3.12 \(\int \frac{\log (e (f (a+b x)^p (c+d x)^q)^r)}{(a+b x)^2} \, dx\)

Optimal. Leaf size=95 \[ -\frac{\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)}+\frac{d q r \log (a+b x)}{b (b c-a d)}-\frac{d q r \log (c+d x)}{b (b c-a d)}-\frac{p r}{b (a+b x)} \]

[Out]

-((p*r)/(b*(a + b*x))) + (d*q*r*Log[a + b*x])/(b*(b*c - a*d)) - (d*q*r*Log[c + d*x])/(b*(b*c - a*d)) - Log[e*(
f*(a + b*x)^p*(c + d*x)^q)^r]/(b*(a + b*x))

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Rubi [A]  time = 0.0374773, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2495, 32, 36, 31} \[ -\frac{\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)}+\frac{d q r \log (a+b x)}{b (b c-a d)}-\frac{d q r \log (c+d x)}{b (b c-a d)}-\frac{p r}{b (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(a + b*x)^2,x]

[Out]

-((p*r)/(b*(a + b*x))) + (d*q*r*Log[a + b*x])/(b*(b*c - a*d)) - (d*q*r*Log[c + d*x])/(b*(b*c - a*d)) - Log[e*(
f*(a + b*x)^p*(c + d*x)^q)^r]/(b*(a + b*x))

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0576862, size = 89, normalized size = 0.94 \[ \frac{r \left (\frac{d q \log (a+b x)}{b c-a d}-\frac{d q \log (c+d x)}{b c-a d}-\frac{p}{a+b x}\right )}{b}-\frac{\log \left (e \left (f (a+b x)^p (c+d x)^q\right )^r\right )}{b (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]/(a + b*x)^2,x]

[Out]

(r*(-(p/(a + b*x)) + (d*q*Log[a + b*x])/(b*c - a*d) - (d*q*Log[c + d*x])/(b*c - a*d)))/b - Log[e*(f*(a + b*x)^
p*(c + d*x)^q)^r]/(b*(a + b*x))

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Maple [F]  time = 0.421, size = 0, normalized size = 0. \begin{align*} \int{\frac{\ln \left ( e \left ( f \left ( bx+a \right ) ^{p} \left ( dx+c \right ) ^{q} \right ) ^{r} \right ) }{ \left ( bx+a \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x)

[Out]

int(ln(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x)

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Maxima [A]  time = 1.26499, size = 134, normalized size = 1.41 \begin{align*} \frac{{\left (d f q{\left (\frac{\log \left (b x + a\right )}{b c - a d} - \frac{\log \left (d x + c\right )}{b c - a d}\right )} - \frac{b f p}{b^{2} x + a b}\right )} r}{b f} - \frac{\log \left (\left ({\left (b x + a\right )}^{p}{\left (d x + c\right )}^{q} f\right )^{r} e\right )}{{\left (b x + a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x, algorithm="maxima")

[Out]

(d*f*q*(log(b*x + a)/(b*c - a*d) - log(d*x + c)/(b*c - a*d)) - b*f*p/(b^2*x + a*b))*r/(b*f) - log(((b*x + a)^p
*(d*x + c)^q*f)^r*e)/((b*x + a)*b)

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Fricas [A]  time = 0.902385, size = 267, normalized size = 2.81 \begin{align*} -\frac{{\left (b c - a d\right )} p r +{\left (b c - a d\right )} r \log \left (f\right ) -{\left (b d q r x +{\left (a d q -{\left (b c - a d\right )} p\right )} r\right )} \log \left (b x + a\right ) +{\left (b d q r x + b c q r\right )} \log \left (d x + c\right ) +{\left (b c - a d\right )} \log \left (e\right )}{a b^{2} c - a^{2} b d +{\left (b^{3} c - a b^{2} d\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x, algorithm="fricas")

[Out]

-((b*c - a*d)*p*r + (b*c - a*d)*r*log(f) - (b*d*q*r*x + (a*d*q - (b*c - a*d)*p)*r)*log(b*x + a) + (b*d*q*r*x +
b*c*q*r)*log(d*x + c) + (b*c - a*d)*log(e))/(a*b^2*c - a^2*b*d + (b^3*c - a*b^2*d)*x)

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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(ln(e*(f*(b*x+a)**p*(d*x+c)**q)**r)/(b*x+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.19727, size = 151, normalized size = 1.59 \begin{align*} \frac{d q r \log \left (b x + a\right )}{b^{2} c - a b d} - \frac{d q r \log \left (d x + c\right )}{b^{2} c - a b d} - \frac{p r \log \left (b x + a\right )}{b^{2} x + a b} - \frac{q r \log \left (d x + c\right )}{b^{2} x + a b} - \frac{p r + r \log \left (f\right ) + 1}{b^{2} x + a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(e*(f*(b*x+a)^p*(d*x+c)^q)^r)/(b*x+a)^2,x, algorithm="giac")

[Out]

d*q*r*log(b*x + a)/(b^2*c - a*b*d) - d*q*r*log(d*x + c)/(b^2*c - a*b*d) - p*r*log(b*x + a)/(b^2*x + a*b) - q*r
*log(d*x + c)/(b^2*x + a*b) - (p*r + r*log(f) + 1)/(b^2*x + a*b)