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

Optimal. Leaf size=112 \[ -\frac {p (b d-a e)^3 \log (a+b x)}{3 b^3 e}-\frac {p x (b d-a e)^2}{3 b^2}+\frac {(d+e x)^3 \log \left (c (a+b x)^p\right )}{3 e}-\frac {p (d+e x)^2 (b d-a e)}{6 b e}-\frac {p (d+e x)^3}{9 e} \]

[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

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Rubi [A]  time = 0.07, antiderivative size = 112, 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, 43} \[ -\frac {p x (b d-a e)^2}{3 b^2}-\frac {p (b d-a e)^3 \log (a+b x)}{3 b^3 e}+\frac {(d+e x)^3 \log \left (c (a+b x)^p\right )}{3 e}-\frac {p (d+e x)^2 (b d-a e)}{6 b e}-\frac {p (d+e x)^3}{9 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*Log[c*(a + b*x)^p],x]

[Out]

-((b*d - a*e)^2*p*x)/(3*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 43

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 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 (d+e x)^2 \log \left (c (a+b x)^p\right ) \, dx &=\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*}

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Mathematica [A]  time = 0.11, size = 121, normalized size = 1.08 \[ \frac {b \left (6 b \left (3 a d^2+b x \left (3 d^2+3 d e x+e^2 x^2\right )\right ) \log \left (c (a+b x)^p\right )-p x \left (6 a^2 e^2-3 a b e (6 d+e x)+b^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )\right )+6 a^2 e p (a e-3 b d) \log (a+b x)}{18 b^3} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.42, size = 172, normalized size = 1.54 \[ -\frac {2 \, b^{3} e^{2} p x^{3} + 3 \, {\left (3 \, b^{3} d e - a b^{2} e^{2}\right )} p x^{2} + 6 \, {\left (3 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2}\right )} p x - 6 \, {\left (b^{3} e^{2} p x^{3} + 3 \, b^{3} d e p x^{2} + 3 \, b^{3} d^{2} p x + {\left (3 \, a b^{2} d^{2} - 3 \, a^{2} b d e + a^{3} e^{2}\right )} p\right )} \log \left (b x + a\right ) - 6 \, {\left (b^{3} e^{2} x^{3} + 3 \, b^{3} d e x^{2} + 3 \, b^{3} d^{2} x\right )} \log \relax (c)}{18 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [B]  time = 0.18, size = 313, normalized size = 2.79 \[ \frac {{\left (b x + a\right )} d^{2} p \log \left (b x + a\right )}{b} + \frac {{\left (b x + a\right )}^{2} d p e \log \left (b x + a\right )}{b^{2}} - \frac {2 \, {\left (b x + a\right )} a d p e \log \left (b x + a\right )}{b^{2}} - \frac {{\left (b x + a\right )} d^{2} p}{b} - \frac {{\left (b x + a\right )}^{2} d p e}{2 \, b^{2}} + \frac {2 \, {\left (b x + a\right )} a d p e}{b^{2}} + \frac {{\left (b x + a\right )}^{3} p e^{2} \log \left (b x + a\right )}{3 \, b^{3}} - \frac {{\left (b x + a\right )}^{2} a p e^{2} \log \left (b x + a\right )}{b^{3}} + \frac {{\left (b x + a\right )} a^{2} p e^{2} \log \left (b x + a\right )}{b^{3}} + \frac {{\left (b x + a\right )} d^{2} \log \relax (c)}{b} + \frac {{\left (b x + a\right )}^{2} d e \log \relax (c)}{b^{2}} - \frac {2 \, {\left (b x + a\right )} a d e \log \relax (c)}{b^{2}} - \frac {{\left (b x + a\right )}^{3} p e^{2}}{9 \, b^{3}} + \frac {{\left (b x + a\right )}^{2} a p e^{2}}{2 \, b^{3}} - \frac {{\left (b x + a\right )} a^{2} p e^{2}}{b^{3}} + \frac {{\left (b x + a\right )}^{3} e^{2} \log \relax (c)}{3 \, b^{3}} - \frac {{\left (b x + a\right )}^{2} a e^{2} \log \relax (c)}{b^{3}} + \frac {{\left (b x + a\right )} a^{2} e^{2} \log \relax (c)}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

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maple [C]  time = 0.45, size = 537, normalized size = 4.79 \[ d e \,x^{2} \ln \relax (c )-\frac {d^{3} p \ln \left (b x +a \right )}{3 e}+\frac {e^{2} x^{3} \ln \relax (c )}{3}+d^{2} x \ln \relax (c )-\frac {e^{2} p \,x^{3}}{9}-d^{2} p x +\frac {\left (e x +d \right )^{3} \ln \left (\left (b x +a \right )^{p}\right )}{3 e}-\frac {d e p \,x^{2}}{2}+\frac {a d e p x}{b}-\frac {a^{2} d e p \ln \left (b x +a \right )}{b^{2}}-\frac {i \pi \,e^{2} x^{3} \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 )}{6}+\frac {i \pi d e \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}+\frac {i \pi d e \,x^{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}-\frac {i \pi \,d^{2} x \,\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}-\frac {i \pi \,e^{2} x^{3} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{6}-\frac {i \pi \,d^{2} x \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \,e^{2} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{6}+\frac {i \pi \,e^{2} x^{3} \mathrm {csgn}\left (i \left (b x +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{6}-\frac {i \pi d e \,x^{2} \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{3}}{2}+\frac {i \pi \,d^{2} x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{p}\right )^{2}}{2}+\frac {i \pi \,d^{2} x \,\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}-\frac {i \pi d e \,x^{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}+\frac {a^{3} e^{2} p \ln \left (b x +a \right )}{3 b^{3}}+\frac {a \,d^{2} p \ln \left (b x +a \right )}{b}-\frac {a^{2} e^{2} p x}{3 b^{2}}+\frac {a \,e^{2} p \,x^{2}}{6 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 136, normalized size = 1.21 \[ -\frac {1}{18} \, b p {\left (\frac {2 \, b^{2} e^{2} x^{3} + 3 \, {\left (3 \, b^{2} d e - a b e^{2}\right )} x^{2} + 6 \, {\left (3 \, b^{2} d^{2} - 3 \, a b d e + a^{2} e^{2}\right )} x}{b^{3}} - \frac {6 \, {\left (3 \, a b^{2} d^{2} - 3 \, a^{2} b d e + a^{3} e^{2}\right )} \log \left (b x + a\right )}{b^{4}}\right )} + \frac {1}{3} \, {\left (e^{2} x^{3} + 3 \, d e x^{2} + 3 \, d^{2} x\right )} \log \left ({\left (b x + a\right )}^{p} c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.27, size = 131, normalized size = 1.17 \[ \ln \left (c\,{\left (a+b\,x\right )}^p\right )\,\left (d^2\,x+d\,e\,x^2+\frac {e^2\,x^3}{3}\right )-x^2\,\left (\frac {d\,e\,p}{2}-\frac {a\,e^2\,p}{6\,b}\right )-x\,\left (d^2\,p-\frac {a\,\left (d\,e\,p-\frac {a\,e^2\,p}{3\,b}\right )}{b}\right )-\frac {e^2\,p\,x^3}{9}+\frac {\ln \left (a+b\,x\right )\,\left (p\,a^3\,e^2-3\,p\,a^2\,b\,d\,e+3\,p\,a\,b^2\,d^2\right )}{3\,b^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 3.02, size = 223, normalized size = 1.99 \[ \begin {cases} \frac {a^{3} e^{2} p \log {\left (a + b x \right )}}{3 b^{3}} - \frac {a^{2} d e p \log {\left (a + b x \right )}}{b^{2}} - \frac {a^{2} e^{2} p x}{3 b^{2}} + \frac {a d^{2} p \log {\left (a + b x \right )}}{b} + \frac {a d e p x}{b} + \frac {a e^{2} p x^{2}}{6 b} + d^{2} p x \log {\left (a + b x \right )} - d^{2} p x + d^{2} x \log {\relax (c )} + d e p x^{2} \log {\left (a + b x \right )} - \frac {d e p x^{2}}{2} + d e x^{2} \log {\relax (c )} + \frac {e^{2} p x^{3} \log {\left (a + b x \right )}}{3} - \frac {e^{2} p x^{3}}{9} + \frac {e^{2} x^{3} \log {\relax (c )}}{3} & \text {for}\: b \neq 0 \\\left (d^{2} x + d e x^{2} + \frac {e^{2} x^{3}}{3}\right ) \log {\left (a^{p} c \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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