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

Optimal. Leaf size=229 \[ \frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b \left (\sqrt {-d}-\sqrt {e} x\right )}{a \sqrt {e}+b \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b \left (\sqrt {-d}+\sqrt {e} x\right )}{b \sqrt {-d}-a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \text {Li}_2\left (-\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}-a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \text {Li}_2\left (\frac {\sqrt {e} (a+b x)}{\sqrt {e} a+b \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2409, 2394, 2393, 2391} \[ -\frac {n \text {PolyLog}\left (2,-\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}-a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \text {PolyLog}\left (2,\frac {\sqrt {e} (a+b x)}{a \sqrt {e}+b \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b \left (\sqrt {-d}-\sqrt {e} x\right )}{a \sqrt {e}+b \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b \left (\sqrt {-d}+\sqrt {e} x\right )}{b \sqrt {-d}-a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x)^n]/(d + e*x^2),x]

[Out]

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

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 2409

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 178, normalized size = 0.78 \[ \frac {\log \left (c (a+b x)^n\right ) \left (\log \left (\frac {b \left (\sqrt {-d}-\sqrt {e} x\right )}{a \sqrt {e}+b \sqrt {-d}}\right )-\log \left (\frac {b \left (\sqrt {-d}+\sqrt {e} x\right )}{b \sqrt {-d}-a \sqrt {e}}\right )\right )-n \text {Li}_2\left (-\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}-a \sqrt {e}}\right )+n \text {Li}_2\left (\frac {\sqrt {e} (a+b x)}{\sqrt {e} a+b \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x + a\right )}^{n} c\right )}{e x^{2} + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.27, size = 419, normalized size = 1.83 \[ -\frac {i \pi \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )}{2 \sqrt {d e}}+\frac {i \pi \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2}}{2 \sqrt {d e}}+\frac {i \pi \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2}}{2 \sqrt {d e}}-\frac {i \pi \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{3}}{2 \sqrt {d e}}+\frac {n \ln \left (\frac {a e +\sqrt {-d e}\, b -\left (b x +a \right ) e}{a e +\sqrt {-d e}\, b}\right ) \ln \left (b x +a \right )}{2 \sqrt {-d e}}-\frac {n \ln \left (\frac {-a e +\sqrt {-d e}\, b +\left (b x +a \right ) e}{-a e +\sqrt {-d e}\, b}\right ) \ln \left (b x +a \right )}{2 \sqrt {-d e}}+\frac {n \dilog \left (\frac {a e +\sqrt {-d e}\, b -\left (b x +a \right ) e}{a e +\sqrt {-d e}\, b}\right )}{2 \sqrt {-d e}}-\frac {n \dilog \left (\frac {-a e +\sqrt {-d e}\, b +\left (b x +a \right ) e}{-a e +\sqrt {-d e}\, b}\right )}{2 \sqrt {-d e}}+\frac {\arctan \left (\frac {e x}{\sqrt {d e}}\right ) \ln \relax (c )}{\sqrt {d e}}+\frac {\left (-n \ln \left (b x +a \right )+\ln \left (\left (b x +a \right )^{n}\right )\right ) \arctan \left (\frac {-2 a e +2 \left (b x +a \right ) e}{2 \sqrt {d e}\, b}\right )}{\sqrt {d e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(ln((b*x+a)^n)-n*ln(b*x+a))/(d*e)^(1/2)*arctan(1/2*(2*(b*x+a)*e-2*a*e)/b/(d*e)^(1/2))+1/2*n*ln(b*x+a)/(-d*e)^(
1/2)*ln((b*(-d*e)^(1/2)-(b*x+a)*e+a*e)/(b*(-d*e)^(1/2)+a*e))-1/2*n*ln(b*x+a)/(-d*e)^(1/2)*ln((b*(-d*e)^(1/2)+(
b*x+a)*e-a*e)/(b*(-d*e)^(1/2)-a*e))+1/2*n/(-d*e)^(1/2)*dilog((b*(-d*e)^(1/2)-(b*x+a)*e+a*e)/(b*(-d*e)^(1/2)+a*
e))-1/2*n/(-d*e)^(1/2)*dilog((b*(-d*e)^(1/2)+(b*x+a)*e-a*e)/(b*(-d*e)^(1/2)-a*e))+1/2*I/(d*e)^(1/2)*arctan(1/(
d*e)^(1/2)*e*x)*Pi*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)^2-1/2*I/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*Pi*csgn
(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)*csgn(I*c)-1/2*I/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*Pi*csgn(I*c*(b*x+a)^n)
^3+1/2*I/(d*e)^(1/2)*arctan(1/(d*e)^(1/2)*e*x)*Pi*csgn(I*c*(b*x+a)^n)^2*csgn(I*c)+1/(d*e)^(1/2)*arctan(1/(d*e)
^(1/2)*e*x)*ln(c)

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maxima [C]  time = 1.24, size = 309, normalized size = 1.35 \[ \frac {b n {\left (\frac {2 \, \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \log \left (b x + a\right )}{b} + \frac {\arctan \left (\frac {{\left (b^{2} x + a b\right )} \sqrt {d} \sqrt {e}}{b^{2} d + a^{2} e}, \frac {a b e x + a^{2} e}{b^{2} d + a^{2} e}\right ) \log \left (e x^{2} + d\right ) - \arctan \left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (\frac {b^{2} e x^{2} + 2 \, a b e x + a^{2} e}{b^{2} d + a^{2} e}\right ) - i \, {\rm Li}_2\left (\frac {a b e x + b^{2} d - {\left (i \, b^{2} x - i \, a b\right )} \sqrt {d} \sqrt {e}}{b^{2} d + 2 i \, a b \sqrt {d} \sqrt {e} - a^{2} e}\right ) + i \, {\rm Li}_2\left (\frac {a b e x + b^{2} d + {\left (i \, b^{2} x - i \, a b\right )} \sqrt {d} \sqrt {e}}{b^{2} d - 2 i \, a b \sqrt {d} \sqrt {e} - a^{2} e}\right )}{b}\right )}}{2 \, \sqrt {d e}} - \frac {n \arctan \left (\frac {e x}{\sqrt {d e}}\right ) \log \left (b x + a\right )}{\sqrt {d e}} + \frac {\arctan \left (\frac {e x}{\sqrt {d e}}\right ) \log \left ({\left (b x + a\right )}^{n} c\right )}{\sqrt {d e}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*n*(2*arctan(e*x/sqrt(d*e))*log(b*x + a)/b + (arctan2((b^2*x + a*b)*sqrt(d)*sqrt(e)/(b^2*d + a^2*e), (a*b
*e*x + a^2*e)/(b^2*d + a^2*e))*log(e*x^2 + d) - arctan(sqrt(e)*x/sqrt(d))*log((b^2*e*x^2 + 2*a*b*e*x + a^2*e)/
(b^2*d + a^2*e)) - I*dilog((a*b*e*x + b^2*d - (I*b^2*x - I*a*b)*sqrt(d)*sqrt(e))/(b^2*d + 2*I*a*b*sqrt(d)*sqrt
(e) - a^2*e)) + I*dilog((a*b*e*x + b^2*d + (I*b^2*x - I*a*b)*sqrt(d)*sqrt(e))/(b^2*d - 2*I*a*b*sqrt(d)*sqrt(e)
- a^2*e)))/b)/sqrt(d*e) - n*arctan(e*x/sqrt(d*e))*log(b*x + a)/sqrt(d*e) + arctan(e*x/sqrt(d*e))*log((b*x + a
)^n*c)/sqrt(d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x)^n)/(d + e*x^2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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