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

Optimal. Leaf size=139 \[ -\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}-\frac {b e^2 n \log \left (f+g x^2\right )}{4 g \left (d^2 g+e^2 f\right )}+\frac {b e^2 n \log (d+e x)}{2 g \left (d^2 g+e^2 f\right )}+\frac {b d e n \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g} \left (d^2 g+e^2 f\right )} \]

[Out]

1/2*b*e^2*n*ln(e*x+d)/g/(d^2*g+e^2*f)+1/2*(-a-b*ln(c*(e*x+d)^n))/g/(g*x^2+f)-1/4*b*e^2*n*ln(g*x^2+f)/g/(d^2*g+
e^2*f)+1/2*b*d*e*n*arctan(x*g^(1/2)/f^(1/2))/(d^2*g+e^2*f)/f^(1/2)/g^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2413, 706, 31, 635, 205, 260} \[ -\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}-\frac {b e^2 n \log \left (f+g x^2\right )}{4 g \left (d^2 g+e^2 f\right )}+\frac {b e^2 n \log (d+e x)}{2 g \left (d^2 g+e^2 f\right )}+\frac {b d e n \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g} \left (d^2 g+e^2 f\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 205

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

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 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 2413

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

Rubi steps

\begin {align*} \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx &=-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}+\frac {(b e n) \int \frac {1}{(d+e x) \left (f+g x^2\right )} \, dx}{2 g}\\ &=-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}+\frac {(b e n) \int \frac {d g-e g x}{f+g x^2} \, dx}{2 g \left (e^2 f+d^2 g\right )}+\frac {\left (b e^3 n\right ) \int \frac {1}{d+e x} \, dx}{2 g \left (e^2 f+d^2 g\right )}\\ &=\frac {b e^2 n \log (d+e x)}{2 g \left (e^2 f+d^2 g\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}+\frac {(b d e n) \int \frac {1}{f+g x^2} \, dx}{2 \left (e^2 f+d^2 g\right )}-\frac {\left (b e^2 n\right ) \int \frac {x}{f+g x^2} \, dx}{2 \left (e^2 f+d^2 g\right )}\\ &=\frac {b d e n \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 \sqrt {f} \sqrt {g} \left (e^2 f+d^2 g\right )}+\frac {b e^2 n \log (d+e x)}{2 g \left (e^2 f+d^2 g\right )}-\frac {a+b \log \left (c (d+e x)^n\right )}{2 g \left (f+g x^2\right )}-\frac {b e^2 n \log \left (f+g x^2\right )}{4 g \left (e^2 f+d^2 g\right )}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 165, normalized size = 1.19 \[ \frac {2 b d e \sqrt {g} n \left (f+g x^2\right ) \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )-\sqrt {f} \left (2 a d^2 g+2 a e^2 f+2 b \left (d^2 g+e^2 f\right ) \log \left (c (d+e x)^n\right )-2 b e^2 n \left (f+g x^2\right ) \log (d+e x)+b e^2 g n x^2 \log \left (f+g x^2\right )+b e^2 f n \log \left (f+g x^2\right )\right )}{4 \sqrt {f} g \left (f+g x^2\right ) \left (d^2 g+e^2 f\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 373, normalized size = 2.68 \[ \left [-\frac {2 \, a e^{2} f^{2} + 2 \, a d^{2} f g + {\left (b d e g n x^{2} + b d e f n\right )} \sqrt {-f g} \log \left (\frac {g x^{2} - 2 \, \sqrt {-f g} x - f}{g x^{2} + f}\right ) + {\left (b e^{2} f g n x^{2} + b e^{2} f^{2} n\right )} \log \left (g x^{2} + f\right ) - 2 \, {\left (b e^{2} f g n x^{2} - b d^{2} f g n\right )} \log \left (e x + d\right ) + 2 \, {\left (b e^{2} f^{2} + b d^{2} f g\right )} \log \relax (c)}{4 \, {\left (e^{2} f^{3} g + d^{2} f^{2} g^{2} + {\left (e^{2} f^{2} g^{2} + d^{2} f g^{3}\right )} x^{2}\right )}}, -\frac {2 \, a e^{2} f^{2} + 2 \, a d^{2} f g - 2 \, {\left (b d e g n x^{2} + b d e f n\right )} \sqrt {f g} \arctan \left (\frac {\sqrt {f g} x}{f}\right ) + {\left (b e^{2} f g n x^{2} + b e^{2} f^{2} n\right )} \log \left (g x^{2} + f\right ) - 2 \, {\left (b e^{2} f g n x^{2} - b d^{2} f g n\right )} \log \left (e x + d\right ) + 2 \, {\left (b e^{2} f^{2} + b d^{2} f g\right )} \log \relax (c)}{4 \, {\left (e^{2} f^{3} g + d^{2} f^{2} g^{2} + {\left (e^{2} f^{2} g^{2} + d^{2} f g^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*a*e^2*f^2 + 2*a*d^2*f*g + (b*d*e*g*n*x^2 + b*d*e*f*n)*sqrt(-f*g)*log((g*x^2 - 2*sqrt(-f*g)*x - f)/(g*
x^2 + f)) + (b*e^2*f*g*n*x^2 + b*e^2*f^2*n)*log(g*x^2 + f) - 2*(b*e^2*f*g*n*x^2 - b*d^2*f*g*n)*log(e*x + d) +
2*(b*e^2*f^2 + b*d^2*f*g)*log(c))/(e^2*f^3*g + d^2*f^2*g^2 + (e^2*f^2*g^2 + d^2*f*g^3)*x^2), -1/4*(2*a*e^2*f^2
+ 2*a*d^2*f*g - 2*(b*d*e*g*n*x^2 + b*d*e*f*n)*sqrt(f*g)*arctan(sqrt(f*g)*x/f) + (b*e^2*f*g*n*x^2 + b*e^2*f^2*
n)*log(g*x^2 + f) - 2*(b*e^2*f*g*n*x^2 - b*d^2*f*g*n)*log(e*x + d) + 2*(b*e^2*f^2 + b*d^2*f*g)*log(c))/(e^2*f^
3*g + d^2*f^2*g^2 + (e^2*f^2*g^2 + d^2*f*g^3)*x^2)]

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giac [A]  time = 0.23, size = 218, normalized size = 1.57 \[ \frac {b d n \arctan \left (\frac {g x}{\sqrt {f g}}\right ) e}{2 \, {\left (d^{2} g + f e^{2}\right )} \sqrt {f g}} - \frac {b n e^{2} \log \left (g x^{2} + f\right )}{4 \, {\left (d^{2} g^{2} + f g e^{2}\right )}} + \frac {b g n x^{2} e^{2} \log \left (x e + d\right ) - b d^{2} g n \log \left (x e + d\right ) - 2 \, b d^{2} g \log \relax (c) - 2 \, a d^{2} g - 2 \, b f e^{2} \log \relax (c) - 2 \, a f e^{2}}{2 \, {\left (d^{2} g^{3} x^{2} + f g^{2} x^{2} e^{2} + d^{2} f g^{2} + f^{2} g e^{2}\right )}} - \frac {b d^{2} g \log \relax (c) + a d^{2} g + b f e^{2} \log \relax (c) + a f e^{2}}{2 \, {\left (d^{2} g + f e^{2}\right )} {\left (g x^{2} + f\right )} g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*d*n*arctan(g*x/sqrt(f*g))*e/((d^2*g + f*e^2)*sqrt(f*g)) - 1/4*b*n*e^2*log(g*x^2 + f)/(d^2*g^2 + f*g*e^2)
+ 1/2*(b*g*n*x^2*e^2*log(x*e + d) - b*d^2*g*n*log(x*e + d) - 2*b*d^2*g*log(c) - 2*a*d^2*g - 2*b*f*e^2*log(c)
- 2*a*f*e^2)/(d^2*g^3*x^2 + f*g^2*x^2*e^2 + d^2*f*g^2 + f^2*g*e^2) - 1/2*(b*d^2*g*log(c) + a*d^2*g + b*f*e^2*l
og(c) + a*f*e^2)/((d^2*g + f*e^2)*(g*x^2 + f)*g)

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maple [C]  time = 0.56, size = 765, normalized size = 5.50 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b/g/(g*x^2+f)*ln((e*x+d)^n)+1/4*(I*Pi*b*e^2*f*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)-I*Pi*b*e^2*
f*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-I*Pi*b*d^2*g*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+I*Pi*b*d^2*g*csgn(I*c*(
e*x+d)^n)^3+I*Pi*b*d^2*g*csgn(I*c)*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)+I*Pi*b*e^2*f*csgn(I*c*(e*x+d)^n)^3-I*
Pi*b*d^2*g*csgn(I*c)*csgn(I*c*(e*x+d)^n)^2-I*Pi*b*e^2*f*csgn(I*(e*x+d)^n)*csgn(I*c*(e*x+d)^n)^2+2*ln(e*x+d)*b*
e^2*g*n*x^2+sum(_R*ln(((-d^2*g^2+3*e^2*f*g)*_R+3*b*e^2*n)*x+4*d*e*f*g*_R+b*d*e*n),_R=RootOf((d^2*f*g^3+e^2*f^2
*g^2)*_Z^2+2*b*e^2*f*g*n*_Z+b^2*e^2*n^2))*d^2*g^3*x^2+sum(_R*ln(((-d^2*g^2+3*e^2*f*g)*_R+3*b*e^2*n)*x+4*d*e*f*
g*_R+b*d*e*n),_R=RootOf((d^2*f*g^3+e^2*f^2*g^2)*_Z^2+2*b*e^2*f*g*n*_Z+b^2*e^2*n^2))*e^2*f*g^2*x^2+2*ln(e*x+d)*
b*e^2*f*n+sum(_R*ln(((-d^2*g^2+3*e^2*f*g)*_R+3*b*e^2*n)*x+4*d*e*f*g*_R+b*d*e*n),_R=RootOf((d^2*f*g^3+e^2*f^2*g
^2)*_Z^2+2*b*e^2*f*g*n*_Z+b^2*e^2*n^2))*d^2*f*g^2+sum(_R*ln(((-d^2*g^2+3*e^2*f*g)*_R+3*b*e^2*n)*x+4*d*e*f*g*_R
+b*d*e*n),_R=RootOf((d^2*f*g^3+e^2*f^2*g^2)*_Z^2+2*b*e^2*f*g*n*_Z+b^2*e^2*n^2))*e^2*f^2*g-2*ln(c)*b*d^2*g-2*ln
(c)*b*e^2*f-2*a*d^2*g-2*a*e^2*f)/(g*x^2+f)/g/(d^2*g+e^2*f)

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maxima [A]  time = 1.05, size = 130, normalized size = 0.94 \[ -\frac {1}{4} \, b e n {\left (\frac {e \log \left (g x^{2} + f\right )}{e^{2} f g + d^{2} g^{2}} - \frac {2 \, e \log \left (e x + d\right )}{e^{2} f g + d^{2} g^{2}} - \frac {2 \, d \arctan \left (\frac {g x}{\sqrt {f g}}\right )}{{\left (e^{2} f + d^{2} g\right )} \sqrt {f g}}\right )} - \frac {b \log \left ({\left (e x + d\right )}^{n} c\right )}{2 \, {\left (g^{2} x^{2} + f g\right )}} - \frac {a}{2 \, {\left (g^{2} x^{2} + f g\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.79, size = 366, normalized size = 2.63 \[ \frac {b\,e^2\,n\,\ln \left (d+e\,x\right )}{2\,d^2\,g^2+2\,f\,e^2\,g}-\frac {\ln \left (\frac {\left (b\,e^2\,f\,g\,n+b\,d\,e\,n\,\sqrt {-f\,g^3}\right )\,\left (x\,\left (2\,d^2\,e\,g^3-6\,e^3\,f\,g^2\right )-8\,d\,e^2\,f\,g^2\right )}{4\,\left (d^2\,f\,g^3+e^2\,f^2\,g^2\right )}+\frac {b\,d\,e^2\,g\,n}{2}+\frac {3\,b\,e^3\,g\,n\,x}{2}\right )\,\left (b\,e^2\,f\,g\,n+b\,d\,e\,n\,\sqrt {-f\,g^3}\right )}{4\,\left (d^2\,f\,g^3+e^2\,f^2\,g^2\right )}-\frac {\ln \left (\frac {\left (b\,e^2\,f\,g\,n-b\,d\,e\,n\,\sqrt {-f\,g^3}\right )\,\left (x\,\left (2\,d^2\,e\,g^3-6\,e^3\,f\,g^2\right )-8\,d\,e^2\,f\,g^2\right )}{4\,\left (d^2\,f\,g^3+e^2\,f^2\,g^2\right )}+\frac {b\,d\,e^2\,g\,n}{2}+\frac {3\,b\,e^3\,g\,n\,x}{2}\right )\,\left (b\,e^2\,f\,g\,n-b\,d\,e\,n\,\sqrt {-f\,g^3}\right )}{4\,\left (d^2\,f\,g^3+e^2\,f^2\,g^2\right )}-\frac {b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )}{2\,g\,\left (g\,x^2+f\right )}-\frac {a}{2\,g^2\,x^2+2\,f\,g} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*e^2*n*log(d + e*x))/(2*d^2*g^2 + 2*e^2*f*g) - (log(((b*e^2*f*g*n + b*d*e*n*(-f*g^3)^(1/2))*(x*(2*d^2*e*g^3
- 6*e^3*f*g^2) - 8*d*e^2*f*g^2))/(4*(d^2*f*g^3 + e^2*f^2*g^2)) + (b*d*e^2*g*n)/2 + (3*b*e^3*g*n*x)/2)*(b*e^2*f
*g*n + b*d*e*n*(-f*g^3)^(1/2)))/(4*(d^2*f*g^3 + e^2*f^2*g^2)) - (log(((b*e^2*f*g*n - b*d*e*n*(-f*g^3)^(1/2))*(
x*(2*d^2*e*g^3 - 6*e^3*f*g^2) - 8*d*e^2*f*g^2))/(4*(d^2*f*g^3 + e^2*f^2*g^2)) + (b*d*e^2*g*n)/2 + (3*b*e^3*g*n
*x)/2)*(b*e^2*f*g*n - b*d*e*n*(-f*g^3)^(1/2)))/(4*(d^2*f*g^3 + e^2*f^2*g^2)) - (b*log(c*(d + e*x)^n))/(2*g*(f
+ g*x^2)) - a/(2*f*g + 2*g^2*x^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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