3.5.46 \(\int \frac {1}{(15+\frac {2}{x^2}+\frac {13}{x}) x^4} \, dx\) [446]

Optimal. Leaf size=34 \[ -\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x) \]

[Out]

-1/2/x-13/4*ln(x)-9/28*ln(2+3*x)+25/7*ln(1+5*x)

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Rubi [A]
time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1368, 723, 814} \begin {gather*} -\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (3 x+2)+\frac {25}{7} \log (5 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((15 + 2/x^2 + 13/x)*x^4),x]

[Out]

-1/2*1/x - (13*Log[x])/4 - (9*Log[2 + 3*x])/28 + (25*Log[1 + 5*x])/7

Rule 723

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m + 1)/((m
+ 1)*(c*d^2 - b*d*e + a*e^2))), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[(d + e*x)^(m + 1)*(Simp[c*d - b*e - c
*e*x, x]/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[m, -1]

Rule 814

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m*((f + g*x)/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[m]

Rule 1368

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + 2*n*p)*(c + b/x^n +
a/x^(2*n))^p, x] /; FreeQ[{a, b, c, m, n}, x] && EqQ[n2, 2*n] && ILtQ[p, 0] && NegQ[n]

Rubi steps

\begin {align*} \int \frac {1}{\left (15+\frac {2}{x^2}+\frac {13}{x}\right ) x^4} \, dx &=\int \frac {1}{x^2 \left (2+13 x+15 x^2\right )} \, dx\\ &=-\frac {1}{2 x}+\frac {1}{2} \int \frac {-13-15 x}{x \left (2+13 x+15 x^2\right )} \, dx\\ &=-\frac {1}{2 x}+\frac {1}{2} \int \left (-\frac {13}{2 x}-\frac {27}{14 (2+3 x)}+\frac {250}{7 (1+5 x)}\right ) \, dx\\ &=-\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 34, normalized size = 1.00 \begin {gather*} -\frac {1}{2 x}-\frac {13 \log (x)}{4}-\frac {9}{28} \log (2+3 x)+\frac {25}{7} \log (1+5 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((15 + 2/x^2 + 13/x)*x^4),x]

[Out]

-1/2*1/x - (13*Log[x])/4 - (9*Log[2 + 3*x])/28 + (25*Log[1 + 5*x])/7

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Maple [A]
time = 0.02, size = 27, normalized size = 0.79

method	result	size
default	\(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (2+3 x \right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\)	\(27\)
norman	\(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (2+3 x \right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\)	\(27\)
risch	\(-\frac {1}{2 x}-\frac {13 \ln \left (x \right )}{4}-\frac {9 \ln \left (2+3 x \right )}{28}+\frac {25 \ln \left (1+5 x \right )}{7}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(15+2/x^2+13/x)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/2/x-13/4*ln(x)-9/28*ln(2+3*x)+25/7*ln(1+5*x)

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Maxima [A]
time = 0.30, size = 26, normalized size = 0.76 \begin {gather*} -\frac {1}{2 \, x} + \frac {25}{7} \, \log \left (5 \, x + 1\right ) - \frac {9}{28} \, \log \left (3 \, x + 2\right ) - \frac {13}{4} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(15+2/x^2+13/x)/x^4,x, algorithm="maxima")

[Out]

-1/2/x + 25/7*log(5*x + 1) - 9/28*log(3*x + 2) - 13/4*log(x)

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Fricas [A]
time = 0.35, size = 30, normalized size = 0.88 \begin {gather*} \frac {100 \, x \log \left (5 \, x + 1\right ) - 9 \, x \log \left (3 \, x + 2\right ) - 91 \, x \log \left (x\right ) - 14}{28 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(15+2/x^2+13/x)/x^4,x, algorithm="fricas")

[Out]

1/28*(100*x*log(5*x + 1) - 9*x*log(3*x + 2) - 91*x*log(x) - 14)/x

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Sympy [A]
time = 0.06, size = 31, normalized size = 0.91 \begin {gather*} - \frac {13 \log {\left (x \right )}}{4} + \frac {25 \log {\left (x + \frac {1}{5} \right )}}{7} - \frac {9 \log {\left (x + \frac {2}{3} \right )}}{28} - \frac {1}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(15+2/x**2+13/x)/x**4,x)

[Out]

-13*log(x)/4 + 25*log(x + 1/5)/7 - 9*log(x + 2/3)/28 - 1/(2*x)

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Giac [A]
time = 4.21, size = 29, normalized size = 0.85 \begin {gather*} -\frac {1}{2 \, x} + \frac {25}{7} \, \log \left ({\left | 5 \, x + 1 \right |}\right ) - \frac {9}{28} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {13}{4} \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(15+2/x^2+13/x)/x^4,x, algorithm="giac")

[Out]

-1/2/x + 25/7*log(abs(5*x + 1)) - 9/28*log(abs(3*x + 2)) - 13/4*log(abs(x))

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Mupad [B]
time = 0.04, size = 22, normalized size = 0.65 \begin {gather*} \frac {25\,\ln \left (x+\frac {1}{5}\right )}{7}-\frac {9\,\ln \left (x+\frac {2}{3}\right )}{28}-\frac {13\,\ln \left (x\right )}{4}-\frac {1}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(13/x + 2/x^2 + 15)),x)

[Out]

(25*log(x + 1/5))/7 - (9*log(x + 2/3))/28 - (13*log(x))/4 - 1/(2*x)

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