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]

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

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Rubi [A]  time = 0.06, antiderivative size = 105, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.10, 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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fricas [B]  time = 0.45, size = 236, normalized size = 2.25 \[ \frac {{\left (b^{2} d e - a b e^{2}\right )} p x + {\left (b^{2} d^{2} - a b d e\right )} p + {\left (b^{2} e^{2} p x^{2} + 2 \, b^{2} d e p x + {\left (2 \, a b d e - a^{2} e^{2}\right )} p\right )} \log \left (b x + a\right ) - {\left (b^{2} e^{2} p x^{2} + 2 \, b^{2} d e p x + b^{2} d^{2} p\right )} \log \left (e x + d\right ) - {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \relax (c)}{2 \, {\left (b^{2} d^{4} e - 2 \, a b d^{3} e^{2} + a^{2} d^{2} e^{3} + {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} x^{2} + 2 \, {\left (b^{2} d^{3} e^{2} - 2 \, a b d^{2} e^{3} + a^{2} d e^{4}\right )} x\right )}} \]

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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giac [B]  time = 0.17, size = 266, normalized size = 2.53 \[ \frac {b^{2} p x^{2} e^{2} \log \left (b x + a\right ) + 2 \, b^{2} d p x e \log \left (b x + a\right ) - b^{2} p x^{2} e^{2} \log \left (x e + d\right ) - 2 \, b^{2} d p x e \log \left (x e + d\right ) + b^{2} d p x e + 2 \, a b d p e \log \left (b x + a\right ) - b^{2} d^{2} p \log \left (x e + d\right ) + b^{2} d^{2} p - a b p x e^{2} - a b d p e - a^{2} p e^{2} \log \left (b x + a\right ) - b^{2} d^{2} \log \relax (c) + 2 \, a b d e \log \relax (c) - a^{2} e^{2} \log \relax (c)}{2 \, {\left (b^{2} d^{2} x^{2} e^{3} + 2 \, b^{2} d^{3} x e^{2} + b^{2} d^{4} e - 2 \, a b d x^{2} e^{4} - 4 \, a b d^{2} x e^{3} - 2 \, a b d^{3} e^{2} + a^{2} x^{2} e^{5} + 2 \, a^{2} d x e^{4} + a^{2} d^{2} e^{3}\right )}} \]

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)

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maple [C]  time = 0.57, size = 582, normalized size = 5.54 \[ -\frac {\ln \left (\left (b x +a \right )^{p}\right )}{2 \left (e x +d \right )^{2} e}-\frac {2 b^{2} d^{2} p \ln \left (e x +d \right )-2 b^{2} d^{2} p \ln \left (-b x -a \right )+2 a^{2} e^{2} \ln \relax (c )+2 a b d e p +2 a b \,e^{2} p x -2 b^{2} d e p x -2 b^{2} d^{2} p +2 b^{2} d^{2} \ln \relax (c )-2 i \pi a b d e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}+i \pi \,a^{2} e^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}+i \pi \,a^{2} e^{2} \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}+i \pi \,b^{2} d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}+i \pi \,b^{2} d^{2} \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}-2 i \pi a b d e \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}+4 b^{2} d e p x \ln \left (e x +d \right )-4 b^{2} d e p x \ln \left (-b x -a \right )+2 b^{2} e^{2} p \,x^{2} \ln \left (e x +d \right )-2 b^{2} e^{2} p \,x^{2} \ln \left (-b x -a \right )-4 a b d e \ln \relax (c )+2 i \pi a b d e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )-i \pi \,a^{2} e^{2} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}-i \pi \,b^{2} d^{2} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}-i \pi \,a^{2} e^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )+2 i \pi a b d e \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}-i \pi \,b^{2} d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )}{4 \left (e x +d \right )^{2} \left (a e -b d \right )^{2} e} \]

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

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maxima [A]  time = 0.45, size = 120, normalized size = 1.14 \[ \frac {b p {\left (\frac {b \log \left (b x + a\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} - \frac {b \log \left (e x + d\right )}{b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}} + \frac {1}{b d^{2} - a d e + {\left (b d e - a e^{2}\right )} x}\right )}}{2 \, e} - \frac {\log \left ({\left (b x + a\right )}^{p} c\right )}{2 \, {\left (e x + d\right )}^{2} e} \]

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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mupad [B]  time = 0.64, size = 96, normalized size = 0.91 \[ -\frac {\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{2\,e\,{\left (d+e\,x\right )}^2}-\frac {b\,p}{2\,e\,\left (a\,e-b\,d\right )\,\left (d+e\,x\right )}-\frac {b^2\,p\,\mathrm {atan}\left (\frac {a\,e\,1{}\mathrm {i}+b\,d\,1{}\mathrm {i}+b\,e\,x\,2{}\mathrm {i}}{a\,e-b\,d}\right )\,1{}\mathrm {i}}{e\,{\left (a\,e-b\,d\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x)^p)/(d + e*x)^3,x)

[Out]

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

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError

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