\(\int (d+e x)^2 \log (c (a+b x)^p) \, dx\) [177]

Optimal result

Integrand size = 18, antiderivative size = 112 \[ \int (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx=-\frac {(b d-a e)^2 p x}{3 b^2}-\frac {(b d-a e) p (d+e x)^2}{6 b e}-\frac {p (d+e x)^3}{9 e}-\frac {(b d-a e)^3 p \log (a+b x)}{3 b^3 e}+\frac {(d+e x)^3 \log \left (c (a+b x)^p\right )}{3 e} \]

[Out]

-1/3*(-a*e+b*d)^2*p*x/b^2-1/6*(-a*e+b*d)*p*(e*x+d)^2/b/e-1/9*p*(e*x+d)^3/e-1/3*(-a*e+b*d)^3*p*ln(b*x+a)/b^3/e+
1/3*(e*x+d)^3*ln(c*(b*x+a)^p)/e

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2442, 45} \[ \int (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx=-\frac {p (b d-a e)^3 \log (a+b x)}{3 b^3 e}-\frac {p x (b d-a e)^2}{3 b^2}+\frac {(d+e x)^3 \log \left (c (a+b x)^p\right )}{3 e}-\frac {p (d+e x)^2 (b d-a e)}{6 b e}-\frac {p (d+e x)^3}{9 e} \]

[In]

Int[(d + e*x)^2*Log[c*(a + b*x)^p],x]

[Out]

-1/3*((b*d - a*e)^2*p*x)/b^2 - ((b*d - a*e)*p*(d + e*x)^2)/(6*b*e) - (p*(d + e*x)^3)/(9*e) - ((b*d - a*e)^3*p*
Log[a + b*x])/(3*b^3*e) + ((d + e*x)^3*Log[c*(a + b*x)^p])/(3*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)^3 \log \left (c (a+b x)^p\right )}{3 e}-\frac {(b p) \int \frac {(d+e x)^3}{a+b x} \, dx}{3 e} \\ & = \frac {(d+e x)^3 \log \left (c (a+b x)^p\right )}{3 e}-\frac {(b p) \int \left (\frac {e (b d-a e)^2}{b^3}+\frac {(b d-a e)^3}{b^3 (a+b x)}+\frac {e (b d-a e) (d+e x)}{b^2}+\frac {e (d+e x)^2}{b}\right ) \, dx}{3 e} \\ & = -\frac {(b d-a e)^2 p x}{3 b^2}-\frac {(b d-a e) p (d+e x)^2}{6 b e}-\frac {p (d+e x)^3}{9 e}-\frac {(b d-a e)^3 p \log (a+b x)}{3 b^3 e}+\frac {(d+e x)^3 \log \left (c (a+b x)^p\right )}{3 e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.08 \[ \int (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx=\frac {6 a^2 e (-3 b d+a e) p \log (a+b x)+b \left (-p x \left (6 a^2 e^2-3 a b e (6 d+e x)+b^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+6 b \left (3 a d^2+b x \left (3 d^2+3 d e x+e^2 x^2\right )\right ) \log \left (c (a+b x)^p\right )\right )}{18 b^3} \]

[In]

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

[Out]

(6*a^2*e*(-3*b*d + a*e)*p*Log[a + b*x] + b*(-(p*x*(6*a^2*e^2 - 3*a*b*e*(6*d + e*x) + b^2*(18*d^2 + 9*d*e*x + 2
*e^2*x^2))) + 6*b*(3*a*d^2 + b*x*(3*d^2 + 3*d*e*x + e^2*x^2))*Log[c*(a + b*x)^p]))/(18*b^3)

Maple [A] (verified)

Time = 0.94 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.64

method	result	size
parts	\(\frac {\ln \left (c \left (b x +a \right )^{p}\right ) e^{2} x^{3}}{3}+\ln \left (c \left (b x +a \right )^{p}\right ) e d \,x^{2}+d^{2} \ln \left (c \left (b x +a \right )^{p}\right ) x +\frac {\ln \left (c \left (b x +a \right )^{p}\right ) d^{3}}{3 e}-\frac {p b \left (\frac {e \left (\frac {1}{3} x^{3} b^{2} e^{2}-\frac {1}{2} a b \,e^{2} x^{2}+\frac {3}{2} d e \,b^{2} x^{2}+x \,a^{2} e^{2}-3 a b d e x +3 x \,b^{2} d^{2}\right )}{b^{3}}+\frac {\left (-a^{3} e^{3}+3 a^{2} b d \,e^{2}-3 a \,b^{2} d^{2} e +b^{3} d^{3}\right ) \ln \left (b x +a \right )}{b^{4}}\right )}{3 e}\)	\(184\)
parallelrisch	\(\frac {6 x^{3} \ln \left (c \left (b x +a \right )^{p}\right ) b^{3} e^{2}-2 x^{3} b^{3} e^{2} p +18 x^{2} \ln \left (c \left (b x +a \right )^{p}\right ) b^{3} d e +3 x^{2} a \,b^{2} e^{2} p -9 x^{2} b^{3} d e p +6 \ln \left (b x +a \right ) a^{3} e^{2} p -18 \ln \left (b x +a \right ) a^{2} b d e p +36 \ln \left (b x +a \right ) a \,b^{2} d^{2} p +18 x \ln \left (c \left (b x +a \right )^{p}\right ) b^{3} d^{2}-6 x \,a^{2} b \,e^{2} p +18 x a \,b^{2} d e p -18 b^{3} d^{2} p x -18 \ln \left (c \left (b x +a \right )^{p}\right ) a \,b^{2} d^{2}+6 a^{3} e^{2} p -18 a^{2} b d e p +18 a \,b^{2} d^{2} p}{18 b^{3}}\)	\(227\)
risch	\(-\frac {i \pi \,d^{2} 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 {\left (e x +d \right )^{3} \ln \left (\left (b x +a \right )^{p}\right )}{3 e}+\frac {i e^{2} \pi \,x^{3} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{6}+\frac {i e^{2} \pi \,x^{3} \operatorname {csgn}\left (i \left (b x +a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{6}-\frac {i e \pi d \,x^{2} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+e \ln \left (c \right ) d \,x^{2}-\frac {\ln \left (b x +a \right ) d^{3} p}{3 e}+\frac {d^{2} p a \ln \left (b x +a \right )}{b}+\frac {e^{2} a p \,x^{2}}{6 b}-\frac {e^{2} a^{2} p x}{3 b^{2}}+\frac {e^{2} \ln \left (b x +a \right ) a^{3} p}{3 b^{3}}+\frac {e a d p x}{b}-\frac {e \ln \left (b x +a \right ) a^{2} d p}{b^{2}}+\frac {i \pi \,d^{2} x \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i \pi \,d^{2} 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 {e^{2} \ln \left (c \right ) x^{3}}{3}+\ln \left (c \right ) d^{2} x -\frac {i \pi \,d^{2} x \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}-\frac {i e \pi d \,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 )}{2}-\frac {i e^{2} \pi \,x^{3} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{6}-\frac {i e^{2} \pi \,x^{3} \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 )}{6}+\frac {i e \pi d \,x^{2} \operatorname {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i e \pi d \,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}}{2}-\frac {d e p \,x^{2}}{2}-\frac {e^{2} p \,x^{3}}{9}-d^{2} p x\)	\(537\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.54 \[ \int (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx=-\frac {2 \, b^{3} e^{2} p x^{3} + 3 \, {\left (3 \, b^{3} d e - a b^{2} e^{2}\right )} p x^{2} + 6 \, {\left (3 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2}\right )} p x - 6 \, {\left (b^{3} e^{2} p x^{3} + 3 \, b^{3} d e p x^{2} + 3 \, b^{3} d^{2} p x + {\left (3 \, a b^{2} d^{2} - 3 \, a^{2} b d e + a^{3} e^{2}\right )} p\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} e^{2} x^{3} + 3 \, b^{3} d e x^{2} + 3 \, b^{3} d^{2} x\right )} \log \left (c\right )}{18 \, b^{3}} \]

[In]

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

[Out]

-1/18*(2*b^3*e^2*p*x^3 + 3*(3*b^3*d*e - a*b^2*e^2)*p*x^2 + 6*(3*b^3*d^2 - 3*a*b^2*d*e + a^2*b*e^2)*p*x - 6*(b^
3*e^2*p*x^3 + 3*b^3*d*e*p*x^2 + 3*b^3*d^2*p*x + (3*a*b^2*d^2 - 3*a^2*b*d*e + a^3*e^2)*p)*log(b*x + a) - 6*(b^3
*e^2*x^3 + 3*b^3*d*e*x^2 + 3*b^3*d^2*x)*log(c))/b^3

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (94) = 188\).

Time = 0.57 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.80 \[ \int (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx=\begin {cases} \frac {a^{3} e^{2} \log {\left (c \left (a + b x\right )^{p} \right )}}{3 b^{3}} - \frac {a^{2} d e \log {\left (c \left (a + b x\right )^{p} \right )}}{b^{2}} - \frac {a^{2} e^{2} p x}{3 b^{2}} + \frac {a d^{2} \log {\left (c \left (a + b x\right )^{p} \right )}}{b} + \frac {a d e p x}{b} + \frac {a e^{2} p x^{2}}{6 b} - d^{2} p x + d^{2} x \log {\left (c \left (a + b x\right )^{p} \right )} - \frac {d e p x^{2}}{2} + d e x^{2} \log {\left (c \left (a + b x\right )^{p} \right )} - \frac {e^{2} p x^{3}}{9} + \frac {e^{2} x^{3} \log {\left (c \left (a + b x\right )^{p} \right )}}{3} & \text {for}\: b \neq 0 \\\left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) \log {\left (a^{p} c \right )} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a**3*e**2*log(c*(a + b*x)**p)/(3*b**3) - a**2*d*e*log(c*(a + b*x)**p)/b**2 - a**2*e**2*p*x/(3*b**2)
+ a*d**2*log(c*(a + b*x)**p)/b + a*d*e*p*x/b + a*e**2*p*x**2/(6*b) - d**2*p*x + d**2*x*log(c*(a + b*x)**p) -
d*e*p*x**2/2 + d*e*x**2*log(c*(a + b*x)**p) - e**2*p*x**3/9 + e**2*x**3*log(c*(a + b*x)**p)/3, Ne(b, 0)), ((d*
*2*x + d*e*x**2 + e**2*x**3/3)*log(a**p*c), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.21 \[ \int (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx=-\frac {1}{18} \, b p {\left (\frac {2 \, b^{2} e^{2} x^{3} + 3 \, {\left (3 \, b^{2} d e - a b e^{2}\right )} x^{2} + 6 \, {\left (3 \, b^{2} d^{2} - 3 \, a b d e + a^{2} e^{2}\right )} x}{b^{3}} - \frac {6 \, {\left (3 \, a b^{2} d^{2} - 3 \, a^{2} b d e + a^{3} e^{2}\right )} \log \left (b x + a\right )}{b^{4}}\right )} + \frac {1}{3} \, {\left (e^{2} x^{3} + 3 \, d e x^{2} + 3 \, d^{2} x\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

[In]

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

[Out]

-1/18*b*p*((2*b^2*e^2*x^3 + 3*(3*b^2*d*e - a*b*e^2)*x^2 + 6*(3*b^2*d^2 - 3*a*b*d*e + a^2*e^2)*x)/b^3 - 6*(3*a*
b^2*d^2 - 3*a^2*b*d*e + a^3*e^2)*log(b*x + a)/b^4) + 1/3*(e^2*x^3 + 3*d*e*x^2 + 3*d^2*x)*log((b*x + a)^p*c)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (102) = 204\).

Time = 0.30 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.82 \[ \int (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx=\frac {{\left (b x + a\right )} d^{2} p \log \left (b x + a\right )}{b} + \frac {{\left (b x + a\right )}^{2} d e p \log \left (b x + a\right )}{b^{2}} - \frac {2 \, {\left (b x + a\right )} a d e p \log \left (b x + a\right )}{b^{2}} + \frac {{\left (b x + a\right )}^{3} e^{2} p \log \left (b x + a\right )}{3 \, b^{3}} - \frac {{\left (b x + a\right )}^{2} a e^{2} p \log \left (b x + a\right )}{b^{3}} + \frac {{\left (b x + a\right )} a^{2} e^{2} p \log \left (b x + a\right )}{b^{3}} - \frac {{\left (b x + a\right )} d^{2} p}{b} - \frac {{\left (b x + a\right )}^{2} d e p}{2 \, b^{2}} + \frac {2 \, {\left (b x + a\right )} a d e p}{b^{2}} - \frac {{\left (b x + a\right )}^{3} e^{2} p}{9 \, b^{3}} + \frac {{\left (b x + a\right )}^{2} a e^{2} p}{2 \, b^{3}} - \frac {{\left (b x + a\right )} a^{2} e^{2} p}{b^{3}} + \frac {{\left (b x + a\right )} d^{2} \log \left (c\right )}{b} + \frac {{\left (b x + a\right )}^{2} d e \log \left (c\right )}{b^{2}} - \frac {2 \, {\left (b x + a\right )} a d e \log \left (c\right )}{b^{2}} + \frac {{\left (b x + a\right )}^{3} e^{2} \log \left (c\right )}{3 \, b^{3}} - \frac {{\left (b x + a\right )}^{2} a e^{2} \log \left (c\right )}{b^{3}} + \frac {{\left (b x + a\right )} a^{2} e^{2} \log \left (c\right )}{b^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.17 \[ \int (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx=\ln \left (c\,{\left (a+b\,x\right )}^p\right )\,\left (d^2\,x+d\,e\,x^2+\frac {e^2\,x^3}{3}\right )-x^2\,\left (\frac {d\,e\,p}{2}-\frac {a\,e^2\,p}{6\,b}\right )-x\,\left (d^2\,p-\frac {a\,\left (d\,e\,p-\frac {a\,e^2\,p}{3\,b}\right )}{b}\right )-\frac {e^2\,p\,x^3}{9}+\frac {\ln \left (a+b\,x\right )\,\left (p\,a^3\,e^2-3\,p\,a^2\,b\,d\,e+3\,p\,a\,b^2\,d^2\right )}{3\,b^3} \]

[In]

int(log(c*(a + b*x)^p)*(d + e*x)^2,x)

[Out]

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