3.4.31 \(\int \frac {\log (c (a+b x)^n)}{d+e x^2} \, dx\) [331]

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

[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 = 0.12, 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 = {2456, 2441, 2440, 2438} \begin {gather*} -\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}} \end {gather*}

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 2438

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 2440

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 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 2456

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 \text {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 \text {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 = 0.07, size = 178, normalized size = 0.78 \begin {gather*} \frac {\log \left (c (a+b x)^n\right ) \left (\log \left (\frac {b \left (\sqrt {-d}-\sqrt {e} x\right )}{b \sqrt {-d}+a \sqrt {e}}\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)}{b \sqrt {-d}+a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}} \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.42, size = 419, normalized size = 1.83

method	result	size
risch	\(\frac {\left (\ln \left (\left (b x +a \right )^{n}\right )-n \ln \left (b x +a \right )\right ) \arctan \left (\frac {2 \left (b x +a \right ) e -2 a e}{2 \sqrt {e d}\, b}\right )}{\sqrt {e d}}+\frac {n \ln \left (b x +a \right ) \ln \left (\frac {b \sqrt {-e d}-\left (b x +a \right ) e +a e}{b \sqrt {-e d}+a e}\right )}{2 \sqrt {-e d}}-\frac {n \ln \left (b x +a \right ) \ln \left (\frac {b \sqrt {-e d}+\left (b x +a \right ) e -a e}{b \sqrt {-e d}-a e}\right )}{2 \sqrt {-e d}}+\frac {n \dilog \left (\frac {b \sqrt {-e d}-\left (b x +a \right ) e +a e}{b \sqrt {-e d}+a e}\right )}{2 \sqrt {-e d}}-\frac {n \dilog \left (\frac {b \sqrt {-e d}+\left (b x +a \right ) e -a e}{b \sqrt {-e d}-a e}\right )}{2 \sqrt {-e d}}-\frac {i \arctan \left (\frac {x e}{\sqrt {e d}}\right ) \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{3}}{2 \sqrt {e d}}+\frac {i \arctan \left (\frac {x e}{\sqrt {e d}}\right ) \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \mathrm {csgn}\left (i c \right )}{2 \sqrt {e d}}+\frac {i \arctan \left (\frac {x e}{\sqrt {e d}}\right ) \pi \mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \mathrm {csgn}\left (i \left (b x +a \right )^{n}\right )}{2 \sqrt {e d}}-\frac {i \arctan \left (\frac {x e}{\sqrt {e d}}\right ) \pi \,\mathrm {csgn}\left (i c \left (b x +a \right )^{n}\right ) \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b x +a \right )^{n}\right )}{2 \sqrt {e d}}+\frac {\arctan \left (\frac {x e}{\sqrt {e d}}\right ) \ln \left (c \right )}{\sqrt {e d}}\)	\(419\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.54, size = 317, normalized size = 1.38 \begin {gather*} \frac {b n {\left (\frac {2 \, \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) \log \left (b x + a\right )}{b} + \frac {\arctan \left (\frac {{\left (b^{2} x + a b\right )} \sqrt {d} e^{\frac {1}{2}}}{b^{2} d + a^{2} e}, \frac {a b x e + a^{2} e}{b^{2} d + a^{2} e}\right ) \log \left (x^{2} e^{2} + d e\right ) - \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) \log \left (\frac {b^{2} x^{2} e + 2 \, a b x e + a^{2} e}{b^{2} d + a^{2} e}\right ) - i \, {\rm Li}_2\left (\frac {a b x e + b^{2} d - {\left (i \, b^{2} x - i \, a b\right )} \sqrt {d} e^{\frac {1}{2}}}{2 i \, a b \sqrt {d} e^{\frac {1}{2}} + b^{2} d - a^{2} e}\right ) + i \, {\rm Li}_2\left (-\frac {a b x e + b^{2} d + {\left (i \, b^{2} x - i \, a b\right )} \sqrt {d} e^{\frac {1}{2}}}{2 i \, a b \sqrt {d} e^{\frac {1}{2}} - b^{2} d + a^{2} e}\right )}{b}\right )} e^{\left (-\frac {1}{2}\right )}}{2 \, \sqrt {d}} - \frac {n \arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )} \log \left (b x + a\right )}{\sqrt {d}} + \frac {\arctan \left (\frac {x e^{\frac {1}{2}}}{\sqrt {d}}\right ) e^{\left (-\frac {1}{2}\right )} \log \left ({\left (b x + a\right )}^{n} c\right )}{\sqrt {d}} \end {gather*}

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

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

[Out]

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

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

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

[Out]

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

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

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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