3.133 \(\int x^{3/2} \log (d (e+f \sqrt{x})^k) (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=367 \[ -\frac{4 b e^5 k n \text{PolyLog}\left (2,\frac{f \sqrt{x}}{e}+1\right )}{5 f^5}+\frac{2}{5} x^{5/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt{x}\right )^k\right )+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}-\frac{2 e^4 k \sqrt{x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac{e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac{2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac{e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac{2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )-\frac{4}{25} b n x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right )+\frac{32 b e^2 k n x^{3/2}}{225 f^2}+\frac{24 b e^4 k n \sqrt{x}}{25 f^4}-\frac{7 b e^3 k n x}{25 f^3}-\frac{4 b e^5 k n \log \left (e+f \sqrt{x}\right )}{25 f^5}-\frac{4 b e^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 f^5}-\frac{9 b e k n x^2}{100 f}+\frac{8}{125} b k n x^{5/2} \]

[Out]

(24*b*e^4*k*n*Sqrt[x])/(25*f^4) - (7*b*e^3*k*n*x)/(25*f^3) + (32*b*e^2*k*n*x^(3/2))/(225*f^2) - (9*b*e*k*n*x^2
)/(100*f) + (8*b*k*n*x^(5/2))/125 - (4*b*e^5*k*n*Log[e + f*Sqrt[x]])/(25*f^5) - (4*b*n*x^(5/2)*Log[d*(e + f*Sq
rt[x])^k])/25 - (4*b*e^5*k*n*Log[e + f*Sqrt[x]]*Log[-((f*Sqrt[x])/e)])/(5*f^5) - (2*e^4*k*Sqrt[x]*(a + b*Log[c
*x^n]))/(5*f^4) + (e^3*k*x*(a + b*Log[c*x^n]))/(5*f^3) - (2*e^2*k*x^(3/2)*(a + b*Log[c*x^n]))/(15*f^2) + (e*k*
x^2*(a + b*Log[c*x^n]))/(10*f) - (2*k*x^(5/2)*(a + b*Log[c*x^n]))/25 + (2*e^5*k*Log[e + f*Sqrt[x]]*(a + b*Log[
c*x^n]))/(5*f^5) + (2*x^(5/2)*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]))/5 - (4*b*e^5*k*n*PolyLog[2, 1 + (f*
Sqrt[x])/e])/(5*f^5)

________________________________________________________________________________________

Rubi [A]  time = 0.298058, antiderivative size = 367, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2454, 2395, 43, 2376, 2394, 2315} \[ -\frac{4 b e^5 k n \text{PolyLog}\left (2,\frac{f \sqrt{x}}{e}+1\right )}{5 f^5}+\frac{2}{5} x^{5/2} \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt{x}\right )^k\right )+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}-\frac{2 e^4 k \sqrt{x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac{e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac{2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac{e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac{2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )-\frac{4}{25} b n x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right )+\frac{32 b e^2 k n x^{3/2}}{225 f^2}+\frac{24 b e^4 k n \sqrt{x}}{25 f^4}-\frac{7 b e^3 k n x}{25 f^3}-\frac{4 b e^5 k n \log \left (e+f \sqrt{x}\right )}{25 f^5}-\frac{4 b e^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 f^5}-\frac{9 b e k n x^2}{100 f}+\frac{8}{125} b k n x^{5/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]),x]

[Out]

(24*b*e^4*k*n*Sqrt[x])/(25*f^4) - (7*b*e^3*k*n*x)/(25*f^3) + (32*b*e^2*k*n*x^(3/2))/(225*f^2) - (9*b*e*k*n*x^2
)/(100*f) + (8*b*k*n*x^(5/2))/125 - (4*b*e^5*k*n*Log[e + f*Sqrt[x]])/(25*f^5) - (4*b*n*x^(5/2)*Log[d*(e + f*Sq
rt[x])^k])/25 - (4*b*e^5*k*n*Log[e + f*Sqrt[x]]*Log[-((f*Sqrt[x])/e)])/(5*f^5) - (2*e^4*k*Sqrt[x]*(a + b*Log[c
*x^n]))/(5*f^4) + (e^3*k*x*(a + b*Log[c*x^n]))/(5*f^3) - (2*e^2*k*x^(3/2)*(a + b*Log[c*x^n]))/(15*f^2) + (e*k*
x^2*(a + b*Log[c*x^n]))/(10*f) - (2*k*x^(5/2)*(a + b*Log[c*x^n]))/25 + (2*e^5*k*Log[e + f*Sqrt[x]]*(a + b*Log[
c*x^n]))/(5*f^5) + (2*x^(5/2)*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]))/5 - (4*b*e^5*k*n*PolyLog[2, 1 + (f*
Sqrt[x])/e])/(5*f^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 43

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 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]

Rubi steps

\begin{align*} \int x^{3/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx &=-\frac{2 e^4 k \sqrt{x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac{e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac{2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac{e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac{2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac{2}{5} x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac{e^3 k}{5 f^3}-\frac{2 e^4 k}{5 f^4 \sqrt{x}}-\frac{2 e^2 k \sqrt{x}}{15 f^2}+\frac{e k x}{10 f}-\frac{2}{25} k x^{3/2}+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right )}{5 f^5 x}+\frac{2}{5} x^{3/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right )\right ) \, dx\\ &=\frac{4 b e^4 k n \sqrt{x}}{5 f^4}-\frac{b e^3 k n x}{5 f^3}+\frac{4 b e^2 k n x^{3/2}}{45 f^2}-\frac{b e k n x^2}{20 f}+\frac{4}{125} b k n x^{5/2}-\frac{2 e^4 k \sqrt{x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac{e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac{2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac{e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac{2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac{2}{5} x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{5} (2 b n) \int x^{3/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \, dx-\frac{\left (2 b e^5 k n\right ) \int \frac{\log \left (e+f \sqrt{x}\right )}{x} \, dx}{5 f^5}\\ &=\frac{4 b e^4 k n \sqrt{x}}{5 f^4}-\frac{b e^3 k n x}{5 f^3}+\frac{4 b e^2 k n x^{3/2}}{45 f^2}-\frac{b e k n x^2}{20 f}+\frac{4}{125} b k n x^{5/2}-\frac{2 e^4 k \sqrt{x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac{e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac{2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac{e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac{2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac{2}{5} x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{5} (4 b n) \operatorname{Subst}\left (\int x^4 \log \left (d (e+f x)^k\right ) \, dx,x,\sqrt{x}\right )-\frac{\left (4 b e^5 k n\right ) \operatorname{Subst}\left (\int \frac{\log (e+f x)}{x} \, dx,x,\sqrt{x}\right )}{5 f^5}\\ &=\frac{4 b e^4 k n \sqrt{x}}{5 f^4}-\frac{b e^3 k n x}{5 f^3}+\frac{4 b e^2 k n x^{3/2}}{45 f^2}-\frac{b e k n x^2}{20 f}+\frac{4}{125} b k n x^{5/2}-\frac{4}{25} b n x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right )-\frac{4 b e^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 f^5}-\frac{2 e^4 k \sqrt{x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac{e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac{2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac{e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac{2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac{2}{5} x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )+\frac{\left (4 b e^5 k n\right ) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{f x}{e}\right )}{e+f x} \, dx,x,\sqrt{x}\right )}{5 f^4}+\frac{1}{25} (4 b f k n) \operatorname{Subst}\left (\int \frac{x^5}{e+f x} \, dx,x,\sqrt{x}\right )\\ &=\frac{4 b e^4 k n \sqrt{x}}{5 f^4}-\frac{b e^3 k n x}{5 f^3}+\frac{4 b e^2 k n x^{3/2}}{45 f^2}-\frac{b e k n x^2}{20 f}+\frac{4}{125} b k n x^{5/2}-\frac{4}{25} b n x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right )-\frac{4 b e^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 f^5}-\frac{2 e^4 k \sqrt{x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac{e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac{2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac{e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac{2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac{2}{5} x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{4 b e^5 k n \text{Li}_2\left (1+\frac{f \sqrt{x}}{e}\right )}{5 f^5}+\frac{1}{25} (4 b f k n) \operatorname{Subst}\left (\int \left (\frac{e^4}{f^5}-\frac{e^3 x}{f^4}+\frac{e^2 x^2}{f^3}-\frac{e x^3}{f^2}+\frac{x^4}{f}-\frac{e^5}{f^5 (e+f x)}\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{24 b e^4 k n \sqrt{x}}{25 f^4}-\frac{7 b e^3 k n x}{25 f^3}+\frac{32 b e^2 k n x^{3/2}}{225 f^2}-\frac{9 b e k n x^2}{100 f}+\frac{8}{125} b k n x^{5/2}-\frac{4 b e^5 k n \log \left (e+f \sqrt{x}\right )}{25 f^5}-\frac{4}{25} b n x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right )-\frac{4 b e^5 k n \log \left (e+f \sqrt{x}\right ) \log \left (-\frac{f \sqrt{x}}{e}\right )}{5 f^5}-\frac{2 e^4 k \sqrt{x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac{e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac{2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac{e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac{2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac{2 e^5 k \log \left (e+f \sqrt{x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac{2}{5} x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{4 b e^5 k n \text{Li}_2\left (1+\frac{f \sqrt{x}}{e}\right )}{5 f^5}\\ \end{align*}

Mathematica [A]  time = 0.41797, size = 394, normalized size = 1.07 \[ \frac{3600 b e^5 k n \text{PolyLog}\left (2,-\frac{f \sqrt{x}}{e}\right )+360 e^5 k \log \left (e+f \sqrt{x}\right ) \left (5 a+5 b \log \left (c x^n\right )-5 b n \log (x)-2 b n\right )+1800 a f^5 x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right )-600 a e^2 f^3 k x^{3/2}+900 a e^3 f^2 k x-1800 a e^4 f k \sqrt{x}+450 a e f^4 k x^2-360 a f^5 k x^{5/2}+1800 b f^5 x^{5/2} \log \left (c x^n\right ) \log \left (d \left (e+f \sqrt{x}\right )^k\right )+900 b e^3 f^2 k x \log \left (c x^n\right )-600 b e^2 f^3 k x^{3/2} \log \left (c x^n\right )-1800 b e^4 f k \sqrt{x} \log \left (c x^n\right )+450 b e f^4 k x^2 \log \left (c x^n\right )-360 b f^5 k x^{5/2} \log \left (c x^n\right )-720 b f^5 n x^{5/2} \log \left (d \left (e+f \sqrt{x}\right )^k\right )+640 b e^2 f^3 k n x^{3/2}-1260 b e^3 f^2 k n x+4320 b e^4 f k n \sqrt{x}+1800 b e^5 k n \log (x) \log \left (\frac{f \sqrt{x}}{e}+1\right )-405 b e f^4 k n x^2+288 b f^5 k n x^{5/2}}{4500 f^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*Log[d*(e + f*Sqrt[x])^k]*(a + b*Log[c*x^n]),x]

[Out]

(-1800*a*e^4*f*k*Sqrt[x] + 4320*b*e^4*f*k*n*Sqrt[x] + 900*a*e^3*f^2*k*x - 1260*b*e^3*f^2*k*n*x - 600*a*e^2*f^3
*k*x^(3/2) + 640*b*e^2*f^3*k*n*x^(3/2) + 450*a*e*f^4*k*x^2 - 405*b*e*f^4*k*n*x^2 - 360*a*f^5*k*x^(5/2) + 288*b
*f^5*k*n*x^(5/2) + 1800*a*f^5*x^(5/2)*Log[d*(e + f*Sqrt[x])^k] - 720*b*f^5*n*x^(5/2)*Log[d*(e + f*Sqrt[x])^k]
+ 1800*b*e^5*k*n*Log[1 + (f*Sqrt[x])/e]*Log[x] - 1800*b*e^4*f*k*Sqrt[x]*Log[c*x^n] + 900*b*e^3*f^2*k*x*Log[c*x
^n] - 600*b*e^2*f^3*k*x^(3/2)*Log[c*x^n] + 450*b*e*f^4*k*x^2*Log[c*x^n] - 360*b*f^5*k*x^(5/2)*Log[c*x^n] + 180
0*b*f^5*x^(5/2)*Log[d*(e + f*Sqrt[x])^k]*Log[c*x^n] + 360*e^5*k*Log[e + f*Sqrt[x]]*(5*a - 2*b*n - 5*b*n*Log[x]
+ 5*b*Log[c*x^n]) + 3600*b*e^5*k*n*PolyLog[2, -((f*Sqrt[x])/e)])/(4500*f^5)

________________________________________________________________________________________

Maple [F]  time = 0.032, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{3}{2}}} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \ln \left ( d \left ( e+f\sqrt{x} \right ) ^{k} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2))^k),x)

[Out]

int(x^(3/2)*(a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2))^k),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{50 \, b e k x^{2} \log \left (x^{n}\right ) + 40 \,{\left (5 \, b f x \log \left (x^{n}\right ) -{\left ({\left (2 \, f n - 5 \, f \log \left (c\right )\right )} b - 5 \, a f\right )} x\right )} x^{\frac{3}{2}} \log \left ({\left (f \sqrt{x} + e\right )}^{k}\right ) + 5 \,{\left (10 \, a e k -{\left (9 \, e k n - 10 \, e k \log \left (c\right )\right )} b\right )} x^{2} + 40 \,{\left (5 \, b f x \log \left (d\right ) \log \left (x^{n}\right ) +{\left (5 \, a f \log \left (d\right ) -{\left (2 \, f n \log \left (d\right ) - 5 \, f \log \left (c\right ) \log \left (d\right )\right )} b\right )} x\right )} x^{\frac{3}{2}} - 8 \,{\left (5 \, b f k x^{2} \log \left (x^{n}\right ) +{\left (5 \, a f k -{\left (4 \, f k n - 5 \, f k \log \left (c\right )\right )} b\right )} x^{2}\right )} \sqrt{x}}{500 \, f} - \int \frac{5 \, b e^{2} k x \log \left (x^{n}\right ) +{\left (5 \, a e^{2} k -{\left (2 \, e^{2} k n - 5 \, e^{2} k \log \left (c\right )\right )} b\right )} x}{25 \,{\left (f^{2} \sqrt{x} + e f\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*log(c*x^n))*log(d*(e+f*x^(1/2))^k),x, algorithm="maxima")

[Out]

1/500*(50*b*e*k*x^2*log(x^n) + 40*(5*b*f*x*log(x^n) - ((2*f*n - 5*f*log(c))*b - 5*a*f)*x)*x^(3/2)*log((f*sqrt(
x) + e)^k) + 5*(10*a*e*k - (9*e*k*n - 10*e*k*log(c))*b)*x^2 + 40*(5*b*f*x*log(d)*log(x^n) + (5*a*f*log(d) - (2
*f*n*log(d) - 5*f*log(c)*log(d))*b)*x)*x^(3/2) - 8*(5*b*f*k*x^2*log(x^n) + (5*a*f*k - (4*f*k*n - 5*f*k*log(c))
*b)*x^2)*sqrt(x))/f - integrate(1/25*(5*b*e^2*k*x*log(x^n) + (5*a*e^2*k - (2*e^2*k*n - 5*e^2*k*log(c))*b)*x)/(
f^2*sqrt(x) + e*f), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b x^{\frac{3}{2}} \log \left (c x^{n}\right ) + a x^{\frac{3}{2}}\right )} \log \left ({\left (f \sqrt{x} + e\right )}^{k} d\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*log(c*x^n))*log(d*(e+f*x^(1/2))^k),x, algorithm="fricas")

[Out]

integral((b*x^(3/2)*log(c*x^n) + a*x^(3/2))*log((f*sqrt(x) + e)^k*d), 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(x**(3/2)*(a+b*ln(c*x**n))*ln(d*(e+f*x**(1/2))**k),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \log \left (c x^{n}\right ) + a\right )} x^{\frac{3}{2}} \log \left ({\left (f \sqrt{x} + e\right )}^{k} d\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(a+b*log(c*x^n))*log(d*(e+f*x^(1/2))^k),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^(3/2)*log((f*sqrt(x) + e)^k*d), x)