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

Optimal. Leaf size=68 \[ -\frac {\log \left (c (a+b x)^p\right )}{e (d+e x)}+\frac {b p \log (a+b x)}{e (b d-a e)}-\frac {b p \log (d+e x)}{e (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2395, 36, 31} \[ -\frac {\log \left (c (a+b x)^p\right )}{e (d+e x)}+\frac {b p \log (a+b x)}{e (b d-a e)}-\frac {b p \log (d+e x)}{e (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x)^p]/(d + e*x)^2,x]

[Out]

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

Rule 31

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

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

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Mathematica [A]  time = 0.05, size = 52, normalized size = 0.76 \[ \frac {\frac {b p (\log (a+b x)-\log (d+e x))}{b d-a e}-\frac {\log \left (c (a+b x)^p\right )}{d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 80, normalized size = 1.18 \[ \frac {{\left (b e p x + a e p\right )} \log \left (b x + a\right ) - {\left (b e p x + b d p\right )} \log \left (e x + d\right ) - {\left (b d - a e\right )} \log \relax (c)}{b d^{2} e - a d e^{2} + {\left (b d e^{2} - a e^{3}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 91, normalized size = 1.34 \[ \frac {b p x e \log \left (b x + a\right ) - b p x e \log \left (x e + d\right ) + a p e \log \left (b x + a\right ) - b d p \log \left (x e + d\right ) - b d \log \relax (c) + a e \log \relax (c)}{b d x e^{2} + b d^{2} e - a x e^{3} - a d e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.45, size = 329, normalized size = 4.84 \[ -\frac {\ln \left (\left (b x +a \right )^{p}\right )}{\left (e x +d \right ) e}-\frac {-i \pi a e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )+i \pi a e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}+i \pi a e \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}-i \pi a e \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}+i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )-i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}-i \pi b d \,\mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}+i \pi b d \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}+2 b e p x \ln \left (b x +a \right )-2 b e p x \ln \left (-e x -d \right )+2 b d p \ln \left (b x +a \right )-2 b d p \ln \left (-e x -d \right )+2 a e \ln \relax (c )-2 b d \ln \relax (c )}{2 \left (e x +d \right ) \left (a e -b d \right ) e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 65, normalized size = 0.96 \[ \frac {b p {\left (\frac {\log \left (b x + a\right )}{b d - a e} - \frac {\log \left (e x + d\right )}{b d - a e}\right )}}{e} - \frac {\log \left ({\left (b x + a\right )}^{p} c\right )}{{\left (e x + d\right )} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.07, size = 70, normalized size = 1.03 \[ -\frac {\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{e\,\left (d+e\,x\right )}+\frac {b\,p\,\mathrm {atan}\left (\frac {a\,e\,1{}\mathrm {i}+b\,d\,1{}\mathrm {i}+b\,e\,x\,2{}\mathrm {i}}{a\,e-b\,d}\right )\,2{}\mathrm {i}}{a\,e^2-b\,d\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x)^p)/(d + e*x)^2,x)

[Out]

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

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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)**2,x)

[Out]

Exception raised: NotImplementedError

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