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

Optimal. Leaf size=50 \[ \frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d \left (d+e x^2\right )}-\frac{b n \log \left (d+e x^2\right )}{4 d e} \]

[Out]

(x^2*(a + b*Log[c*x^n]))/(2*d*(d + e*x^2)) - (b*n*Log[d + e*x^2])/(4*d*e)

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Rubi [A]  time = 0.041939, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2335, 260} \[ \frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d \left (d+e x^2\right )}-\frac{b n \log \left (d+e x^2\right )}{4 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(a + b*Log[c*x^n]))/(2*d*(d + e*x^2)) - (b*n*Log[d + e*x^2])/(4*d*e)

Rule 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

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]

Rubi steps

\begin{align*} \int \frac{x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx &=\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d \left (d+e x^2\right )}-\frac{(b n) \int \frac{x}{d+e x^2} \, dx}{2 d}\\ &=\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 d \left (d+e x^2\right )}-\frac{b n \log \left (d+e x^2\right )}{4 d e}\\ \end{align*}

Mathematica [A]  time = 0.0648603, size = 74, normalized size = 1.48 \[ -\frac{2 a d+2 b d \log \left (c x^n\right )+b e n x^2 \log \left (d+e x^2\right )-2 b n \log (x) \left (d+e x^2\right )+b d n \log \left (d+e x^2\right )}{4 d e \left (d+e x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*a*d - 2*b*n*(d + e*x^2)*Log[x] + 2*b*d*Log[c*x^n] + b*d*n*Log[d + e*x^2] + b*e*n*x^2*Log[d + e*x^2])/(4*d*
e*(d + e*x^2))

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Maple [C]  time = 0.096, size = 179, normalized size = 3.6 \begin{align*} -{\frac{b\ln \left ({x}^{n} \right ) }{2\, \left ( e{x}^{2}+d \right ) e}}-{\frac{i\pi \,bd{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-i\pi \,bd{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -i\pi \,bd \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+i\pi \,bd \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -2\,\ln \left ( x \right ) ben{x}^{2}+\ln \left ( e{x}^{2}+d \right ) ben{x}^{2}-2\,\ln \left ( x \right ) bdn+\ln \left ( e{x}^{2}+d \right ) bdn+2\,\ln \left ( c \right ) bd+2\,ad}{ \left ( 4\,e{x}^{2}+4\,d \right ) ed}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/e/(e*x^2+d)*ln(x^n)-1/4*(I*Pi*b*d*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*d*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I
*c)-I*Pi*b*d*csgn(I*c*x^n)^3+I*Pi*b*d*csgn(I*c*x^n)^2*csgn(I*c)-2*ln(x)*b*e*n*x^2+ln(e*x^2+d)*b*e*n*x^2-2*ln(x
)*b*d*n+ln(e*x^2+d)*b*d*n+2*ln(c)*b*d+2*a*d)/(e*x^2+d)/e/d

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Maxima [A]  time = 1.19873, size = 96, normalized size = 1.92 \begin{align*} -\frac{1}{4} \, b n{\left (\frac{\log \left (e x^{2} + d\right )}{d e} - \frac{\log \left (x^{2}\right )}{d e}\right )} - \frac{b \log \left (c x^{n}\right )}{2 \,{\left (e^{2} x^{2} + d e\right )}} - \frac{a}{2 \,{\left (e^{2} x^{2} + d e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*b*n*(log(e*x^2 + d)/(d*e) - log(x^2)/(d*e)) - 1/2*b*log(c*x^n)/(e^2*x^2 + d*e) - 1/2*a/(e^2*x^2 + d*e)

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Fricas [A]  time = 1.34862, size = 143, normalized size = 2.86 \begin{align*} \frac{2 \, b e n x^{2} \log \left (x\right ) - 2 \, b d \log \left (c\right ) - 2 \, a d -{\left (b e n x^{2} + b d n\right )} \log \left (e x^{2} + d\right )}{4 \,{\left (d e^{2} x^{2} + d^{2} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b*e*n*x^2*log(x) - 2*b*d*log(c) - 2*a*d - (b*e*n*x^2 + b*d*n)*log(e*x^2 + d))/(d*e^2*x^2 + d^2*e)

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Sympy [A]  time = 78.7155, size = 366, normalized size = 7.32 \begin{align*} \begin{cases} \tilde{\infty } \left (- \frac{a}{2 x^{2}} - \frac{b n \log{\left (x \right )}}{2 x^{2}} - \frac{b n}{4 x^{2}} - \frac{b \log{\left (c \right )}}{2 x^{2}}\right ) & \text{for}\: d = 0 \wedge e = 0 \\\frac{\frac{a x^{2}}{2} + \frac{b n x^{2} \log{\left (x \right )}}{2} - \frac{b n x^{2}}{4} + \frac{b x^{2} \log{\left (c \right )}}{2}}{d^{2}} & \text{for}\: e = 0 \\\frac{- \frac{a}{2 x^{2}} - \frac{b n \log{\left (x \right )}}{2 x^{2}} - \frac{b n}{4 x^{2}} - \frac{b \log{\left (c \right )}}{2 x^{2}}}{e^{2}} & \text{for}\: d = 0 \\- \frac{2 a d}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac{b d n \log{\left (- i \sqrt{d} \sqrt{\frac{1}{e}} + x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac{b d n \log{\left (i \sqrt{d} \sqrt{\frac{1}{e}} + x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} + \frac{2 b e n x^{2} \log{\left (x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac{b e n x^{2} \log{\left (- i \sqrt{d} \sqrt{\frac{1}{e}} + x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} - \frac{b e n x^{2} \log{\left (i \sqrt{d} \sqrt{\frac{1}{e}} + x \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} + \frac{2 b e x^{2} \log{\left (c \right )}}{4 d^{2} e + 4 d e^{2} x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*(-a/(2*x**2) - b*n*log(x)/(2*x**2) - b*n/(4*x**2) - b*log(c)/(2*x**2)), Eq(d, 0) & Eq(e, 0)), (
(a*x**2/2 + b*n*x**2*log(x)/2 - b*n*x**2/4 + b*x**2*log(c)/2)/d**2, Eq(e, 0)), ((-a/(2*x**2) - b*n*log(x)/(2*x
**2) - b*n/(4*x**2) - b*log(c)/(2*x**2))/e**2, Eq(d, 0)), (-2*a*d/(4*d**2*e + 4*d*e**2*x**2) - b*d*n*log(-I*sq
rt(d)*sqrt(1/e) + x)/(4*d**2*e + 4*d*e**2*x**2) - b*d*n*log(I*sqrt(d)*sqrt(1/e) + x)/(4*d**2*e + 4*d*e**2*x**2
) + 2*b*e*n*x**2*log(x)/(4*d**2*e + 4*d*e**2*x**2) - b*e*n*x**2*log(-I*sqrt(d)*sqrt(1/e) + x)/(4*d**2*e + 4*d*
e**2*x**2) - b*e*n*x**2*log(I*sqrt(d)*sqrt(1/e) + x)/(4*d**2*e + 4*d*e**2*x**2) + 2*b*e*x**2*log(c)/(4*d**2*e
+ 4*d*e**2*x**2), True))

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Giac [A]  time = 1.33931, size = 95, normalized size = 1.9 \begin{align*} -\frac{b n x^{2} e \log \left (x^{2} e + d\right ) - 2 \, b n x^{2} e \log \left (x\right ) + b d n \log \left (x^{2} e + d\right ) + 2 \, b d \log \left (c\right ) + 2 \, a d}{4 \,{\left (d x^{2} e^{2} + d^{2} e\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*n*x^2*e*log(x^2*e + d) - 2*b*n*x^2*e*log(x) + b*d*n*log(x^2*e + d) + 2*b*d*log(c) + 2*a*d)/(d*x^2*e^2
+ d^2*e)