3.266 \(\int \frac {x^5 (a+b \log (c (d+e x)^n))}{(f+g x^2)^2} \, dx\)

Optimal. Leaf size=417 \[ -\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}-\frac {f \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}+\frac {b d e f^{3/2} n \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 g^{5/2} \left (d^2 g+e^2 f\right )}-\frac {b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (d^2 g+e^2 f\right )}+\frac {b e^2 f^2 n \log (d+e x)}{2 g^3 \left (d^2 g+e^2 f\right )}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}-\frac {b f n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}-\frac {b f n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{\sqrt {g} d+e \sqrt {-f}}\right )}{g^3}+\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2} \]

[Out]

1/2*b*d*n*x/e/g^2-1/4*b*n*x^2/g^2+1/2*b*d*e*f^(3/2)*n*arctan(x*g^(1/2)/f^(1/2))/g^(5/2)/(d^2*g+e^2*f)-1/2*b*d^
2*n*ln(e*x+d)/e^2/g^2+1/2*b*e^2*f^2*n*ln(e*x+d)/g^3/(d^2*g+e^2*f)+1/2*x^2*(a+b*ln(c*(e*x+d)^n))/g^2-1/2*f^2*(a
+b*ln(c*(e*x+d)^n))/g^3/(g*x^2+f)-1/4*b*e^2*f^2*n*ln(g*x^2+f)/g^3/(d^2*g+e^2*f)-f*(a+b*ln(c*(e*x+d)^n))*ln(e*(
(-f)^(1/2)-x*g^(1/2))/(e*(-f)^(1/2)+d*g^(1/2)))/g^3-f*(a+b*ln(c*(e*x+d)^n))*ln(e*((-f)^(1/2)+x*g^(1/2))/(e*(-f
)^(1/2)-d*g^(1/2)))/g^3-b*f*n*polylog(2,-(e*x+d)*g^(1/2)/(e*(-f)^(1/2)-d*g^(1/2)))/g^3-b*f*n*polylog(2,(e*x+d)
*g^(1/2)/(e*(-f)^(1/2)+d*g^(1/2)))/g^3

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Rubi [A]  time = 0.49, antiderivative size = 417, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.482, Rules used = {266, 43, 2416, 2395, 2413, 706, 31, 635, 205, 260, 2394, 2393, 2391} \[ -\frac {b f n \text {PolyLog}\left (2,-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}-\frac {b f n \text {PolyLog}\left (2,\frac {\sqrt {g} (d+e x)}{d \sqrt {g}+e \sqrt {-f}}\right )}{g^3}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{d \sqrt {g}+e \sqrt {-f}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}-\frac {f \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right ) \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^3}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (d^2 g+e^2 f\right )}+\frac {b e^2 f^2 n \log (d+e x)}{2 g^3 \left (d^2 g+e^2 f\right )}+\frac {b d e f^{3/2} n \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 g^{5/2} \left (d^2 g+e^2 f\right )}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^5*(a + b*Log[c*(d + e*x)^n]))/(f + g*x^2)^2,x]

[Out]

(b*d*n*x)/(2*e*g^2) - (b*n*x^2)/(4*g^2) + (b*d*e*f^(3/2)*n*ArcTan[(Sqrt[g]*x)/Sqrt[f]])/(2*g^(5/2)*(e^2*f + d^
2*g)) - (b*d^2*n*Log[d + e*x])/(2*e^2*g^2) + (b*e^2*f^2*n*Log[d + e*x])/(2*g^3*(e^2*f + d^2*g)) + (x^2*(a + b*
Log[c*(d + e*x)^n]))/(2*g^2) - (f^2*(a + b*Log[c*(d + e*x)^n]))/(2*g^3*(f + g*x^2)) - (f*(a + b*Log[c*(d + e*x
)^n])*Log[(e*(Sqrt[-f] - Sqrt[g]*x))/(e*Sqrt[-f] + d*Sqrt[g])])/g^3 - (f*(a + b*Log[c*(d + e*x)^n])*Log[(e*(Sq
rt[-f] + Sqrt[g]*x))/(e*Sqrt[-f] - d*Sqrt[g])])/g^3 - (b*e^2*f^2*n*Log[f + g*x^2])/(4*g^3*(e^2*f + d^2*g)) - (
b*f*n*PolyLog[2, -((Sqrt[g]*(d + e*x))/(e*Sqrt[-f] - d*Sqrt[g]))])/g^3 - (b*f*n*PolyLog[2, (Sqrt[g]*(d + e*x))
/(e*Sqrt[-f] + d*Sqrt[g])])/g^3

Rule 31

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 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 2395

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

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]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {x^5 \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx &=\int \left (\frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2}+\frac {f^2 x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 \left (f+g x^2\right )^2}-\frac {2 f x \left (a+b \log \left (c (d+e x)^n\right )\right )}{g^2 \left (f+g x^2\right )}\right ) \, dx\\ &=\frac {\int x \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx}{g^2}-\frac {(2 f) \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{f+g x^2} \, dx}{g^2}+\frac {f^2 \int \frac {x \left (a+b \log \left (c (d+e x)^n\right )\right )}{\left (f+g x^2\right )^2} \, dx}{g^2}\\ &=\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac {(2 f) \int \left (-\frac {a+b \log \left (c (d+e x)^n\right )}{2 \sqrt {g} \left (\sqrt {-f}-\sqrt {g} x\right )}+\frac {a+b \log \left (c (d+e x)^n\right )}{2 \sqrt {g} \left (\sqrt {-f}+\sqrt {g} x\right )}\right ) \, dx}{g^2}+\frac {\left (b e f^2 n\right ) \int \frac {1}{(d+e x) \left (f+g x^2\right )} \, dx}{2 g^3}-\frac {(b e n) \int \frac {x^2}{d+e x} \, dx}{2 g^2}\\ &=\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}+\frac {f \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {-f}-\sqrt {g} x} \, dx}{g^{5/2}}-\frac {f \int \frac {a+b \log \left (c (d+e x)^n\right )}{\sqrt {-f}+\sqrt {g} x} \, dx}{g^{5/2}}-\frac {(b e n) \int \left (-\frac {d}{e^2}+\frac {x}{e}+\frac {d^2}{e^2 (d+e x)}\right ) \, dx}{2 g^2}+\frac {\left (b e f^2 n\right ) \int \frac {d g-e g x}{f+g x^2} \, dx}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac {\left (b e^3 f^2 n\right ) \int \frac {1}{d+e x} \, dx}{2 g^3 \left (e^2 f+d^2 g\right )}\\ &=\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac {b e^2 f^2 n \log (d+e x)}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{g^3}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}+\frac {(b e f n) \int \frac {\log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{d+e x} \, dx}{g^3}+\frac {(b e f n) \int \frac {\log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{d+e x} \, dx}{g^3}+\frac {\left (b d e f^2 n\right ) \int \frac {1}{f+g x^2} \, dx}{2 g^2 \left (e^2 f+d^2 g\right )}-\frac {\left (b e^2 f^2 n\right ) \int \frac {x}{f+g x^2} \, dx}{2 g^2 \left (e^2 f+d^2 g\right )}\\ &=\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2}+\frac {b d e f^{3/2} n \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 g^{5/2} \left (e^2 f+d^2 g\right )}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac {b e^2 f^2 n \log (d+e x)}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{g^3}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}-\frac {b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (e^2 f+d^2 g\right )}+\frac {(b f n) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {\sqrt {g} x}{e \sqrt {-f}-d \sqrt {g}}\right )}{x} \, dx,x,d+e x\right )}{g^3}+\frac {(b f n) \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {\sqrt {g} x}{e \sqrt {-f}+d \sqrt {g}}\right )}{x} \, dx,x,d+e x\right )}{g^3}\\ &=\frac {b d n x}{2 e g^2}-\frac {b n x^2}{4 g^2}+\frac {b d e f^{3/2} n \tan ^{-1}\left (\frac {\sqrt {g} x}{\sqrt {f}}\right )}{2 g^{5/2} \left (e^2 f+d^2 g\right )}-\frac {b d^2 n \log (d+e x)}{2 e^2 g^2}+\frac {b e^2 f^2 n \log (d+e x)}{2 g^3 \left (e^2 f+d^2 g\right )}+\frac {x^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^2}-\frac {f^2 \left (a+b \log \left (c (d+e x)^n\right )\right )}{2 g^3 \left (f+g x^2\right )}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}-\sqrt {g} x\right )}{e \sqrt {-f}+d \sqrt {g}}\right )}{g^3}-\frac {f \left (a+b \log \left (c (d+e x)^n\right )\right ) \log \left (\frac {e \left (\sqrt {-f}+\sqrt {g} x\right )}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}-\frac {b e^2 f^2 n \log \left (f+g x^2\right )}{4 g^3 \left (e^2 f+d^2 g\right )}-\frac {b f n \text {Li}_2\left (-\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}-d \sqrt {g}}\right )}{g^3}-\frac {b f n \text {Li}_2\left (\frac {\sqrt {g} (d+e x)}{e \sqrt {-f}+d \sqrt {g}}\right )}{g^3}\\ \end {align*}

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Mathematica [C]  time = 1.48, size = 530, normalized size = 1.27 \[ \frac {-\frac {2 f^2 \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )}{f+g x^2}-4 f \log \left (f+g x^2\right ) \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )+2 g x^2 \left (a+b \log \left (c (d+e x)^n\right )-b n \log (d+e x)\right )+b n \left (\frac {g \left (e x (2 d-e x)-2 \left (d^2-e^2 x^2\right ) \log (d+e x)\right )}{e^2}+\frac {f^{3/2} \left (i \sqrt {g} (d+e x) \log (d+e x)-e \left (\sqrt {f}+i \sqrt {g} x\right ) \log \left (-\sqrt {g} x+i \sqrt {f}\right )\right )}{\left (\sqrt {f}+i \sqrt {g} x\right ) \left (e \sqrt {f}-i d \sqrt {g}\right )}+\frac {i f^{3/2} \left (-\sqrt {g} (d+e x) \log (d+e x)+e \left (\sqrt {g} x+i \sqrt {f}\right ) \log \left (\sqrt {g} x+i \sqrt {f}\right )\right )}{\left (\sqrt {f}-i \sqrt {g} x\right ) \left (e \sqrt {f}+i d \sqrt {g}\right )}-4 f \left (\text {Li}_2\left (-\frac {i \sqrt {g} (d+e x)}{e \sqrt {f}-i d \sqrt {g}}\right )+\log (d+e x) \log \left (\frac {e \left (\sqrt {f}+i \sqrt {g} x\right )}{e \sqrt {f}-i d \sqrt {g}}\right )\right )-4 f \left (\text {Li}_2\left (\frac {i \sqrt {g} (d+e x)}{i \sqrt {g} d+e \sqrt {f}}\right )+\log (d+e x) \log \left (\frac {e \left (\sqrt {f}-i \sqrt {g} x\right )}{e \sqrt {f}+i d \sqrt {g}}\right )\right )\right )}{4 g^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(a + b*Log[c*(d + e*x)^n]))/(f + g*x^2)^2,x]

[Out]

(2*g*x^2*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]) - (2*f^2*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n]))
/(f + g*x^2) - 4*f*(a - b*n*Log[d + e*x] + b*Log[c*(d + e*x)^n])*Log[f + g*x^2] + b*n*((g*(e*x*(2*d - e*x) - 2
*(d^2 - e^2*x^2)*Log[d + e*x]))/e^2 + (f^(3/2)*(I*Sqrt[g]*(d + e*x)*Log[d + e*x] - e*(Sqrt[f] + I*Sqrt[g]*x)*L
og[I*Sqrt[f] - Sqrt[g]*x]))/((e*Sqrt[f] - I*d*Sqrt[g])*(Sqrt[f] + I*Sqrt[g]*x)) + (I*f^(3/2)*(-(Sqrt[g]*(d + e
*x)*Log[d + e*x]) + e*(I*Sqrt[f] + Sqrt[g]*x)*Log[I*Sqrt[f] + Sqrt[g]*x]))/((e*Sqrt[f] + I*d*Sqrt[g])*(Sqrt[f]
- I*Sqrt[g]*x)) - 4*f*(Log[d + e*x]*Log[(e*(Sqrt[f] + I*Sqrt[g]*x))/(e*Sqrt[f] - I*d*Sqrt[g])] + PolyLog[2, (
(-I)*Sqrt[g]*(d + e*x))/(e*Sqrt[f] - I*d*Sqrt[g])]) - 4*f*(Log[d + e*x]*Log[(e*(Sqrt[f] - I*Sqrt[g]*x))/(e*Sqr
t[f] + I*d*Sqrt[g])] + PolyLog[2, (I*Sqrt[g]*(d + e*x))/(e*Sqrt[f] + I*d*Sqrt[g])])))/(4*g^3)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{5} \log \left ({\left (e x + d\right )}^{n} c\right ) + a x^{5}}{g^{2} x^{4} + 2 \, f g x^{2} + f^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*(e*x+d)^n))/(g*x^2+f)^2,x, algorithm="fricas")

[Out]

integral((b*x^5*log((e*x + d)^n*c) + a*x^5)/(g^2*x^4 + 2*f*g*x^2 + f^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left ({\left (e x + d\right )}^{n} c\right ) + a\right )} x^{5}}{{\left (g x^{2} + f\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*(e*x+d)^n))/(g*x^2+f)^2,x, algorithm="giac")

[Out]

integrate((b*log((e*x + d)^n*c) + a)*x^5/(g*x^2 + f)^2, x)

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maple [C]  time = 0.28, size = 1008, normalized size = 2.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(b*ln(c*(e*x+d)^n)+a)/(g*x^2+f)^2,x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a {\left (\frac {f^{2}}{g^{4} x^{2} + f g^{3}} - \frac {x^{2}}{g^{2}} + \frac {2 \, f \log \left (g x^{2} + f\right )}{g^{3}}\right )} + b \int \frac {x^{5} \log \left ({\left (e x + d\right )}^{n}\right ) + x^{5} \log \relax (c)}{g^{2} x^{4} + 2 \, f g x^{2} + f^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*log(c*(e*x+d)^n))/(g*x^2+f)^2,x, algorithm="maxima")

[Out]

-1/2*a*(f^2/(g^4*x^2 + f*g^3) - x^2/g^2 + 2*f*log(g*x^2 + f)/g^3) + b*integrate((x^5*log((e*x + d)^n) + x^5*lo
g(c))/(g^2*x^4 + 2*f*g*x^2 + f^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5\,\left (a+b\,\ln \left (c\,{\left (d+e\,x\right )}^n\right )\right )}{{\left (g\,x^2+f\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(a + b*log(c*(d + e*x)^n)))/(f + g*x^2)^2,x)

[Out]

int((x^5*(a + b*log(c*(d + e*x)^n)))/(f + g*x^2)^2, x)

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

[Out]

Timed out

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