\(\int (f+g x) (a+b \log (c (d+e x)^n)) \, dx\) [38]

Optimal result

Integrand size = 20, antiderivative size = 91 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {b (e f-d g) n x}{2 e}-\frac {b n (f+g x)^2}{4 g}-\frac {b (e f-d g)^2 n \log (d+e x)}{2 e^2 g}+\frac {(f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2442, 45} \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {(f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {b n (e f-d g)^2 \log (d+e x)}{2 e^2 g}-\frac {b n x (e f-d g)}{2 e}-\frac {b n (f+g x)^2}{4 g} \]

[In]

Int[(f + g*x)*(a + b*Log[c*(d + e*x)^n]),x]

[Out]

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

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 {(f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {(b e n) \int \frac {(f+g x)^2}{d+e x} \, dx}{2 g} \\ & = \frac {(f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g}-\frac {(b e n) \int \left (\frac {g (e f-d g)}{e^2}+\frac {(e f-d g)^2}{e^2 (d+e x)}+\frac {g (f+g x)}{e}\right ) \, dx}{2 g} \\ & = -\frac {b (e f-d g) n x}{2 e}-\frac {b n (f+g x)^2}{4 g}-\frac {b (e f-d g)^2 n \log (d+e x)}{2 e^2 g}+\frac {(f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.04 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=a f x-b f n x+\frac {1}{2} a g x^2-\frac {1}{2} b g n \left (-\frac {d x}{e}+\frac {x^2}{2}+\frac {d^2 \log (d+e x)}{e^2}\right )+\frac {1}{2} b g x^2 \log \left (c (d+e x)^n\right )+\frac {b f (d+e x) \log \left (c (d+e x)^n\right )}{e} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.08

method	result	size
parts	\(a \left (\frac {1}{2} g \,x^{2}+f x \right )+b \left (f \ln \left (c \left (e x +d \right )^{n}\right ) x -f n x +\frac {f n d \ln \left (e x +d \right )}{e}+\frac {x^{2} g \ln \left (c \,{\mathrm e}^{n \ln \left (e x +d \right )}\right )}{2}-\frac {g n \,x^{2}}{4}-\frac {n \,d^{2} g \ln \left (e x +d \right )}{2 e^{2}}+\frac {d g n x}{2 e}\right )\)	\(98\)
norman	\(\left (-\frac {1}{4} b g n +\frac {1}{2} a g \right ) x^{2}+b f x \ln \left (c \,{\mathrm e}^{n \ln \left (e x +d \right )}\right )+\frac {\left (d g b n -2 b e f n +2 a e f \right ) x}{2 e}+\frac {b g \,x^{2} \ln \left (c \,{\mathrm e}^{n \ln \left (e x +d \right )}\right )}{2}-\frac {n \left (b \,d^{2} g -2 b d e f \right ) \ln \left (e x +d \right )}{2 e^{2}}\)	\(99\)
default	\(a f x +\frac {a g \,x^{2}}{2}+b f \ln \left (c \left (e x +d \right )^{n}\right ) x -b f n x +\frac {b f n d \ln \left (e x +d \right )}{e}+\frac {b g \,x^{2} \ln \left (c \,{\mathrm e}^{n \ln \left (e x +d \right )}\right )}{2}-\frac {b g n \,x^{2}}{4}-\frac {n b \,d^{2} g \ln \left (e x +d \right )}{2 e^{2}}+\frac {d g b n x}{2 e}\)	\(101\)
parallelrisch	\(-\frac {-2 x^{2} \ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{2} g +b \,e^{2} g n \,x^{2}+2 \ln \left (e x +d \right ) b \,d^{2} g n -8 \ln \left (e x +d \right ) b d e f n -2 a \,e^{2} g \,x^{2}-4 x \ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{2} f -2 b d e g n x +4 b \,e^{2} f n x -4 a \,e^{2} f x +4 \ln \left (c \left (e x +d \right )^{n}\right ) b d e f +2 d^{2} g b n -4 b d e f n +4 a d e f}{4 e^{2}}\)	\(154\)
risch	\(\frac {b x \left (g x +2 f \right ) \ln \left (\left (e x +d \right )^{n}\right )}{2}-\frac {i \pi b f x \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )}{2}-\frac {i \pi b f x \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{2}-\frac {i \pi b g \,x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )}{4}-\frac {i \pi b g \,x^{2} \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{3}}{4}+\frac {i \pi b f x \,\operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}+\frac {i \pi b g \,x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{4}+\frac {i \pi b f x \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{2}+\frac {i \pi b g \,x^{2} \operatorname {csgn}\left (i \left (e x +d \right )^{n}\right ) \operatorname {csgn}\left (i c \left (e x +d \right )^{n}\right )^{2}}{4}+\frac {\ln \left (c \right ) b g \,x^{2}}{2}-\frac {b g n \,x^{2}}{4}+\ln \left (c \right ) b f x -\frac {n b \,d^{2} g \ln \left (e x +d \right )}{2 e^{2}}+\frac {b f n d \ln \left (e x +d \right )}{e}+\frac {a g \,x^{2}}{2}+\frac {d g b n x}{2 e}-b f n x +a f x\)	\(338\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.31 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {{\left (b e^{2} g n - 2 \, a e^{2} g\right )} x^{2} - 2 \, {\left (2 \, a e^{2} f - {\left (2 \, b e^{2} f - b d e g\right )} n\right )} x - 2 \, {\left (b e^{2} g n x^{2} + 2 \, b e^{2} f n x + {\left (2 \, b d e f - b d^{2} g\right )} n\right )} \log \left (e x + d\right ) - 2 \, {\left (b e^{2} g x^{2} + 2 \, b e^{2} f x\right )} \log \left (c\right )}{4 \, e^{2}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.47 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\begin {cases} a f x + \frac {a g x^{2}}{2} - \frac {b d^{2} g \log {\left (c \left (d + e x\right )^{n} \right )}}{2 e^{2}} + \frac {b d f \log {\left (c \left (d + e x\right )^{n} \right )}}{e} + \frac {b d g n x}{2 e} - b f n x + b f x \log {\left (c \left (d + e x\right )^{n} \right )} - \frac {b g n x^{2}}{4} + \frac {b g x^{2} \log {\left (c \left (d + e x\right )^{n} \right )}}{2} & \text {for}\: e \neq 0 \\\left (a + b \log {\left (c d^{n} \right )}\right ) \left (f x + \frac {g x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((g*x+f)*(a+b*ln(c*(e*x+d)**n)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.12 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-b e f n {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} - \frac {1}{4} \, b e g n {\left (\frac {2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {e x^{2} - 2 \, d x}{e^{2}}\right )} + \frac {1}{2} \, b g x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac {1}{2} \, a g x^{2} + b f x \log \left ({\left (e x + d\right )}^{n} c\right ) + a f x \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (83) = 166\).

Time = 0.31 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.01 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {{\left (e x + d\right )} b f n \log \left (e x + d\right )}{e} + \frac {{\left (e x + d\right )}^{2} b g n \log \left (e x + d\right )}{2 \, e^{2}} - \frac {{\left (e x + d\right )} b d g n \log \left (e x + d\right )}{e^{2}} - \frac {{\left (e x + d\right )} b f n}{e} - \frac {{\left (e x + d\right )}^{2} b g n}{4 \, e^{2}} + \frac {{\left (e x + d\right )} b d g n}{e^{2}} + \frac {{\left (e x + d\right )} b f \log \left (c\right )}{e} + \frac {{\left (e x + d\right )}^{2} b g \log \left (c\right )}{2 \, e^{2}} - \frac {{\left (e x + d\right )} b d g \log \left (c\right )}{e^{2}} + \frac {{\left (e x + d\right )} a f}{e} + \frac {{\left (e x + d\right )}^{2} a g}{2 \, e^{2}} - \frac {{\left (e x + d\right )} a d g}{e^{2}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.14 \[ \int (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=x\,\left (\frac {2\,a\,d\,g+2\,a\,e\,f-2\,b\,e\,f\,n}{2\,e}-\frac {d\,g\,\left (2\,a-b\,n\right )}{2\,e}\right )+\ln \left (c\,{\left (d+e\,x\right )}^n\right )\,\left (\frac {b\,g\,x^2}{2}+b\,f\,x\right )-\frac {\ln \left (d+e\,x\right )\,\left (b\,d^2\,g\,n-2\,b\,d\,e\,f\,n\right )}{2\,e^2}+\frac {g\,x^2\,\left (2\,a-b\,n\right )}{4} \]

[In]

int((f + g*x)*(a + b*log(c*(d + e*x)^n)),x)

[Out]

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