3.35 \(\int (f+g x)^4 (a+b \log (c (d+e x)^n)) \, dx\)

Optimal. Leaf size=178 \[ \frac{(f+g x)^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac{b n x (e f-d g)^4}{5 e^4}-\frac{b n (f+g x)^2 (e f-d g)^3}{10 e^3 g}-\frac{b n (f+g x)^3 (e f-d g)^2}{15 e^2 g}-\frac{b n (e f-d g)^5 \log (d+e x)}{5 e^5 g}-\frac{b n (f+g x)^4 (e f-d g)}{20 e g}-\frac{b n (f+g x)^5}{25 g} \]

[Out]

-(b*(e*f - d*g)^4*n*x)/(5*e^4) - (b*(e*f - d*g)^3*n*(f + g*x)^2)/(10*e^3*g) - (b*(e*f - d*g)^2*n*(f + g*x)^3)/
(15*e^2*g) - (b*(e*f - d*g)*n*(f + g*x)^4)/(20*e*g) - (b*n*(f + g*x)^5)/(25*g) - (b*(e*f - d*g)^5*n*Log[d + e*
x])/(5*e^5*g) + ((f + g*x)^5*(a + b*Log[c*(d + e*x)^n]))/(5*g)

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Rubi [A]  time = 0.100471, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {2395, 43} \[ \frac{(f+g x)^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac{b n x (e f-d g)^4}{5 e^4}-\frac{b n (f+g x)^2 (e f-d g)^3}{10 e^3 g}-\frac{b n (f+g x)^3 (e f-d g)^2}{15 e^2 g}-\frac{b n (e f-d g)^5 \log (d+e x)}{5 e^5 g}-\frac{b n (f+g x)^4 (e f-d g)}{20 e g}-\frac{b n (f+g x)^5}{25 g} \]

Antiderivative was successfully verified.

[In]

Int[(f + g*x)^4*(a + b*Log[c*(d + e*x)^n]),x]

[Out]

-(b*(e*f - d*g)^4*n*x)/(5*e^4) - (b*(e*f - d*g)^3*n*(f + g*x)^2)/(10*e^3*g) - (b*(e*f - d*g)^2*n*(f + g*x)^3)/
(15*e^2*g) - (b*(e*f - d*g)*n*(f + g*x)^4)/(20*e*g) - (b*n*(f + g*x)^5)/(25*g) - (b*(e*f - d*g)^5*n*Log[d + e*
x])/(5*e^5*g) + ((f + g*x)^5*(a + b*Log[c*(d + e*x)^n]))/(5*g)

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 (f+g x)^4 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx &=\frac{(f+g x)^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac{(b e n) \int \frac{(f+g x)^5}{d+e x} \, dx}{5 g}\\ &=\frac{(f+g x)^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}-\frac{(b e n) \int \left (\frac{g (e f-d g)^4}{e^5}+\frac{(e f-d g)^5}{e^5 (d+e x)}+\frac{g (e f-d g)^3 (f+g x)}{e^4}+\frac{g (e f-d g)^2 (f+g x)^2}{e^3}+\frac{g (e f-d g) (f+g x)^3}{e^2}+\frac{g (f+g x)^4}{e}\right ) \, dx}{5 g}\\ &=-\frac{b (e f-d g)^4 n x}{5 e^4}-\frac{b (e f-d g)^3 n (f+g x)^2}{10 e^3 g}-\frac{b (e f-d g)^2 n (f+g x)^3}{15 e^2 g}-\frac{b (e f-d g) n (f+g x)^4}{20 e g}-\frac{b n (f+g x)^5}{25 g}-\frac{b (e f-d g)^5 n \log (d+e x)}{5 e^5 g}+\frac{(f+g x)^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g}\\ \end{align*}

Mathematica [A]  time = 0.301256, size = 315, normalized size = 1.77 \[ \frac{e x \left (60 a e^4 \left (10 f^2 g^2 x^2+10 f^3 g x+5 f^4+5 f g^3 x^3+g^4 x^4\right )-b n \left (10 d^2 e^2 g^2 \left (60 f^2+15 f g x+2 g^2 x^2\right )-30 d^3 e g^3 (10 f+g x)+60 d^4 g^4-5 d e^3 g \left (60 f^2 g x+120 f^3+20 f g^2 x^2+3 g^3 x^3\right )+e^4 \left (200 f^2 g^2 x^2+300 f^3 g x+300 f^4+75 f g^3 x^3+12 g^4 x^4\right )\right )\right )+60 b e^4 \left (5 d f^4+e x \left (10 f^2 g^2 x^2+10 f^3 g x+5 f^4+5 f g^3 x^3+g^4 x^4\right )\right ) \log \left (c (d+e x)^n\right )+60 b d^2 g n \left (-5 d^2 e f g^2+d^3 g^3+10 d e^2 f^2 g-10 e^3 f^3\right ) \log (d+e x)}{300 e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(f + g*x)^4*(a + b*Log[c*(d + e*x)^n]),x]

[Out]

(e*x*(60*a*e^4*(5*f^4 + 10*f^3*g*x + 10*f^2*g^2*x^2 + 5*f*g^3*x^3 + g^4*x^4) - b*n*(60*d^4*g^4 - 30*d^3*e*g^3*
(10*f + g*x) + 10*d^2*e^2*g^2*(60*f^2 + 15*f*g*x + 2*g^2*x^2) - 5*d*e^3*g*(120*f^3 + 60*f^2*g*x + 20*f*g^2*x^2
+ 3*g^3*x^3) + e^4*(300*f^4 + 300*f^3*g*x + 200*f^2*g^2*x^2 + 75*f*g^3*x^3 + 12*g^4*x^4))) + 60*b*d^2*g*(-10*
e^3*f^3 + 10*d*e^2*f^2*g - 5*d^2*e*f*g^2 + d^3*g^3)*n*Log[d + e*x] + 60*b*e^4*(5*d*f^4 + e*x*(5*f^4 + 10*f^3*g
*x + 10*f^2*g^2*x^2 + 5*f*g^3*x^3 + g^4*x^4))*Log[c*(d + e*x)^n])/(300*e^5)

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Maple [C]  time = 0.514, size = 1105, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x+f)^4*(a+b*ln(c*(e*x+d)^n)),x)

[Out]

1/5*(g*x+f)^5*b/g*ln((e*x+d)^n)+a*f^4*x-1/25*g^4*b*n*x^5+1/5*g^4*a*x^5+1/5*g^4*ln(c)*b*x^5+ln(c)*b*f^4*x+g^3*a
*f*x^4+2*g^2*a*f^2*x^3+2*g*a*f^3*x^2+2*g^2*ln(c)*b*f^2*x^3+2*g*ln(c)*b*f^3*x^2+g^3*ln(c)*b*f*x^4-1/5/g*ln(e*x+
d)*b*f^5*n-1/2*I*Pi*b*f^4*x*csgn(I*c*(e*x+d)^n)^3-1/10*I*g^4*Pi*b*x^5*csgn(I*c*(e*x+d)^n)^3-1/15/e^2*g^4*b*d^2
*n*x^3-2/3*g^2*b*f^2*n*x^3+1/10/e^3*g^4*b*d^3*n*x^2-g*b*f^3*n*x^2-1/5/e^4*g^4*b*d^4*n*x-b*f^4*n*x+1/5/e^5*g^4*
ln(e*x+d)*b*d^5*n+1/e*ln(e*x+d)*b*d*f^4*n+1/20/e*g^4*b*d*n*x^4-1/4*g^3*b*f*n*x^4+I*g^2*Pi*b*f^2*x^3*csgn(I*(e*
x+d)^n)*csgn(I*c*(e*x+d)^n)^2+I*g*Pi*b*f^3*x^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*g^2*Pi*b*f^2*x^3*csgn(I*c)*cs
gn(I*c*(e*x+d)^n)^2+I*g*Pi*b*f^3*x^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*I*Pi*b*f^4*x*csgn(I*c)*csgn(I
*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I*g^3*Pi*b*f*x^4*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/2*I*g^3*Pi*b*f*x^4*csgn
(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/10*I*g^4*Pi*b*x^5*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-2/e^2*
g^2*b*d^2*f^2*n*x+2/e*g*b*d*f^3*n*x+1/3/e*g^3*b*d*f*n*x^3-1/2/e^2*g^3*b*d^2*f*n*x^2+1/e*g^2*b*d*f^2*n*x^2-1/e^
4*g^3*ln(e*x+d)*b*d^4*f*n+2/e^3*g^2*ln(e*x+d)*b*d^3*f^2*n-2/e^2*g*ln(e*x+d)*b*d^2*f^3*n-I*g^2*Pi*b*f^2*x^3*csg
n(I*c*(e*x+d)^n)^3-I*g*Pi*b*f^3*x^2*csgn(I*c*(e*x+d)^n)^3+1/10*I*g^4*Pi*b*x^5*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+
d)^n)^2-1/2*I*g^3*Pi*b*f*x^4*csgn(I*c*(e*x+d)^n)^3+1/10*I*g^4*Pi*b*x^5*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/2*I*P
i*b*f^4*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/2*I*Pi*b*f^4*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/e^3*g^3*b
*d^3*f*n*x-I*g^2*Pi*b*f^2*x^3*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-I*g*Pi*b*f^3*x^2*csgn(I*c)*csgn(
I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/2*I*g^3*Pi*b*f*x^4*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)

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Maxima [B]  time = 1.21324, size = 531, normalized size = 2.98 \begin{align*} \frac{1}{5} \, b g^{4} x^{5} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac{1}{5} \, a g^{4} x^{5} + b f g^{3} x^{4} \log \left ({\left (e x + d\right )}^{n} c\right ) + a f g^{3} x^{4} + 2 \, b f^{2} g^{2} x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + 2 \, a f^{2} g^{2} x^{3} - b e f^{4} n{\left (\frac{x}{e} - \frac{d \log \left (e x + d\right )}{e^{2}}\right )} + \frac{1}{300} \, b e g^{4} n{\left (\frac{60 \, d^{5} \log \left (e x + d\right )}{e^{6}} - \frac{12 \, e^{4} x^{5} - 15 \, d e^{3} x^{4} + 20 \, d^{2} e^{2} x^{3} - 30 \, d^{3} e x^{2} + 60 \, d^{4} x}{e^{5}}\right )} - \frac{1}{12} \, b e f g^{3} n{\left (\frac{12 \, d^{4} \log \left (e x + d\right )}{e^{5}} + \frac{3 \, e^{3} x^{4} - 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} - 12 \, d^{3} x}{e^{4}}\right )} + \frac{1}{3} \, b e f^{2} g^{2} n{\left (\frac{6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac{2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} - b e f^{3} 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 )} + 2 \, b f^{3} g x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) + 2 \, a f^{3} g x^{2} + b f^{4} x \log \left ({\left (e x + d\right )}^{n} c\right ) + a f^{4} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^4*(a+b*log(c*(e*x+d)^n)),x, algorithm="maxima")

[Out]

1/5*b*g^4*x^5*log((e*x + d)^n*c) + 1/5*a*g^4*x^5 + b*f*g^3*x^4*log((e*x + d)^n*c) + a*f*g^3*x^4 + 2*b*f^2*g^2*
x^3*log((e*x + d)^n*c) + 2*a*f^2*g^2*x^3 - b*e*f^4*n*(x/e - d*log(e*x + d)/e^2) + 1/300*b*e*g^4*n*(60*d^5*log(
e*x + d)/e^6 - (12*e^4*x^5 - 15*d*e^3*x^4 + 20*d^2*e^2*x^3 - 30*d^3*e*x^2 + 60*d^4*x)/e^5) - 1/12*b*e*f*g^3*n*
(12*d^4*log(e*x + d)/e^5 + (3*e^3*x^4 - 4*d*e^2*x^3 + 6*d^2*e*x^2 - 12*d^3*x)/e^4) + 1/3*b*e*f^2*g^2*n*(6*d^3*
log(e*x + d)/e^4 - (2*e^2*x^3 - 3*d*e*x^2 + 6*d^2*x)/e^3) - b*e*f^3*g*n*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d
*x)/e^2) + 2*b*f^3*g*x^2*log((e*x + d)^n*c) + 2*a*f^3*g*x^2 + b*f^4*x*log((e*x + d)^n*c) + a*f^4*x

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Fricas [B]  time = 2.21632, size = 998, normalized size = 5.61 \begin{align*} -\frac{12 \,{\left (b e^{5} g^{4} n - 5 \, a e^{5} g^{4}\right )} x^{5} - 15 \,{\left (20 \, a e^{5} f g^{3} -{\left (5 \, b e^{5} f g^{3} - b d e^{4} g^{4}\right )} n\right )} x^{4} - 20 \,{\left (30 \, a e^{5} f^{2} g^{2} -{\left (10 \, b e^{5} f^{2} g^{2} - 5 \, b d e^{4} f g^{3} + b d^{2} e^{3} g^{4}\right )} n\right )} x^{3} - 30 \,{\left (20 \, a e^{5} f^{3} g -{\left (10 \, b e^{5} f^{3} g - 10 \, b d e^{4} f^{2} g^{2} + 5 \, b d^{2} e^{3} f g^{3} - b d^{3} e^{2} g^{4}\right )} n\right )} x^{2} - 60 \,{\left (5 \, a e^{5} f^{4} -{\left (5 \, b e^{5} f^{4} - 10 \, b d e^{4} f^{3} g + 10 \, b d^{2} e^{3} f^{2} g^{2} - 5 \, b d^{3} e^{2} f g^{3} + b d^{4} e g^{4}\right )} n\right )} x - 60 \,{\left (b e^{5} g^{4} n x^{5} + 5 \, b e^{5} f g^{3} n x^{4} + 10 \, b e^{5} f^{2} g^{2} n x^{3} + 10 \, b e^{5} f^{3} g n x^{2} + 5 \, b e^{5} f^{4} n x +{\left (5 \, b d e^{4} f^{4} - 10 \, b d^{2} e^{3} f^{3} g + 10 \, b d^{3} e^{2} f^{2} g^{2} - 5 \, b d^{4} e f g^{3} + b d^{5} g^{4}\right )} n\right )} \log \left (e x + d\right ) - 60 \,{\left (b e^{5} g^{4} x^{5} + 5 \, b e^{5} f g^{3} x^{4} + 10 \, b e^{5} f^{2} g^{2} x^{3} + 10 \, b e^{5} f^{3} g x^{2} + 5 \, b e^{5} f^{4} x\right )} \log \left (c\right )}{300 \, e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^4*(a+b*log(c*(e*x+d)^n)),x, algorithm="fricas")

[Out]

-1/300*(12*(b*e^5*g^4*n - 5*a*e^5*g^4)*x^5 - 15*(20*a*e^5*f*g^3 - (5*b*e^5*f*g^3 - b*d*e^4*g^4)*n)*x^4 - 20*(3
0*a*e^5*f^2*g^2 - (10*b*e^5*f^2*g^2 - 5*b*d*e^4*f*g^3 + b*d^2*e^3*g^4)*n)*x^3 - 30*(20*a*e^5*f^3*g - (10*b*e^5
*f^3*g - 10*b*d*e^4*f^2*g^2 + 5*b*d^2*e^3*f*g^3 - b*d^3*e^2*g^4)*n)*x^2 - 60*(5*a*e^5*f^4 - (5*b*e^5*f^4 - 10*
b*d*e^4*f^3*g + 10*b*d^2*e^3*f^2*g^2 - 5*b*d^3*e^2*f*g^3 + b*d^4*e*g^4)*n)*x - 60*(b*e^5*g^4*n*x^5 + 5*b*e^5*f
*g^3*n*x^4 + 10*b*e^5*f^2*g^2*n*x^3 + 10*b*e^5*f^3*g*n*x^2 + 5*b*e^5*f^4*n*x + (5*b*d*e^4*f^4 - 10*b*d^2*e^3*f
^3*g + 10*b*d^3*e^2*f^2*g^2 - 5*b*d^4*e*f*g^3 + b*d^5*g^4)*n)*log(e*x + d) - 60*(b*e^5*g^4*x^5 + 5*b*e^5*f*g^3
*x^4 + 10*b*e^5*f^2*g^2*x^3 + 10*b*e^5*f^3*g*x^2 + 5*b*e^5*f^4*x)*log(c))/e^5

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Sympy [A]  time = 15.9663, size = 620, normalized size = 3.48 \begin{align*} \begin{cases} a f^{4} x + 2 a f^{3} g x^{2} + 2 a f^{2} g^{2} x^{3} + a f g^{3} x^{4} + \frac{a g^{4} x^{5}}{5} + \frac{b d^{5} g^{4} n \log{\left (d + e x \right )}}{5 e^{5}} - \frac{b d^{4} f g^{3} n \log{\left (d + e x \right )}}{e^{4}} - \frac{b d^{4} g^{4} n x}{5 e^{4}} + \frac{2 b d^{3} f^{2} g^{2} n \log{\left (d + e x \right )}}{e^{3}} + \frac{b d^{3} f g^{3} n x}{e^{3}} + \frac{b d^{3} g^{4} n x^{2}}{10 e^{3}} - \frac{2 b d^{2} f^{3} g n \log{\left (d + e x \right )}}{e^{2}} - \frac{2 b d^{2} f^{2} g^{2} n x}{e^{2}} - \frac{b d^{2} f g^{3} n x^{2}}{2 e^{2}} - \frac{b d^{2} g^{4} n x^{3}}{15 e^{2}} + \frac{b d f^{4} n \log{\left (d + e x \right )}}{e} + \frac{2 b d f^{3} g n x}{e} + \frac{b d f^{2} g^{2} n x^{2}}{e} + \frac{b d f g^{3} n x^{3}}{3 e} + \frac{b d g^{4} n x^{4}}{20 e} + b f^{4} n x \log{\left (d + e x \right )} - b f^{4} n x + b f^{4} x \log{\left (c \right )} + 2 b f^{3} g n x^{2} \log{\left (d + e x \right )} - b f^{3} g n x^{2} + 2 b f^{3} g x^{2} \log{\left (c \right )} + 2 b f^{2} g^{2} n x^{3} \log{\left (d + e x \right )} - \frac{2 b f^{2} g^{2} n x^{3}}{3} + 2 b f^{2} g^{2} x^{3} \log{\left (c \right )} + b f g^{3} n x^{4} \log{\left (d + e x \right )} - \frac{b f g^{3} n x^{4}}{4} + b f g^{3} x^{4} \log{\left (c \right )} + \frac{b g^{4} n x^{5} \log{\left (d + e x \right )}}{5} - \frac{b g^{4} n x^{5}}{25} + \frac{b g^{4} x^{5} \log{\left (c \right )}}{5} & \text{for}\: e \neq 0 \\\left (a + b \log{\left (c d^{n} \right )}\right ) \left (f^{4} x + 2 f^{3} g x^{2} + 2 f^{2} g^{2} x^{3} + f g^{3} x^{4} + \frac{g^{4} x^{5}}{5}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)**4*(a+b*ln(c*(e*x+d)**n)),x)

[Out]

Piecewise((a*f**4*x + 2*a*f**3*g*x**2 + 2*a*f**2*g**2*x**3 + a*f*g**3*x**4 + a*g**4*x**5/5 + b*d**5*g**4*n*log
(d + e*x)/(5*e**5) - b*d**4*f*g**3*n*log(d + e*x)/e**4 - b*d**4*g**4*n*x/(5*e**4) + 2*b*d**3*f**2*g**2*n*log(d
+ e*x)/e**3 + b*d**3*f*g**3*n*x/e**3 + b*d**3*g**4*n*x**2/(10*e**3) - 2*b*d**2*f**3*g*n*log(d + e*x)/e**2 - 2
*b*d**2*f**2*g**2*n*x/e**2 - b*d**2*f*g**3*n*x**2/(2*e**2) - b*d**2*g**4*n*x**3/(15*e**2) + b*d*f**4*n*log(d +
e*x)/e + 2*b*d*f**3*g*n*x/e + b*d*f**2*g**2*n*x**2/e + b*d*f*g**3*n*x**3/(3*e) + b*d*g**4*n*x**4/(20*e) + b*f
**4*n*x*log(d + e*x) - b*f**4*n*x + b*f**4*x*log(c) + 2*b*f**3*g*n*x**2*log(d + e*x) - b*f**3*g*n*x**2 + 2*b*f
**3*g*x**2*log(c) + 2*b*f**2*g**2*n*x**3*log(d + e*x) - 2*b*f**2*g**2*n*x**3/3 + 2*b*f**2*g**2*x**3*log(c) + b
*f*g**3*n*x**4*log(d + e*x) - b*f*g**3*n*x**4/4 + b*f*g**3*x**4*log(c) + b*g**4*n*x**5*log(d + e*x)/5 - b*g**4
*n*x**5/25 + b*g**4*x**5*log(c)/5, Ne(e, 0)), ((a + b*log(c*d**n))*(f**4*x + 2*f**3*g*x**2 + 2*f**2*g**2*x**3
+ f*g**3*x**4 + g**4*x**5/5), True))

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Giac [B]  time = 1.292, size = 1652, normalized size = 9.28 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x+f)^4*(a+b*log(c*(e*x+d)^n)),x, algorithm="giac")

[Out]

1/5*(x*e + d)^5*b*g^4*n*e^(-5)*log(x*e + d) - (x*e + d)^4*b*d*g^4*n*e^(-5)*log(x*e + d) + 2*(x*e + d)^3*b*d^2*
g^4*n*e^(-5)*log(x*e + d) - 2*(x*e + d)^2*b*d^3*g^4*n*e^(-5)*log(x*e + d) + (x*e + d)*b*d^4*g^4*n*e^(-5)*log(x
*e + d) - 1/25*(x*e + d)^5*b*g^4*n*e^(-5) + 1/4*(x*e + d)^4*b*d*g^4*n*e^(-5) - 2/3*(x*e + d)^3*b*d^2*g^4*n*e^(
-5) + (x*e + d)^2*b*d^3*g^4*n*e^(-5) - (x*e + d)*b*d^4*g^4*n*e^(-5) + (x*e + d)^4*b*f*g^3*n*e^(-4)*log(x*e + d
) - 4*(x*e + d)^3*b*d*f*g^3*n*e^(-4)*log(x*e + d) + 6*(x*e + d)^2*b*d^2*f*g^3*n*e^(-4)*log(x*e + d) - 4*(x*e +
d)*b*d^3*f*g^3*n*e^(-4)*log(x*e + d) + 1/5*(x*e + d)^5*b*g^4*e^(-5)*log(c) - (x*e + d)^4*b*d*g^4*e^(-5)*log(c
) + 2*(x*e + d)^3*b*d^2*g^4*e^(-5)*log(c) - 2*(x*e + d)^2*b*d^3*g^4*e^(-5)*log(c) + (x*e + d)*b*d^4*g^4*e^(-5)
*log(c) - 1/4*(x*e + d)^4*b*f*g^3*n*e^(-4) + 4/3*(x*e + d)^3*b*d*f*g^3*n*e^(-4) - 3*(x*e + d)^2*b*d^2*f*g^3*n*
e^(-4) + 4*(x*e + d)*b*d^3*f*g^3*n*e^(-4) + 1/5*(x*e + d)^5*a*g^4*e^(-5) - (x*e + d)^4*a*d*g^4*e^(-5) + 2*(x*e
+ d)^3*a*d^2*g^4*e^(-5) - 2*(x*e + d)^2*a*d^3*g^4*e^(-5) + (x*e + d)*a*d^4*g^4*e^(-5) + 2*(x*e + d)^3*b*f^2*g
^2*n*e^(-3)*log(x*e + d) - 6*(x*e + d)^2*b*d*f^2*g^2*n*e^(-3)*log(x*e + d) + 6*(x*e + d)*b*d^2*f^2*g^2*n*e^(-3
)*log(x*e + d) + (x*e + d)^4*b*f*g^3*e^(-4)*log(c) - 4*(x*e + d)^3*b*d*f*g^3*e^(-4)*log(c) + 6*(x*e + d)^2*b*d
^2*f*g^3*e^(-4)*log(c) - 4*(x*e + d)*b*d^3*f*g^3*e^(-4)*log(c) - 2/3*(x*e + d)^3*b*f^2*g^2*n*e^(-3) + 3*(x*e +
d)^2*b*d*f^2*g^2*n*e^(-3) - 6*(x*e + d)*b*d^2*f^2*g^2*n*e^(-3) + (x*e + d)^4*a*f*g^3*e^(-4) - 4*(x*e + d)^3*a
*d*f*g^3*e^(-4) + 6*(x*e + d)^2*a*d^2*f*g^3*e^(-4) - 4*(x*e + d)*a*d^3*f*g^3*e^(-4) + 2*(x*e + d)^2*b*f^3*g*n*
e^(-2)*log(x*e + d) - 4*(x*e + d)*b*d*f^3*g*n*e^(-2)*log(x*e + d) + 2*(x*e + d)^3*b*f^2*g^2*e^(-3)*log(c) - 6*
(x*e + d)^2*b*d*f^2*g^2*e^(-3)*log(c) + 6*(x*e + d)*b*d^2*f^2*g^2*e^(-3)*log(c) - (x*e + d)^2*b*f^3*g*n*e^(-2)
+ 4*(x*e + d)*b*d*f^3*g*n*e^(-2) + 2*(x*e + d)^3*a*f^2*g^2*e^(-3) - 6*(x*e + d)^2*a*d*f^2*g^2*e^(-3) + 6*(x*e
+ d)*a*d^2*f^2*g^2*e^(-3) + (x*e + d)*b*f^4*n*e^(-1)*log(x*e + d) + 2*(x*e + d)^2*b*f^3*g*e^(-2)*log(c) - 4*(
x*e + d)*b*d*f^3*g*e^(-2)*log(c) - (x*e + d)*b*f^4*n*e^(-1) + 2*(x*e + d)^2*a*f^3*g*e^(-2) - 4*(x*e + d)*a*d*f
^3*g*e^(-2) + (x*e + d)*b*f^4*e^(-1)*log(c) + (x*e + d)*a*f^4*e^(-1)