3.1.51 \(\int \frac {(a+b \log (c (d+e x)^n))^2}{(f+g x)^4} \, dx\) [51]

Optimal. Leaf size=317 \[ -\frac {b^2 e^2 n^2}{3 g (e f-d g)^2 (f+g x)}-\frac {b^2 e^3 n^2 \log (d+e x)}{3 g (e f-d g)^3}+\frac {b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (e f-d g) (f+g x)^2}-\frac {2 b e^2 n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g)^3 (f+g x)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}+\frac {b^2 e^3 n^2 \log (f+g x)}{g (e f-d g)^3}-\frac {2 b e^3 n \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (1+\frac {e f-d g}{g (d+e x)}\right )}{3 g (e f-d g)^3}+\frac {2 b^2 e^3 n^2 \text {Li}_2\left (-\frac {e f-d g}{g (d+e x)}\right )}{3 g (e f-d g)^3} \]

[Out]

-1/3*b^2*e^2*n^2/g/(-d*g+e*f)^2/(g*x+f)-1/3*b^2*e^3*n^2*ln(e*x+d)/g/(-d*g+e*f)^3+1/3*b*e*n*(a+b*ln(c*(e*x+d)^n
))/g/(-d*g+e*f)/(g*x+f)^2-2/3*b*e^2*n*(e*x+d)*(a+b*ln(c*(e*x+d)^n))/(-d*g+e*f)^3/(g*x+f)-1/3*(a+b*ln(c*(e*x+d)
^n))^2/g/(g*x+f)^3+b^2*e^3*n^2*ln(g*x+f)/g/(-d*g+e*f)^3-2/3*b*e^3*n*(a+b*ln(c*(e*x+d)^n))*ln(1+(-d*g+e*f)/g/(e
*x+d))/g/(-d*g+e*f)^3+2/3*b^2*e^3*n^2*polylog(2,(d*g-e*f)/g/(e*x+d))/g/(-d*g+e*f)^3

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Rubi [A]
time = 0.37, antiderivative size = 317, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2445, 2458, 2389, 2379, 2438, 2351, 31, 2356, 46} \begin {gather*} \frac {2 b^2 e^3 n^2 \text {PolyLog}\left (2,-\frac {e f-d g}{g (d+e x)}\right )}{3 g (e f-d g)^3}-\frac {2 b e^3 n \log \left (\frac {e f-d g}{g (d+e x)}+1\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (e f-d g)^3}-\frac {2 b e^2 n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (f+g x) (e f-d g)^3}+\frac {b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (f+g x)^2 (e f-d g)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}-\frac {b^2 e^3 n^2 \log (d+e x)}{3 g (e f-d g)^3}+\frac {b^2 e^3 n^2 \log (f+g x)}{g (e f-d g)^3}-\frac {b^2 e^2 n^2}{3 g (f+g x) (e f-d g)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b^2*e^2*n^2)/(g*(e*f - d*g)^2*(f + g*x)) - (b^2*e^3*n^2*Log[d + e*x])/(3*g*(e*f - d*g)^3) + (b*e*n*(a +
b*Log[c*(d + e*x)^n]))/(3*g*(e*f - d*g)*(f + g*x)^2) - (2*b*e^2*n*(d + e*x)*(a + b*Log[c*(d + e*x)^n]))/(3*(e*
f - d*g)^3*(f + g*x)) - (a + b*Log[c*(d + e*x)^n])^2/(3*g*(f + g*x)^3) + (b^2*e^3*n^2*Log[f + g*x])/(g*(e*f -
d*g)^3) - (2*b*e^3*n*(a + b*Log[c*(d + e*x)^n])*Log[1 + (e*f - d*g)/(g*(d + e*x))])/(3*g*(e*f - d*g)^3) + (2*b
^2*e^3*n^2*PolyLog[2, -((e*f - d*g)/(g*(d + e*x)))])/(3*g*(e*f - d*g)^3)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[x*(d + e*x^r)^(q +
1)*((a + b*Log[c*x^n])/d), x] - Dist[b*(n/d), Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2356

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)
*((a + b*Log[c*x^n])^p/(e*(q + 1))), x] - Dist[b*n*(p/(e*(q + 1))), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2379

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^(r_.))), x_Symbol] :> Simp[(-Log[1 +
d/(e*x^r)])*((a + b*Log[c*x^n])^p/(d*r)), x] + Dist[b*n*(p/(d*r)), Int[Log[1 + d/(e*x^r)]*((a + b*Log[c*x^n])^
(p - 1)/x), x], x] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[p, 0]

Rule 2389

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[(d
+ e*x)^(q + 1)*((a + b*Log[c*x^n])^p/x), x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2445

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[(f
+ g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])^p/(g*(q + 1))), x] - Dist[b*e*n*(p/(g*(q + 1))), Int[(f + g*x)^(q
+ 1)*((a + b*Log[c*(d + e*x)^n])^(p - 1)/(d + e*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^4} \, dx &=-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}+\frac {(2 b e n) \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) (f+g x)^3} \, dx}{3 g}\\ &=-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}+\frac {(2 b n) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{x \left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^3} \, dx,x,d+e x\right )}{3 g}\\ &=-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}-\frac {(2 b n) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{\left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^3} \, dx,x,d+e x\right )}{3 (e f-d g)}+\frac {(2 b e n) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{x \left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^2} \, dx,x,d+e x\right )}{3 g (e f-d g)}\\ &=\frac {b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (e f-d g) (f+g x)^2}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}-\frac {(2 b e n) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{\left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^2} \, dx,x,d+e x\right )}{3 (e f-d g)^2}+\frac {\left (2 b e^2 n\right ) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{x \left (\frac {e f-d g}{e}+\frac {g x}{e}\right )} \, dx,x,d+e x\right )}{3 g (e f-d g)^2}-\frac {\left (b^2 e n^2\right ) \text {Subst}\left (\int \frac {1}{x \left (\frac {e f-d g}{e}+\frac {g x}{e}\right )^2} \, dx,x,d+e x\right )}{3 g (e f-d g)}\\ &=\frac {b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (e f-d g) (f+g x)^2}-\frac {2 b e^2 n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g)^3 (f+g x)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}-\frac {\left (2 b e^2 n\right ) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{\frac {e f-d g}{e}+\frac {g x}{e}} \, dx,x,d+e x\right )}{3 (e f-d g)^3}+\frac {\left (2 b e^3 n\right ) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{x} \, dx,x,d+e x\right )}{3 g (e f-d g)^3}+\frac {\left (2 b^2 e^2 n^2\right ) \text {Subst}\left (\int \frac {1}{\frac {e f-d g}{e}+\frac {g x}{e}} \, dx,x,d+e x\right )}{3 (e f-d g)^3}-\frac {\left (b^2 e n^2\right ) \text {Subst}\left (\int \left (\frac {e^2}{(e f-d g)^2 x}-\frac {e^2 g}{(e f-d g) (e f-d g+g x)^2}-\frac {e^2 g}{(e f-d g)^2 (e f-d g+g x)}\right ) \, dx,x,d+e x\right )}{3 g (e f-d g)}\\ &=-\frac {b^2 e^2 n^2}{3 g (e f-d g)^2 (f+g x)}-\frac {b^2 e^3 n^2 \log (d+e x)}{3 g (e f-d g)^3}+\frac {b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (e f-d g) (f+g x)^2}-\frac {2 b e^2 n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g)^3 (f+g x)}+\frac {e^3 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (e f-d g)^3}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}+\frac {b^2 e^3 n^2 \log (f+g x)}{g (e f-d g)^3}-\frac {2 b e^3 n \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{3 g (e f-d g)^3}+\frac {\left (2 b^2 e^3 n^2\right ) \text {Subst}\left (\int \frac {\log \left (1+\frac {g x}{e f-d g}\right )}{x} \, dx,x,d+e x\right )}{3 g (e f-d g)^3}\\ &=-\frac {b^2 e^2 n^2}{3 g (e f-d g)^2 (f+g x)}-\frac {b^2 e^3 n^2 \log (d+e x)}{3 g (e f-d g)^3}+\frac {b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (e f-d g) (f+g x)^2}-\frac {2 b e^2 n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (e f-d g)^3 (f+g x)}+\frac {e^3 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (e f-d g)^3}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}+\frac {b^2 e^3 n^2 \log (f+g x)}{g (e f-d g)^3}-\frac {2 b e^3 n \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )}{3 g (e f-d g)^3}-\frac {2 b^2 e^3 n^2 \text {Li}_2\left (-\frac {g (d+e x)}{e f-d g}\right )}{3 g (e f-d g)^3}\\ \end {align*}

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Mathematica [A]
time = 0.24, size = 302, normalized size = 0.95 \begin {gather*} \frac {-\left (a+b \log \left (c (d+e x)^n\right )\right )^2+\frac {e (f+g x) \left (b (e f-d g)^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )+2 b e (e f-d g) n (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )+e^2 (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2-2 b^2 e^2 n^2 (f+g x)^2 (\log (d+e x)-\log (f+g x))-b^2 e n^2 (f+g x) (e f-d g+e (f+g x) \log (d+e x)-e (f+g x) \log (f+g x))-2 b e^2 n (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e (f+g x)}{e f-d g}\right )-2 b^2 e^2 n^2 (f+g x)^2 \text {Li}_2\left (\frac {g (d+e x)}{-e f+d g}\right )\right )}{(e f-d g)^3}}{3 g (f+g x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.47, size = 1815, normalized size = 5.73

method	result	size
risch	\(\text {Expression too large to display}\)	\(1815\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(e*x+d)^n))^2/(g*x+f)^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*I/(g*x+f)^3/g*ln((e*x+d)^n)*Pi*b^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-2/3*b/g*n*e^3/(d*g-e*f)^3*ln(e*x+d)*a+
2/3/g*n*e^3/(d*g-e*f)^3*ln(g*x+f)*b^2*ln(c)-1/3/g*n*e/(d*g-e*f)/(g*x+f)^2*b^2*ln(c)+2/3/g*n*e^2/(d*g-e*f)^2/(g
*x+f)*b^2*ln(c)-2/3/g*n*e^3/(d*g-e*f)^3*ln(e*x+d)*b^2*ln(c)-1/3*I/g*n*e^3/(d*g-e*f)^3*ln(g*x+f)*Pi*b^2*csgn(I*
c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/6*I/g*n*e/(d*g-e*f)/(g*x+f)^2*Pi*b^2*csgn(I*c)*csgn(I*(e*x+d)^n)*cs
gn(I*c*(e*x+d)^n)+1/6*I/g*n*e/(d*g-e*f)/(g*x+f)^2*Pi*b^2*csgn(I*c*(e*x+d)^n)^3-1/3*I/(g*x+f)^3/g*ln((e*x+d)^n)
*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-2/3*b^2/g*n^2*e^3/(d*g-e*f)^3*ln(g*x+f)*ln(((g*x+f)*e+d*g-e*f)
/(d*g-e*f))-1/6*I/g*n*e/(d*g-e*f)/(g*x+f)^2*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-2/3/(g*x+f)^3/g*ln(
(e*x+d)^n)*b^2*ln(c)-1/3*b^2/(g*x+f)^3/g*ln((e*x+d)^n)^2-2/3*b/(g*x+f)^3/g*ln((e*x+d)^n)*a-1/3*I/g*n*e^2/(d*g-
e*f)^2/(g*x+f)*Pi*b^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/3*I/g*n*e^3/(d*g-e*f)^3*ln(g*x+f)*Pi*b
^2*csgn(I*c*(e*x+d)^n)^3+1/3*I/(g*x+f)^3/g*ln((e*x+d)^n)*Pi*b^2*csgn(I*c*(e*x+d)^n)^3-1/3*b^2/g*n*e*ln((e*x+d)
^n)/(d*g-e*f)/(g*x+f)^2+2/3*b^2/g*n*e^3*ln((e*x+d)^n)/(d*g-e*f)^3*ln(g*x+f)+2/3*b^2/g*n*e^2*ln((e*x+d)^n)/(d*g
-e*f)^2/(g*x+f)-2/3*b^2/g*n*e^3*ln((e*x+d)^n)/(d*g-e*f)^3*ln(e*x+d)-1/3*b/g*n*e/(d*g-e*f)/(g*x+f)^2*a+2/3*b/g*
n*e^2/(d*g-e*f)^2/(g*x+f)*a+2/3*b/g*n*e^3/(d*g-e*f)^3*ln(g*x+f)*a+1/3*I/g*n*e^3/(d*g-e*f)^3*ln(e*x+d)*Pi*b^2*c
sgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+1/3*I/g*n*e^3/(d*g-e*f)^3*ln(e*x+d)*Pi*b^2*csgn(I*c*(e*x+d)^n)^
3+1/3*I/(g*x+f)^3/g*ln((e*x+d)^n)*Pi*b^2*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-1/3*I/g*n*e^2/(d*g-e*
f)^2/(g*x+f)*Pi*b^2*csgn(I*c*(e*x+d)^n)^3-1/3*b^2/g*n^2*e^2/(d*g-e*f)^2/(g*x+f)-b^2/g*n^2*e^3/(d*g-e*f)^3*ln(g
*x+f)+b^2/g*n^2*e^3/(d*g-e*f)^3*ln(e*x+d)+1/3*b^2/g*n^2*e^3/(d*g-e*f)^3*ln(e*x+d)^2-2/3*b^2/g*n^2*e^3/(d*g-e*f
)^3*dilog(((g*x+f)*e+d*g-e*f)/(d*g-e*f))+1/3*I/g*n*e^3/(d*g-e*f)^3*ln(g*x+f)*Pi*b^2*csgn(I*c)*csgn(I*c*(e*x+d)
^n)^2-1/12*(-I*b*Pi*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*b*Pi*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+I*b
*Pi*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-I*b*Pi*csgn(I*c*(e*x+d)^n)^3+2*b*ln(c)+2*a)^2/(g*x+f)^3/g-1/3*I/g*
n*e^3/(d*g-e*f)^3*ln(e*x+d)*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/6*I/g*n*e/(d*g-e*f)/(g*x+f)^2*Pi*
b^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/3*I/g*n*e^3/(d*g-e*f)^3*ln(g*x+f)*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x
+d)^n)^2+1/3*I/g*n*e^2/(d*g-e*f)^2/(g*x+f)*Pi*b^2*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2+1/3*I/g*n*e^2/(d*g-e*f)^2/(g
*x+f)*Pi*b^2*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2-1/3*I/g*n*e^3/(d*g-e*f)^3*ln(e*x+d)*Pi*b^2*csgn(I*c)*csgn
(I*c*(e*x+d)^n)^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^4,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^4,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \log {\left (c \left (d + e x\right )^{n} \right )}\right )^{2}}{\left (f + g x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(e*x+d)**n))**2/(g*x+f)**4,x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^4,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}{{\left (f+g\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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