3.15.58 \(\int \frac {1}{x^5 (a+b x^8)} \, dx\) [1458]

Optimal. Leaf size=40 \[ -\frac {1}{4 a x^4}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x^4}{\sqrt {a}}\right )}{4 a^{3/2}} \]

[Out]

-1/4/a/x^4-1/4*arctan(x^4*b^(1/2)/a^(1/2))*b^(1/2)/a^(3/2)

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Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {281, 331, 211} \begin {gather*} -\frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b} x^4}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {1}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^8)),x]

[Out]

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

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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^5 \left (a+b x^8\right )} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )} \, dx,x,x^4\right )\\ &=-\frac {1}{4 a x^4}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,x^4\right )}{4 a}\\ &=-\frac {1}{4 a x^4}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x^4}{\sqrt {a}}\right )}{4 a^{3/2}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(164\) vs. \(2(40)=80\).
time = 0.02, size = 164, normalized size = 4.10 \begin {gather*} \frac {-\sqrt {a}+\sqrt {b} x^4 \tan ^{-1}\left (\cot \left (\frac {\pi }{8}\right )-\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}\right )+\sqrt {b} x^4 \tan ^{-1}\left (\cot \left (\frac {\pi }{8}\right )+\frac {\sqrt [8]{b} x \csc \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}\right )+\sqrt {b} x^4 \tan ^{-1}\left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}-\tan \left (\frac {\pi }{8}\right )\right )-\sqrt {b} x^4 \tan ^{-1}\left (\frac {\sqrt [8]{b} x \sec \left (\frac {\pi }{8}\right )}{\sqrt [8]{a}}+\tan \left (\frac {\pi }{8}\right )\right )}{4 a^{3/2} x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^8)),x]

[Out]

(-Sqrt[a] + Sqrt[b]*x^4*ArcTan[Cot[Pi/8] - (b^(1/8)*x*Csc[Pi/8])/a^(1/8)] + Sqrt[b]*x^4*ArcTan[Cot[Pi/8] + (b^
(1/8)*x*Csc[Pi/8])/a^(1/8)] + Sqrt[b]*x^4*ArcTan[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) - Tan[Pi/8]] - Sqrt[b]*x^4*ArcT
an[(b^(1/8)*x*Sec[Pi/8])/a^(1/8) + Tan[Pi/8]])/(4*a^(3/2)*x^4)

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Maple [A]
time = 0.20, size = 32, normalized size = 0.80

method	result	size
default	\(-\frac {1}{4 a \,x^{4}}-\frac {b \arctan \left (\frac {b \,x^{4}}{\sqrt {a b}}\right )}{4 a \sqrt {a b}}\)	\(32\)
risch	\(-\frac {1}{4 a \,x^{4}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a^{3} \textit {\_Z}^{2}+b \right )}{\sum }\textit {\_R} \ln \left (\left (-9 a^{3} \textit {\_R}^{2}-8 b \right ) x^{4}-a^{2} \textit {\_R} \right )\right )}{8}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^8+a),x,method=_RETURNVERBOSE)

[Out]

-1/4/a/x^4-1/4*b/a/(a*b)^(1/2)*arctan(b*x^4/(a*b)^(1/2))

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Maxima [A]
time = 0.49, size = 31, normalized size = 0.78 \begin {gather*} -\frac {b \arctan \left (\frac {b x^{4}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a} - \frac {1}{4 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^8+a),x, algorithm="maxima")

[Out]

-1/4*b*arctan(b*x^4/sqrt(a*b))/(sqrt(a*b)*a) - 1/4/(a*x^4)

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Fricas [A]
time = 0.41, size = 94, normalized size = 2.35 \begin {gather*} \left [\frac {x^{4} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{8} - 2 \, a x^{4} \sqrt {-\frac {b}{a}} - a}{b x^{8} + a}\right ) - 2}{8 \, a x^{4}}, \frac {x^{4} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b x^{4}}\right ) - 1}{4 \, a x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^8+a),x, algorithm="fricas")

[Out]

[1/8*(x^4*sqrt(-b/a)*log((b*x^8 - 2*a*x^4*sqrt(-b/a) - a)/(b*x^8 + a)) - 2)/(a*x^4), 1/4*(x^4*sqrt(b/a)*arctan
(a*sqrt(b/a)/(b*x^4)) - 1)/(a*x^4)]

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Sympy [A]
time = 0.15, size = 71, normalized size = 1.78 \begin {gather*} \frac {\sqrt {- \frac {b}{a^{3}}} \log {\left (- \frac {a^{2} \sqrt {- \frac {b}{a^{3}}}}{b} + x^{4} \right )}}{8} - \frac {\sqrt {- \frac {b}{a^{3}}} \log {\left (\frac {a^{2} \sqrt {- \frac {b}{a^{3}}}}{b} + x^{4} \right )}}{8} - \frac {1}{4 a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**8+a),x)

[Out]

sqrt(-b/a**3)*log(-a**2*sqrt(-b/a**3)/b + x**4)/8 - sqrt(-b/a**3)*log(a**2*sqrt(-b/a**3)/b + x**4)/8 - 1/(4*a*
x**4)

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Giac [A]
time = 0.99, size = 31, normalized size = 0.78 \begin {gather*} -\frac {b \arctan \left (\frac {b x^{4}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a} - \frac {1}{4 \, a x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^8+a),x, algorithm="giac")

[Out]

-1/4*b*arctan(b*x^4/sqrt(a*b))/(sqrt(a*b)*a) - 1/4/(a*x^4)

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Mupad [B]
time = 0.05, size = 28, normalized size = 0.70 \begin {gather*} -\frac {1}{4\,a\,x^4}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x^4}{\sqrt {a}}\right )}{4\,a^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^8)),x)

[Out]

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

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