3.2.17 \(\int \log (d (e+f \sqrt {x})^k) (a+b \log (c x^n)) \, dx\) [117]

Optimal. Leaf size=209 \[ -\frac {3 b e k n \sqrt {x}}{f}+b k n x+\frac {b e^2 k n \log \left (e+f \sqrt {x}\right )}{f^2}-b n x \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {2 b e^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{f^2}+\frac {e k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{f}-\frac {1}{2} k x \left (a+b \log \left (c x^n\right )\right )-\frac {e^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{f^2}+x \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \left (a+b \log \left (c x^n\right )\right )+\frac {2 b e^2 k n \text {Li}_2\left (1+\frac {f \sqrt {x}}{e}\right )}{f^2} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2498, 272, 45, 2417, 2504, 2441, 2352} \begin {gather*} \frac {2 b e^2 k n \text {PolyLog}\left (2,\frac {f \sqrt {x}}{e}+1\right )}{f^2}+x \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {e^2 k \log \left (e+f \sqrt {x}\right ) \left (a+b \log \left (c x^n\right )\right )}{f^2}+\frac {e k \sqrt {x} \left (a+b \log \left (c x^n\right )\right )}{f}-\frac {1}{2} k x \left (a+b \log \left (c x^n\right )\right )-b n x \log \left (d \left (e+f \sqrt {x}\right )^k\right )+\frac {b e^2 k n \log \left (e+f \sqrt {x}\right )}{f^2}+\frac {2 b e^2 k n \log \left (e+f \sqrt {x}\right ) \log \left (-\frac {f \sqrt {x}}{e}\right )}{f^2}-\frac {3 b e k n \sqrt {x}}{f}+b k n x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 2417

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> With[
{u = IntHide[Log[d*(e + f*x^m)^r], x]}, Dist[(a + b*Log[c*x^n])^p, u, x] - Dist[b*n*p, Int[Dist[(a + b*Log[c*x
^n])^(p - 1)/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] && RationalQ[m] && (EqQ[
p, 1] || (FractionQ[m] && IntegerQ[1/m]) || (EqQ[r, 1] && EqQ[m, 1] && EqQ[d*e, 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 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

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

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Mathematica [A]
time = 0.16, size = 218, normalized size = 1.04 \begin {gather*} \frac {a e k \sqrt {x}}{f}-\frac {3 b e k n \sqrt {x}}{f}-\frac {a k x}{2}+b k n x+a x \log \left (d \left (e+f \sqrt {x}\right )^k\right )-b n x \log \left (d \left (e+f \sqrt {x}\right )^k\right )-\frac {b e^2 k n \log \left (1+\frac {f \sqrt {x}}{e}\right ) \log (x)}{f^2}+\frac {b e k \sqrt {x} \log \left (c x^n\right )}{f}-\frac {1}{2} b k x \log \left (c x^n\right )+b x \log \left (d \left (e+f \sqrt {x}\right )^k\right ) \log \left (c x^n\right )-\frac {e^2 k \log \left (e+f \sqrt {x}\right ) \left (a-b n-b n \log (x)+b \log \left (c x^n\right )\right )}{f^2}-\frac {2 b e^2 k n \text {Li}_2\left (-\frac {f \sqrt {x}}{e}\right )}{f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \left (a +b \ln \left (c \,x^{n}\right )\right ) \ln \left (d \left (e +f \sqrt {x}\right )^{k}\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

1/9*(9*b*x*e*log(d)*log(x^n) - 9*((n*log(d) - log(c)*log(d))*b - a*log(d))*x*e + 9*(b*x*e*log(x^n) - (b*(n - l
og(c)) - a)*x*e)*k*log(f*sqrt(x) + e) - (3*b*f*k*x^2*log(x^n) + (3*a*f*k - (5*f*k*n - 3*f*k*log(c))*b)*x^2)/sq
rt(x))*e^(-1) + integrate(1/2*(b*f^2*k*x*log(x^n) + (a*f^2*k - (f^2*k*n - f^2*k*log(c))*b)*x)/(f*e^(1/2*log(x)
+ 1) + 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, algorithm="fricas")

[Out]

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

[Out]

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

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