3.183 \(\int \frac{\log (c (a+b x)^p)}{(d+e x)^4} \, dx\)

Optimal. Leaf size=133 \[ \frac{b^2 p}{3 e (d+e x) (b d-a e)^2}+\frac{b^3 p \log (a+b x)}{3 e (b d-a e)^3}-\frac{b^3 p \log (d+e x)}{3 e (b d-a e)^3}-\frac{\log \left (c (a+b x)^p\right )}{3 e (d+e x)^3}+\frac{b p}{6 e (d+e x)^2 (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0773629, antiderivative size = 133, 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} \[ \frac{b^2 p}{3 e (d+e x) (b d-a e)^2}+\frac{b^3 p \log (a+b x)}{3 e (b d-a e)^3}-\frac{b^3 p \log (d+e x)}{3 e (b d-a e)^3}-\frac{\log \left (c (a+b x)^p\right )}{3 e (d+e x)^3}+\frac{b p}{6 e (d+e x)^2 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x)^p]/(d + e*x)^4,x]

[Out]

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

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

Mathematica [A]  time = 0.131363, size = 105, normalized size = 0.79 \[ \frac{\frac{b p (d+e x) \left (2 b^2 (d+e x)^2 \log (a+b x)+(b d-a e) (-a e+3 b d+2 b e x)-2 b^2 (d+e x)^2 \log (d+e x)\right )}{(b d-a e)^3}-2 \log \left (c (a+b x)^p\right )}{6 e (d+e x)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x)^p]/(d + e*x)^4,x]

[Out]

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

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Maple [C]  time = 0.393, size = 873, normalized size = 6.6 \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)^4,x)

[Out]

-1/3/e/(e*x+d)^3*ln((b*x+a)^p)+1/6*(-3*I*Pi*a^2*b*d*e^2*csgn(I*c)*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)+3*I*Pi
*a*b^2*d^2*e*csgn(I*c)*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)-2*ln(b*x+a)*b^3*d^3*p+6*a*b^2*d*e^2*p*x+3*I*Pi*a^
2*b*d*e^2*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2+2*ln(-e*x-d)*b^3*d^3*p+2*ln(c)*b^3*d^3-2*ln(c)*a^3*e^3-a^2*b
*d*e^2*p-3*b^3*d^3*p+4*a*b^2*d^2*p*e+2*a*b^2*e^3*p*x^2-2*b^3*d*e^2*p*x^2-a^2*b*e^3*p*x-2*ln(b*x+a)*b^3*e^3*p*x
^3+2*ln(-e*x-d)*b^3*e^3*p*x^3+6*ln(c)*a^2*b*d*e^2-6*ln(c)*a*b^2*d^2*e-5*b^3*d^2*e*p*x-3*I*Pi*a*b^2*d^2*e*csgn(
I*c)*csgn(I*c*(b*x+a)^p)^2-3*I*Pi*a*b^2*d^2*e*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2+3*I*Pi*a^2*b*d*e^2*csgn(
I*c)*csgn(I*c*(b*x+a)^p)^2+3*I*Pi*a*b^2*d^2*e*csgn(I*c*(b*x+a)^p)^3-I*Pi*b^3*d^3*csgn(I*c)*csgn(I*(b*x+a)^p)*c
sgn(I*c*(b*x+a)^p)+I*Pi*a^3*e^3*csgn(I*c)*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)-I*Pi*a^3*e^3*csgn(I*(b*x+a)^p)
*csgn(I*c*(b*x+a)^p)^2+I*Pi*a^3*e^3*csgn(I*c*(b*x+a)^p)^3-I*Pi*b^3*d^3*csgn(I*c*(b*x+a)^p)^3-3*I*Pi*a^2*b*d*e^
2*csgn(I*c*(b*x+a)^p)^3-I*Pi*a^3*e^3*csgn(I*c)*csgn(I*c*(b*x+a)^p)^2-6*ln(b*x+a)*b^3*d*e^2*p*x^2+6*ln(-e*x-d)*
b^3*d*e^2*p*x^2-6*ln(b*x+a)*b^3*d^2*e*p*x+6*ln(-e*x-d)*b^3*d^2*e*p*x+I*Pi*b^3*d^3*csgn(I*c)*csgn(I*c*(b*x+a)^p
)^2+I*Pi*b^3*d^3*csgn(I*(b*x+a)^p)*csgn(I*c*(b*x+a)^p)^2)/(e*x+d)^3/(a^2*e^2-2*a*b*d*e+b^2*d^2)/(a*e-b*d)/e

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Maxima [A]  time = 1.10358, size = 313, normalized size = 2.35 \begin{align*} \frac{{\left (\frac{2 \, b^{2} \log \left (b x + a\right )}{b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}} - \frac{2 \, b^{2} \log \left (e x + d\right )}{b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}} + \frac{2 \, b e x + 3 \, b d - a e}{b^{2} d^{4} - 2 \, a b d^{3} e + a^{2} d^{2} e^{2} +{\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x^{2} + 2 \,{\left (b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3}\right )} x}\right )} b p}{6 \, e} - \frac{\log \left ({\left (b x + a\right )}^{p} c\right )}{3 \,{\left (e x + d\right )}^{3} e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x+a)^p)/(e*x+d)^4,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.37855, size = 894, normalized size = 6.72 \begin{align*} \frac{2 \,{\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} p x^{2} +{\left (5 \, b^{3} d^{2} e - 6 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} p x +{\left (3 \, b^{3} d^{3} - 4 \, a b^{2} d^{2} e + a^{2} b d e^{2}\right )} p + 2 \,{\left (b^{3} e^{3} p x^{3} + 3 \, b^{3} d e^{2} p x^{2} + 3 \, b^{3} d^{2} e p x +{\left (3 \, a b^{2} d^{2} e - 3 \, a^{2} b d e^{2} + a^{3} e^{3}\right )} p\right )} \log \left (b x + a\right ) - 2 \,{\left (b^{3} e^{3} p x^{3} + 3 \, b^{3} d e^{2} p x^{2} + 3 \, b^{3} d^{2} e p x + b^{3} d^{3} p\right )} \log \left (e x + d\right ) - 2 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \log \left (c\right )}{6 \,{\left (b^{3} d^{6} e - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d^{4} e^{3} - a^{3} d^{3} e^{4} +{\left (b^{3} d^{3} e^{4} - 3 \, a b^{2} d^{2} e^{5} + 3 \, a^{2} b d e^{6} - a^{3} e^{7}\right )} x^{3} + 3 \,{\left (b^{3} d^{4} e^{3} - 3 \, a b^{2} d^{3} e^{4} + 3 \, a^{2} b d^{2} e^{5} - a^{3} d e^{6}\right )} x^{2} + 3 \,{\left (b^{3} d^{5} e^{2} - 3 \, a b^{2} d^{4} e^{3} + 3 \, a^{2} b d^{3} e^{4} - a^{3} d^{2} e^{5}\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)^4,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.22556, size = 668, normalized size = 5.02 \begin{align*} \frac{2 \, b^{3} p x^{3} e^{3} \log \left (b x + a\right ) + 6 \, b^{3} d p x^{2} e^{2} \log \left (b x + a\right ) + 6 \, b^{3} d^{2} p x e \log \left (b x + a\right ) - 2 \, b^{3} p x^{3} e^{3} \log \left (x e + d\right ) - 6 \, b^{3} d p x^{2} e^{2} \log \left (x e + d\right ) - 6 \, b^{3} d^{2} p x e \log \left (x e + d\right ) + 2 \, b^{3} d p x^{2} e^{2} + 5 \, b^{3} d^{2} p x e + 6 \, a b^{2} d^{2} p e \log \left (b x + a\right ) - 2 \, b^{3} d^{3} p \log \left (x e + d\right ) + 3 \, b^{3} d^{3} p - 2 \, a b^{2} p x^{2} e^{3} - 6 \, a b^{2} d p x e^{2} - 4 \, a b^{2} d^{2} p e - 6 \, a^{2} b d p e^{2} \log \left (b x + a\right ) - 2 \, b^{3} d^{3} \log \left (c\right ) + 6 \, a b^{2} d^{2} e \log \left (c\right ) + a^{2} b p x e^{3} + a^{2} b d p e^{2} + 2 \, a^{3} p e^{3} \log \left (b x + a\right ) - 6 \, a^{2} b d e^{2} \log \left (c\right ) + 2 \, a^{3} e^{3} \log \left (c\right )}{6 \,{\left (b^{3} d^{3} x^{3} e^{4} + 3 \, b^{3} d^{4} x^{2} e^{3} + 3 \, b^{3} d^{5} x e^{2} + b^{3} d^{6} e - 3 \, a b^{2} d^{2} x^{3} e^{5} - 9 \, a b^{2} d^{3} x^{2} e^{4} - 9 \, a b^{2} d^{4} x e^{3} - 3 \, a b^{2} d^{5} e^{2} + 3 \, a^{2} b d x^{3} e^{6} + 9 \, a^{2} b d^{2} x^{2} e^{5} + 9 \, a^{2} b d^{3} x e^{4} + 3 \, a^{2} b d^{4} e^{3} - a^{3} x^{3} e^{7} - 3 \, a^{3} d x^{2} e^{6} - 3 \, a^{3} d^{2} x e^{5} - a^{3} d^{3} e^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x+a)^p)/(e*x+d)^4,x, algorithm="giac")

[Out]

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