3.182 \(\int \frac{\log (c (a+b x)^p)}{(d+e x)^3} \, dx\)

Optimal. Leaf size=105 \[ \frac{b^2 p \log (a+b x)}{2 e (b d-a e)^2}-\frac{b^2 p \log (d+e x)}{2 e (b d-a e)^2}-\frac{\log \left (c (a+b x)^p\right )}{2 e (d+e x)^2}+\frac{b p}{2 e (d+e x) (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0588471, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2395, 44} \[ \frac{b^2 p \log (a+b x)}{2 e (b d-a e)^2}-\frac{b^2 p \log (d+e x)}{2 e (b d-a e)^2}-\frac{\log \left (c (a+b x)^p\right )}{2 e (d+e x)^2}+\frac{b p}{2 e (d+e x) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x)^p]/(d + e*x)^3,x]

[Out]

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

Rule 2395

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0818907, size = 80, normalized size = 0.76 \[ \frac{\frac{b p (d+e x) (b (d+e x) \log (a+b x)-a e-b (d+e x) \log (d+e x)+b d)}{(b d-a e)^2}-\log \left (c (a+b x)^p\right )}{2 e (d+e x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x)^p]/(d + e*x)^3,x]

[Out]

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

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Maple [C]  time = 0.375, size = 582, normalized size = 5.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x+a)^p)/(e*x+d)^3,x)

[Out]

-1/2/e/(e*x+d)^2*ln((b*x+a)^p)-1/4*(2*I*Pi*a*b*d*e*csgn(I*c)*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)-I*Pi*b^2*d^
2*csgn(I*c*(b*x+a)^p)^3-I*Pi*a^2*e^2*csgn(I*c*(b*x+a)^p)^3-2*ln(-b*x-a)*b^2*e^2*p*x^2+2*ln(e*x+d)*b^2*e^2*p*x^
2-4*ln(c)*a*b*d*e-2*I*Pi*a*b*d*e*csgn(I*c)*csgn(I*c*(b*x+a)^p)^2+2*ln(c)*a^2*e^2-2*ln(-b*x-a)*b^2*d^2*p+2*ln(e
*x+d)*b^2*d^2*p-2*b^2*d^2*p+2*a*b*d*p*e-2*I*Pi*a*b*d*e*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2+2*a*b*e^2*p*x-2
*b^2*d*e*p*x+2*I*Pi*a*b*d*e*csgn(I*c*(b*x+a)^p)^3-I*Pi*b^2*d^2*csgn(I*c)*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)
+I*Pi*a^2*e^2*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2+2*ln(c)*b^2*d^2+I*Pi*b^2*d^2*csgn(I*c)*csgn(I*c*(b*x+a)^
p)^2-I*Pi*a^2*e^2*csgn(I*c)*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)-4*ln(-b*x-a)*b^2*d*e*p*x+4*ln(e*x+d)*b^2*d*e
*p*x+I*Pi*b^2*d^2*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2+I*Pi*a^2*e^2*csgn(I*c)*csgn(I*c*(b*x+a)^p)^2)/(e*x+d
)^2/(a*e-b*d)^2/e

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Maxima [A]  time = 1.07574, size = 162, normalized size = 1.54 \begin{align*} \frac{b p{\left (\frac{b \log \left (b x + a\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} - \frac{b \log \left (e x + d\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} + \frac{1}{b d^{2} - a d e +{\left (b d e - a e^{2}\right )} x}\right )}}{2 \, e} - \frac{\log \left ({\left (b x + a\right )}^{p} c\right )}{2 \,{\left (e x + d\right )}^{2} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x+a)^p)/(e*x+d)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.99638, size = 489, normalized size = 4.66 \begin{align*} \frac{{\left (b^{2} d e - a b e^{2}\right )} p x +{\left (b^{2} d^{2} - a b d e\right )} p +{\left (b^{2} e^{2} p x^{2} + 2 \, b^{2} d e p x +{\left (2 \, a b d e - a^{2} e^{2}\right )} p\right )} \log \left (b x + a\right ) -{\left (b^{2} e^{2} p x^{2} + 2 \, b^{2} d e p x + b^{2} d^{2} p\right )} \log \left (e x + d\right ) -{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (c\right )}{2 \,{\left (b^{2} d^{4} e - 2 \, a b d^{3} e^{2} + a^{2} d^{2} e^{3} +{\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} x^{2} + 2 \,{\left (b^{2} d^{3} e^{2} - 2 \, a b d^{2} e^{3} + a^{2} d e^{4}\right )} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x+a)^p)/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/2*((b^2*d*e - a*b*e^2)*p*x + (b^2*d^2 - a*b*d*e)*p + (b^2*e^2*p*x^2 + 2*b^2*d*e*p*x + (2*a*b*d*e - a^2*e^2)*
p)*log(b*x + a) - (b^2*e^2*p*x^2 + 2*b^2*d*e*p*x + b^2*d^2*p)*log(e*x + d) - (b^2*d^2 - 2*a*b*d*e + a^2*e^2)*l
og(c))/(b^2*d^4*e - 2*a*b*d^3*e^2 + a^2*d^2*e^3 + (b^2*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5)*x^2 + 2*(b^2*d^3*e^2 -
2*a*b*d^2*e^3 + a^2*d*e^4)*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(c*(b*x+a)**p)/(e*x+d)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.1903, size = 359, normalized size = 3.42 \begin{align*} \frac{b^{2} p x^{2} e^{2} \log \left (b x + a\right ) + 2 \, b^{2} d p x e \log \left (b x + a\right ) - b^{2} p x^{2} e^{2} \log \left (x e + d\right ) - 2 \, b^{2} d p x e \log \left (x e + d\right ) + b^{2} d p x e + 2 \, a b d p e \log \left (b x + a\right ) - b^{2} d^{2} p \log \left (x e + d\right ) + b^{2} d^{2} p - a b p x e^{2} - a b d p e - a^{2} p e^{2} \log \left (b x + a\right ) - b^{2} d^{2} \log \left (c\right ) + 2 \, a b d e \log \left (c\right ) - a^{2} e^{2} \log \left (c\right )}{2 \,{\left (b^{2} d^{2} x^{2} e^{3} + 2 \, b^{2} d^{3} x e^{2} + b^{2} d^{4} e - 2 \, a b d x^{2} e^{4} - 4 \, a b d^{2} x e^{3} - 2 \, a b d^{3} e^{2} + a^{2} x^{2} e^{5} + 2 \, a^{2} d x e^{4} + a^{2} d^{2} e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x+a)^p)/(e*x+d)^3,x, algorithm="giac")

[Out]

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