3.3.38 \(\int \frac {1}{x^5 (a+b x^2) (c+d x^2)} \, dx\)

Optimal. Leaf size=119 \[ -\frac {b^3 \log \left (a+b x^2\right )}{2 a^3 (b c-a d)}+\frac {a d+b c}{2 a^2 c^2 x^2}+\frac {\log (x) \left (a^2 d^2+a b c d+b^2 c^2\right )}{a^3 c^3}+\frac {d^3 \log \left (c+d x^2\right )}{2 c^3 (b c-a d)}-\frac {1}{4 a c x^4} \]

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Rubi [A]  time = 0.13, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {446, 72} \begin {gather*} \frac {\log (x) \left (a^2 d^2+a b c d+b^2 c^2\right )}{a^3 c^3}-\frac {b^3 \log \left (a+b x^2\right )}{2 a^3 (b c-a d)}+\frac {a d+b c}{2 a^2 c^2 x^2}+\frac {d^3 \log \left (c+d x^2\right )}{2 c^3 (b c-a d)}-\frac {1}{4 a c x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^2)*(c + d*x^2)),x]

[Out]

-1/(4*a*c*x^4) + (b*c + a*d)/(2*a^2*c^2*x^2) + ((b^2*c^2 + a*b*c*d + a^2*d^2)*Log[x])/(a^3*c^3) - (b^3*Log[a +
b*x^2])/(2*a^3*(b*c - a*d)) + (d^3*Log[c + d*x^2])/(2*c^3*(b*c - a*d))

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {1}{x^5 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 (a+b x) (c+d x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {1}{a c x^3}+\frac {-b c-a d}{a^2 c^2 x^2}+\frac {b^2 c^2+a b c d+a^2 d^2}{a^3 c^3 x}+\frac {b^4}{a^3 (-b c+a d) (a+b x)}+\frac {d^4}{c^3 (b c-a d) (c+d x)}\right ) \, dx,x,x^2\right )\\ &=-\frac {1}{4 a c x^4}+\frac {b c+a d}{2 a^2 c^2 x^2}+\frac {\left (b^2 c^2+a b c d+a^2 d^2\right ) \log (x)}{a^3 c^3}-\frac {b^3 \log \left (a+b x^2\right )}{2 a^3 (b c-a d)}+\frac {d^3 \log \left (c+d x^2\right )}{2 c^3 (b c-a d)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 119, normalized size = 1.00 \begin {gather*} \frac {b^3 \log \left (a+b x^2\right )}{2 a^3 (a d-b c)}+\frac {a d+b c}{2 a^2 c^2 x^2}+\frac {\log (x) \left (a^2 d^2+a b c d+b^2 c^2\right )}{a^3 c^3}+\frac {d^3 \log \left (c+d x^2\right )}{2 c^3 (b c-a d)}-\frac {1}{4 a c x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*1/(a*c*x^4) + (b*c + a*d)/(2*a^2*c^2*x^2) + ((b^2*c^2 + a*b*c*d + a^2*d^2)*Log[x])/(a^3*c^3) + (b^3*Log[a
+ b*x^2])/(2*a^3*(-(b*c) + a*d)) + (d^3*Log[c + d*x^2])/(2*c^3*(b*c - a*d))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^5 \left (a+b x^2\right ) \left (c+d x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/(x^5*(a + b*x^2)*(c + d*x^2)),x]

[Out]

IntegrateAlgebraic[1/(x^5*(a + b*x^2)*(c + d*x^2)), x]

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fricas [A]  time = 3.09, size = 127, normalized size = 1.07 \begin {gather*} -\frac {2 \, b^{3} c^{3} x^{4} \log \left (b x^{2} + a\right ) - 2 \, a^{3} d^{3} x^{4} \log \left (d x^{2} + c\right ) + a^{2} b c^{3} - a^{3} c^{2} d - 4 \, {\left (b^{3} c^{3} - a^{3} d^{3}\right )} x^{4} \log \relax (x) - 2 \, {\left (a b^{2} c^{3} - a^{3} c d^{2}\right )} x^{2}}{4 \, {\left (a^{3} b c^{4} - a^{4} c^{3} d\right )} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*b^3*c^3*x^4*log(b*x^2 + a) - 2*a^3*d^3*x^4*log(d*x^2 + c) + a^2*b*c^3 - a^3*c^2*d - 4*(b^3*c^3 - a^3*d
^3)*x^4*log(x) - 2*(a*b^2*c^3 - a^3*c*d^2)*x^2)/((a^3*b*c^4 - a^4*c^3*d)*x^4)

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giac [A]  time = 0.31, size = 167, normalized size = 1.40 \begin {gather*} -\frac {b^{4} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \, {\left (a^{3} b^{2} c - a^{4} b d\right )}} + \frac {d^{4} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \, {\left (b c^{4} d - a c^{3} d^{2}\right )}} + \frac {{\left (b^{2} c^{2} + a b c d + a^{2} d^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{3} c^{3}} - \frac {3 \, b^{2} c^{2} x^{4} + 3 \, a b c d x^{4} + 3 \, a^{2} d^{2} x^{4} - 2 \, a b c^{2} x^{2} - 2 \, a^{2} c d x^{2} + a^{2} c^{2}}{4 \, a^{3} c^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^4*log(abs(b*x^2 + a))/(a^3*b^2*c - a^4*b*d) + 1/2*d^4*log(abs(d*x^2 + c))/(b*c^4*d - a*c^3*d^2) + 1/2*(
b^2*c^2 + a*b*c*d + a^2*d^2)*log(x^2)/(a^3*c^3) - 1/4*(3*b^2*c^2*x^4 + 3*a*b*c*d*x^4 + 3*a^2*d^2*x^4 - 2*a*b*c
^2*x^2 - 2*a^2*c*d*x^2 + a^2*c^2)/(a^3*c^3*x^4)

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maple [A]  time = 0.02, size = 124, normalized size = 1.04 \begin {gather*} \frac {b^{3} \ln \left (b \,x^{2}+a \right )}{2 \left (a d -b c \right ) a^{3}}-\frac {d^{3} \ln \left (d \,x^{2}+c \right )}{2 \left (a d -b c \right ) c^{3}}+\frac {d^{2} \ln \relax (x )}{a \,c^{3}}+\frac {b d \ln \relax (x )}{a^{2} c^{2}}+\frac {b^{2} \ln \relax (x )}{a^{3} c}+\frac {d}{2 a \,c^{2} x^{2}}+\frac {b}{2 a^{2} c \,x^{2}}-\frac {1}{4 a c \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^2+a)/(d*x^2+c),x)

[Out]

1/2*b^3/a^3/(a*d-b*c)*ln(b*x^2+a)-1/2*d^3/c^3/(a*d-b*c)*ln(d*x^2+c)-1/4/a/c/x^4+1/2/a/c^2/x^2*d+1/2/a^2/c/x^2*
b+1/a/c^3*ln(x)*d^2+1/a^2/c^2*ln(x)*b*d+1/a^3/c*ln(x)*b^2

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maxima [A]  time = 1.11, size = 117, normalized size = 0.98 \begin {gather*} -\frac {b^{3} \log \left (b x^{2} + a\right )}{2 \, {\left (a^{3} b c - a^{4} d\right )}} + \frac {d^{3} \log \left (d x^{2} + c\right )}{2 \, {\left (b c^{4} - a c^{3} d\right )}} + \frac {{\left (b^{2} c^{2} + a b c d + a^{2} d^{2}\right )} \log \left (x^{2}\right )}{2 \, a^{3} c^{3}} + \frac {2 \, {\left (b c + a d\right )} x^{2} - a c}{4 \, a^{2} c^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b^3*log(b*x^2 + a)/(a^3*b*c - a^4*d) + 1/2*d^3*log(d*x^2 + c)/(b*c^4 - a*c^3*d) + 1/2*(b^2*c^2 + a*b*c*d
+ a^2*d^2)*log(x^2)/(a^3*c^3) + 1/4*(2*(b*c + a*d)*x^2 - a*c)/(a^2*c^2*x^4)

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mupad [B]  time = 0.43, size = 118, normalized size = 0.99 \begin {gather*} \frac {b^3\,\ln \left (b\,x^2+a\right )}{2\,a^4\,d-2\,a^3\,b\,c}-\frac {\frac {1}{4\,a\,c}-\frac {x^2\,\left (a\,d+b\,c\right )}{2\,a^2\,c^2}}{x^4}+\frac {d^3\,\ln \left (d\,x^2+c\right )}{2\,b\,c^4-2\,a\,c^3\,d}+\frac {\ln \relax (x)\,\left (a^2\,d^2+a\,b\,c\,d+b^2\,c^2\right )}{a^3\,c^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^2)*(c + d*x^2)),x)

[Out]

(b^3*log(a + b*x^2))/(2*a^4*d - 2*a^3*b*c) - (1/(4*a*c) - (x^2*(a*d + b*c))/(2*a^2*c^2))/x^4 + (d^3*log(c + d*
x^2))/(2*b*c^4 - 2*a*c^3*d) + (log(x)*(a^2*d^2 + b^2*c^2 + a*b*c*d))/(a^3*c^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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