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

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.60, antiderivative size = 347, normalized size of antiderivative = 1.09, number of steps used = 13, number of rules used = 11, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.458, Rules used = {2398, 2411, 2347, 2344, 2301, 2317, 2391, 2314, 31, 2319, 44} \[ -\frac {2 b^2 e^3 n^2 \text {PolyLog}\left (2,-\frac {g (d+e x)}{e f-d g}\right )}{3 g (e f-d g)^3}+\frac {e^3 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (e f-d g)^3}-\frac {2 b e^3 n \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (e f-d g)^3}-\frac {2 b e^2 n (d+e x) \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 (f+g x) (e f-d g)^3}+\frac {b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{3 g (f+g x)^2 (e f-d g)}-\frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3}-\frac {b^2 e^2 n^2}{3 g (f+g x) (e f-d g)^2}-\frac {b^2 e^3 n^2 \log (d+e x)}{3 g (e f-d g)^3}+\frac {b^2 e^3 n^2 \log (f+g x)}{g (e f-d g)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*e^2*n^2)/(3*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)) + (e^3*(a + b*Log[c*(d + e*x)^n])^2)/(3*g*(e*f - d*g)^3) - (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[(e*
(f + g*x))/(e*f - d*g)])/(3*g*(e*f - d*g)^3) - (2*b^2*e^3*n^2*PolyLog[2, -((g*(d + e*x))/(e*f - d*g))])/(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 44

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] && L
tQ[m + n + 2, 0])

Rule 2301

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 2314

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 2317

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

Rule 2319

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 2344

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

Rule 2347

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 2391

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 2398

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 2411

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) \operatorname {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) \operatorname {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) \operatorname {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) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.34, size = 302, normalized size = 0.95 \[ \frac {\frac {e (f+g x) \left (e^2 (f+g x)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2-2 b e^2 n (f+g x)^2 \log \left (\frac {e (f+g x)}{e f-d g}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )+b n (e f-d g)^2 \left (a+b \log \left (c (d+e x)^n\right )\right )+2 b e n (f+g x) (e f-d g) \left (a+b \log \left (c (d+e x)^n\right )\right )-2 b^2 e^2 n^2 (f+g x)^2 \text {Li}_2\left (\frac {g (d+e x)}{d g-e f}\right )-2 b^2 e^2 n^2 (f+g x)^2 (\log (d+e x)-\log (f+g x))-b^2 e n^2 (f+g x) (e (f+g x) \log (d+e x)-d g-e (f+g x) \log (f+g x)+e f)\right )}{(e f-d g)^3}-\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{3 g (f+g x)^3} \]

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)

________________________________________________________________________________________

fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \log \left ({\left (e x + d\right )}^{n} c\right )^{2} + 2 \, a b \log \left ({\left (e x + d\right )}^{n} c\right ) + a^{2}}{g^{4} x^{4} + 4 \, f g^{3} x^{3} + 6 \, f^{2} g^{2} x^{2} + 4 \, f^{3} g x + f^{4}}, x\right ) \]

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((e*x + d)^n*c)^2 + 2*a*b*log((e*x + 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)

________________________________________________________________________________________

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

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

________________________________________________________________________________________

maple [C]  time = 0.54, size = 1815, normalized size = 5.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (\frac {2 \, e^{2} \log \left (e x + d\right )}{e^{3} f^{3} g - 3 \, d e^{2} f^{2} g^{2} + 3 \, d^{2} e f g^{3} - d^{3} g^{4}} - \frac {2 \, e^{2} \log \left (g x + f\right )}{e^{3} f^{3} g - 3 \, d e^{2} f^{2} g^{2} + 3 \, d^{2} e f g^{3} - d^{3} g^{4}} + \frac {2 \, e g x + 3 \, e f - d g}{e^{2} f^{4} g - 2 \, d e f^{3} g^{2} + d^{2} f^{2} g^{3} + {\left (e^{2} f^{2} g^{3} - 2 \, d e f g^{4} + d^{2} g^{5}\right )} x^{2} + 2 \, {\left (e^{2} f^{3} g^{2} - 2 \, d e f^{2} g^{3} + d^{2} f g^{4}\right )} x}\right )} a b e n - \frac {1}{3} \, b^{2} {\left (\frac {\log \left ({\left (e x + d\right )}^{n}\right )^{2}}{g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g} - 3 \, \int \frac {3 \, e g x \log \relax (c)^{2} + 3 \, d g \log \relax (c)^{2} + 2 \, {\left (e f n + 3 \, d g \log \relax (c) + {\left (e g n + 3 \, e g \log \relax (c)\right )} x\right )} \log \left ({\left (e x + d\right )}^{n}\right )}{3 \, {\left (e g^{5} x^{5} + d f^{4} g + {\left (4 \, e f g^{4} + d g^{5}\right )} x^{4} + 2 \, {\left (3 \, e f^{2} g^{3} + 2 \, d f g^{4}\right )} x^{3} + 2 \, {\left (2 \, e f^{3} g^{2} + 3 \, d f^{2} g^{3}\right )} x^{2} + {\left (e f^{4} g + 4 \, d f^{3} g^{2}\right )} x\right )}}\,{d x}\right )} - \frac {2 \, a b \log \left ({\left (e x + d\right )}^{n} c\right )}{3 \, {\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} - \frac {a^{2}}{3 \, {\left (g^{4} x^{3} + 3 \, f g^{3} x^{2} + 3 \, f^{2} g^{2} x + f^{3} g\right )}} \]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________