\(\int (d+e x) \log (c (a+b x)^p) \, dx\) [178]

Optimal result

Integrand size = 16, antiderivative size = 84 \[ \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx=-\frac {(b d-a e) p x}{2 b}-\frac {p (d+e x)^2}{4 e}-\frac {(b d-a e)^2 p \log (a+b x)}{2 b^2 e}+\frac {(d+e x)^2 \log \left (c (a+b x)^p\right )}{2 e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2442, 45} \[ \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx=-\frac {p (b d-a e)^2 \log (a+b x)}{2 b^2 e}+\frac {(d+e x)^2 \log \left (c (a+b x)^p\right )}{2 e}-\frac {p x (b d-a e)}{2 b}-\frac {p (d+e x)^2}{4 e} \]

[In]

Int[(d + e*x)*Log[c*(a + b*x)^p],x]

[Out]

-1/2*((b*d - a*e)*p*x)/b - (p*(d + e*x)^2)/(4*e) - ((b*d - a*e)^2*p*Log[a + b*x])/(2*b^2*e) + ((d + e*x)^2*Log
[c*(a + b*x)^p])/(2*e)

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 {(d+e x)^2 \log \left (c (a+b x)^p\right )}{2 e}-\frac {(b p) \int \frac {(d+e x)^2}{a+b x} \, dx}{2 e} \\ & = \frac {(d+e x)^2 \log \left (c (a+b x)^p\right )}{2 e}-\frac {(b p) \int \left (\frac {e (b d-a e)}{b^2}+\frac {(b d-a e)^2}{b^2 (a+b x)}+\frac {e (d+e x)}{b}\right ) \, dx}{2 e} \\ & = -\frac {(b d-a e) p x}{2 b}-\frac {p (d+e x)^2}{4 e}-\frac {(b d-a e)^2 p \log (a+b x)}{2 b^2 e}+\frac {(d+e x)^2 \log \left (c (a+b x)^p\right )}{2 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93 \[ \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx=-d p x-\frac {1}{2} e p \left (-\frac {a x}{b}+\frac {x^2}{2}+\frac {a^2 \log (a+b x)}{b^2}\right )+\frac {1}{2} e x^2 \log \left (c (a+b x)^p\right )+\frac {d (a+b x) \log \left (c (a+b x)^p\right )}{b} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92

method	result	size
parts	\(\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e \,x^{2}}{2}+d \ln \left (c \left (b x +a \right )^{p}\right ) x -\frac {p b \left (-\frac {-\frac {1}{2} b e \,x^{2}+a e x -2 b d x}{b^{2}}+\frac {a \left (a e -2 b d \right ) \ln \left (b x +a \right )}{b^{3}}\right )}{2}\)	\(77\)
norman	\(d x \ln \left (c \,{\mathrm e}^{p \ln \left (b x +a \right )}\right )-\frac {e p \,x^{2}}{4}+\frac {e \,x^{2} \ln \left (c \,{\mathrm e}^{p \ln \left (b x +a \right )}\right )}{2}+\frac {p \left (a e -2 b d \right ) x}{2 b}-\frac {p \left (a^{2} e -2 a b d \right ) \ln \left (b x +a \right )}{2 b^{2}}\)	\(80\)
default	\(d \ln \left (c \left (b x +a \right )^{p}\right ) x -d p x +\frac {d p a \ln \left (b x +a \right )}{b}+\frac {e \,x^{2} \ln \left (c \,{\mathrm e}^{p \ln \left (b x +a \right )}\right )}{2}-\frac {e p \,x^{2}}{4}-\frac {p \,a^{2} e \ln \left (b x +a \right )}{2 b^{2}}+\frac {a e p x}{2 b}\)	\(83\)
parallelrisch	\(-\frac {-2 x^{2} \ln \left (c \left (b x +a \right )^{p}\right ) b^{2} e +b^{2} e p \,x^{2}+2 \ln \left (b x +a \right ) a^{2} e p -8 \ln \left (b x +a \right ) a b d p -4 x \ln \left (c \left (b x +a \right )^{p}\right ) b^{2} d -2 a b e p x +4 b^{2} d p x +4 \ln \left (c \left (b x +a \right )^{p}\right ) a b d +2 a^{2} e p -4 a b d p}{4 b^{2}}\)	\(120\)
risch	\(\left (\frac {1}{2} e \,x^{2}+d x \right ) \ln \left (\left (b x +a \right )^{p}\right )+\frac {i \pi d x \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}-\frac {i \pi e \,x^{2} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{4}+\frac {i \pi e \,x^{2} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{4}-\frac {i \pi d x \,\operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}+\frac {i \pi d x \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi e \,x^{2} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{4}+\frac {i \pi e \,x^{2} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{4}-\frac {i \pi d x \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {\ln \left (c \right ) e \,x^{2}}{2}-\frac {e p \,x^{2}}{4}+\ln \left (c \right ) d x -\frac {p \,a^{2} e \ln \left (b x +a \right )}{2 b^{2}}+\frac {d p a \ln \left (b x +a \right )}{b}+\frac {a e p x}{2 b}-d p x\)	\(312\)

[In]

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

[Out]

1/2*ln(c*(b*x+a)^p)*e*x^2+d*ln(c*(b*x+a)^p)*x-1/2*p*b*(-1/b^2*(-1/2*b*e*x^2+a*e*x-2*b*d*x)+a*(a*e-2*b*d)/b^3*l
n(b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.08 \[ \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx=-\frac {b^{2} e p x^{2} + 2 \, {\left (2 \, b^{2} d - a b e\right )} p x - 2 \, {\left (b^{2} e p x^{2} + 2 \, b^{2} d p x + {\left (2 \, a b d - a^{2} e\right )} p\right )} \log \left (b x + a\right ) - 2 \, {\left (b^{2} e x^{2} + 2 \, b^{2} d x\right )} \log \left (c\right )}{4 \, b^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.25 \[ \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx=\begin {cases} - \frac {a^{2} e \log {\left (c \left (a + b x\right )^{p} \right )}}{2 b^{2}} + \frac {a d \log {\left (c \left (a + b x\right )^{p} \right )}}{b} + \frac {a e p x}{2 b} - d p x + d x \log {\left (c \left (a + b x\right )^{p} \right )} - \frac {e p x^{2}}{4} + \frac {e x^{2} \log {\left (c \left (a + b x\right )^{p} \right )}}{2} & \text {for}\: b \neq 0 \\\left (d x + \frac {e x^{2}}{2}\right ) \log {\left (a^{p} c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a**2*e*log(c*(a + b*x)**p)/(2*b**2) + a*d*log(c*(a + b*x)**p)/b + a*e*p*x/(2*b) - d*p*x + d*x*log(
c*(a + b*x)**p) - e*p*x**2/4 + e*x**2*log(c*(a + b*x)**p)/2, Ne(b, 0)), ((d*x + e*x**2/2)*log(a**p*c), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88 \[ \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx=-\frac {1}{4} \, b p {\left (\frac {b e x^{2} + 2 \, {\left (2 \, b d - a e\right )} x}{b^{2}} - \frac {2 \, {\left (2 \, a b d - a^{2} e\right )} \log \left (b x + a\right )}{b^{3}}\right )} + \frac {1}{2} \, {\left (e x^{2} + 2 \, d x\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.62 \[ \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx=\frac {{\left (b x + a\right )} d p \log \left (b x + a\right )}{b} + \frac {{\left (b x + a\right )}^{2} e p \log \left (b x + a\right )}{2 \, b^{2}} - \frac {{\left (b x + a\right )} a e p \log \left (b x + a\right )}{b^{2}} - \frac {{\left (b x + a\right )} d p}{b} - \frac {{\left (b x + a\right )}^{2} e p}{4 \, b^{2}} + \frac {{\left (b x + a\right )} a e p}{b^{2}} + \frac {{\left (b x + a\right )} d \log \left (c\right )}{b} + \frac {{\left (b x + a\right )}^{2} e \log \left (c\right )}{2 \, b^{2}} - \frac {{\left (b x + a\right )} a e \log \left (c\right )}{b^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx=\ln \left (c\,{\left (a+b\,x\right )}^p\right )\,\left (\frac {e\,x^2}{2}+d\,x\right )-x\,\left (d\,p-\frac {a\,e\,p}{2\,b}\right )-\frac {e\,p\,x^2}{4}-\frac {\ln \left (a+b\,x\right )\,\left (a^2\,e\,p-2\,a\,b\,d\,p\right )}{2\,b^2} \]

[In]

int(log(c*(a + b*x)^p)*(d + e*x),x)

[Out]

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