3.250 \(\int \frac {\log (c (a+\frac {b}{x^2})^p)}{d+e x} \, dx\)

Optimal. Leaf size=241 \[ \frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}-\frac {p \text {Li}_2\left (\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e}-\frac {p \text {Li}_2\left (\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e}-\frac {p \log (d+e x) \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right )}{e}-\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a} x+\sqrt {b}\right )}{\sqrt {-a} d-\sqrt {b} e}\right )}{e}+\frac {2 p \text {Li}_2\left (\frac {e x}{d}+1\right )}{e}+\frac {2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2462, 260, 2416, 2394, 2315, 2393, 2391} \[ -\frac {p \text {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e}-\frac {p \text {PolyLog}\left (2,\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e}+\frac {2 p \text {PolyLog}\left (2,\frac {e x}{d}+1\right )}{e}+\frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}-\frac {p \log (d+e x) \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right )}{e}-\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a} x+\sqrt {b}\right )}{\sqrt {-a} d-\sqrt {b} e}\right )}{e}+\frac {2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x^2)^p]/(d + e*x),x]

[Out]

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

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 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2462

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[f +
g*x]*(a + b*Log[c*(d + e*x^n)^p]))/g, x] - Dist[(b*e*n*p)/g, Int[(x^(n - 1)*Log[f + g*x])/(d + e*x^n), x], x]
/; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && RationalQ[n]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 242, normalized size = 1.00 \[ \frac {\log (d+e x) \log \left (c \left (a+\frac {b}{x^2}\right )^p\right )}{e}-\frac {p \text {Li}_2\left (\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d-\sqrt {b} e}\right )}{e}-\frac {p \text {Li}_2\left (\frac {\sqrt {-a} (d+e x)}{\sqrt {-a} d+\sqrt {b} e}\right )}{e}-\frac {p \log (d+e x) \log \left (\frac {e \left (\sqrt {b}-\sqrt {-a} x\right )}{\sqrt {-a} d+\sqrt {b} e}\right )}{e}-\frac {p \log (d+e x) \log \left (-\frac {e \left (\sqrt {-a} x+\sqrt {b}\right )}{\sqrt {-a} d-\sqrt {b} e}\right )}{e}+\frac {2 p \text {Li}_2\left (\frac {d+e x}{d}\right )}{e}+\frac {2 p \log \left (-\frac {e x}{d}\right ) \log (d+e x)}{e} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x^2)^p]/(d + e*x),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left (c \left (\frac {a x^{2} + b}{x^{2}}\right )^{p}\right )}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(log(c*((a*x^2 + b)/x^2)^p)/(e*x + d), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((a + b/x^2)^p*c)/(e*x + d), x)

________________________________________________________________________________________

maple [F]  time = 0.42, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (a +\frac {b}{x^{2}}\right )^{p}\right )}{e x +d}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(a+b/x^2)^p)/(e*x+d),x)

[Out]

int(ln(c*(a+b/x^2)^p)/(e*x+d),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (a + \frac {b}{x^{2}}\right )}^{p} c\right )}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(log((a + b/x^2)^p*c)/(e*x + d), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\ln \left (c\,{\left (a+\frac {b}{x^2}\right )}^p\right )}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b/x^2)^p)/(d + e*x),x)

[Out]

int(log(c*(a + b/x^2)^p)/(d + e*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (c \left (a + \frac {b}{x^{2}}\right )^{p} \right )}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(log(c*(a + b/x**2)**p)/(d + e*x), x)

________________________________________________________________________________________