\(\int \frac {\log (c (a+b x^2)^p)}{x} \, dx\) [6]

Optimal result

Integrand size = 16, antiderivative size = 44 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx=\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+\frac {1}{2} p \operatorname {PolyLog}\left (2,1+\frac {b x^2}{a}\right ) \]

[Out]

1/2*ln(-b*x^2/a)*ln(c*(b*x^2+a)^p)+1/2*p*polylog(2,1+b*x^2/a)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2504, 2441, 2352} \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx=\frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+\frac {1}{2} p \operatorname {PolyLog}\left (2,\frac {b x^2}{a}+1\right ) \]

[In]

Int[Log[c*(a + b*x^2)^p]/x,x]

[Out]

(Log[-((b*x^2)/a)]*Log[c*(a + b*x^2)^p])/2 + (p*PolyLog[2, 1 + (b*x^2)/a])/2

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 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 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*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {\log \left (c (a+b x)^p\right )}{x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )-\frac {1}{2} (b p) \text {Subst}\left (\int \frac {\log \left (-\frac {b x}{a}\right )}{a+b x} \, dx,x,x^2\right ) \\ & = \frac {1}{2} \log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+\frac {1}{2} p \text {Li}_2\left (1+\frac {b x^2}{a}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx=\frac {1}{2} \left (\log \left (-\frac {b x^2}{a}\right ) \log \left (c \left (a+b x^2\right )^p\right )+p \operatorname {PolyLog}\left (2,\frac {a+b x^2}{a}\right )\right ) \]

[In]

Integrate[Log[c*(a + b*x^2)^p]/x,x]

[Out]

(Log[-((b*x^2)/a)]*Log[c*(a + b*x^2)^p] + p*PolyLog[2, (a + b*x^2)/a])/2

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(109\) vs. \(2(40)=80\).

Time = 0.23 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.50

method	result	size
parts	\(\ln \left (c \left (b \,x^{2}+a \right )^{p}\right ) \ln \left (x \right )-2 p b \left (\frac {\ln \left (x \right ) \left (\ln \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\ln \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )\right )}{2 b}+\frac {\operatorname {dilog}\left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\operatorname {dilog}\left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )}{2 b}\right )\)	\(110\)
risch	\(\ln \left (\left (b \,x^{2}+a \right )^{p}\right ) \ln \left (x \right )-p \ln \left (x \right ) \ln \left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )-p \ln \left (x \right ) \ln \left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )-p \operatorname {dilog}\left (\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}\right )-p \operatorname {dilog}\left (\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}\right )+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \ln \left (x \right )\)	\(225\)

[In]

int(ln(c*(b*x^2+a)^p)/x,x,method=_RETURNVERBOSE)

[Out]

ln(c*(b*x^2+a)^p)*ln(x)-2*p*b*(1/2*ln(x)*(ln((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+ln((b*x+(-a*b)^(1/2))/(-a*b)^(1
/2)))/b+1/2*(dilog((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))+dilog((b*x+(-a*b)^(1/2))/(-a*b)^(1/2)))/b)

Fricas [F]

\[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{x} \,d x } \]

[In]

integrate(log(c*(b*x^2+a)^p)/x,x, algorithm="fricas")

[Out]

integral(log((b*x^2 + a)^p*c)/x, x)

Sympy [F]

\[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx=\int \frac {\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}{x}\, dx \]

[In]

integrate(ln(c*(b*x**2+a)**p)/x,x)

[Out]

Integral(log(c*(a + b*x**2)**p)/x, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (39) = 78\).

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.82 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx=\frac {1}{2} \, b p {\left (\frac {2 \, \log \left (b x^{2} + a\right ) \log \left (x\right )}{b} - \frac {2 \, \log \left (\frac {b x^{2}}{a} + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-\frac {b x^{2}}{a}\right )}{b}\right )} - p \log \left (b x^{2} + a\right ) \log \left (x\right ) + \log \left ({\left (b x^{2} + a\right )}^{p} c\right ) \log \left (x\right ) \]

[In]

integrate(log(c*(b*x^2+a)^p)/x,x, algorithm="maxima")

[Out]

1/2*b*p*(2*log(b*x^2 + a)*log(x)/b - (2*log(b*x^2/a + 1)*log(x) + dilog(-b*x^2/a))/b) - p*log(b*x^2 + a)*log(x
) + log((b*x^2 + a)^p*c)*log(x)

Giac [F]

\[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx=\int { \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{x} \,d x } \]

[In]

integrate(log(c*(b*x^2+a)^p)/x,x, algorithm="giac")

[Out]

integrate(log((b*x^2 + a)^p*c)/x, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{x} \, dx=\int \frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{x} \,d x \]

[In]

int(log(c*(a + b*x^2)^p)/x,x)

[Out]

int(log(c*(a + b*x^2)^p)/x, x)