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

Optimal. Leaf size=84 \[ -\frac{p (b d-a e)^2 \log (a+b x)}{2 b^2 e}+\frac{(d+e x)^2 \log \left (c (a+b x)^p\right )}{2 e}-\frac{p x (b d-a e)}{2 b}-\frac{p (d+e x)^2}{4 e} \]

[Out]

-((b*d - a*e)*p*x)/(2*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)

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Rubi [A]  time = 0.0371142, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2395, 43} \[ -\frac{p (b d-a e)^2 \log (a+b x)}{2 b^2 e}+\frac{(d+e x)^2 \log \left (c (a+b x)^p\right )}{2 e}-\frac{p x (b d-a e)}{2 b}-\frac{p (d+e x)^2}{4 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*Log[c*(a + b*x)^p],x]

[Out]

-((b*d - a*e)*p*x)/(2*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 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]

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])

Rubi steps

\begin{align*} \int (d+e x) \log \left (c (a+b x)^p\right ) \, dx &=\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]  time = 0.0455267, size = 82, normalized size = 0.98 \[ -\frac{a^2 e p \log (a+b x)}{2 b^2}+\frac{d (a+b x) \log \left (c (a+b x)^p\right )}{b}+\frac{1}{2} e x^2 \log \left (c (a+b x)^p\right )+\frac{a e p x}{2 b}-d p x-\frac{1}{4} e p x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.091, size = 83, normalized size = 1. \begin{align*} d\ln \left ( c \left ( bx+a \right ) ^{p} \right ) x-dpx+{\frac{dpa\ln \left ( bx+a \right ) }{b}}+{\frac{{x}^{2}e\ln \left ( c{{\rm e}^{p\ln \left ( bx+a \right ) }} \right ) }{2}}-{\frac{ep{x}^{2}}{4}}-{\frac{{a}^{2}pe\ln \left ( bx+a \right ) }{2\,{b}^{2}}}+{\frac{apex}{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.10274, size = 100, normalized size = 1.19 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.02251, size = 205, normalized size = 2.44 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 1.42708, size = 116, normalized size = 1.38 \begin{align*} \begin{cases} - \frac{a^{2} e p \log{\left (a + b x \right )}}{2 b^{2}} + \frac{a d p \log{\left (a + b x \right )}}{b} + \frac{a e p x}{2 b} + d p x \log{\left (a + b x \right )} - d p x + d x \log{\left (c \right )} + \frac{e p x^{2} \log{\left (a + b x \right )}}{2} - \frac{e p x^{2}}{4} + \frac{e x^{2} \log{\left (c \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27378, size = 192, normalized size = 2.29 \begin{align*} \frac{{\left (b x + a\right )} d p \log \left (b x + a\right )}{b} + \frac{{\left (b x + a\right )}^{2} p e \log \left (b x + a\right )}{2 \, b^{2}} - \frac{{\left (b x + a\right )} a p e \log \left (b x + a\right )}{b^{2}} - \frac{{\left (b x + a\right )} d p}{b} - \frac{{\left (b x + a\right )}^{2} p e}{4 \, b^{2}} + \frac{{\left (b x + a\right )} a p e}{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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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