\(\int (f+g x)^3 (a+b \log (c (d+e x)^n)) \, dx\) [36]

Optimal result

Integrand size = 22, antiderivative size = 149 \[ \int (f+g x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {b (e f-d g)^3 n x}{4 e^3}-\frac {b (e f-d g)^2 n (f+g x)^2}{8 e^2 g}-\frac {b (e f-d g) n (f+g x)^3}{12 e g}-\frac {b n (f+g x)^4}{16 g}-\frac {b (e f-d g)^4 n \log (d+e x)}{4 e^4 g}+\frac {(f+g x)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 g} \]

[Out]

-1/4*b*(-d*g+e*f)^3*n*x/e^3-1/8*b*(-d*g+e*f)^2*n*(g*x+f)^2/e^2/g-1/12*b*(-d*g+e*f)*n*(g*x+f)^3/e/g-1/16*b*n*(g
*x+f)^4/g-1/4*b*(-d*g+e*f)^4*n*ln(e*x+d)/e^4/g+1/4*(g*x+f)^4*(a+b*ln(c*(e*x+d)^n))/g

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2442, 45} \[ \int (f+g x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {(f+g x)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 g}-\frac {b n (e f-d g)^4 \log (d+e x)}{4 e^4 g}-\frac {b n x (e f-d g)^3}{4 e^3}-\frac {b n (f+g x)^2 (e f-d g)^2}{8 e^2 g}-\frac {b n (f+g x)^3 (e f-d g)}{12 e g}-\frac {b n (f+g x)^4}{16 g} \]

[In]

Int[(f + g*x)^3*(a + b*Log[c*(d + e*x)^n]),x]

[Out]

-1/4*(b*(e*f - d*g)^3*n*x)/e^3 - (b*(e*f - d*g)^2*n*(f + g*x)^2)/(8*e^2*g) - (b*(e*f - d*g)*n*(f + g*x)^3)/(12
*e*g) - (b*n*(f + g*x)^4)/(16*g) - (b*(e*f - d*g)^4*n*Log[d + e*x])/(4*e^4*g) + ((f + g*x)^4*(a + b*Log[c*(d +
e*x)^n]))/(4*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)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 g}-\frac {(b e n) \int \frac {(f+g x)^4}{d+e x} \, dx}{4 g} \\ & = \frac {(f+g x)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 g}-\frac {(b e n) \int \left (\frac {g (e f-d g)^3}{e^4}+\frac {(e f-d g)^4}{e^4 (d+e x)}+\frac {g (e f-d g)^2 (f+g x)}{e^3}+\frac {g (e f-d g) (f+g x)^2}{e^2}+\frac {g (f+g x)^3}{e}\right ) \, dx}{4 g} \\ & = -\frac {b (e f-d g)^3 n x}{4 e^3}-\frac {b (e f-d g)^2 n (f+g x)^2}{8 e^2 g}-\frac {b (e f-d g) n (f+g x)^3}{12 e g}-\frac {b n (f+g x)^4}{16 g}-\frac {b (e f-d g)^4 n \log (d+e x)}{4 e^4 g}+\frac {(f+g x)^4 \left (a+b \log \left (c (d+e x)^n\right )\right )}{4 g} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.52 \[ \int (f+g x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {e x \left (12 a e^3 \left (4 f^3+6 f^2 g x+4 f g^2 x^2+g^3 x^3\right )-b n \left (-12 d^3 g^3+6 d^2 e g^2 (8 f+g x)-4 d e^2 g \left (18 f^2+6 f g x+g^2 x^2\right )+e^3 \left (48 f^3+36 f^2 g x+16 f g^2 x^2+3 g^3 x^3\right )\right )\right )-12 b d^2 g \left (6 e^2 f^2-4 d e f g+d^2 g^2\right ) n \log (d+e x)+12 b e^3 \left (4 d f^3+e x \left (4 f^3+6 f^2 g x+4 f g^2 x^2+g^3 x^3\right )\right ) \log \left (c (d+e x)^n\right )}{48 e^4} \]

[In]

Integrate[(f + g*x)^3*(a + b*Log[c*(d + e*x)^n]),x]

[Out]

(e*x*(12*a*e^3*(4*f^3 + 6*f^2*g*x + 4*f*g^2*x^2 + g^3*x^3) - b*n*(-12*d^3*g^3 + 6*d^2*e*g^2*(8*f + g*x) - 4*d*
e^2*g*(18*f^2 + 6*f*g*x + g^2*x^2) + e^3*(48*f^3 + 36*f^2*g*x + 16*f*g^2*x^2 + 3*g^3*x^3))) - 12*b*d^2*g*(6*e^
2*f^2 - 4*d*e*f*g + d^2*g^2)*n*Log[d + e*x] + 12*b*e^3*(4*d*f^3 + e*x*(4*f^3 + 6*f^2*g*x + 4*f*g^2*x^2 + g^3*x
^3))*Log[c*(d + e*x)^n])/(48*e^4)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(427\) vs. \(2(137)=274\).

Time = 1.07 (sec) , antiderivative size = 428, normalized size of antiderivative = 2.87

method	result	size
parallelrisch	\(-\frac {-48 b \,d^{3} e f \,g^{2} n +72 b \,d^{2} e^{2} f^{2} g n -72 x^{2} a \,e^{4} f^{2} g -48 x \ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{4} f^{3}+48 x b \,e^{4} f^{3} n +48 \ln \left (c \left (e x +d \right )^{n}\right ) b d \,e^{3} f^{3}+12 \ln \left (e x +d \right ) b \,d^{4} g^{3} n -12 x^{4} \ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{4} g^{3}+3 x^{4} b \,e^{4} g^{3} n -48 x^{3} a \,e^{4} f \,g^{2}+48 a d \,f^{3} e^{3}-48 x^{3} \ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{4} f \,g^{2}-4 x^{3} b d \,e^{3} g^{3} n +16 x^{3} b \,e^{4} f \,g^{2} n -72 x^{2} \ln \left (c \left (e x +d \right )^{n}\right ) b \,e^{4} f^{2} g +6 x^{2} b \,d^{2} e^{2} g^{3} n +36 x^{2} b \,e^{4} f^{2} g n -12 x b \,d^{3} e \,g^{3} n -96 \ln \left (e x +d \right ) b d \,e^{3} f^{3} n -48 b d \,e^{3} f^{3} n +12 b \,d^{4} g^{3} n -48 \ln \left (e x +d \right ) b \,d^{3} e f \,g^{2} n +72 \ln \left (e x +d \right ) b \,d^{2} e^{2} f^{2} g n +48 x b \,d^{2} e^{2} f \,g^{2} n -72 x b d \,e^{3} f^{2} g n -24 x^{2} b d \,e^{3} f \,g^{2} n -12 x^{4} a \,e^{4} g^{3}-48 x a \,e^{4} f^{3}}{48 e^{4}}\)	\(428\)
risch	\(\text {Expression too large to display}\)	\(836\)

[In]

int((g*x+f)^3*(a+b*ln(c*(e*x+d)^n)),x,method=_RETURNVERBOSE)

[Out]

-1/48*(-48*b*d^3*e*f*g^2*n+72*b*d^2*e^2*f^2*g*n-72*x^2*a*e^4*f^2*g-48*x*ln(c*(e*x+d)^n)*b*e^4*f^3+48*x*b*e^4*f
^3*n+48*ln(c*(e*x+d)^n)*b*d*e^3*f^3+12*ln(e*x+d)*b*d^4*g^3*n-12*x^4*ln(c*(e*x+d)^n)*b*e^4*g^3+3*x^4*b*e^4*g^3*
n-48*x^3*a*e^4*f*g^2+48*a*d*f^3*e^3-48*x^3*ln(c*(e*x+d)^n)*b*e^4*f*g^2-4*x^3*b*d*e^3*g^3*n+16*x^3*b*e^4*f*g^2*
n-72*x^2*ln(c*(e*x+d)^n)*b*e^4*f^2*g+6*x^2*b*d^2*e^2*g^3*n+36*x^2*b*e^4*f^2*g*n-12*x*b*d^3*e*g^3*n-96*ln(e*x+d
)*b*d*e^3*f^3*n-48*b*d*e^3*f^3*n+12*b*d^4*g^3*n-48*ln(e*x+d)*b*d^3*e*f*g^2*n+72*ln(e*x+d)*b*d^2*e^2*f^2*g*n+48
*x*b*d^2*e^2*f*g^2*n-72*x*b*d*e^3*f^2*g*n-24*x^2*b*d*e^3*f*g^2*n-12*x^4*a*e^4*g^3-48*x*a*e^4*f^3)/e^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (137) = 274\).

Time = 0.29 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.28 \[ \int (f+g x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=-\frac {3 \, {\left (b e^{4} g^{3} n - 4 \, a e^{4} g^{3}\right )} x^{4} - 4 \, {\left (12 \, a e^{4} f g^{2} - {\left (4 \, b e^{4} f g^{2} - b d e^{3} g^{3}\right )} n\right )} x^{3} - 6 \, {\left (12 \, a e^{4} f^{2} g - {\left (6 \, b e^{4} f^{2} g - 4 \, b d e^{3} f g^{2} + b d^{2} e^{2} g^{3}\right )} n\right )} x^{2} - 12 \, {\left (4 \, a e^{4} f^{3} - {\left (4 \, b e^{4} f^{3} - 6 \, b d e^{3} f^{2} g + 4 \, b d^{2} e^{2} f g^{2} - b d^{3} e g^{3}\right )} n\right )} x - 12 \, {\left (b e^{4} g^{3} n x^{4} + 4 \, b e^{4} f g^{2} n x^{3} + 6 \, b e^{4} f^{2} g n x^{2} + 4 \, b e^{4} f^{3} n x + {\left (4 \, b d e^{3} f^{3} - 6 \, b d^{2} e^{2} f^{2} g + 4 \, b d^{3} e f g^{2} - b d^{4} g^{3}\right )} n\right )} \log \left (e x + d\right ) - 12 \, {\left (b e^{4} g^{3} x^{4} + 4 \, b e^{4} f g^{2} x^{3} + 6 \, b e^{4} f^{2} g x^{2} + 4 \, b e^{4} f^{3} x\right )} \log \left (c\right )}{48 \, e^{4}} \]

[In]

integrate((g*x+f)^3*(a+b*log(c*(e*x+d)^n)),x, algorithm="fricas")

[Out]

-1/48*(3*(b*e^4*g^3*n - 4*a*e^4*g^3)*x^4 - 4*(12*a*e^4*f*g^2 - (4*b*e^4*f*g^2 - b*d*e^3*g^3)*n)*x^3 - 6*(12*a*
e^4*f^2*g - (6*b*e^4*f^2*g - 4*b*d*e^3*f*g^2 + b*d^2*e^2*g^3)*n)*x^2 - 12*(4*a*e^4*f^3 - (4*b*e^4*f^3 - 6*b*d*
e^3*f^2*g + 4*b*d^2*e^2*f*g^2 - b*d^3*e*g^3)*n)*x - 12*(b*e^4*g^3*n*x^4 + 4*b*e^4*f*g^2*n*x^3 + 6*b*e^4*f^2*g*
n*x^2 + 4*b*e^4*f^3*n*x + (4*b*d*e^3*f^3 - 6*b*d^2*e^2*f^2*g + 4*b*d^3*e*f*g^2 - b*d^4*g^3)*n)*log(e*x + d) -
12*(b*e^4*g^3*x^4 + 4*b*e^4*f*g^2*x^3 + 6*b*e^4*f^2*g*x^2 + 4*b*e^4*f^3*x)*log(c))/e^4

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 410 vs. \(2 (128) = 256\).

Time = 1.16 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.75 \[ \int (f+g x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\begin {cases} a f^{3} x + \frac {3 a f^{2} g x^{2}}{2} + a f g^{2} x^{3} + \frac {a g^{3} x^{4}}{4} - \frac {b d^{4} g^{3} \log {\left (c \left (d + e x\right )^{n} \right )}}{4 e^{4}} + \frac {b d^{3} f g^{2} \log {\left (c \left (d + e x\right )^{n} \right )}}{e^{3}} + \frac {b d^{3} g^{3} n x}{4 e^{3}} - \frac {3 b d^{2} f^{2} g \log {\left (c \left (d + e x\right )^{n} \right )}}{2 e^{2}} - \frac {b d^{2} f g^{2} n x}{e^{2}} - \frac {b d^{2} g^{3} n x^{2}}{8 e^{2}} + \frac {b d f^{3} \log {\left (c \left (d + e x\right )^{n} \right )}}{e} + \frac {3 b d f^{2} g n x}{2 e} + \frac {b d f g^{2} n x^{2}}{2 e} + \frac {b d g^{3} n x^{3}}{12 e} - b f^{3} n x + b f^{3} x \log {\left (c \left (d + e x\right )^{n} \right )} - \frac {3 b f^{2} g n x^{2}}{4} + \frac {3 b f^{2} g x^{2} \log {\left (c \left (d + e x\right )^{n} \right )}}{2} - \frac {b f g^{2} n x^{3}}{3} + b f g^{2} x^{3} \log {\left (c \left (d + e x\right )^{n} \right )} - \frac {b g^{3} n x^{4}}{16} + \frac {b g^{3} x^{4} \log {\left (c \left (d + e x\right )^{n} \right )}}{4} & \text {for}\: e \neq 0 \\\left (a + b \log {\left (c d^{n} \right )}\right ) \left (f^{3} x + \frac {3 f^{2} g x^{2}}{2} + f g^{2} x^{3} + \frac {g^{3} x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((g*x+f)**3*(a+b*ln(c*(e*x+d)**n)),x)

[Out]

Piecewise((a*f**3*x + 3*a*f**2*g*x**2/2 + a*f*g**2*x**3 + a*g**3*x**4/4 - b*d**4*g**3*log(c*(d + e*x)**n)/(4*e
**4) + b*d**3*f*g**2*log(c*(d + e*x)**n)/e**3 + b*d**3*g**3*n*x/(4*e**3) - 3*b*d**2*f**2*g*log(c*(d + e*x)**n)
/(2*e**2) - b*d**2*f*g**2*n*x/e**2 - b*d**2*g**3*n*x**2/(8*e**2) + b*d*f**3*log(c*(d + e*x)**n)/e + 3*b*d*f**2
*g*n*x/(2*e) + b*d*f*g**2*n*x**2/(2*e) + b*d*g**3*n*x**3/(12*e) - b*f**3*n*x + b*f**3*x*log(c*(d + e*x)**n) -
3*b*f**2*g*n*x**2/4 + 3*b*f**2*g*x**2*log(c*(d + e*x)**n)/2 - b*f*g**2*n*x**3/3 + b*f*g**2*x**3*log(c*(d + e*x
)**n) - b*g**3*n*x**4/16 + b*g**3*x**4*log(c*(d + e*x)**n)/4, Ne(e, 0)), ((a + b*log(c*d**n))*(f**3*x + 3*f**2
*g*x**2/2 + f*g**2*x**3 + g**3*x**4/4), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (137) = 274\).

Time = 0.20 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.91 \[ \int (f+g x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {1}{4} \, b g^{3} x^{4} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac {1}{4} \, a g^{3} x^{4} + b f g^{2} x^{3} \log \left ({\left (e x + d\right )}^{n} c\right ) + a f g^{2} x^{3} - b e f^{3} n {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} - \frac {1}{48} \, b e g^{3} n {\left (\frac {12 \, d^{4} \log \left (e x + d\right )}{e^{5}} + \frac {3 \, e^{3} x^{4} - 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} - 12 \, d^{3} x}{e^{4}}\right )} + \frac {1}{6} \, b e f g^{2} n {\left (\frac {6 \, d^{3} \log \left (e x + d\right )}{e^{4}} - \frac {2 \, e^{2} x^{3} - 3 \, d e x^{2} + 6 \, d^{2} x}{e^{3}}\right )} - \frac {3}{4} \, b e f^{2} g n {\left (\frac {2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {e x^{2} - 2 \, d x}{e^{2}}\right )} + \frac {3}{2} \, b f^{2} g x^{2} \log \left ({\left (e x + d\right )}^{n} c\right ) + \frac {3}{2} \, a f^{2} g x^{2} + b f^{3} x \log \left ({\left (e x + d\right )}^{n} c\right ) + a f^{3} x \]

[In]

integrate((g*x+f)^3*(a+b*log(c*(e*x+d)^n)),x, algorithm="maxima")

[Out]

1/4*b*g^3*x^4*log((e*x + d)^n*c) + 1/4*a*g^3*x^4 + b*f*g^2*x^3*log((e*x + d)^n*c) + a*f*g^2*x^3 - b*e*f^3*n*(x
/e - d*log(e*x + d)/e^2) - 1/48*b*e*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) + 1/6*b*e*f*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) - 3/4*b*e*f^
2*g*n*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + 3/2*b*f^2*g*x^2*log((e*x + d)^n*c) + 3/2*a*f^2*g*x^2 +
b*f^3*x*log((e*x + d)^n*c) + a*f^3*x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 770 vs. \(2 (137) = 274\).

Time = 0.32 (sec) , antiderivative size = 770, normalized size of antiderivative = 5.17 \[ \int (f+g x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=\frac {{\left (e x + d\right )} b f^{3} n \log \left (e x + d\right )}{e} + \frac {3 \, {\left (e x + d\right )}^{2} b f^{2} g n \log \left (e x + d\right )}{2 \, e^{2}} - \frac {3 \, {\left (e x + d\right )} b d f^{2} g n \log \left (e x + d\right )}{e^{2}} + \frac {{\left (e x + d\right )}^{3} b f g^{2} n \log \left (e x + d\right )}{e^{3}} - \frac {3 \, {\left (e x + d\right )}^{2} b d f g^{2} n \log \left (e x + d\right )}{e^{3}} + \frac {3 \, {\left (e x + d\right )} b d^{2} f g^{2} n \log \left (e x + d\right )}{e^{3}} + \frac {{\left (e x + d\right )}^{4} b g^{3} n \log \left (e x + d\right )}{4 \, e^{4}} - \frac {{\left (e x + d\right )}^{3} b d g^{3} n \log \left (e x + d\right )}{e^{4}} + \frac {3 \, {\left (e x + d\right )}^{2} b d^{2} g^{3} n \log \left (e x + d\right )}{2 \, e^{4}} - \frac {{\left (e x + d\right )} b d^{3} g^{3} n \log \left (e x + d\right )}{e^{4}} - \frac {{\left (e x + d\right )} b f^{3} n}{e} - \frac {3 \, {\left (e x + d\right )}^{2} b f^{2} g n}{4 \, e^{2}} + \frac {3 \, {\left (e x + d\right )} b d f^{2} g n}{e^{2}} - \frac {{\left (e x + d\right )}^{3} b f g^{2} n}{3 \, e^{3}} + \frac {3 \, {\left (e x + d\right )}^{2} b d f g^{2} n}{2 \, e^{3}} - \frac {3 \, {\left (e x + d\right )} b d^{2} f g^{2} n}{e^{3}} - \frac {{\left (e x + d\right )}^{4} b g^{3} n}{16 \, e^{4}} + \frac {{\left (e x + d\right )}^{3} b d g^{3} n}{3 \, e^{4}} - \frac {3 \, {\left (e x + d\right )}^{2} b d^{2} g^{3} n}{4 \, e^{4}} + \frac {{\left (e x + d\right )} b d^{3} g^{3} n}{e^{4}} + \frac {{\left (e x + d\right )} b f^{3} \log \left (c\right )}{e} + \frac {3 \, {\left (e x + d\right )}^{2} b f^{2} g \log \left (c\right )}{2 \, e^{2}} - \frac {3 \, {\left (e x + d\right )} b d f^{2} g \log \left (c\right )}{e^{2}} + \frac {{\left (e x + d\right )}^{3} b f g^{2} \log \left (c\right )}{e^{3}} - \frac {3 \, {\left (e x + d\right )}^{2} b d f g^{2} \log \left (c\right )}{e^{3}} + \frac {3 \, {\left (e x + d\right )} b d^{2} f g^{2} \log \left (c\right )}{e^{3}} + \frac {{\left (e x + d\right )}^{4} b g^{3} \log \left (c\right )}{4 \, e^{4}} - \frac {{\left (e x + d\right )}^{3} b d g^{3} \log \left (c\right )}{e^{4}} + \frac {3 \, {\left (e x + d\right )}^{2} b d^{2} g^{3} \log \left (c\right )}{2 \, e^{4}} - \frac {{\left (e x + d\right )} b d^{3} g^{3} \log \left (c\right )}{e^{4}} + \frac {{\left (e x + d\right )} a f^{3}}{e} + \frac {3 \, {\left (e x + d\right )}^{2} a f^{2} g}{2 \, e^{2}} - \frac {3 \, {\left (e x + d\right )} a d f^{2} g}{e^{2}} + \frac {{\left (e x + d\right )}^{3} a f g^{2}}{e^{3}} - \frac {3 \, {\left (e x + d\right )}^{2} a d f g^{2}}{e^{3}} + \frac {3 \, {\left (e x + d\right )} a d^{2} f g^{2}}{e^{3}} + \frac {{\left (e x + d\right )}^{4} a g^{3}}{4 \, e^{4}} - \frac {{\left (e x + d\right )}^{3} a d g^{3}}{e^{4}} + \frac {3 \, {\left (e x + d\right )}^{2} a d^{2} g^{3}}{2 \, e^{4}} - \frac {{\left (e x + d\right )} a d^{3} g^{3}}{e^{4}} \]

[In]

integrate((g*x+f)^3*(a+b*log(c*(e*x+d)^n)),x, algorithm="giac")

[Out]

(e*x + d)*b*f^3*n*log(e*x + d)/e + 3/2*(e*x + d)^2*b*f^2*g*n*log(e*x + d)/e^2 - 3*(e*x + d)*b*d*f^2*g*n*log(e*
x + d)/e^2 + (e*x + d)^3*b*f*g^2*n*log(e*x + d)/e^3 - 3*(e*x + d)^2*b*d*f*g^2*n*log(e*x + d)/e^3 + 3*(e*x + d)
*b*d^2*f*g^2*n*log(e*x + d)/e^3 + 1/4*(e*x + d)^4*b*g^3*n*log(e*x + d)/e^4 - (e*x + d)^3*b*d*g^3*n*log(e*x + d
)/e^4 + 3/2*(e*x + d)^2*b*d^2*g^3*n*log(e*x + d)/e^4 - (e*x + d)*b*d^3*g^3*n*log(e*x + d)/e^4 - (e*x + d)*b*f^
3*n/e - 3/4*(e*x + d)^2*b*f^2*g*n/e^2 + 3*(e*x + d)*b*d*f^2*g*n/e^2 - 1/3*(e*x + d)^3*b*f*g^2*n/e^3 + 3/2*(e*x
+ d)^2*b*d*f*g^2*n/e^3 - 3*(e*x + d)*b*d^2*f*g^2*n/e^3 - 1/16*(e*x + d)^4*b*g^3*n/e^4 + 1/3*(e*x + d)^3*b*d*g
^3*n/e^4 - 3/4*(e*x + d)^2*b*d^2*g^3*n/e^4 + (e*x + d)*b*d^3*g^3*n/e^4 + (e*x + d)*b*f^3*log(c)/e + 3/2*(e*x +
d)^2*b*f^2*g*log(c)/e^2 - 3*(e*x + d)*b*d*f^2*g*log(c)/e^2 + (e*x + d)^3*b*f*g^2*log(c)/e^3 - 3*(e*x + d)^2*b
*d*f*g^2*log(c)/e^3 + 3*(e*x + d)*b*d^2*f*g^2*log(c)/e^3 + 1/4*(e*x + d)^4*b*g^3*log(c)/e^4 - (e*x + d)^3*b*d*
g^3*log(c)/e^4 + 3/2*(e*x + d)^2*b*d^2*g^3*log(c)/e^4 - (e*x + d)*b*d^3*g^3*log(c)/e^4 + (e*x + d)*a*f^3/e + 3
/2*(e*x + d)^2*a*f^2*g/e^2 - 3*(e*x + d)*a*d*f^2*g/e^2 + (e*x + d)^3*a*f*g^2/e^3 - 3*(e*x + d)^2*a*d*f*g^2/e^3
+ 3*(e*x + d)*a*d^2*f*g^2/e^3 + 1/4*(e*x + d)^4*a*g^3/e^4 - (e*x + d)^3*a*d*g^3/e^4 + 3/2*(e*x + d)^2*a*d^2*g
^3/e^4 - (e*x + d)*a*d^3*g^3/e^4

Mupad [B] (verification not implemented)

Time = 0.82 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.36 \[ \int (f+g x)^3 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx=x\,\left (\frac {4\,a\,e\,f^3+12\,a\,d\,f^2\,g-4\,b\,e\,f^3\,n}{4\,e}+\frac {d\,\left (\frac {d\,\left (\frac {g^2\,\left (a\,d\,g+3\,a\,e\,f-b\,e\,f\,n\right )}{e}-\frac {d\,g^3\,\left (4\,a-b\,n\right )}{4\,e}\right )}{e}-\frac {3\,f\,g\,\left (2\,a\,d\,g+2\,a\,e\,f-b\,e\,f\,n\right )}{2\,e}\right )}{e}\right )+x^3\,\left (\frac {g^2\,\left (a\,d\,g+3\,a\,e\,f-b\,e\,f\,n\right )}{3\,e}-\frac {d\,g^3\,\left (4\,a-b\,n\right )}{12\,e}\right )+\ln \left (c\,{\left (d+e\,x\right )}^n\right )\,\left (b\,f^3\,x+\frac {3\,b\,f^2\,g\,x^2}{2}+b\,f\,g^2\,x^3+\frac {b\,g^3\,x^4}{4}\right )-x^2\,\left (\frac {d\,\left (\frac {g^2\,\left (a\,d\,g+3\,a\,e\,f-b\,e\,f\,n\right )}{e}-\frac {d\,g^3\,\left (4\,a-b\,n\right )}{4\,e}\right )}{2\,e}-\frac {3\,f\,g\,\left (2\,a\,d\,g+2\,a\,e\,f-b\,e\,f\,n\right )}{4\,e}\right )-\frac {\ln \left (d+e\,x\right )\,\left (b\,n\,d^4\,g^3-4\,b\,n\,d^3\,e\,f\,g^2+6\,b\,n\,d^2\,e^2\,f^2\,g-4\,b\,n\,d\,e^3\,f^3\right )}{4\,e^4}+\frac {g^3\,x^4\,\left (4\,a-b\,n\right )}{16} \]

[In]

int((f + g*x)^3*(a + b*log(c*(d + e*x)^n)),x)

[Out]

x*((4*a*e*f^3 + 12*a*d*f^2*g - 4*b*e*f^3*n)/(4*e) + (d*((d*((g^2*(a*d*g + 3*a*e*f - b*e*f*n))/e - (d*g^3*(4*a
- b*n))/(4*e)))/e - (3*f*g*(2*a*d*g + 2*a*e*f - b*e*f*n))/(2*e)))/e) + x^3*((g^2*(a*d*g + 3*a*e*f - b*e*f*n))/
(3*e) - (d*g^3*(4*a - b*n))/(12*e)) + log(c*(d + e*x)^n)*((b*g^3*x^4)/4 + b*f^3*x + (3*b*f^2*g*x^2)/2 + b*f*g^
2*x^3) - x^2*((d*((g^2*(a*d*g + 3*a*e*f - b*e*f*n))/e - (d*g^3*(4*a - b*n))/(4*e)))/(2*e) - (3*f*g*(2*a*d*g +
2*a*e*f - b*e*f*n))/(4*e)) - (log(d + e*x)*(b*d^4*g^3*n - 4*b*d*e^3*f^3*n - 4*b*d^3*e*f*g^2*n + 6*b*d^2*e^2*f^
2*g*n))/(4*e^4) + (g^3*x^4*(4*a - b*n))/16