3.3.49 \(\int \frac {a+b \log (c (d x^m)^n)}{e+f x^2} \, dx\) [249]

Optimal. Leaf size=111 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{\sqrt {e} \sqrt {f}}-\frac {i b m n \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{2 \sqrt {e} \sqrt {f}}+\frac {i b m n \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{2 \sqrt {e} \sqrt {f}} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {211, 2361, 12, 4940, 2438, 2495} \begin {gather*} -\frac {i b m n \text {PolyLog}\left (2,-\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{2 \sqrt {e} \sqrt {f}}+\frac {i b m n \text {PolyLog}\left (2,\frac {i \sqrt {f} x}{\sqrt {e}}\right )}{2 \sqrt {e} \sqrt {f}}+\frac {\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c \left (d x^m\right )^n\right )\right )}{\sqrt {e} \sqrt {f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[(Sqrt[f]*x)/Sqrt[e]]*(a + b*Log[c*(d*x^m)^n]))/(Sqrt[e]*Sqrt[f]) - ((I/2)*b*m*n*PolyLog[2, ((-I)*Sqrt[
f]*x)/Sqrt[e]])/(Sqrt[e]*Sqrt[f]) + ((I/2)*b*m*n*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]])/(Sqrt[e]*Sqrt[f])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

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 2495

Int[((a_.) + Log[(c_.)*((d_.)*((e_.) + (f_.)*(x_))^(m_.))^(n_)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Subst[Int[u*(
a + b*Log[c*d^n*(e + f*x)^(m*n)])^p, x], c*d^n*(e + f*x)^(m*n), c*(d*(e + f*x)^m)^n] /; FreeQ[{a, b, c, d, e,
f, m, n, p}, x] &&  !IntegerQ[n] &&  !(EqQ[d, 1] && EqQ[m, 1]) && IntegralFreeQ[IntHide[u*(a + b*Log[c*d^n*(e
+ f*x)^(m*n)])^p, x]]

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[I*(b/2), Int[Log[1 - I*c*x
]/x, x], x] - Dist[I*(b/2), Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 113, normalized size = 1.02 \begin {gather*} \frac {-\left (\left (a+b \log \left (c \left (d x^m\right )^n\right )\right ) \left (\log \left (1+\frac {\sqrt {f} x}{\sqrt {-e}}\right )-\log \left (1+\frac {e \sqrt {f} x}{(-e)^{3/2}}\right )\right )\right )+b m n \text {Li}_2\left (\frac {\sqrt {f} x}{\sqrt {-e}}\right )-b m n \text {Li}_2\left (\frac {e \sqrt {f} x}{(-e)^{3/2}}\right )}{2 \sqrt {-e} \sqrt {f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*Log[c*(d*x^m)^n])*(Log[1 + (Sqrt[f]*x)/Sqrt[-e]] - Log[1 + (e*Sqrt[f]*x)/(-e)^(3/2)])) + b*m*n*PolyL
og[2, (Sqrt[f]*x)/Sqrt[-e]] - b*m*n*PolyLog[2, (e*Sqrt[f]*x)/(-e)^(3/2)])/(2*Sqrt[-e]*Sqrt[f])

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d \,x^{m}\right )^{n}\right )}{f \,x^{2}+e}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*arctan(sqrt(f)*x*e^(-1/2))*e^(-1/2)/sqrt(f) + b*integrate((n*log(d) + log(c) + log((x^m)^n))/(f*x^2 + e), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d\,x^m\right )}^n\right )}{f\,x^2+e} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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