3.2.19 \(\int \frac {\log (d (e+f \sqrt {x})^k) (a+b \log (c x^n))}{x^2} \, dx\) [119]

Optimal. Leaf size=248 \[ -\frac {3 b f k n}{e \sqrt {x}}+\frac {b f^2 k n \log \left (e+f \sqrt {x}\right )}{e^2}-\frac {b n \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}-\frac {2 b f^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{e^2}-\frac {b f^2 k n \log (x)}{2 e^2}+\frac {b f^2 k n \log ^2(x)}{4 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {2 b f^2 k n \text {Li}_2\left (1+\frac {f \sqrt {x}}{e}\right )}{e^2} \]

[Out]

-1/2*b*f^2*k*n*ln(x)/e^2+1/4*b*f^2*k*n*ln(x)^2/e^2-1/2*f^2*k*ln(x)*(a+b*ln(c*x^n))/e^2+b*f^2*k*n*ln(e+f*x^(1/2
))/e^2+f^2*k*(a+b*ln(c*x^n))*ln(e+f*x^(1/2))/e^2-2*b*f^2*k*n*ln(-f*x^(1/2)/e)*ln(e+f*x^(1/2))/e^2-b*n*ln(d*(e+
f*x^(1/2))^k)/x-(a+b*ln(c*x^n))*ln(d*(e+f*x^(1/2))^k)/x-2*b*f^2*k*n*polylog(2,1+f*x^(1/2)/e)/e^2-3*b*f*k*n/e/x
^(1/2)-f*k*(a+b*ln(c*x^n))/e/x^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2504, 2442, 46, 2423, 2441, 2352, 2338} \begin {gather*} -\frac {2 b f^2 k n \text {PolyLog}\left (2,\frac {f \sqrt {x}}{e}+1\right )}{e^2}-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}-\frac {b n \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}+\frac {b f^2 k n \log ^2(x)}{4 e^2}+\frac {b f^2 k n \log \left (e+f \sqrt {x}\right )}{e^2}-\frac {2 b f^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{e^2}-\frac {b f^2 k n \log (x)}{2 e^2}-\frac {3 b f k n}{e \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*f*k*n)/(e*Sqrt[x]) + (b*f^2*k*n*Log[e + f*Sqrt[x]])/e^2 - (b*n*Log[d*(e + f*Sqrt[x])^k])/x - (2*b*f^2*k*
n*Log[e + f*Sqrt[x]]*Log[-((f*Sqrt[x])/e)])/e^2 - (b*f^2*k*n*Log[x])/(2*e^2) + (b*f^2*k*n*Log[x]^2)/(4*e^2) -
(f*k*(a + b*Log[c*x^n]))/(e*Sqrt[x]) + (f^2*k*Log[e + f*Sqrt[x]]*(a + b*Log[c*x^n]))/e^2 - (Log[d*(e + f*Sqrt[
x])^k]*(a + b*Log[c*x^n]))/x - (f^2*k*Log[x]*(a + b*Log[c*x^n]))/(2*e^2) - (2*b*f^2*k*n*PolyLog[2, 1 + (f*Sqrt
[x])/e])/e^2

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2338

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 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 \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx &=-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-(b n) \int \left (-\frac {f k}{e x^{3/2}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right )}{e^2 x}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x^2}-\frac {f^2 k \log (x)}{2 e^2 x}\right ) \, dx\\ &=-\frac {2 b f k n}{e \sqrt {x}}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+(b n) \int \frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x^2} \, dx+\frac {\left (b f^2 k n\right ) \int \frac {\log (x)}{x} \, dx}{2 e^2}-\frac {\left (b f^2 k n\right ) \int \frac {\log \left (e+f \sqrt {x}\right )}{x} \, dx}{e^2}\\ &=-\frac {2 b f k n}{e \sqrt {x}}+\frac {b f^2 k n \log ^2(x)}{4 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+(2 b n) \text {Subst}\left (\int \frac {\log \left (d (e+f x)^k\right )}{x^3} \, dx,x,\sqrt {x}\right )-\frac {\left (2 b f^2 k n\right ) \text {Subst}\left (\int \frac {\log (e+f x)}{x} \, dx,x,\sqrt {x}\right )}{e^2}\\ &=-\frac {2 b f k n}{e \sqrt {x}}-\frac {b n \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}-\frac {2 b f^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{e^2}+\frac {b f^2 k n \log ^2(x)}{4 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+(b f k n) \text {Subst}\left (\int \frac {1}{x^2 (e+f x)} \, dx,x,\sqrt {x}\right )+\frac {\left (2 b f^3 k n\right ) \text {Subst}\left (\int \frac {\log \left (-\frac {f x}{e}\right )}{e+f x} \, dx,x,\sqrt {x}\right )}{e^2}\\ &=-\frac {2 b f k n}{e \sqrt {x}}-\frac {b n \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}-\frac {2 b f^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{e^2}+\frac {b f^2 k n \log ^2(x)}{4 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {2 b f^2 k n \text {Li}_2\left (1+\frac {f \sqrt {x}}{e}\right )}{e^2}+(b f k n) \text {Subst}\left (\int \left (\frac {1}{e x^2}-\frac {f}{e^2 x}+\frac {f^2}{e^2 (e+f x)}\right ) \, dx,x,\sqrt {x}\right )\\ &=-\frac {3 b f k n}{e \sqrt {x}}+\frac {b f^2 k n \log \left (e+f \sqrt {x}\right )}{e^2}-\frac {b n \log \left (d \left (e+f \sqrt {x}\right )^k\right )}{x}-\frac {2 b f^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{e^2}-\frac {b f^2 k n \log (x)}{2 e^2}+\frac {b f^2 k n \log ^2(x)}{4 e^2}-\frac {f k \left (a+b \log \left (c x^n\right )\right )}{e \sqrt {x}}+\frac {f^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^2}-\frac {\log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {f^2 k \log (x) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {2 b f^2 k n \text {Li}_2\left (1+\frac {f \sqrt {x}}{e}\right )}{e^2}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 250, normalized size = 1.01 \begin {gather*} -\frac {4 a e f k \sqrt {x}+12 b e f k n \sqrt {x}+4 a e^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right )+4 b e^2 n \log \left (d \left (e+f \sqrt {x}\right )^k\right )+2 a f^2 k x \log (x)+2 b f^2 k n x \log (x)-4 b f^2 k n x \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)-b f^2 k n x \log ^2(x)+4 b e f k \sqrt {x} \log \left (c x^n\right )+4 b e^2 \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )+2 b f^2 k x \log (x) \log \left (c x^n\right )-4 f^2 k x \log \left (e+f \sqrt {x}\right ) \left (a+b n-b n \log (x)+b \log \left (c x^n\right )\right )-8 b f^2 k n x \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )}{4 e^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(4*a*e*f*k*Sqrt[x] + 12*b*e*f*k*n*Sqrt[x] + 4*a*e^2*Log[d*(e + f*Sqrt[x])^k] + 4*b*e^2*n*Log[d*(e + f*Sqr
t[x])^k] + 2*a*f^2*k*x*Log[x] + 2*b*f^2*k*n*x*Log[x] - 4*b*f^2*k*n*x*Log[1 + (f*Sqrt[x])/e]*Log[x] - b*f^2*k*n
*x*Log[x]^2 + 4*b*e*f*k*Sqrt[x]*Log[c*x^n] + 4*b*e^2*Log[d*(e + f*Sqrt[x])^k]*Log[c*x^n] + 2*b*f^2*k*x*Log[x]*
Log[c*x^n] - 4*f^2*k*x*Log[e + f*Sqrt[x]]*(a + b*n - b*n*Log[x] + b*Log[c*x^n]) - 8*b*f^2*k*n*x*PolyLog[2, -((
f*Sqrt[x])/e)])/(e^2*x)

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

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^2,x)

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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**2,x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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

[Out]

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

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

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^2,x)

[Out]

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

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