Integrand size = 18, antiderivative size = 46 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx=2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \log \left (-\frac {b \sqrt {x}}{a}\right )+2 p \operatorname {PolyLog}\left (2,1+\frac {b \sqrt {x}}{a}\right ) \] Output:
2*ln(c*(a+b*x^(1/2))^p)*ln(-b*x^(1/2)/a)+2*p*polylog(2,1+b*x^(1/2)/a)
Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.02 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx=2 \log \left (c \left (a+b \sqrt {x}\right )^p\right ) \log \left (-\frac {b \sqrt {x}}{a}\right )+2 p \operatorname {PolyLog}\left (2,\frac {a+b \sqrt {x}}{a}\right ) \] Input:
Integrate[Log[c*(a + b*Sqrt[x])^p]/x,x]
Output:
2*Log[c*(a + b*Sqrt[x])^p]*Log[-((b*Sqrt[x])/a)] + 2*p*PolyLog[2, (a + b*S qrt[x])/a]
Time = 0.40 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2904, 2841, 2752}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle 2 \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{\sqrt {x}}d\sqrt {x}\) |
\(\Big \downarrow \) 2841 |
\(\displaystyle 2 \left (\log \left (-\frac {b \sqrt {x}}{a}\right ) \log \left (c \left (a+b \sqrt {x}\right )^p\right )-b p \int \frac {\log \left (-\frac {b \sqrt {x}}{a}\right )}{a+b \sqrt {x}}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 2752 |
\(\displaystyle 2 \left (\log \left (-\frac {b \sqrt {x}}{a}\right ) \log \left (c \left (a+b \sqrt {x}\right )^p\right )+p \operatorname {PolyLog}\left (2,\frac {\sqrt {x} b}{a}+1\right )\right )\) |
Input:
Int[Log[c*(a + b*Sqrt[x])^p]/x,x]
Output:
2*(Log[c*(a + b*Sqrt[x])^p]*Log[-((b*Sqrt[x])/a)] + p*PolyLog[2, 1 + (b*Sq rt[x])/a])
Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLo g[2, 1 - c*x], x] /; FreeQ[{c, d, e}, x] && EqQ[e + c*d, 0]
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] - Simp[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]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[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])
Time = 0.64 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.26
method | result | size |
parts | \(\ln \left (c \left (a +b \sqrt {x}\right )^{p}\right ) \ln \left (x \right )-\frac {p b \left (\frac {4 \operatorname {dilog}\left (\frac {a +b \sqrt {x}}{a}\right )}{b}+\frac {2 \ln \left (x \right ) \ln \left (\frac {a +b \sqrt {x}}{a}\right )}{b}\right )}{2}\) | \(58\) |
Input:
int(ln(c*(a+b*x^(1/2))^p)/x,x,method=_RETURNVERBOSE)
Output:
ln(c*(a+b*x^(1/2))^p)*ln(x)-1/2*p*b*(4*dilog((a+b*x^(1/2))/a)/b+2*ln(x)*ln ((a+b*x^(1/2))/a)/b)
\[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx=\int { \frac {\log \left ({\left (b \sqrt {x} + a\right )}^{p} c\right )}{x} \,d x } \] Input:
integrate(log(c*(a+b*x^(1/2))^p)/x,x, algorithm="fricas")
Output:
integral(log((b*sqrt(x) + a)^p*c)/x, x)
\[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx=\int \frac {\log {\left (c \left (a + b \sqrt {x}\right )^{p} \right )}}{x}\, dx \] Input:
integrate(ln(c*(a+b*x**(1/2))**p)/x,x)
Output:
Integral(log(c*(a + b*sqrt(x))**p)/x, x)
Leaf count of result is larger than twice the leaf count of optimal. 79 vs. \(2 (39) = 78\).
Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.72 \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx=b p {\left (\frac {\log \left (b \sqrt {x} + a\right ) \log \left (x\right )}{b} - \frac {\log \left (x\right ) \log \left (\frac {b \sqrt {x}}{a} + 1\right ) + 2 \, {\rm Li}_2\left (-\frac {b \sqrt {x}}{a}\right )}{b}\right )} - p \log \left (b \sqrt {x} + a\right ) \log \left (x\right ) + \log \left ({\left (b \sqrt {x} + a\right )}^{p} c\right ) \log \left (x\right ) \] Input:
integrate(log(c*(a+b*x^(1/2))^p)/x,x, algorithm="maxima")
Output:
b*p*(log(b*sqrt(x) + a)*log(x)/b - (log(x)*log(b*sqrt(x)/a + 1) + 2*dilog( -b*sqrt(x)/a))/b) - p*log(b*sqrt(x) + a)*log(x) + log((b*sqrt(x) + a)^p*c) *log(x)
\[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx=\int { \frac {\log \left ({\left (b \sqrt {x} + a\right )}^{p} c\right )}{x} \,d x } \] Input:
integrate(log(c*(a+b*x^(1/2))^p)/x,x, algorithm="giac")
Output:
integrate(log((b*sqrt(x) + a)^p*c)/x, x)
Timed out. \[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (a+b\,\sqrt {x}\right )}^p\right )}{x} \,d x \] Input:
int(log(c*(a + b*x^(1/2))^p)/x,x)
Output:
int(log(c*(a + b*x^(1/2))^p)/x, x)
\[ \int \frac {\log \left (c \left (a+b \sqrt {x}\right )^p\right )}{x} \, dx=\frac {\left (\int \frac {\mathrm {log}\left (\left (\sqrt {x}\, b +a \right )^{p} c \right )}{-b^{2} x^{2}+a^{2} x}d x \right ) a^{2} p -\left (\int \frac {\sqrt {x}\, \mathrm {log}\left (\left (\sqrt {x}\, b +a \right )^{p} c \right )}{-b^{2} x^{2}+a^{2} x}d x \right ) a b p +\mathrm {log}\left (\left (\sqrt {x}\, b +a \right )^{p} c \right )^{2}}{p} \] Input:
int(log(c*(a+b*x^(1/2))^p)/x,x)
Output:
(int(log((sqrt(x)*b + a)**p*c)/(a**2*x - b**2*x**2),x)*a**2*p - int((sqrt( x)*log((sqrt(x)*b + a)**p*c))/(a**2*x - b**2*x**2),x)*a*b*p + log((sqrt(x) *b + a)**p*c)**2)/p