3.4.80 \(\int x (a+b \log (c (d+e x)^n)) (f+g \log (c (d+e x)^n)) \, dx\) [380]

Optimal. Leaf size=196 \[ -\frac {2 b d g n^2 x}{e}+\frac {b g n^2 (d+e x)^2}{4 e^2}+\frac {b d^2 g n^2 \log ^2(d+e x)}{2 e^2}+\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {d n (d+e x) \left (b f+a g+2 b g \log \left (c (d+e x)^n\right )\right )}{e^2}-\frac {n (d+e x)^2 \left (b f+a g+2 b g \log \left (c (d+e x)^n\right )\right )}{4 e^2}-\frac {d^2 n \log (d+e x) \left (b f+a g+2 b g \log \left (c (d+e x)^n\right )\right )}{2 e^2} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2483, 2458, 45, 2372, 12, 14, 2338} \begin {gather*} -\frac {d^2 n \log (d+e x) \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{2 e^2}+\frac {d n (d+e x) \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{e^2}-\frac {n (d+e x)^2 \left (a g+2 b g \log \left (c (d+e x)^n\right )+b f\right )}{4 e^2}+\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (g \log \left (c (d+e x)^n\right )+f\right )+\frac {b d^2 g n^2 \log ^2(d+e x)}{2 e^2}+\frac {b g n^2 (d+e x)^2}{4 e^2}-\frac {2 b d g n^2 x}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2372

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]]
/; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2483

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(g_.))*
(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*(a + b*Log[c*(d + e*x)^n])*((f + g*Log[c*(d + e*x)^n])/(m + 1)), x] -
Dist[e*(n/(m + 1)), Int[(x^(m + 1)*(b*f + a*g + 2*b*g*Log[c*(d + e*x)^n]))/(d + e*x), x], x] /; FreeQ[{a, b, c
, d, e, f, g, n, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right ) \, dx &=\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} (b e n) \int \frac {x^2 \left (f+g \log \left (c (d+e x)^n\right )\right )}{d+e x} \, dx-\frac {1}{2} (e g n) \int \frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{d+e x} \, dx\\ &=\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )-\frac {1}{2} (b n) \text {Subst}\left (\int \frac {\left (-\frac {d}{e}+\frac {x}{e}\right )^2 \left (f+g \log \left (c x^n\right )\right )}{x} \, dx,x,d+e x\right )-\frac {1}{2} (g n) \text {Subst}\left (\int \frac {\left (-\frac {d}{e}+\frac {x}{e}\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x} \, dx,x,d+e x\right )\\ &=\frac {1}{4} g n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+2 \left (\frac {1}{2} \left (b g n^2\right ) \text {Subst}\left (\int \frac {x (-4 d+x)+2 d^2 \log (x)}{2 e^2 x} \, dx,x,d+e x\right )\right )\\ &=\frac {1}{4} g n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+2 \frac {\left (b g n^2\right ) \text {Subst}\left (\int \frac {x (-4 d+x)+2 d^2 \log (x)}{x} \, dx,x,d+e x\right )}{4 e^2}\\ &=\frac {1}{4} g n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+2 \frac {\left (b g n^2\right ) \text {Subst}\left (\int \left (-4 d+x+\frac {2 d^2 \log (x)}{x}\right ) \, dx,x,d+e x\right )}{4 e^2}\\ &=\frac {1}{4} g n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+2 \left (-\frac {b d g n^2 x}{e}+\frac {b g n^2 (d+e x)^2}{8 e^2}+\frac {\left (b d^2 g n^2\right ) \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,d+e x\right )}{2 e^2}\right )\\ &=2 \left (-\frac {b d g n^2 x}{e}+\frac {b g n^2 (d+e x)^2}{8 e^2}+\frac {b d^2 g n^2 \log ^2(d+e x)}{4 e^2}\right )+\frac {1}{4} g n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+\frac {1}{4} b n \left (\frac {4 d (d+e x)}{e^2}-\frac {(d+e x)^2}{e^2}-\frac {2 d^2 \log (d+e x)}{e^2}\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )+\frac {1}{2} x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \left (f+g \log \left (c (d+e x)^n\right )\right )\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 263, normalized size = 1.34 \begin {gather*} \frac {b d f n x}{2 e}+\frac {a d g n x}{2 e}-\frac {3 b d g n^2 x}{2 e}+\frac {1}{2} a f x^2-\frac {1}{4} b f n x^2-\frac {1}{4} a g n x^2+\frac {1}{4} b g n^2 x^2-\frac {b d^2 f n \log (d+e x)}{2 e^2}-\frac {a d^2 g n \log (d+e x)}{2 e^2}+\frac {3 b d^2 g n \log \left (c (d+e x)^n\right )}{2 e^2}+\frac {b d g n x \log \left (c (d+e x)^n\right )}{e}+\frac {1}{2} b f x^2 \log \left (c (d+e x)^n\right )+\frac {1}{2} a g x^2 \log \left (c (d+e x)^n\right )-\frac {1}{2} b g n x^2 \log \left (c (d+e x)^n\right )-\frac {b d^2 g \log ^2\left (c (d+e x)^n\right )}{2 e^2}+\frac {1}{2} b g x^2 \log ^2\left (c (d+e x)^n\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.55, size = 1558, normalized size = 7.95

method	result	size
risch	\(\text {Expression too large to display}\)	\(1558\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*b*d*g*n^2*x/e+1/4*x^2*b*g*n^2+1/2*x^2*a*f-1/4*n*b*f*x^2+1/2*I/e^2*Pi*ln(e*x+d)*b*d^2*g*n*csgn(I*c)*csgn(I
*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/4*n*a*g*x^2+1/2*ln(c)*a*g*x^2+1/2*ln(c)*b*f*x^2+1/2*ln(c)^2*b*g*x^2+1/2*x^2*
b*g*ln((e*x+d)^n)^2-1/8*Pi^2*b*g*x^2*csgn(I*c)^2*csgn(I*c*(e*x+d)^n)^4-1/2*I/e*Pi*b*d*g*n*x*csgn(I*c)*csgn(I*(
e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/2*I*Pi*ln(c)*b*g*x^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/4*I*n*P
i*b*g*x^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I/e^2*Pi*ln(e*x+d)*b*d^2*g*n*csgn(I*c*(e*x+d)^n)
^3-1/2*I/e*Pi*b*d*g*n*x*csgn(I*c*(e*x+d)^n)^3+1/2*I/e*Pi*b*d*g*n*x*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/2*I/e^2*P
i*ln(e*x+d)*b*d^2*g*n*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/4*Pi^2*b*g*x^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^5
-1/8*Pi^2*b*g*x^2*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^4+1/4*Pi^2*b*g*x^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d
)^n)^5-1/2*I/e^2*Pi*ln(e*x+d)*b*d^2*g*n*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/2*n*ln(c)*b*g*x^2-1/8*Pi^2*b*g*x^2*c
sgn(I*c*(e*x+d)^n)^6+1/4*Pi^2*b*g*x^2*csgn(I*c)^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^3+1/4*Pi^2*b*g*x^2*csg
n(I*c)*csgn(I*(e*x+d)^n)^2*csgn(I*c*(e*x+d)^n)^3-1/2*Pi^2*b*g*x^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)
^n)^4-1/2*I*Pi*ln(c)*b*g*x^2*csgn(I*c*(e*x+d)^n)^3+1/4*I*n*Pi*b*g*x^2*csgn(I*c*(e*x+d)^n)^3+1/4*I*Pi*a*g*x^2*c
sgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/4*I*Pi*b*f*x^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/4*I*Pi*a*g*x^2*csgn(I*(e*x+d
)^n)*csgn(I*c*(e*x+d)^n)^2+1/4*I*Pi*b*f*x^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/2*n*b*d^2*f/e^2*ln(e*x+d
)-1/e^2*ln(c)*ln(e*x+d)*b*d^2*g*n+1/e*ln(c)*b*d*g*n*x-1/8*Pi^2*b*g*x^2*csgn(I*c)^2*csgn(I*(e*x+d)^n)^2*csgn(I*
c*(e*x+d)^n)^2+1/2*I*Pi*ln(c)*b*g*x^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-1/4*I*n*Pi*b*g*x^2*csgn(I*c)*csgn(I*c*(e
*x+d)^n)^2+1/2*I*Pi*ln(c)*b*g*x^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/4*I*n*Pi*b*g*x^2*csgn(I*(e*x+d)^n)
*csgn(I*c*(e*x+d)^n)^2-1/4*I*Pi*a*g*x^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/4*I*Pi*b*f*x^2*csgn(
I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/2*I/e*Pi*b*d*g*n*x*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+1/2*(-
I*Pi*b*e^2*g*x^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*Pi*b*e^2*g*x^2*csgn(I*c)*csgn(I*c*(e*x+d)^n
)^2+I*Pi*b*e^2*g*x^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*Pi*b*e^2*g*x^2*csgn(I*c*(e*x+d)^n)^3+2*ln(c)*b*
e^2*g*x^2-b*e^2*g*n*x^2+a*e^2*g*x^2-2*ln(e*x+d)*b*d^2*g*n+2*b*d*e*g*n*x+b*e^2*f*x^2)/e^2*ln((e*x+d)^n)+3/2*b*d
^2*g*n^2/e^2*ln(e*x+d)-1/2*n*a*d^2*g/e^2*ln(e*x+d)+1/2*d*n*a*g/e*x-1/4*I*Pi*a*g*x^2*csgn(I*c*(e*x+d)^n)^3-1/4*
I*Pi*b*f*x^2*csgn(I*c*(e*x+d)^n)^3+1/2*b*d^2*g*n^2*ln(e*x+d)^2/e^2+1/2*b*d*f*n*x/e

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Maxima [A]
time = 0.30, size = 231, normalized size = 1.18 \begin {gather*} \frac {1}{2} \, b g x^{2} \log \left ({\left (x e + d\right )}^{n} c\right )^{2} - \frac {1}{4} \, {\left (2 \, d^{2} e^{\left (-3\right )} \log \left (x e + d\right ) + {\left (x^{2} e - 2 \, d x\right )} e^{\left (-2\right )}\right )} b f n e - \frac {1}{4} \, {\left (2 \, d^{2} e^{\left (-3\right )} \log \left (x e + d\right ) + {\left (x^{2} e - 2 \, d x\right )} e^{\left (-2\right )}\right )} a g n e + \frac {1}{2} \, b f x^{2} \log \left ({\left (x e + d\right )}^{n} c\right ) + \frac {1}{2} \, a g x^{2} \log \left ({\left (x e + d\right )}^{n} c\right ) + \frac {1}{2} \, a f x^{2} + \frac {1}{4} \, {\left ({\left (2 \, d^{2} \log \left (x e + d\right )^{2} + x^{2} e^{2} - 6 \, d x e + 6 \, d^{2} \log \left (x e + d\right )\right )} n^{2} e^{\left (-2\right )} - 2 \, {\left (2 \, d^{2} e^{\left (-3\right )} \log \left (x e + d\right ) + {\left (x^{2} e - 2 \, d x\right )} e^{\left (-2\right )}\right )} n e \log \left ({\left (x e + d\right )}^{n} c\right )\right )} b g \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 231, normalized size = 1.18 \begin {gather*} \frac {1}{4} \, {\left (2 \, b g x^{2} e^{2} \log \left (c\right )^{2} + {\left (b g n^{2} + 2 \, a f - {\left (b f + a g\right )} n\right )} x^{2} e^{2} - 2 \, {\left (3 \, b d g n^{2} - {\left (b d f + a d g\right )} n\right )} x e + 2 \, {\left (b g n^{2} x^{2} e^{2} - b d^{2} g n^{2}\right )} \log \left (x e + d\right )^{2} + 2 \, {\left (2 \, b d g n^{2} x e + 3 \, b d^{2} g n^{2} - {\left (b g n^{2} - {\left (b f + a g\right )} n\right )} x^{2} e^{2} - {\left (b d^{2} f + a d^{2} g\right )} n + 2 \, {\left (b g n x^{2} e^{2} - b d^{2} g n\right )} \log \left (c\right )\right )} \log \left (x e + d\right ) + 2 \, {\left (2 \, b d g n x e - {\left (b g n - b f - a g\right )} x^{2} e^{2}\right )} \log \left (c\right )\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.64, size = 296, normalized size = 1.51 \begin {gather*} \begin {cases} - \frac {a d^{2} g \log {\left (c \left (d + e x\right )^{n} \right )}}{2 e^{2}} + \frac {a d g n x}{2 e} + \frac {a f x^{2}}{2} - \frac {a g n x^{2}}{4} + \frac {a g x^{2} \log {\left (c \left (d + e x\right )^{n} \right )}}{2} - \frac {b d^{2} f \log {\left (c \left (d + e x\right )^{n} \right )}}{2 e^{2}} + \frac {3 b d^{2} g n \log {\left (c \left (d + e x\right )^{n} \right )}}{2 e^{2}} - \frac {b d^{2} g \log {\left (c \left (d + e x\right )^{n} \right )}^{2}}{2 e^{2}} + \frac {b d f n x}{2 e} - \frac {3 b d g n^{2} x}{2 e} + \frac {b d g n x \log {\left (c \left (d + e x\right )^{n} \right )}}{e} - \frac {b f n x^{2}}{4} + \frac {b f x^{2} \log {\left (c \left (d + e x\right )^{n} \right )}}{2} + \frac {b g n^{2} x^{2}}{4} - \frac {b g n x^{2} \log {\left (c \left (d + e x\right )^{n} \right )}}{2} + \frac {b g x^{2} \log {\left (c \left (d + e x\right )^{n} \right )}^{2}}{2} & \text {for}\: e \neq 0 \\\frac {x^{2} \left (a + b \log {\left (c d^{n} \right )}\right ) \left (f + g \log {\left (c d^{n} \right )}\right )}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 477 vs. \(2 (190) = 380\).
time = 4.10, size = 477, normalized size = 2.43 \begin {gather*} \frac {1}{2} \, {\left (x e + d\right )}^{2} b g n^{2} e^{\left (-2\right )} \log \left (x e + d\right )^{2} - {\left (x e + d\right )} b d g n^{2} e^{\left (-2\right )} \log \left (x e + d\right )^{2} - \frac {1}{2} \, {\left (x e + d\right )}^{2} b g n^{2} e^{\left (-2\right )} \log \left (x e + d\right ) + 2 \, {\left (x e + d\right )} b d g n^{2} e^{\left (-2\right )} \log \left (x e + d\right ) + {\left (x e + d\right )}^{2} b g n e^{\left (-2\right )} \log \left (x e + d\right ) \log \left (c\right ) - 2 \, {\left (x e + d\right )} b d g n e^{\left (-2\right )} \log \left (x e + d\right ) \log \left (c\right ) + \frac {1}{4} \, {\left (x e + d\right )}^{2} b g n^{2} e^{\left (-2\right )} - 2 \, {\left (x e + d\right )} b d g n^{2} e^{\left (-2\right )} + \frac {1}{2} \, {\left (x e + d\right )}^{2} b f n e^{\left (-2\right )} \log \left (x e + d\right ) - {\left (x e + d\right )} b d f n e^{\left (-2\right )} \log \left (x e + d\right ) + \frac {1}{2} \, {\left (x e + d\right )}^{2} a g n e^{\left (-2\right )} \log \left (x e + d\right ) - {\left (x e + d\right )} a d g n e^{\left (-2\right )} \log \left (x e + d\right ) - \frac {1}{2} \, {\left (x e + d\right )}^{2} b g n e^{\left (-2\right )} \log \left (c\right ) + 2 \, {\left (x e + d\right )} b d g n e^{\left (-2\right )} \log \left (c\right ) + \frac {1}{2} \, {\left (x e + d\right )}^{2} b g e^{\left (-2\right )} \log \left (c\right )^{2} - {\left (x e + d\right )} b d g e^{\left (-2\right )} \log \left (c\right )^{2} - \frac {1}{4} \, {\left (x e + d\right )}^{2} b f n e^{\left (-2\right )} + {\left (x e + d\right )} b d f n e^{\left (-2\right )} - \frac {1}{4} \, {\left (x e + d\right )}^{2} a g n e^{\left (-2\right )} + {\left (x e + d\right )} a d g n e^{\left (-2\right )} + \frac {1}{2} \, {\left (x e + d\right )}^{2} b f e^{\left (-2\right )} \log \left (c\right ) - {\left (x e + d\right )} b d f e^{\left (-2\right )} \log \left (c\right ) + \frac {1}{2} \, {\left (x e + d\right )}^{2} a g e^{\left (-2\right )} \log \left (c\right ) - {\left (x e + d\right )} a d g e^{\left (-2\right )} \log \left (c\right ) + \frac {1}{2} \, {\left (x e + d\right )}^{2} a f e^{\left (-2\right )} - {\left (x e + d\right )} a d f e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.35, size = 203, normalized size = 1.04 \begin {gather*} x\,\left (\frac {d\,\left (a\,f-b\,g\,n^2\right )}{e}-\frac {d\,\left (a\,f-\frac {a\,g\,n}{2}-\frac {b\,f\,n}{2}+\frac {b\,g\,n^2}{2}\right )}{e}\right )+\ln \left (c\,{\left (d+e\,x\right )}^n\right )\,\left (\left (\frac {a\,g}{2}+\frac {b\,f}{2}-\frac {b\,g\,n}{2}\right )\,x^2+\left (\frac {d\,\left (a\,g+b\,f\right )}{e}-\frac {d\,\left (a\,g+b\,f-b\,g\,n\right )}{e}\right )\,x\right )+{\ln \left (c\,{\left (d+e\,x\right )}^n\right )}^2\,\left (\frac {b\,g\,x^2}{2}-\frac {b\,d^2\,g}{2\,e^2}\right )+x^2\,\left (\frac {a\,f}{2}-\frac {a\,g\,n}{4}-\frac {b\,f\,n}{4}+\frac {b\,g\,n^2}{4}\right )-\frac {\ln \left (d+e\,x\right )\,\left (a\,d^2\,g\,n+b\,d^2\,f\,n-3\,b\,d^2\,g\,n^2\right )}{2\,e^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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