\(\int \frac {x^5 (a+b \log (c x^n))}{d+e x^2} \, dx\) [210]

Optimal result

Integrand size = 23, antiderivative size = 121 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {b d n x^2}{4 e^2}-\frac {b n x^4}{16 e}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^3}+\frac {b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^3} \]

[Out]

1/4*b*d*n*x^2/e^2-1/16*b*n*x^4/e-1/2*d*x^2*(a+b*ln(c*x^n))/e^2+1/4*x^4*(a+b*ln(c*x^n))/e+1/2*d^2*(a+b*ln(c*x^n
))*ln(1+e*x^2/d)/e^3+1/4*b*d^2*n*polylog(2,-e*x^2/d)/e^3

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {272, 45, 2393, 2341, 2375, 2438} \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {d^2 \log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {b d^2 n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^3}+\frac {b d n x^2}{4 e^2}-\frac {b n x^4}{16 e} \]

[In]

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

[Out]

(b*d*n*x^2)/(4*e^2) - (b*n*x^4)/(16*e) - (d*x^2*(a + b*Log[c*x^n]))/(2*e^2) + (x^4*(a + b*Log[c*x^n]))/(4*e) +
(d^2*(a + b*Log[c*x^n])*Log[1 + (e*x^2)/d])/(2*e^3) + (b*d^2*n*PolyLog[2, -((e*x^2)/d)])/(4*e^3)

Rule 45

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 272

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 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2375

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

Rule 2393

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {d x \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{e}+\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}\right ) \, dx \\ & = -\frac {d \int x \left (a+b \log \left (c x^n\right )\right ) \, dx}{e^2}+\frac {d^2 \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx}{e^2}+\frac {\int x^3 \left (a+b \log \left (c x^n\right )\right ) \, dx}{e} \\ & = \frac {b d n x^2}{4 e^2}-\frac {b n x^4}{16 e}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^3}-\frac {\left (b d^2 n\right ) \int \frac {\log \left (1+\frac {e x^2}{d}\right )}{x} \, dx}{2 e^3} \\ & = \frac {b d n x^2}{4 e^2}-\frac {b n x^4}{16 e}-\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 e}+\frac {d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^3}+\frac {b d^2 n \text {Li}_2\left (-\frac {e x^2}{d}\right )}{4 e^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.44 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {4 b d e n x^2-b e^2 n x^4-8 d e x^2 \left (a+b \log \left (c x^n\right )\right )+4 e^2 x^4 \left (a+b \log \left (c x^n\right )\right )+8 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 d^2 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )+8 b d^2 n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} x}{\sqrt {-d}}\right )+8 b d^2 n \operatorname {PolyLog}\left (2,\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{16 e^3} \]

[In]

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

[Out]

(4*b*d*e*n*x^2 - b*e^2*n*x^4 - 8*d*e*x^2*(a + b*Log[c*x^n]) + 4*e^2*x^4*(a + b*Log[c*x^n]) + 8*d^2*(a + b*Log[
c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]] + 8*d^2*(a + b*Log[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)] + 8*b*d^2*n
*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]] + 8*b*d^2*n*PolyLog[2, (d*Sqrt[e]*x)/(-d)^(3/2)])/(16*e^3)

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.62 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.83

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) x^{4}}{4 e}-\frac {b \ln \left (x^{n}\right ) d \,x^{2}}{2 e^{2}}+\frac {b \ln \left (x^{n}\right ) d^{2} \ln \left (e \,x^{2}+d \right )}{2 e^{3}}-\frac {b n \,d^{2} \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{3}}+\frac {b n \,d^{2} \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \,d^{2} \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \,d^{2} \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}+\frac {b n \,d^{2} \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{3}}-\frac {b n \,x^{4}}{16 e}+\frac {b d n \,x^{2}}{4 e^{2}}-\frac {b n \,d^{2}}{4 e^{3}}+\left (-\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+b \ln \left (c \right )+a \right ) \left (\frac {\frac {1}{2} e \,x^{4}-d \,x^{2}}{2 e^{2}}+\frac {d^{2} \ln \left (e \,x^{2}+d \right )}{2 e^{3}}\right )\)	\(343\)

[In]

int(x^5*(a+b*ln(c*x^n))/(e*x^2+d),x,method=_RETURNVERBOSE)

[Out]

1/4*b*ln(x^n)/e*x^4-1/2*b*ln(x^n)/e^2*d*x^2+1/2*b*ln(x^n)*d^2/e^3*ln(e*x^2+d)-1/2*b*n*d^2/e^3*ln(x)*ln(e*x^2+d
)+1/2*b*n*d^2/e^3*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n*d^2/e^3*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e
)^(1/2))+1/2*b*n*d^2/e^3*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n*d^2/e^3*dilog((e*x+(-d*e)^(1/2))/(-d*
e)^(1/2))-1/16*b*n*x^4/e+1/4*b*d*n*x^2/e^2-1/4*b*n*d^2/e^3+(-1/2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1/
2*I*b*Pi*csgn(I*c)*csgn(I*c*x^n)^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*b*Pi*csgn(I*c*x^n)^3+b*ln(c)+a
)*(1/2/e^2*(1/2*e*x^4-d*x^2)+1/2*d^2/e^3*ln(e*x^2+d))

Fricas [F]

\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{e x^{2} + d} \,d x } \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 35.17 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.12 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\frac {a d^{2} \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} - \frac {a d x^{2}}{2 e^{2}} + \frac {a x^{4}}{4 e} - \frac {b d^{2} n \left (\begin {cases} \frac {x^{2}}{2 d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{2 e^{2}} + \frac {b d n x^{2}}{4 e^{2}} - \frac {b d x^{2} \log {\left (c x^{n} \right )}}{2 e^{2}} - \frac {b n x^{4}}{16 e} + \frac {b x^{4} \log {\left (c x^{n} \right )}}{4 e} \]

[In]

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

[Out]

a*d**2*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))/(2*e**2) - a*d*x**2/(2*e**2) + a*x**4/(4*e) -
b*d**2*n*Piecewise((x**2/(2*d), Eq(e, 0)), (Piecewise((-polylog(2, e*x**2*exp_polar(I*pi)/d)/2, (Abs(x) < 1) &
(1/Abs(x) < 1)), (log(d)*log(x) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, Abs(x) < 1), (-log(d)*log(1/x) - po
lylog(2, e*x**2*exp_polar(I*pi)/d)/2, 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) + meijerg
(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, True))/e, True))/(2*e**2) + b
*d**2*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))*log(c*x**n)/(2*e**2) + b*d*n*x**2/(4*e**2) - b*
d*x**2*log(c*x**n)/(2*e**2) - b*n*x**4/(16*e) + b*x**4*log(c*x**n)/(4*e)

Maxima [F]

\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{e x^{2} + d} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{e x^{2} + d} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{e\,x^2+d} \,d x \]

[In]

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

[Out]

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