\(\int \frac {(a+b \log (c (d+e x)^n))^2}{(f+g x)^{7/2}} \, dx\) [150]

Optimal result

Integrand size = 26, antiderivative size = 503 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{7/2}} \, dx=-\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {64 b^2 e^{5/2} n^2 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{15 g (e f-d g)^{5/2}}+\frac {8 b^2 e^{5/2} n^2 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{5 g (e f-d g)^{5/2}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}-\frac {16 b^2 e^{5/2} n^2 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}}-\frac {8 b^2 e^{5/2} n^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}} \]

[Out]

64/15*b^2*e^(5/2)*n^2*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))/g/(-d*g+e*f)^(5/2)+8/5*b^2*e^(5/2)*n^2*a
rctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))^2/g/(-d*g+e*f)^(5/2)+8/15*b*e*n*(a+b*ln(c*(e*x+d)^n))/g/(-d*g+e
*f)/(g*x+f)^(3/2)-8/5*b*e^(5/2)*n*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2))*(a+b*ln(c*(e*x+d)^n))/g/(-d*
g+e*f)^(5/2)-2/5*(a+b*ln(c*(e*x+d)^n))^2/g/(g*x+f)^(5/2)-16/5*b^2*e^(5/2)*n^2*arctanh(e^(1/2)*(g*x+f)^(1/2)/(-
d*g+e*f)^(1/2))*ln(2/(1-e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2)))/g/(-d*g+e*f)^(5/2)-8/5*b^2*e^(5/2)*n^2*polylo
g(2,1-2/(1-e^(1/2)*(g*x+f)^(1/2)/(-d*g+e*f)^(1/2)))/g/(-d*g+e*f)^(5/2)-16/15*b^2*e^2*n^2/g/(-d*g+e*f)^2/(g*x+f
)^(1/2)+8/5*b*e^2*n*(a+b*ln(c*(e*x+d)^n))/g/(-d*g+e*f)^2/(g*x+f)^(1/2)

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 503, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {2445, 2458, 2389, 65, 214, 2390, 12, 1601, 6873, 6131, 6055, 2449, 2352, 2356, 53} \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{7/2}} \, dx=-\frac {8 b e^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g \sqrt {f+g x} (e f-d g)^2}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (f+g x)^{3/2} (e f-d g)}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}+\frac {8 b^2 e^{5/2} n^2 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{5 g (e f-d g)^{5/2}}+\frac {64 b^2 e^{5/2} n^2 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{15 g (e f-d g)^{5/2}}-\frac {16 b^2 e^{5/2} n^2 \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}}-\frac {8 b^2 e^{5/2} n^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}}-\frac {16 b^2 e^2 n^2}{15 g \sqrt {f+g x} (e f-d g)^2} \]

[In]

Int[(a + b*Log[c*(d + e*x)^n])^2/(f + g*x)^(7/2),x]

[Out]

(-16*b^2*e^2*n^2)/(15*g*(e*f - d*g)^2*Sqrt[f + g*x]) + (64*b^2*e^(5/2)*n^2*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqr
t[e*f - d*g]])/(15*g*(e*f - d*g)^(5/2)) + (8*b^2*e^(5/2)*n^2*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]^
2)/(5*g*(e*f - d*g)^(5/2)) + (8*b*e*n*(a + b*Log[c*(d + e*x)^n]))/(15*g*(e*f - d*g)*(f + g*x)^(3/2)) + (8*b*e^
2*n*(a + b*Log[c*(d + e*x)^n]))/(5*g*(e*f - d*g)^2*Sqrt[f + g*x]) - (8*b*e^(5/2)*n*ArcTanh[(Sqrt[e]*Sqrt[f + g
*x])/Sqrt[e*f - d*g]]*(a + b*Log[c*(d + e*x)^n]))/(5*g*(e*f - d*g)^(5/2)) - (2*(a + b*Log[c*(d + e*x)^n])^2)/(
5*g*(f + g*x)^(5/2)) - (16*b^2*e^(5/2)*n^2*ArcTanh[(Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g]]*Log[2/(1 - (Sqrt[e
]*Sqrt[f + g*x])/Sqrt[e*f - d*g])])/(5*g*(e*f - d*g)^(5/2)) - (8*b^2*e^(5/2)*n^2*PolyLog[2, 1 - 2/(1 - (Sqrt[e
]*Sqrt[f + g*x])/Sqrt[e*f - d*g])])/(5*g*(e*f - d*g)^(5/2))

Rule 12

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

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 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 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 2390

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.))/(x_), x_Symbol] :> With[{u = IntHi
de[(d + e*x^r)^q/x, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[Dist[1/x, u, x], x], x]] /; FreeQ[{a, b
, c, d, e, n, r}, x] && IntegerQ[q - 1/2]

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 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

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 6055

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)
*(Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 - c^
2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0]

Rule 6131

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}+\frac {(4 b e n) \int \frac {a+b \log \left (c (d+e x)^n\right )}{(d+e x) (f+g x)^{5/2}} \, dx}{5 g} \\ & = -\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}+\frac {(4 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 )^{5/2}} \, dx,x,d+e x\right )}{5 g} \\ & = -\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}-\frac {(4 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 )^{5/2}} \, dx,x,d+e x\right )}{5 (e f-d g)}+\frac {(4 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 )^{3/2}} \, dx,x,d+e x\right )}{5 g (e f-d g)} \\ & = \frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}-\frac {(4 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 )^{3/2}} \, dx,x,d+e x\right )}{5 (e f-d g)^2}+\frac {\left (4 b e^2 n\right ) \text {Subst}\left (\int \frac {a+b \log \left (c x^n\right )}{x \sqrt {\frac {e f-d g}{e}+\frac {g x}{e}}} \, dx,x,d+e x\right )}{5 g (e f-d g)^2}-\frac {\left (8 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 )^{3/2}} \, dx,x,d+e x\right )}{15 g (e f-d g)} \\ & = -\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}-\frac {\left (8 b^2 e^2 n^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {\frac {e f-d g}{e}+\frac {g x}{e}}} \, dx,x,d+e x\right )}{15 g (e f-d g)^2}-\frac {\left (4 b^2 e^2 n^2\right ) \text {Subst}\left (\int -\frac {2 \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f-\frac {d g}{e}+\frac {g x}{e}}}{\sqrt {e f-d g}}\right )}{\sqrt {e f-d g} x} \, dx,x,d+e x\right )}{5 g (e f-d g)^2}-\frac {\left (8 b^2 e^2 n^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {\frac {e f-d g}{e}+\frac {g x}{e}}} \, dx,x,d+e x\right )}{5 g (e f-d g)^2} \\ & = -\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}+\frac {\left (8 b^2 e^{5/2} n^2\right ) \text {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f-\frac {d g}{e}+\frac {g x}{e}}}{\sqrt {e f-d g}}\right )}{x} \, dx,x,d+e x\right )}{5 g (e f-d g)^{5/2}}-\frac {\left (16 b^2 e^3 n^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {e f-d g}{g}+\frac {e x^2}{g}} \, dx,x,\sqrt {f+g x}\right )}{15 g^2 (e f-d g)^2}-\frac {\left (16 b^2 e^3 n^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {e f-d g}{g}+\frac {e x^2}{g}} \, dx,x,\sqrt {f+g x}\right )}{5 g^2 (e f-d g)^2} \\ & = -\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {64 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{15 g (e f-d g)^{5/2}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}+\frac {\left (16 b^2 e^{7/2} n^2\right ) \text {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {e f-d g}}\right )}{d g+e \left (-f+x^2\right )} \, dx,x,\sqrt {f+g x}\right )}{5 g (e f-d g)^{5/2}} \\ & = -\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {64 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{15 g (e f-d g)^{5/2}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}+\frac {\left (16 b^2 e^{7/2} n^2\right ) \text {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {e f-d g}}\right )}{-e f+d g+e x^2} \, dx,x,\sqrt {f+g x}\right )}{5 g (e f-d g)^{5/2}} \\ & = -\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {64 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{15 g (e f-d g)^{5/2}}+\frac {8 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{5 g (e f-d g)^{5/2}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}-\frac {\left (16 b^2 e^3 n^2\right ) \text {Subst}\left (\int \frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {e f-d g}}\right )}{1-\frac {\sqrt {e} x}{\sqrt {e f-d g}}} \, dx,x,\sqrt {f+g x}\right )}{5 g (e f-d g)^3} \\ & = -\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {64 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{15 g (e f-d g)^{5/2}}+\frac {8 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{5 g (e f-d g)^{5/2}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}-\frac {16 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}}+\frac {\left (16 b^2 e^3 n^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {2}{1-\frac {\sqrt {e} x}{\sqrt {e f-d g}}}\right )}{1-\frac {e x^2}{e f-d g}} \, dx,x,\sqrt {f+g x}\right )}{5 g (e f-d g)^3} \\ & = -\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {64 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{15 g (e f-d g)^{5/2}}+\frac {8 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{5 g (e f-d g)^{5/2}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}-\frac {16 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}}-\frac {\left (16 b^2 e^{5/2} n^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}} \\ & = -\frac {16 b^2 e^2 n^2}{15 g (e f-d g)^2 \sqrt {f+g x}}+\frac {64 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )}{15 g (e f-d g)^{5/2}}+\frac {8 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )^2}{5 g (e f-d g)^{5/2}}+\frac {8 b e n \left (a+b \log \left (c (d+e x)^n\right )\right )}{15 g (e f-d g) (f+g x)^{3/2}}+\frac {8 b e^2 n \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^2 \sqrt {f+g x}}-\frac {8 b e^{5/2} n \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{5 g (e f-d g)^{5/2}}-\frac {2 \left (a+b \log \left (c (d+e x)^n\right )\right )^2}{5 g (f+g x)^{5/2}}-\frac {16 b^2 e^{5/2} n^2 \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}}-\frac {8 b^2 e^{5/2} n^2 \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}}\right )}{5 g (e f-d g)^{5/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.22 (sec) , antiderivative size = 639, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{7/2}} \, dx=\frac {2 \left (-3 \left (a+b \log \left (c (d+e x)^n\right )\right )^2+\frac {b e n (f+g x) \left (24 b e^{3/2} n (f+g x)^{3/2} \text {arctanh}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )-8 b e \sqrt {e f-d g} n (f+g x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {e (f+g x)}{e f-d g}\right )+4 (e f-d g)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right )+12 e \sqrt {e f-d g} (f+g x) \left (a+b \log \left (c (d+e x)^n\right )\right )+6 e^{3/2} (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right )-6 e^{3/2} (f+g x)^{3/2} \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right )-3 b e^{3/2} n (f+g x)^{3/2} \left (\log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right ) \left (\log \left (\sqrt {e f-d g}-\sqrt {e} \sqrt {f+g x}\right )+2 \log \left (\frac {1}{2} \left (1+\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {1}{2}-\frac {\sqrt {e} \sqrt {f+g x}}{2 \sqrt {e f-d g}}\right )\right )+3 b e^{3/2} n (f+g x)^{3/2} \left (\log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right ) \left (\log \left (\sqrt {e f-d g}+\sqrt {e} \sqrt {f+g x}\right )+2 \log \left (\frac {1}{2}-\frac {\sqrt {e} \sqrt {f+g x}}{2 \sqrt {e f-d g}}\right )\right )+2 \operatorname {PolyLog}\left (2,\frac {1}{2} \left (1+\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {e f-d g}}\right )\right )\right )\right )}{(e f-d g)^{5/2}}\right )}{15 g (f+g x)^{5/2}} \]

[In]

Integrate[(a + b*Log[c*(d + e*x)^n])^2/(f + g*x)^(7/2),x]

[Out]

(2*(-3*(a + b*Log[c*(d + e*x)^n])^2 + (b*e*n*(f + g*x)*(24*b*e^(3/2)*n*(f + g*x)^(3/2)*ArcTanh[(Sqrt[e]*Sqrt[f
+ g*x])/Sqrt[e*f - d*g]] - 8*b*e*Sqrt[e*f - d*g]*n*(f + g*x)*Hypergeometric2F1[-1/2, 1, 1/2, (e*(f + g*x))/(e
*f - d*g)] + 4*(e*f - d*g)^(3/2)*(a + b*Log[c*(d + e*x)^n]) + 12*e*Sqrt[e*f - d*g]*(f + g*x)*(a + b*Log[c*(d +
e*x)^n]) + 6*e^(3/2)*(f + g*x)^(3/2)*(a + b*Log[c*(d + e*x)^n])*Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqrt[f + g*x]]
- 6*e^(3/2)*(f + g*x)^(3/2)*(a + b*Log[c*(d + e*x)^n])*Log[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]] - 3*b*e^(3
/2)*n*(f + g*x)^(3/2)*(Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqrt[f + g*x]]*(Log[Sqrt[e*f - d*g] - Sqrt[e]*Sqrt[f + g*
x]] + 2*Log[(1 + (Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g])/2]) + 2*PolyLog[2, 1/2 - (Sqrt[e]*Sqrt[f + g*x])/(2*
Sqrt[e*f - d*g])]) + 3*b*e^(3/2)*n*(f + g*x)^(3/2)*(Log[Sqrt[e*f - d*g] + Sqrt[e]*Sqrt[f + g*x]]*(Log[Sqrt[e*f
- d*g] + Sqrt[e]*Sqrt[f + g*x]] + 2*Log[1/2 - (Sqrt[e]*Sqrt[f + g*x])/(2*Sqrt[e*f - d*g])]) + 2*PolyLog[2, (1
+ (Sqrt[e]*Sqrt[f + g*x])/Sqrt[e*f - d*g])/2])))/(e*f - d*g)^(5/2)))/(15*g*(f + g*x)^(5/2))

Maple [F]

\[\int \frac {{\left (a +b \ln \left (c \left (e x +d \right )^{n}\right )\right )}^{2}}{\left (g x +f \right )^{\frac {7}{2}}}d x\]

[In]

int((a+b*ln(c*(e*x+d)^n))^2/(g*x+f)^(7/2),x)

[Out]

int((a+b*ln(c*(e*x+d)^n))^2/(g*x+f)^(7/2),x)

Fricas [F]

\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{7/2}} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}{{\left (g x + f\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^(7/2),x, algorithm="fricas")

[Out]

integral((sqrt(g*x + f)*b^2*log((e*x + d)^n*c)^2 + 2*sqrt(g*x + f)*a*b*log((e*x + d)^n*c) + sqrt(g*x + f)*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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+b*ln(c*(e*x+d)**n))**2/(g*x+f)**(7/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{7/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e*(d*g-e*f)>0)', see `assume?`
for more de

Giac [F]

\[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{7/2}} \, dx=\int { \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )}^{2}}{{\left (g x + f\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+b*log(c*(e*x+d)^n))^2/(g*x+f)^(7/2),x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)^2/(g*x + f)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c (d+e x)^n\right )\right )^2}{(f+g x)^{7/2}} \, dx=\int \frac {{\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}^2}{{\left (f+g\,x\right )}^{7/2}} \,d x \]

[In]

int((a + b*log(c*(d + e*x)^n))^2/(f + g*x)^(7/2),x)

[Out]

int((a + b*log(c*(d + e*x)^n))^2/(f + g*x)^(7/2), x)