\(\int \frac {\log (c (a+b x)^p)}{(d+e x)^2} \, dx\) [181]

Optimal result

Integrand size = 18, antiderivative size = 68 \[ \int \frac {\log \left (c (a+b x)^p\right )}{(d+e x)^2} \, dx=\frac {b p \log (a+b x)}{e (b d-a e)}-\frac {\log \left (c (a+b x)^p\right )}{e (d+e x)}-\frac {b p \log (d+e x)}{e (b d-a e)} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2442, 36, 31} \[ \int \frac {\log \left (c (a+b x)^p\right )}{(d+e x)^2} \, dx=-\frac {\log \left (c (a+b x)^p\right )}{e (d+e x)}+\frac {b p \log (a+b x)}{e (b d-a e)}-\frac {b p \log (d+e x)}{e (b d-a e)} \]

[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 2442

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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c (a+b x)^p\right )}{(d+e x)^2} \, dx=\frac {\frac {b p \log (a+b x)}{b d-a e}-\frac {\log \left (c (a+b x)^p\right )}{d+e x}+\frac {b p \log (d+e x)}{-b d+a e}}{e} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97

method	result	size
parts	\(-\frac {\ln \left (c \left (b x +a \right )^{p}\right )}{e \left (e x +d \right )}+\frac {p b \left (\frac {\ln \left (e x +d \right )}{a e -b d}-\frac {\ln \left (b x +a \right )}{a e -b d}\right )}{e}\)	\(66\)
parallelrisch	\(-\frac {\ln \left (b x +a \right ) x \,b^{2} e p -\ln \left (e x +d \right ) x \,b^{2} e p +\ln \left (b x +a \right ) b^{2} d p -\ln \left (e x +d \right ) b^{2} d p +\ln \left (c \left (b x +a \right )^{p}\right ) a b e -\ln \left (c \left (b x +a \right )^{p}\right ) b^{2} d}{\left (a e -b d \right ) \left (e x +d \right ) b e}\)	\(109\)
risch	\(-\frac {\ln \left (\left (b x +a \right )^{p}\right )}{e \left (e x +d \right )}-\frac {i \pi a e \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}-i \pi a e \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi a e \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}+i \pi a e \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )-i \pi b d \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}+i \pi b d \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+i \pi b d \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}-i \pi b d \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (b x +a \right ) b e p x -2 \ln \left (-e x -d \right ) b e p x +2 \ln \left (b x +a \right ) b d p -2 \ln \left (-e x -d \right ) b d p +2 \ln \left (c \right ) a e -2 d b \ln \left (c \right )}{2 \left (e x +d \right ) e \left (a e -b d \right )}\)	\(329\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.18 \[ \int \frac {\log \left (c (a+b x)^p\right )}{(d+e x)^2} \, dx=\frac {{\left (b e p x + a e p\right )} \log \left (b x + a\right ) - {\left (b e p x + b d p\right )} \log \left (e x + d\right ) - {\left (b d - a e\right )} \log \left (c\right )}{b d^{2} e - a d e^{2} + {\left (b d e^{2} - a e^{3}\right )} x} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 236 vs. \(2 (53) = 106\).

Time = 1.47 (sec) , antiderivative size = 236, normalized size of antiderivative = 3.47 \[ \int \frac {\log \left (c (a+b x)^p\right )}{(d+e x)^2} \, dx=\begin {cases} \frac {\frac {a \log {\left (c \left (a + b x\right )^{p} \right )}}{b} - p x + x \log {\left (c \left (a + b x\right )^{p} \right )}}{d^{2}} & \text {for}\: e = 0 \\- \frac {p}{d e + e^{2} x} - \frac {\log {\left (c \left (\frac {b d}{e} + b x\right )^{p} \right )}}{d e + e^{2} x} & \text {for}\: a = \frac {b d}{e} \\- \frac {a e \log {\left (c \left (a + b x\right )^{p} \right )}}{a d e^{2} + a e^{3} x - b d^{2} e - b d e^{2} x} + \frac {b d p \log {\left (\frac {d}{e} + x \right )}}{a d e^{2} + a e^{3} x - b d^{2} e - b d e^{2} x} + \frac {b e p x \log {\left (\frac {d}{e} + x \right )}}{a d e^{2} + a e^{3} x - b d^{2} e - b d e^{2} x} - \frac {b e x \log {\left (c \left (a + b x\right )^{p} \right )}}{a d e^{2} + a e^{3} x - b d^{2} e - b d e^{2} x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((a*log(c*(a + b*x)**p)/b - p*x + x*log(c*(a + b*x)**p))/d**2, Eq(e, 0)), (-p/(d*e + e**2*x) - log(c
*(b*d/e + b*x)**p)/(d*e + e**2*x), Eq(a, b*d/e)), (-a*e*log(c*(a + b*x)**p)/(a*d*e**2 + a*e**3*x - b*d**2*e -
b*d*e**2*x) + b*d*p*log(d/e + x)/(a*d*e**2 + a*e**3*x - b*d**2*e - b*d*e**2*x) + b*e*p*x*log(d/e + x)/(a*d*e**
2 + a*e**3*x - b*d**2*e - b*d*e**2*x) - b*e*x*log(c*(a + b*x)**p)/(a*d*e**2 + a*e**3*x - b*d**2*e - b*d*e**2*x
), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.96 \[ \int \frac {\log \left (c (a+b x)^p\right )}{(d+e x)^2} \, dx=\frac {b p {\left (\frac {\log \left (b x + a\right )}{b d - a e} - \frac {\log \left (e x + d\right )}{b d - a e}\right )}}{e} - \frac {\log \left ({\left (b x + a\right )}^{p} c\right )}{{\left (e x + d\right )} e} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.19 \[ \int \frac {\log \left (c (a+b x)^p\right )}{(d+e x)^2} \, dx=\frac {b p \log \left (b x + a\right )}{b d e - a e^{2}} - \frac {b p \log \left (e x + d\right )}{b d e - a e^{2}} - \frac {p \log \left (b x + a\right )}{e^{2} x + d e} - \frac {\log \left (c\right )}{e^{2} x + d e} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.03 \[ \int \frac {\log \left (c (a+b x)^p\right )}{(d+e x)^2} \, dx=-\frac {\ln \left (c\,{\left (a+b\,x\right )}^p\right )}{e\,\left (d+e\,x\right )}+\frac {b\,p\,\mathrm {atan}\left (\frac {a\,e\,1{}\mathrm {i}+b\,d\,1{}\mathrm {i}+b\,e\,x\,2{}\mathrm {i}}{a\,e-b\,d}\right )\,2{}\mathrm {i}}{a\,e^2-b\,d\,e} \]

[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))