3.5 \(\int \frac{1}{(a+b x) (c+d x) (e+f x) (g+h x)} \, dx\)

Optimal. Leaf size=163 \[ \frac{b^2 \log (a+b x)}{(b c-a d) (b e-a f) (b g-a h)}-\frac{d^2 \log (c+d x)}{(b c-a d) (d e-c f) (d g-c h)}+\frac{f^2 \log (e+f x)}{(b e-a f) (d e-c f) (f g-e h)}-\frac{h^2 \log (g+h x)}{(b g-a h) (d g-c h) (f g-e h)} \]

[Out]

(b^2*Log[a + b*x])/((b*c - a*d)*(b*e - a*f)*(b*g - a*h)) - (d^2*Log[c + d*x])/((b*c - a*d)*(d*e - c*f)*(d*g -
c*h)) + (f^2*Log[e + f*x])/((b*e - a*f)*(d*e - c*f)*(f*g - e*h)) - (h^2*Log[g + h*x])/((b*g - a*h)*(d*g - c*h)
*(f*g - e*h))

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Rubi [A]  time = 0.211911, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.034, Rules used = {180} \[ \frac{b^2 \log (a+b x)}{(b c-a d) (b e-a f) (b g-a h)}-\frac{d^2 \log (c+d x)}{(b c-a d) (d e-c f) (d g-c h)}+\frac{f^2 \log (e+f x)}{(b e-a f) (d e-c f) (f g-e h)}-\frac{h^2 \log (g+h x)}{(b g-a h) (d g-c h) (f g-e h)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)*(c + d*x)*(e + f*x)*(g + h*x)),x]

[Out]

(b^2*Log[a + b*x])/((b*c - a*d)*(b*e - a*f)*(b*g - a*h)) - (d^2*Log[c + d*x])/((b*c - a*d)*(d*e - c*f)*(d*g -
c*h)) + (f^2*Log[e + f*x])/((b*e - a*f)*(d*e - c*f)*(f*g - e*h)) - (h^2*Log[g + h*x])/((b*g - a*h)*(d*g - c*h)
*(f*g - e*h))

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rubi steps

\begin{align*} \int \frac{1}{(a+b x) (c+d x) (e+f x) (g+h x)} \, dx &=\int \left (\frac{b^3}{(b c-a d) (b e-a f) (b g-a h) (a+b x)}-\frac{d^3}{(b c-a d) (-d e+c f) (-d g+c h) (c+d x)}-\frac{f^3}{(b e-a f) (d e-c f) (-f g+e h) (e+f x)}-\frac{h^3}{(b g-a h) (d g-c h) (f g-e h) (g+h x)}\right ) \, dx\\ &=\frac{b^2 \log (a+b x)}{(b c-a d) (b e-a f) (b g-a h)}-\frac{d^2 \log (c+d x)}{(b c-a d) (d e-c f) (d g-c h)}+\frac{f^2 \log (e+f x)}{(b e-a f) (d e-c f) (f g-e h)}-\frac{h^2 \log (g+h x)}{(b g-a h) (d g-c h) (f g-e h)}\\ \end{align*}

Mathematica [A]  time = 0.246551, size = 164, normalized size = 1.01 \[ \frac{b^2 \log (a+b x)}{(b c-a d) (b e-a f) (b g-a h)}-\frac{d^2 \log (c+d x)}{(b c-a d) (c f-d e) (c h-d g)}-\frac{f^2 \log (e+f x)}{(b e-a f) (d e-c f) (e h-f g)}-\frac{h^2 \log (g+h x)}{(b g-a h) (d g-c h) (f g-e h)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)*(c + d*x)*(e + f*x)*(g + h*x)),x]

[Out]

(b^2*Log[a + b*x])/((b*c - a*d)*(b*e - a*f)*(b*g - a*h)) - (d^2*Log[c + d*x])/((b*c - a*d)*(-(d*e) + c*f)*(-(d
*g) + c*h)) - (f^2*Log[e + f*x])/((b*e - a*f)*(d*e - c*f)*(-(f*g) + e*h)) - (h^2*Log[g + h*x])/((b*g - a*h)*(d
*g - c*h)*(f*g - e*h))

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Maple [A]  time = 0.007, size = 164, normalized size = 1. \begin{align*}{\frac{{d}^{2}\ln \left ( dx+c \right ) }{ \left ( ad-bc \right ) \left ( cf-de \right ) \left ( ch-dg \right ) }}-{\frac{{f}^{2}\ln \left ( fx+e \right ) }{ \left ( af-be \right ) \left ( cf-de \right ) \left ( eh-fg \right ) }}+{\frac{{h}^{2}\ln \left ( hx+g \right ) }{ \left ( ch-dg \right ) \left ( ah-bg \right ) \left ( eh-fg \right ) }}-{\frac{{b}^{2}\ln \left ( bx+a \right ) }{ \left ( ad-bc \right ) \left ( af-be \right ) \left ( ah-bg \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)/(d*x+c)/(f*x+e)/(h*x+g),x)

[Out]

d^2/(a*d-b*c)/(c*f-d*e)/(c*h-d*g)*ln(d*x+c)-f^2/(a*f-b*e)/(c*f-d*e)/(e*h-f*g)*ln(f*x+e)+h^2/(c*h-d*g)/(a*h-b*g
)/(e*h-f*g)*ln(h*x+g)-b^2/(a*d-b*c)/(a*f-b*e)/(a*h-b*g)*ln(b*x+a)

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Maxima [A]  time = 1.28156, size = 419, normalized size = 2.57 \begin{align*} \frac{b^{2} \log \left (b x + a\right )}{{\left ({\left (b^{3} c - a b^{2} d\right )} e -{\left (a b^{2} c - a^{2} b d\right )} f\right )} g -{\left ({\left (a b^{2} c - a^{2} b d\right )} e -{\left (a^{2} b c - a^{3} d\right )} f\right )} h} - \frac{d^{2} \log \left (d x + c\right )}{{\left ({\left (b c d^{2} - a d^{3}\right )} e -{\left (b c^{2} d - a c d^{2}\right )} f\right )} g -{\left ({\left (b c^{2} d - a c d^{2}\right )} e -{\left (b c^{3} - a c^{2} d\right )} f\right )} h} + \frac{f^{2} \log \left (f x + e\right )}{{\left (b d e^{2} f + a c f^{3} -{\left (b c + a d\right )} e f^{2}\right )} g -{\left (b d e^{3} + a c e f^{2} -{\left (b c + a d\right )} e^{2} f\right )} h} - \frac{h^{2} \log \left (h x + g\right )}{b d f g^{3} - a c e h^{3} -{\left (b d e +{\left (b c + a d\right )} f\right )} g^{2} h +{\left (a c f +{\left (b c + a d\right )} e\right )} g h^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)/(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="maxima")

[Out]

b^2*log(b*x + a)/(((b^3*c - a*b^2*d)*e - (a*b^2*c - a^2*b*d)*f)*g - ((a*b^2*c - a^2*b*d)*e - (a^2*b*c - a^3*d)
*f)*h) - d^2*log(d*x + c)/(((b*c*d^2 - a*d^3)*e - (b*c^2*d - a*c*d^2)*f)*g - ((b*c^2*d - a*c*d^2)*e - (b*c^3 -
a*c^2*d)*f)*h) + f^2*log(f*x + e)/((b*d*e^2*f + a*c*f^3 - (b*c + a*d)*e*f^2)*g - (b*d*e^3 + a*c*e*f^2 - (b*c
+ a*d)*e^2*f)*h) - h^2*log(h*x + g)/(b*d*f*g^3 - a*c*e*h^3 - (b*d*e + (b*c + a*d)*f)*g^2*h + (a*c*f + (b*c + a
*d)*e)*g*h^2)

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Fricas [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/(b*x+a)/(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="fricas")

[Out]

Timed out

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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/(b*x+a)/(d*x+c)/(f*x+e)/(h*x+g),x)

[Out]

Timed out

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Giac [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/(b*x+a)/(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="giac")

[Out]

Timed out