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 \[ -\frac {2 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{5 x^{5/2}}-\frac {2 f^5 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac {f^5 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{5 e^5}+\frac {2 f^4 k \left (a+b \log \left (c x^n\right )\right )}{5 e^4 \sqrt {x}}-\frac {f^3 k \left (a+b \log \left (c x^n\right )\right )}{5 e^3 x}+\frac {2 f^2 k \left (a+b \log \left (c x^n\right )\right )}{15 e^2 x^{3/2}}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{10 e x^2}-\frac {4 b n \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{25 x^{5/2}}+\frac {4 b f^5 k n \text {Li}_2\left (\frac {\sqrt {x} f}{e}+1\right )}{5 e^5}-\frac {b f^5 k n \log ^2(x)}{10 e^5}-\frac {4 b f^5 k n \log \left (e+f \sqrt {x}\right )}{25 e^5}+\frac {4 b f^5 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{5 e^5}+\frac {2 b f^5 k n \log (x)}{25 e^5}+\frac {24 b f^4 k n}{25 e^4 \sqrt {x}}-\frac {7 b f^3 k n}{25 e^3 x}+\frac {32 b f^2 k n}{225 e^2 x^{3/2}}-\frac {9 b f k n}{100 e x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.30, antiderivative size = 394, normalized size of antiderivative = 1.00, 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 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 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 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 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 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])

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 = 0.45, size = 422, normalized size = 1.07 \[ \frac {-72 f^5 k x^{5/2} \log \left (e+f \sqrt {x}\right ) \left (5 a+5 b \log \left (c x^n\right )-5 b n \log (x)+2 b n\right )-360 a e^5 \log \left (d \left (e+f \sqrt {x}\right )^k\right )-90 a e^4 f k \sqrt {x}+120 a e^3 f^2 k x-180 a e^2 f^3 k x^{3/2}+360 a e f^4 k x^2+180 a f^5 k x^{5/2} \log (x)-360 b e^5 \log \left (c x^n\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )-90 b e^4 f k \sqrt {x} \log \left (c x^n\right )+120 b e^3 f^2 k x \log \left (c x^n\right )-180 b e^2 f^3 k x^{3/2} \log \left (c x^n\right )+360 b e f^4 k x^2 \log \left (c x^n\right )+180 b f^5 k x^{5/2} \log (x) \log \left (c x^n\right )-144 b e^5 n \log \left (d \left (e+f \sqrt {x}\right )^k\right )-81 b e^4 f k n \sqrt {x}+128 b e^3 f^2 k n x-252 b e^2 f^3 k n x^{3/2}-720 b f^5 k n x^{5/2} \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )-360 b f^5 k n x^{5/2} \log (x) \log \left (\frac {f \sqrt {x}}{e}+1\right )+864 b e f^4 k n x^2-90 b f^5 k n x^{5/2} \log ^2(x)+72 b f^5 k n x^{5/2} \log (x)}{900 e^5 x^{5/2}} \]

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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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b \sqrt {x} \log \left (c x^{n}\right ) + a \sqrt {x}\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x^{4}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left ({\left (f \sqrt {x} + e\right )}^{k} d\right )}{x^{\frac {7}{2}}}\,{d x} \]

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)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right ) \ln \left (d \left (f \sqrt {x}+e \right )^{k}\right )}{x^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {-\frac {9 \, b f k n}{4 \, x^{2}} - \frac {5 \, b f k \log \relax (c)}{2 \, x^{2}} - \frac {5 \, b f k \log \left (x^{n}\right )}{2 \, x^{2}} - \frac {5 \, a f k}{2 \, x^{2}}}{25 \, e} + \frac {-\frac {7 \, b f^{3} k n}{x} - \frac {5 \, b f^{3} k \log \relax (c)}{x} - \frac {5 \, b f^{3} k \log \left (x^{n}\right )}{x} - \frac {5 \, a f^{3} k}{x}}{25 \, e^{3}} + \frac {2 \, b f^{5} k n \log \relax (x) + 5 \, b f^{5} k \log \relax (c) \log \relax (x) + 5 \, a f^{5} k \log \relax (x) + \frac {5 \, b f^{5} k \log \left (x^{n}\right )^{2}}{2 \, n}}{25 \, e^{5}} - \frac {\frac {2 \, {\left (15 \, b f^{8} k x^{2} \log \left (x^{n}\right ) + {\left (15 \, a f^{8} k - {\left (4 \, f^{8} k n - 15 \, f^{8} k \log \relax (c)\right )} b\right )} x^{2}\right )}}{\sqrt {x}} - \frac {9 \, {\left (5 \, b e f^{7} k x^{2} \log \left (x^{n}\right ) + {\left (5 \, a e f^{7} k - {\left (3 \, e f^{7} k n - 5 \, e f^{7} k \log \relax (c)\right )} b\right )} x^{2}\right )}}{x} + \frac {18 \, {\left (5 \, b e^{2} f^{6} k x^{2} \log \left (x^{n}\right ) + {\left (5 \, a e^{2} f^{6} k - {\left (8 \, e^{2} f^{6} k n - 5 \, e^{2} f^{6} k \log \relax (c)\right )} b\right )} x^{2}\right )}}{x^{\frac {3}{2}}} + \frac {18 \, {\left (5 \, b e^{8} x \log \left (x^{n}\right ) + {\left (5 \, a e^{8} + {\left (2 \, e^{8} n + 5 \, e^{8} \log \relax (c)\right )} b\right )} x\right )} k \log \left (f \sqrt {x} + e\right )}{x^{\frac {7}{2}}} - \frac {18 \, {\left (5 \, b e^{4} f^{4} k x^{2} \log \left (x^{n}\right ) + {\left (5 \, a e^{4} f^{4} k + {\left (12 \, e^{4} f^{4} k n + 5 \, e^{4} f^{4} k \log \relax (c)\right )} b\right )} x^{2}\right )}}{x^{\frac {5}{2}}} - \frac {2 \, {\left ({\left (15 \, a e^{6} f^{2} k + {\left (16 \, e^{6} f^{2} k n + 15 \, e^{6} f^{2} k \log \relax (c)\right )} b\right )} x^{2} - 9 \, {\left (5 \, a e^{8} \log \relax (d) + {\left (2 \, e^{8} n \log \relax (d) + 5 \, e^{8} \log \relax (c) \log \relax (d)\right )} b\right )} x + 15 \, {\left (b e^{6} f^{2} k x^{2} - 3 \, b e^{8} x \log \relax (d)\right )} \log \left (x^{n}\right )\right )}}{x^{\frac {7}{2}}}}{225 \, e^{8}} + \int \frac {5 \, b f^{9} k x \log \left (x^{n}\right ) + {\left (5 \, a f^{9} k + {\left (2 \, f^{9} k n + 5 \, f^{9} k \log \relax (c)\right )} b\right )} x}{25 \, {\left (e^{8} f \sqrt {x} + e^{9}\right )}}\,{d x} \]

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^8*x*log(x^n) + (5*a*e^8 + (2*e^8*n + 5*e^8*log(c))*b)*x)*k*log(f
*sqrt(x) + e)/x^(7/2) - 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) - 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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