3.2.33 \(\int x^{3/2} \log (d (e+f \sqrt {x})^k) (a+b \log (c x^n)) \, dx\) [133]

Optimal. Leaf size=367 \[ \frac {24 b e^4 k n \sqrt {x}}{25 f^4}-\frac {7 b e^3 k n x}{25 f^3}+\frac {32 b e^2 k n x^{3/2}}{225 f^2}-\frac {9 b e k n x^2}{100 f}+\frac {8}{125} b k n x^{5/2}-\frac {4 b e^5 k n \log \left (e+f \sqrt {x}\right )}{25 f^5}-\frac {4}{25} b n x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {4 b e^5 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{5 f^5}-\frac {2 e^4 k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{5 f^4}+\frac {e^3 k x \left (a+b \log \left (c x^n\right )\right )}{5 f^3}-\frac {2 e^2 k x^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 f^2}+\frac {e k x^2 \left (a+b \log \left (c x^n\right )\right )}{10 f}-\frac {2}{25} k x^{5/2} \left (a+b \log \left (c x^n\right )\right )+\frac {2 e^5 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{5 f^5}+\frac {2}{5} x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b e^5 k n \text {Li}_2\left (1+\frac {f \sqrt {x}}{e}\right )}{5 f^5} \]

[Out]

-7/25*b*e^3*k*n*x/f^3+32/225*b*e^2*k*n*x^(3/2)/f^2-9/100*b*e*k*n*x^2/f+8/125*b*k*n*x^(5/2)+1/5*e^3*k*x*(a+b*ln
(c*x^n))/f^3-2/15*e^2*k*x^(3/2)*(a+b*ln(c*x^n))/f^2+1/10*e*k*x^2*(a+b*ln(c*x^n))/f-2/25*k*x^(5/2)*(a+b*ln(c*x^
n))-4/25*b*e^5*k*n*ln(e+f*x^(1/2))/f^5+2/5*e^5*k*(a+b*ln(c*x^n))*ln(e+f*x^(1/2))/f^5-4/5*b*e^5*k*n*ln(-f*x^(1/
2)/e)*ln(e+f*x^(1/2))/f^5-4/25*b*n*x^(5/2)*ln(d*(e+f*x^(1/2))^k)+2/5*x^(5/2)*(a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2)
)^k)-4/5*b*e^5*k*n*polylog(2,1+f*x^(1/2)/e)/f^5+24/25*b*e^4*k*n*x^(1/2)/f^4-2/5*e^4*k*(a+b*ln(c*x^n))*x^(1/2)/
f^4

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Rubi [A]
time = 0.20, antiderivative size = 367, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2504, 2442, 45, 2423, 2441, 2352} \begin {gather*} -\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 {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 {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} \end {gather*}

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 45

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 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 2423

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 2441

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 2442

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 2504

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 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) \text {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 ) \text {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 ) \text {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) \text {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) \text {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*}

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Mathematica [A]
time = 0.26, size = 394, normalized size = 1.07 \begin {gather*} \frac {-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 \left (d \left (e+f \sqrt {x}\right )^k\right )-720 b f^5 n x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right )+1800 b e^5 k n \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)-1800 b e^4 f k \sqrt {x} \log \left (c x^n\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 )+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 )+1800 b f^5 x^{5/2} \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )+360 e^5 k \log \left (e+f \sqrt {x}\right ) \left (5 a-2 b n-5 b n \log (x)+5 b \log \left (c x^n\right )\right )+3600 b e^5 k n \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )}{4500 f^5} \end {gather*}

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)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int x^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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*k*x^2*e*log(x^n) + 40*(5*b*f*x*log(x^n) - ((2*f*n - 5*f*log(c))*b - 5*a*f)*x)*k*x^(3/2)*log(f*sqrt
(x) + e) - 5*((9*k*n - 10*k*log(c))*b - 10*a*k)*x^2*e + 40*(5*b*f*x*log(d)*log(x^n) + (5*a*f*log(d) - (2*f*n*l
og(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*k*x*e^2*log(x^n) - ((2*k*n - 5*k*log(c))*b - 5*a*k)*x*e^2)/(f^2*sqrt(x) +
f*e), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^{3/2}\,\ln \left (d\,{\left (e+f\,\sqrt {x}\right )}^k\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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