*b^2*x^4 + 10*a*b*x^2 - 6*a^2
)/(a^3*x^5), -1/15*(15*b^2*x^5*sqrt(b/a)*arctan(x*sqrt(b/a)) + 15*b^2*x^4 - 5*a*b*x^2 + 3*a^2)/(a^3*x^5)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (49) = 98\).
time = 0.10, size = 100, normalized size = 1.72 \begin {gather*} \frac {\sqrt {- \frac {b^{5}}{a^{7}}} \log {\left (- \frac {a^{4} \sqrt {- \frac {b^{5}}{a^{7}}}}{b^{3}} + x \right )}}{2} - \frac {\sqrt {- \frac {b^{5}}{a^{7}}} \log {\left (\frac {a^{4} \sqrt {- \frac {b^{5}}{a^{7}}}}{b^{3}} + x \right )}}{2} + \frac {- 3 a^{2} + 5 a b x^{2} - 15 b^{2} x^{4}}{15 a^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(b*x**2+a),x)

[Out]

sqrt(-b**5/a**7)*log(-a**4*sqrt(-b**5/a**7)/b**3 + x)/2 - sqrt(-b**5/a**7)*log(a**4*sqrt(-b**5/a**7)/b**3 + x)
/2 + (-3*a**2 + 5*a*b*x**2 - 15*b**2*x**4)/(15*a**3*x**5)

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Giac [A]
time = 0.45, size = 52, normalized size = 0.90 \begin {gather*} -\frac {b^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {15 \, b^{2} x^{4} - 5 \, a b x^{2} + 3 \, a^{2}}{15 \, a^{3} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^2+a),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.06, size = 48, normalized size = 0.83 \begin {gather*} -\frac {\frac {1}{5\,a}-\frac {b\,x^2}{3\,a^2}+\frac {b^2\,x^4}{a^3}}{x^5}-\frac {b^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{a^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b*x^2)),x)

[Out]

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

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