3.315 \(\int \frac{1}{x (a+b x^2)^2 (c+d x^2)^3} \, dx\)

Optimal. Leaf size=192 \[ -\frac{d^2 \left (a^2 d^2-4 a b c d+6 b^2 c^2\right ) \log \left (c+d x^2\right )}{2 c^3 (b c-a d)^4}-\frac{b^3 (b c-4 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^4}+\frac{\log (x)}{a^2 c^3}+\frac{b^3}{2 a \left (a+b x^2\right ) (b c-a d)^3}+\frac{d^2 (3 b c-a d)}{2 c^2 \left (c+d x^2\right ) (b c-a d)^3}+\frac{d^2}{4 c \left (c+d x^2\right )^2 (b c-a d)^2} \]

[Out]

b^3/(2*a*(b*c - a*d)^3*(a + b*x^2)) + d^2/(4*c*(b*c - a*d)^2*(c + d*x^2)^2) + (d^2*(3*b*c - a*d))/(2*c^2*(b*c
- a*d)^3*(c + d*x^2)) + Log[x]/(a^2*c^3) - (b^3*(b*c - 4*a*d)*Log[a + b*x^2])/(2*a^2*(b*c - a*d)^4) - (d^2*(6*
b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Log[c + d*x^2])/(2*c^3*(b*c - a*d)^4)

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Rubi [A]  time = 0.240651, antiderivative size = 192, normalized size of antiderivative = 1., 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, 88} \[ -\frac{d^2 \left (a^2 d^2-4 a b c d+6 b^2 c^2\right ) \log \left (c+d x^2\right )}{2 c^3 (b c-a d)^4}-\frac{b^3 (b c-4 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^4}+\frac{\log (x)}{a^2 c^3}+\frac{b^3}{2 a \left (a+b x^2\right ) (b c-a d)^3}+\frac{d^2 (3 b c-a d)}{2 c^2 \left (c+d x^2\right ) (b c-a d)^3}+\frac{d^2}{4 c \left (c+d x^2\right )^2 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*x^2)^2*(c + d*x^2)^3),x]

[Out]

b^3/(2*a*(b*c - a*d)^3*(a + b*x^2)) + d^2/(4*c*(b*c - a*d)^2*(c + d*x^2)^2) + (d^2*(3*b*c - a*d))/(2*c^2*(b*c
- a*d)^3*(c + d*x^2)) + Log[x]/(a^2*c^3) - (b^3*(b*c - 4*a*d)*Log[a + b*x^2])/(2*a^2*(b*c - a*d)^4) - (d^2*(6*
b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Log[c + d*x^2])/(2*c^3*(b*c - a*d)^4)

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]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin{align*} \int \frac{1}{x \left (a+b x^2\right )^2 \left (c+d x^2\right )^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x (a+b x)^2 (c+d x)^3} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{1}{a^2 c^3 x}+\frac{b^4}{a (-b c+a d)^3 (a+b x)^2}+\frac{b^4 (-b c+4 a d)}{a^2 (-b c+a d)^4 (a+b x)}-\frac{d^3}{c (b c-a d)^2 (c+d x)^3}-\frac{d^3 (3 b c-a d)}{c^2 (b c-a d)^3 (c+d x)^2}-\frac{d^3 \left (6 b^2 c^2-4 a b c d+a^2 d^2\right )}{c^3 (b c-a d)^4 (c+d x)}\right ) \, dx,x,x^2\right )\\ &=\frac{b^3}{2 a (b c-a d)^3 \left (a+b x^2\right )}+\frac{d^2}{4 c (b c-a d)^2 \left (c+d x^2\right )^2}+\frac{d^2 (3 b c-a d)}{2 c^2 (b c-a d)^3 \left (c+d x^2\right )}+\frac{\log (x)}{a^2 c^3}-\frac{b^3 (b c-4 a d) \log \left (a+b x^2\right )}{2 a^2 (b c-a d)^4}-\frac{d^2 \left (6 b^2 c^2-4 a b c d+a^2 d^2\right ) \log \left (c+d x^2\right )}{2 c^3 (b c-a d)^4}\\ \end{align*}

Mathematica [A]  time = 0.302499, size = 187, normalized size = 0.97 \[ \frac{1}{4} \left (-\frac{2 d^2 \left (a^2 d^2-4 a b c d+6 b^2 c^2\right ) \log \left (c+d x^2\right )}{c^3 (b c-a d)^4}+\frac{2 b^3 (4 a d-b c) \log \left (a+b x^2\right )}{a^2 (b c-a d)^4}+\frac{4 \log (x)}{a^2 c^3}-\frac{2 b^3}{a \left (a+b x^2\right ) (a d-b c)^3}+\frac{2 d^2 (3 b c-a d)}{c^2 \left (c+d x^2\right ) (b c-a d)^3}+\frac{d^2}{c \left (c+d x^2\right )^2 (b c-a d)^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*x^2)^2*(c + d*x^2)^3),x]

[Out]

((-2*b^3)/(a*(-(b*c) + a*d)^3*(a + b*x^2)) + d^2/(c*(b*c - a*d)^2*(c + d*x^2)^2) + (2*d^2*(3*b*c - a*d))/(c^2*
(b*c - a*d)^3*(c + d*x^2)) + (4*Log[x])/(a^2*c^3) + (2*b^3*(-(b*c) + 4*a*d)*Log[a + b*x^2])/(a^2*(b*c - a*d)^4
) - (2*d^2*(6*b^2*c^2 - 4*a*b*c*d + a^2*d^2)*Log[c + d*x^2])/(c^3*(b*c - a*d)^4))/4

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Maple [B]  time = 0.023, size = 374, normalized size = 2. \begin{align*} -{\frac{{d}^{4}\ln \left ( d{x}^{2}+c \right ){a}^{2}}{2\,{c}^{3} \left ( ad-bc \right ) ^{4}}}+2\,{\frac{{d}^{3}\ln \left ( d{x}^{2}+c \right ) ab}{{c}^{2} \left ( ad-bc \right ) ^{4}}}-3\,{\frac{{d}^{2}\ln \left ( d{x}^{2}+c \right ){b}^{2}}{c \left ( ad-bc \right ) ^{4}}}+{\frac{{a}^{2}{d}^{4}}{4\,c \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}-{\frac{ab{d}^{3}}{2\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{{b}^{2}c{d}^{2}}{4\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) ^{2}}}+{\frac{{a}^{2}{d}^{4}}{2\,{c}^{2} \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) }}-2\,{\frac{ab{d}^{3}}{c \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) }}+{\frac{3\,{b}^{2}{d}^{2}}{2\, \left ( ad-bc \right ) ^{4} \left ( d{x}^{2}+c \right ) }}+{\frac{\ln \left ( x \right ) }{{a}^{2}{c}^{3}}}+2\,{\frac{{b}^{3}\ln \left ( b{x}^{2}+a \right ) d}{a \left ( ad-bc \right ) ^{4}}}-{\frac{{b}^{4}\ln \left ( b{x}^{2}+a \right ) c}{2\,{a}^{2} \left ( ad-bc \right ) ^{4}}}-{\frac{{b}^{3}d}{2\, \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) }}+{\frac{{b}^{4}c}{2\,a \left ( ad-bc \right ) ^{4} \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^2+a)^2/(d*x^2+c)^3,x)

[Out]

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

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Maxima [B]  time = 1.08827, size = 711, normalized size = 3.7 \begin{align*} -\frac{{\left (b^{4} c - 4 \, a b^{3} d\right )} \log \left (b x^{2} + a\right )}{2 \,{\left (a^{2} b^{4} c^{4} - 4 \, a^{3} b^{3} c^{3} d + 6 \, a^{4} b^{2} c^{2} d^{2} - 4 \, a^{5} b c d^{3} + a^{6} d^{4}\right )}} - \frac{{\left (6 \, b^{2} c^{2} d^{2} - 4 \, a b c d^{3} + a^{2} d^{4}\right )} \log \left (d x^{2} + c\right )}{2 \,{\left (b^{4} c^{7} - 4 \, a b^{3} c^{6} d + 6 \, a^{2} b^{2} c^{5} d^{2} - 4 \, a^{3} b c^{4} d^{3} + a^{4} c^{3} d^{4}\right )}} + \frac{2 \, b^{3} c^{4} + 7 \, a^{2} b c^{2} d^{2} - 3 \, a^{3} c d^{3} + 2 \,{\left (b^{3} c^{2} d^{2} + 3 \, a b^{2} c d^{3} - a^{2} b d^{4}\right )} x^{4} +{\left (4 \, b^{3} c^{3} d + 7 \, a b^{2} c^{2} d^{2} + 3 \, a^{2} b c d^{3} - 2 \, a^{3} d^{4}\right )} x^{2}}{4 \,{\left (a^{2} b^{3} c^{7} - 3 \, a^{3} b^{2} c^{6} d + 3 \, a^{4} b c^{5} d^{2} - a^{5} c^{4} d^{3} +{\left (a b^{4} c^{5} d^{2} - 3 \, a^{2} b^{3} c^{4} d^{3} + 3 \, a^{3} b^{2} c^{3} d^{4} - a^{4} b c^{2} d^{5}\right )} x^{6} +{\left (2 \, a b^{4} c^{6} d - 5 \, a^{2} b^{3} c^{5} d^{2} + 3 \, a^{3} b^{2} c^{4} d^{3} + a^{4} b c^{3} d^{4} - a^{5} c^{2} d^{5}\right )} x^{4} +{\left (a b^{4} c^{7} - a^{2} b^{3} c^{6} d - 3 \, a^{3} b^{2} c^{5} d^{2} + 5 \, a^{4} b c^{4} d^{3} - 2 \, a^{5} c^{3} d^{4}\right )} x^{2}\right )}} + \frac{\log \left (x^{2}\right )}{2 \, a^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="maxima")

[Out]

-1/2*(b^4*c - 4*a*b^3*d)*log(b*x^2 + a)/(a^2*b^4*c^4 - 4*a^3*b^3*c^3*d + 6*a^4*b^2*c^2*d^2 - 4*a^5*b*c*d^3 + a
^6*d^4) - 1/2*(6*b^2*c^2*d^2 - 4*a*b*c*d^3 + a^2*d^4)*log(d*x^2 + c)/(b^4*c^7 - 4*a*b^3*c^6*d + 6*a^2*b^2*c^5*
d^2 - 4*a^3*b*c^4*d^3 + a^4*c^3*d^4) + 1/4*(2*b^3*c^4 + 7*a^2*b*c^2*d^2 - 3*a^3*c*d^3 + 2*(b^3*c^2*d^2 + 3*a*b
^2*c*d^3 - a^2*b*d^4)*x^4 + (4*b^3*c^3*d + 7*a*b^2*c^2*d^2 + 3*a^2*b*c*d^3 - 2*a^3*d^4)*x^2)/(a^2*b^3*c^7 - 3*
a^3*b^2*c^6*d + 3*a^4*b*c^5*d^2 - a^5*c^4*d^3 + (a*b^4*c^5*d^2 - 3*a^2*b^3*c^4*d^3 + 3*a^3*b^2*c^3*d^4 - a^4*b
*c^2*d^5)*x^6 + (2*a*b^4*c^6*d - 5*a^2*b^3*c^5*d^2 + 3*a^3*b^2*c^4*d^3 + a^4*b*c^3*d^4 - a^5*c^2*d^5)*x^4 + (a
*b^4*c^7 - a^2*b^3*c^6*d - 3*a^3*b^2*c^5*d^2 + 5*a^4*b*c^4*d^3 - 2*a^5*c^3*d^4)*x^2) + 1/2*log(x^2)/(a^2*c^3)

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Fricas [B]  time = 60.7381, size = 2079, normalized size = 10.83 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="fricas")

[Out]

1/4*(2*a*b^4*c^6 - 2*a^2*b^3*c^5*d + 7*a^3*b^2*c^4*d^2 - 10*a^4*b*c^3*d^3 + 3*a^5*c^2*d^4 + 2*(a*b^4*c^4*d^2 +
2*a^2*b^3*c^3*d^3 - 4*a^3*b^2*c^2*d^4 + a^4*b*c*d^5)*x^4 + (4*a*b^4*c^5*d + 3*a^2*b^3*c^4*d^2 - 4*a^3*b^2*c^3
*d^3 - 5*a^4*b*c^2*d^4 + 2*a^5*c*d^5)*x^2 - 2*(a*b^4*c^6 - 4*a^2*b^3*c^5*d + (b^5*c^4*d^2 - 4*a*b^4*c^3*d^3)*x
^6 + (2*b^5*c^5*d - 7*a*b^4*c^4*d^2 - 4*a^2*b^3*c^3*d^3)*x^4 + (b^5*c^6 - 2*a*b^4*c^5*d - 8*a^2*b^3*c^4*d^2)*x
^2)*log(b*x^2 + a) - 2*(6*a^3*b^2*c^4*d^2 - 4*a^4*b*c^3*d^3 + a^5*c^2*d^4 + (6*a^2*b^3*c^2*d^4 - 4*a^3*b^2*c*d
^5 + a^4*b*d^6)*x^6 + (12*a^2*b^3*c^3*d^3 - 2*a^3*b^2*c^2*d^4 - 2*a^4*b*c*d^5 + a^5*d^6)*x^4 + (6*a^2*b^3*c^4*
d^2 + 8*a^3*b^2*c^3*d^3 - 7*a^4*b*c^2*d^4 + 2*a^5*c*d^5)*x^2)*log(d*x^2 + c) + 4*(a*b^4*c^6 - 4*a^2*b^3*c^5*d
+ 6*a^3*b^2*c^4*d^2 - 4*a^4*b*c^3*d^3 + a^5*c^2*d^4 + (b^5*c^4*d^2 - 4*a*b^4*c^3*d^3 + 6*a^2*b^3*c^2*d^4 - 4*a
^3*b^2*c*d^5 + a^4*b*d^6)*x^6 + (2*b^5*c^5*d - 7*a*b^4*c^4*d^2 + 8*a^2*b^3*c^3*d^3 - 2*a^3*b^2*c^2*d^4 - 2*a^4
*b*c*d^5 + a^5*d^6)*x^4 + (b^5*c^6 - 2*a*b^4*c^5*d - 2*a^2*b^3*c^4*d^2 + 8*a^3*b^2*c^3*d^3 - 7*a^4*b*c^2*d^4 +
2*a^5*c*d^5)*x^2)*log(x))/(a^3*b^4*c^9 - 4*a^4*b^3*c^8*d + 6*a^5*b^2*c^7*d^2 - 4*a^6*b*c^6*d^3 + a^7*c^5*d^4
+ (a^2*b^5*c^7*d^2 - 4*a^3*b^4*c^6*d^3 + 6*a^4*b^3*c^5*d^4 - 4*a^5*b^2*c^4*d^5 + a^6*b*c^3*d^6)*x^6 + (2*a^2*b
^5*c^8*d - 7*a^3*b^4*c^7*d^2 + 8*a^4*b^3*c^6*d^3 - 2*a^5*b^2*c^5*d^4 - 2*a^6*b*c^4*d^5 + a^7*c^3*d^6)*x^4 + (a
^2*b^5*c^9 - 2*a^3*b^4*c^8*d - 2*a^4*b^3*c^7*d^2 + 8*a^5*b^2*c^6*d^3 - 7*a^6*b*c^5*d^4 + 2*a^7*c^4*d^5)*x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x**2+a)**2/(d*x**2+c)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.18703, size = 635, normalized size = 3.31 \begin{align*} -\frac{{\left (b^{5} c - 4 \, a b^{4} d\right )} \log \left ({\left | b x^{2} + a \right |}\right )}{2 \,{\left (a^{2} b^{5} c^{4} - 4 \, a^{3} b^{4} c^{3} d + 6 \, a^{4} b^{3} c^{2} d^{2} - 4 \, a^{5} b^{2} c d^{3} + a^{6} b d^{4}\right )}} - \frac{{\left (6 \, b^{2} c^{2} d^{3} - 4 \, a b c d^{4} + a^{2} d^{5}\right )} \log \left ({\left | d x^{2} + c \right |}\right )}{2 \,{\left (b^{4} c^{7} d - 4 \, a b^{3} c^{6} d^{2} + 6 \, a^{2} b^{2} c^{5} d^{3} - 4 \, a^{3} b c^{4} d^{4} + a^{4} c^{3} d^{5}\right )}} + \frac{b^{5} c x^{2} - 4 \, a b^{4} d x^{2} + 2 \, a b^{4} c - 5 \, a^{2} b^{3} d}{2 \,{\left (a^{2} b^{4} c^{4} - 4 \, a^{3} b^{3} c^{3} d + 6 \, a^{4} b^{2} c^{2} d^{2} - 4 \, a^{5} b c d^{3} + a^{6} d^{4}\right )}{\left (b x^{2} + a\right )}} + \frac{18 \, b^{2} c^{2} d^{4} x^{4} - 12 \, a b c d^{5} x^{4} + 3 \, a^{2} d^{6} x^{4} + 42 \, b^{2} c^{3} d^{3} x^{2} - 32 \, a b c^{2} d^{4} x^{2} + 8 \, a^{2} c d^{5} x^{2} + 25 \, b^{2} c^{4} d^{2} - 22 \, a b c^{3} d^{3} + 6 \, a^{2} c^{2} d^{4}}{4 \,{\left (b^{4} c^{7} - 4 \, a b^{3} c^{6} d + 6 \, a^{2} b^{2} c^{5} d^{2} - 4 \, a^{3} b c^{4} d^{3} + a^{4} c^{3} d^{4}\right )}{\left (d x^{2} + c\right )}^{2}} + \frac{\log \left (x^{2}\right )}{2 \, a^{2} c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b*x^2+a)^2/(d*x^2+c)^3,x, algorithm="giac")

[Out]

-1/2*(b^5*c - 4*a*b^4*d)*log(abs(b*x^2 + a))/(a^2*b^5*c^4 - 4*a^3*b^4*c^3*d + 6*a^4*b^3*c^2*d^2 - 4*a^5*b^2*c*
d^3 + a^6*b*d^4) - 1/2*(6*b^2*c^2*d^3 - 4*a*b*c*d^4 + a^2*d^5)*log(abs(d*x^2 + c))/(b^4*c^7*d - 4*a*b^3*c^6*d^
2 + 6*a^2*b^2*c^5*d^3 - 4*a^3*b*c^4*d^4 + a^4*c^3*d^5) + 1/2*(b^5*c*x^2 - 4*a*b^4*d*x^2 + 2*a*b^4*c - 5*a^2*b^
3*d)/((a^2*b^4*c^4 - 4*a^3*b^3*c^3*d + 6*a^4*b^2*c^2*d^2 - 4*a^5*b*c*d^3 + a^6*d^4)*(b*x^2 + a)) + 1/4*(18*b^2
*c^2*d^4*x^4 - 12*a*b*c*d^5*x^4 + 3*a^2*d^6*x^4 + 42*b^2*c^3*d^3*x^2 - 32*a*b*c^2*d^4*x^2 + 8*a^2*c*d^5*x^2 +
25*b^2*c^4*d^2 - 22*a*b*c^3*d^3 + 6*a^2*c^2*d^4)/((b^4*c^7 - 4*a*b^3*c^6*d + 6*a^2*b^2*c^5*d^2 - 4*a^3*b*c^4*d
^3 + a^4*c^3*d^4)*(d*x^2 + c)^2) + 1/2*log(x^2)/(a^2*c^3)