3.137 \(\int \frac{\log (d (e+f \sqrt{x})^k) (a+b \log (c x^n))}{x^{7/2}} \, dx\)

Optimal. Leaf size=394 \[ \frac{4 b f^5 k n \text{PolyLog}\left (2,\frac{f \sqrt{x}}{e}+1\right )}{5 e^5}-\frac{2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt{x}\right )^k\right )}{5 x^{5/2}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt{x}}-\frac{f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac{2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac{f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}-\frac{4 b n \log \left (d \left (e+f \sqrt{x}\right )^k\right )}{25 x^{5/2}}+\frac{32 b f^2 k n}{225 e^2 x^{3/2}}+\frac{24 b f^4 k n}{25 e^4 \sqrt{x}}-\frac{7 b f^3 k n}{25 e^3 x}-\frac{b f^5 k n \log ^2(x)}{10 e^5}-\frac{4 b f^5 k n \log \left (e+f \sqrt{x}\right )}{25 e^5}+\frac{4 b f^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 e^5}+\frac{2 b f^5 k n \log (x)}{25 e^5}-\frac{9 b f k n}{100 e x^2} \]

[Out]

(-9*b*f*k*n)/(100*e*x^2) + (32*b*f^2*k*n)/(225*e^2*x^(3/2)) - (7*b*f^3*k*n)/(25*e^3*x) + (24*b*f^4*k*n)/(25*e^
4*Sqrt[x]) - (4*b*f^5*k*n*Log[e + f*Sqrt[x]])/(25*e^5) - (4*b*n*Log[d*(e + f*Sqrt[x])^k])/(25*x^(5/2)) + (4*b*
f^5*k*n*Log[e + f*Sqrt[x]]*Log[-((f*Sqrt[x])/e)])/(5*e^5) + (2*b*f^5*k*n*Log[x])/(25*e^5) - (b*f^5*k*n*Log[x]^
2)/(10*e^5) - (f*k*(a + b*Log[c*x^n]))/(10*e*x^2) + (2*f^2*k*(a + b*Log[c*x^n]))/(15*e^2*x^(3/2)) - (f^3*k*(a
+ b*Log[c*x^n]))/(5*e^3*x) + (2*f^4*k*(a + b*Log[c*x^n]))/(5*e^4*Sqrt[x]) - (2*f^5*k*Log[e + f*Sqrt[x]]*(a + b
*Log[c*x^n]))/(5*e^5) - (2*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]))/(5*x^(5/2)) + (f^5*k*Log[x]*(a + b*Log
[c*x^n]))/(5*e^5) + (4*b*f^5*k*n*PolyLog[2, 1 + (f*Sqrt[x])/e])/(5*e^5)

________________________________________________________________________________________

Rubi [A]  time = 0.30474, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {2454, 2395, 44, 2376, 2394, 2315, 2301} \[ \frac{4 b f^5 k n \text{PolyLog}\left (2,\frac{f \sqrt{x}}{e}+1\right )}{5 e^5}-\frac{2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt{x}\right )^k\right )}{5 x^{5/2}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt{x}}-\frac{f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac{2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac{f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}-\frac{4 b n \log \left (d \left (e+f \sqrt{x}\right )^k\right )}{25 x^{5/2}}+\frac{32 b f^2 k n}{225 e^2 x^{3/2}}+\frac{24 b f^4 k n}{25 e^4 \sqrt{x}}-\frac{7 b f^3 k n}{25 e^3 x}-\frac{b f^5 k n \log ^2(x)}{10 e^5}-\frac{4 b f^5 k n \log \left (e+f \sqrt{x}\right )}{25 e^5}+\frac{4 b f^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 e^5}+\frac{2 b f^5 k n \log (x)}{25 e^5}-\frac{9 b f k n}{100 e x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*b*f*k*n)/(100*e*x^2) + (32*b*f^2*k*n)/(225*e^2*x^(3/2)) - (7*b*f^3*k*n)/(25*e^3*x) + (24*b*f^4*k*n)/(25*e^
4*Sqrt[x]) - (4*b*f^5*k*n*Log[e + f*Sqrt[x]])/(25*e^5) - (4*b*n*Log[d*(e + f*Sqrt[x])^k])/(25*x^(5/2)) + (4*b*
f^5*k*n*Log[e + f*Sqrt[x]]*Log[-((f*Sqrt[x])/e)])/(5*e^5) + (2*b*f^5*k*n*Log[x])/(25*e^5) - (b*f^5*k*n*Log[x]^
2)/(10*e^5) - (f*k*(a + b*Log[c*x^n]))/(10*e*x^2) + (2*f^2*k*(a + b*Log[c*x^n]))/(15*e^2*x^(3/2)) - (f^3*k*(a
+ b*Log[c*x^n]))/(5*e^3*x) + (2*f^4*k*(a + b*Log[c*x^n]))/(5*e^4*Sqrt[x]) - (2*f^5*k*Log[e + f*Sqrt[x]]*(a + b
*Log[c*x^n]))/(5*e^5) - (2*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]))/(5*x^(5/2)) + (f^5*k*Log[x]*(a + b*Log
[c*x^n]))/(5*e^5) + (4*b*f^5*k*n*PolyLog[2, 1 + (f*Sqrt[x])/e])/(5*e^5)

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2395

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]

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 2376

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] &
& RationalQ[q])) && NeQ[q, -1]

Rule 2394

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

Rule 2315

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

Rubi steps

\begin{align*} \int \frac{\log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^{7/2}} \, dx &=-\frac{f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}+\frac{2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac{f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac{2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt{x}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}-\frac{2 \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{5 x^{5/2}}+\frac{f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}-(b n) \int \left (-\frac{f k}{10 e x^3}+\frac{2 f^2 k}{15 e^2 x^{5/2}}-\frac{f^3 k}{5 e^3 x^2}+\frac{2 f^4 k}{5 e^4 x^{3/2}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right )}{5 e^5 x}-\frac{2 \log \left (d \left (e+f \sqrt{x}\right )^k\right )}{5 x^{7/2}}+\frac{f^5 k \log (x)}{5 e^5 x}\right ) \, dx\\ &=-\frac{b f k n}{20 e x^2}+\frac{4 b f^2 k n}{45 e^2 x^{3/2}}-\frac{b f^3 k n}{5 e^3 x}+\frac{4 b f^4 k n}{5 e^4 \sqrt{x}}-\frac{f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}+\frac{2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac{f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac{2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt{x}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}-\frac{2 \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{5 x^{5/2}}+\frac{f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{1}{5} (2 b n) \int \frac{\log \left (d \left (e+f \sqrt{x}\right )^k\right )}{x^{7/2}} \, dx-\frac{\left (b f^5 k n\right ) \int \frac{\log (x)}{x} \, dx}{5 e^5}+\frac{\left (2 b f^5 k n\right ) \int \frac{\log \left (e+f \sqrt{x}\right )}{x} \, dx}{5 e^5}\\ &=-\frac{b f k n}{20 e x^2}+\frac{4 b f^2 k n}{45 e^2 x^{3/2}}-\frac{b f^3 k n}{5 e^3 x}+\frac{4 b f^4 k n}{5 e^4 \sqrt{x}}-\frac{b f^5 k n \log ^2(x)}{10 e^5}-\frac{f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}+\frac{2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac{f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac{2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt{x}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}-\frac{2 \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{5 x^{5/2}}+\frac{f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{1}{5} (4 b n) \operatorname{Subst}\left (\int \frac{\log \left (d (e+f x)^k\right )}{x^6} \, dx,x,\sqrt{x}\right )+\frac{\left (4 b f^5 k n\right ) \operatorname{Subst}\left (\int \frac{\log (e+f x)}{x} \, dx,x,\sqrt{x}\right )}{5 e^5}\\ &=-\frac{b f k n}{20 e x^2}+\frac{4 b f^2 k n}{45 e^2 x^{3/2}}-\frac{b f^3 k n}{5 e^3 x}+\frac{4 b f^4 k n}{5 e^4 \sqrt{x}}-\frac{4 b n \log \left (d \left (e+f \sqrt{x}\right )^k\right )}{25 x^{5/2}}+\frac{4 b f^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 e^5}-\frac{b f^5 k n \log ^2(x)}{10 e^5}-\frac{f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}+\frac{2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac{f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac{2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt{x}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}-\frac{2 \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{5 x^{5/2}}+\frac{f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{1}{25} (4 b f k n) \operatorname{Subst}\left (\int \frac{1}{x^5 (e+f x)} \, dx,x,\sqrt{x}\right )-\frac{\left (4 b f^6 k n\right ) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{f x}{e}\right )}{e+f x} \, dx,x,\sqrt{x}\right )}{5 e^5}\\ &=-\frac{b f k n}{20 e x^2}+\frac{4 b f^2 k n}{45 e^2 x^{3/2}}-\frac{b f^3 k n}{5 e^3 x}+\frac{4 b f^4 k n}{5 e^4 \sqrt{x}}-\frac{4 b n \log \left (d \left (e+f \sqrt{x}\right )^k\right )}{25 x^{5/2}}+\frac{4 b f^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 e^5}-\frac{b f^5 k n \log ^2(x)}{10 e^5}-\frac{f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}+\frac{2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac{f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac{2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt{x}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}-\frac{2 \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{5 x^{5/2}}+\frac{f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{4 b f^5 k n \text{Li}_2\left (1+\frac{f \sqrt{x}}{e}\right )}{5 e^5}+\frac{1}{25} (4 b f k n) \operatorname{Subst}\left (\int \left (\frac{1}{e x^5}-\frac{f}{e^2 x^4}+\frac{f^2}{e^3 x^3}-\frac{f^3}{e^4 x^2}+\frac{f^4}{e^5 x}-\frac{f^5}{e^5 (e+f x)}\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{9 b f k n}{100 e x^2}+\frac{32 b f^2 k n}{225 e^2 x^{3/2}}-\frac{7 b f^3 k n}{25 e^3 x}+\frac{24 b f^4 k n}{25 e^4 \sqrt{x}}-\frac{4 b f^5 k n \log \left (e+f \sqrt{x}\right )}{25 e^5}-\frac{4 b n \log \left (d \left (e+f \sqrt{x}\right )^k\right )}{25 x^{5/2}}+\frac{4 b f^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 e^5}+\frac{2 b f^5 k n \log (x)}{25 e^5}-\frac{b f^5 k n \log ^2(x)}{10 e^5}-\frac{f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}+\frac{2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac{f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac{2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt{x}}-\frac{2 f^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}-\frac{2 \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{5 x^{5/2}}+\frac{f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac{4 b f^5 k n \text{Li}_2\left (1+\frac{f \sqrt{x}}{e}\right )}{5 e^5}\\ \end{align*}

Mathematica [A]  time = 0.451916, size = 422, normalized size = 1.07 \[ \frac{-720 b f^5 k n x^{5/2} \text{PolyLog}\left (2,-\frac{f \sqrt{x}}{e}\right )-72 f^5 k x^{5/2} \log \left (e+f \sqrt{x}\right ) \left (5 a+5 b \log \left (c x^n\right )-5 b n \log (x)+2 b n\right )-360 a e^5 \log \left (d \left (e+f \sqrt{x}\right )^k\right )-180 a e^2 f^3 k x^{3/2}+120 a e^3 f^2 k x-90 a e^4 f k \sqrt{x}+360 a e f^4 k x^2+180 a f^5 k x^{5/2} \log (x)-360 b e^5 \log \left (c x^n\right ) \log \left (d \left (e+f \sqrt{x}\right )^k\right )+120 b e^3 f^2 k x \log \left (c x^n\right )-180 b e^2 f^3 k x^{3/2} \log \left (c x^n\right )-90 b e^4 f k \sqrt{x} \log \left (c x^n\right )+360 b e f^4 k x^2 \log \left (c x^n\right )+180 b f^5 k x^{5/2} \log (x) \log \left (c x^n\right )-144 b e^5 n \log \left (d \left (e+f \sqrt{x}\right )^k\right )-252 b e^2 f^3 k n x^{3/2}+128 b e^3 f^2 k n x-81 b e^4 f k n \sqrt{x}+864 b e f^4 k n x^2-360 b f^5 k n x^{5/2} \log (x) \log \left (\frac{f \sqrt{x}}{e}+1\right )-90 b f^5 k n x^{5/2} \log ^2(x)+72 b f^5 k n x^{5/2} \log (x)}{900 e^5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-90*a*e^4*f*k*Sqrt[x] - 81*b*e^4*f*k*n*Sqrt[x] + 120*a*e^3*f^2*k*x + 128*b*e^3*f^2*k*n*x - 180*a*e^2*f^3*k*x^
(3/2) - 252*b*e^2*f^3*k*n*x^(3/2) + 360*a*e*f^4*k*x^2 + 864*b*e*f^4*k*n*x^2 - 360*a*e^5*Log[d*(e + f*Sqrt[x])^
k] - 144*b*e^5*n*Log[d*(e + f*Sqrt[x])^k] + 180*a*f^5*k*x^(5/2)*Log[x] + 72*b*f^5*k*n*x^(5/2)*Log[x] - 360*b*f
^5*k*n*x^(5/2)*Log[1 + (f*Sqrt[x])/e]*Log[x] - 90*b*f^5*k*n*x^(5/2)*Log[x]^2 - 90*b*e^4*f*k*Sqrt[x]*Log[c*x^n]
+ 120*b*e^3*f^2*k*x*Log[c*x^n] - 180*b*e^2*f^3*k*x^(3/2)*Log[c*x^n] + 360*b*e*f^4*k*x^2*Log[c*x^n] - 360*b*e^
5*Log[d*(e + f*Sqrt[x])^k]*Log[c*x^n] + 180*b*f^5*k*x^(5/2)*Log[x]*Log[c*x^n] - 72*f^5*k*x^(5/2)*Log[e + f*Sqr
t[x]]*(5*a + 2*b*n - 5*b*n*Log[x] + 5*b*Log[c*x^n]) - 720*b*f^5*k*n*x^(5/2)*PolyLog[2, -((f*Sqrt[x])/e)])/(900
*e^5*x^(5/2))

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Maple [F]  time = 0.023, size = 0, normalized size = 0. \begin{align*} \int{(a+b\ln \left ( c{x}^{n} \right ) )\ln \left ( d \left ( e+f\sqrt{x} \right ) ^{k} \right ){x}^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2))^k)/x^(7/2),x)

[Out]

int((a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2))^k)/x^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(d*(e+f*x^(1/2))^k)/x^(7/2),x, algorithm="maxima")

[Out]

1/25*integrate((5*b*f*k*x*log(x^n) + (5*a*f*k + (2*f*k*n + 5*f*k*log(c))*b)*x)/x^4, x)/e + 1/25*integrate((5*b
*f^3*k*x*log(x^n) + (5*a*f^3*k + (2*f^3*k*n + 5*f^3*k*log(c))*b)*x)/x^3, x)/e^3 + 1/25*integrate((5*b*f^5*k*x*
log(x^n) + (5*a*f^5*k + (2*f^5*k*n + 5*f^5*k*log(c))*b)*x)/x^2, x)/e^5 - 1/225*(2*(15*b*f^8*k*x^2*log(x^n) + (
15*a*f^8*k - (4*f^8*k*n - 15*f^8*k*log(c))*b)*x^2)/sqrt(x) - 9*(5*b*e*f^7*k*x^2*log(x^n) + (5*a*e*f^7*k - (3*e
*f^7*k*n - 5*e*f^7*k*log(c))*b)*x^2)/x + 18*(5*b*e^2*f^6*k*x^2*log(x^n) + (5*a*e^2*f^6*k - (8*e^2*f^6*k*n - 5*
e^2*f^6*k*log(c))*b)*x^2)/x^(3/2) - 18*(5*b*e^4*f^4*k*x^2*log(x^n) + (5*a*e^4*f^4*k + (12*e^4*f^4*k*n + 5*e^4*
f^4*k*log(c))*b)*x^2)/x^(5/2) + 18*(5*b*e^8*x*log(x^n) + (5*a*e^8 + (2*e^8*n + 5*e^8*log(c))*b)*x)*log((f*sqrt
(x) + e)^k)/x^(7/2) - 2*((15*a*e^6*f^2*k + (16*e^6*f^2*k*n + 15*e^6*f^2*k*log(c))*b)*x^2 - 9*(5*a*e^8*log(d) +
(2*e^8*n*log(d) + 5*e^8*log(c)*log(d))*b)*x + 15*(b*e^6*f^2*k*x^2 - 3*b*e^8*x*log(d))*log(x^n))/x^(7/2))/e^8
+ integrate(1/25*(5*b*f^9*k*x*log(x^n) + (5*a*f^9*k + (2*f^9*k*n + 5*f^9*k*log(c))*b)*x)/(e^8*f*sqrt(x) + e^9)
, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b \sqrt{x} \log \left (c x^{n}\right ) + a \sqrt{x}\right )} \log \left ({\left (f \sqrt{x} + e\right )}^{k} d\right )}{x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(d*(e+f*x^(1/2))^k)/x^(7/2),x, algorithm="fricas")

[Out]

integral((b*sqrt(x)*log(c*x^n) + a*sqrt(x))*log((f*sqrt(x) + e)^k*d)/x^4, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))*ln(d*(e+f*x**(1/2))**k)/x**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt{x} + e\right )}^{k} d\right )}{x^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(d*(e+f*x^(1/2))^k)/x^(7/2),x, algorithm="giac")

[Out]

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