3.3.11 \(\int \frac {x^3 (a+b \log (c x^n))}{d+e x^2} \, dx\) [211]

Optimal. Leaf size=83 \[ -\frac {b n x^2}{4 e}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^2}-\frac {b d n \text {Li}_2\left (-\frac {e x^2}{d}\right )}{4 e^2} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {272, 45, 2393, 2341, 2375, 2438} \begin {gather*} -\frac {b d n \text {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^2}-\frac {d \log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {b n x^2}{4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2),x]

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 135, normalized size = 1.63 \begin {gather*} -\frac {b e n x^2-2 e x^2 \left (a+b \log \left (c x^n\right )\right )+2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {\sqrt {e} x}{\sqrt {-d}}\right )+2 d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )+2 b d n \text {Li}_2\left (\frac {\sqrt {e} x}{\sqrt {-d}}\right )+2 b d n \text {Li}_2\left (\frac {d \sqrt {e} x}{(-d)^{3/2}}\right )}{4 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2),x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 460, normalized size = 5.54

method	result	size
risch	\(\frac {b \ln \left (x^{n}\right ) x^{2}}{2 e}-\frac {b \ln \left (x^{n}\right ) d \ln \left (e \,x^{2}+d \right )}{2 e^{2}}-\frac {b n \,x^{2}}{4 e}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e^{2}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e^{2}}+\frac {b n d \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{2 e^{2}}-\frac {b n d \dilog \left (\frac {-e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e^{2}}-\frac {b n d \dilog \left (\frac {e x +\sqrt {-e d}}{\sqrt {-e d}}\right )}{2 e^{2}}+\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} d \ln \left (e \,x^{2}+d \right )}{4 e^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (e \,x^{2}+d \right )}{4 e^{2}}-\frac {i b \pi \mathrm {csgn}\left (i c \,x^{n}\right )^{3} x^{2}}{4 e}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) d \ln \left (e \,x^{2}+d \right )}{4 e^{2}}-\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) x^{2}}{4 e}-\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} d \ln \left (e \,x^{2}+d \right )}{4 e^{2}}+\frac {i b \pi \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x^{2}}{4 e}+\frac {i b \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x^{2}}{4 e}+\frac {b \ln \left (c \right ) x^{2}}{2 e}-\frac {b \ln \left (c \right ) d \ln \left (e \,x^{2}+d \right )}{2 e^{2}}+\frac {a \,x^{2}}{2 e}-\frac {a d \ln \left (e \,x^{2}+d \right )}{2 e^{2}}\)	\(460\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*ln(c*x^n))/(e*x^2+d),x,method=_RETURNVERBOSE)

[Out]

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

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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(x^3*(a+b*log(c*x^n))/(e*x^2+d),x, algorithm="maxima")

[Out]

1/2*(x^2*e^(-1) - d*e^(-2)*log(x^2*e + d))*a + b*integrate((x^3*log(c) + x^3*log(x^n))/(x^2*e + d), 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(x^3*(a+b*log(c*x^n))/(e*x^2+d),x, algorithm="fricas")

[Out]

integral((b*x^3*log(c*x^n) + a*x^3)/(x^2*e + d), x)

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Sympy [A]
time = 21.60, size = 202, normalized size = 2.43 \begin {gather*} - \frac {a d \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} + \frac {a x^{2}}{2 e} + \frac {b d 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} - \frac {b d \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} - \frac {b n x^{2}}{4 e} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*ln(c*x**n))/(e*x**2+d),x)

[Out]

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

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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(x^3*(a+b*log(c*x^n))/(e*x^2+d),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{e\,x^2+d} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*log(c*x^n)))/(d + e*x^2),x)

[Out]

int((x^3*(a + b*log(c*x^n)))/(d + e*x^2), x)

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