3.2.73 \(\int x (d+e x^2) (a+b \log (c x^n)) \, dx\) [173]
Optimal. Leaf size=47 \[ -\frac {1}{4} b d n x^2-\frac {1}{16} b e n x^4+\frac {1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right ) \]
[Out]
-1/4*b*d*n*x^2-1/16*b*e*n*x^4+1/4*(e*x^4+2*d*x^2)*(a+b*ln(c*x^n))
________________________________________________________________________________________
Rubi [A]
time = 0.03, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {14, 2371, 12}
\begin {gather*} \frac {1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b d n x^2-\frac {1}{16} b e n x^4 \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x*(d + e*x^2)*(a + b*Log[c*x^n]),x]
[Out]
-1/4*(b*d*n*x^2) - (b*e*n*x^4)/16 + ((2*d*x^2 + e*x^4)*(a + b*Log[c*x^n]))/4
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
Rule 14
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&& !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Rule 2371
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IGtQ[m, 0]
Rubi steps
\begin {align*} \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac {1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {1}{4} x \left (2 d+e x^2\right ) \, dx\\ &=\frac {1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int x \left (2 d+e x^2\right ) \, dx\\ &=\frac {1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \left (2 d x+e x^3\right ) \, dx\\ &=-\frac {1}{4} b d n x^2-\frac {1}{16} b e n x^4+\frac {1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 0.01, size = 69, normalized size = 1.47 \begin {gather*} \frac {1}{2} a d x^2-\frac {1}{4} b d n x^2+\frac {1}{4} a e x^4-\frac {1}{16} b e n x^4+\frac {1}{2} b d x^2 \log \left (c x^n\right )+\frac {1}{4} b e x^4 \log \left (c x^n\right ) \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x*(d + e*x^2)*(a + b*Log[c*x^n]),x]
[Out]
(a*d*x^2)/2 - (b*d*n*x^2)/4 + (a*e*x^4)/4 - (b*e*n*x^4)/16 + (b*d*x^2*Log[c*x^n])/2 + (b*e*x^4*Log[c*x^n])/4
________________________________________________________________________________________
Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 0.16, size = 2346, normalized size = 49.91
| | |
method |
result |
size |
| | |
risch |
\(\text {Expression too large to display}\) |
\(2346\) |
| | |
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(e*x^2+d)*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
[Out]
1/4*(e*x^2+d)^2*b/e*ln(x^n)+1/16*(16*x^4*a^2*e^2-8*Pi^2*b^2*d*e*x^2*csgn(I*c*x^n)^6+4*I*Pi*b^2*e^2*n*x^4*csgn(
I*c*x^n)^3-16*I*Pi*a*b*e^2*x^4*csgn(I*c*x^n)^3+16*I*Pi*a*b*d^2*csgn(I*c)*csgn(I*c*x^n)^2+16*I*Pi*a*b*d^2*csgn(
I*x^n)*csgn(I*c*x^n)^2-12*I*Pi*b^2*d*e*n*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+32*I*Pi*a*b*d*e*x^2*csgn(I*c)*csgn(I*
c*x^n)^2+32*I*Pi*a*b*d*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-32*Pi^2*b^2*d*e*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^
n)^4-4*I*Pi*b^2*e^2*n*x^4*csgn(I*c)*csgn(I*c*x^n)^2+16*I*Pi*a*b*e^2*x^4*csgn(I*c)*csgn(I*c*x^n)^2+32*a^2*d*e*x
^2-16*b^2*d^2*ln(c)*n-4*Pi^2*b^2*e^2*x^4*csgn(I*c*x^n)^6+8*I*ln(x)*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*x^n)*csgn(I*c
*x^n)-16*I*Pi*ln(c)*b^2*e^2*x^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+4*I*Pi*b^2*e^2*n*x^4*csgn(I*c)*csgn(I*x^n)
*csgn(I*c*x^n)+16*d^2*b^2*ln(c)^2+16*a^2*d^2+16*I*Pi*ln(c)*b^2*d^2*csgn(I*c)*csgn(I*c*x^n)^2+16*I*Pi*ln(c)*b^2
*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2+16*I*Pi*ln(c)*b^2*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2+8*I*Pi*b^2*d^2*n*csgn(I
*c)*csgn(I*x^n)*csgn(I*c*x^n)+4*ln(x)*b^2*d^2*n^2+32*d^2*a*b*ln(c)+16*I*Pi*ln(c)*b^2*e^2*x^4*csgn(I*c)*csgn(I*
c*x^n)^2-8*Pi^2*b^2*d*e*x^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2+16*Pi^2*b^2*d*e*x^2*csgn(I*c)^2*csgn(I*x
^n)*csgn(I*c*x^n)^3+16*Pi^2*b^2*d*e*x^2*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3-16*I*Pi*ln(c)*b^2*e^2*x^4*csgn
(I*c*x^n)^3-4*Pi^2*b^2*d^2*csgn(I*x^n)^2*csgn(I*c*x^n)^4+4*b^2*d^2*n^2+32*ln(c)^2*b^2*d*e*x^2-4*I*Pi*b^2*e^2*n
*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2+16*I*Pi*a*b*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)^2-32*I*Pi*ln(c)*b^2*d*e*x^2*csg
n(I*c*x^n)^3-16*I*Pi*ln(c)*b^2*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-8*I*ln(x)*Pi*b^2*d^2*n*csgn(I*c)*csgn(I
*c*x^n)^2-8*I*ln(x)*Pi*b^2*d^2*n*csgn(I*x^n)*csgn(I*c*x^n)^2-16*ln(x)*a*b*d^2*n+b^2*e^2*n^2*x^4-4*Pi^2*b^2*d^2
*csgn(I*c*x^n)^6-16*I*Pi*a*b*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+12*I*Pi*b^2*d*e*n*x^2*csgn(I*c*x^n)^3-16*
b*d^2*n*a-32*I*Pi*a*b*d*e*x^2*csgn(I*c*x^n)^3-16*ln(x)*ln(c)*b^2*d^2*n+16*ln(c)^2*b^2*e^2*x^4-12*I*Pi*b^2*d*e*
n*x^2*csgn(I*c)*csgn(I*c*x^n)^2-24*b*n*a*d*e*x^2-32*I*Pi*a*b*d*e*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-32*I*
Pi*ln(c)*b^2*d*e*x^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-4*Pi^2*b^2*e^2*x^4*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c
*x^n)^2+8*Pi^2*b^2*e^2*x^4*csgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3+8*Pi^2*b^2*e^2*x^4*csgn(I*c)*csgn(I*x^n)^2*
csgn(I*c*x^n)^3-16*Pi^2*b^2*e^2*x^4*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4-8*a*b*e^2*n*x^4-16*I*Pi*a*b*e^2*x^4*
csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+32*I*Pi*ln(c)*b^2*d*e*x^2*csgn(I*c)*csgn(I*c*x^n)^2+32*I*Pi*ln(c)*b^2*d*e*
x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+8*I*ln(x)*Pi*b^2*d^2*n*csgn(I*c*x^n)^3-8*Pi^2*b^2*d*e*x^2*csgn(I*x^n)^2*csgn(I
*c*x^n)^4+16*Pi^2*b^2*d*e*x^2*csgn(I*x^n)*csgn(I*c*x^n)^5+4*b^2*d*e*n^2*x^2-16*I*Pi*a*b*d^2*csgn(I*c*x^n)^3+8*
Pi^2*b^2*e^2*x^4*csgn(I*x^n)*csgn(I*c*x^n)^5-4*Pi^2*b^2*e^2*x^4*csgn(I*c)^2*csgn(I*c*x^n)^4-16*I*Pi*ln(c)*b^2*
d^2*csgn(I*c*x^n)^3+8*I*Pi*b^2*d^2*n*csgn(I*c*x^n)^3+8*Pi^2*b^2*e^2*x^4*csgn(I*c)*csgn(I*c*x^n)^5-4*Pi^2*b^2*e
^2*x^4*csgn(I*x^n)^2*csgn(I*c*x^n)^4-4*Pi^2*b^2*d^2*csgn(I*c)^2*csgn(I*x^n)^2*csgn(I*c*x^n)^2+8*Pi^2*b^2*d^2*c
sgn(I*c)^2*csgn(I*x^n)*csgn(I*c*x^n)^3+8*Pi^2*b^2*d^2*csgn(I*c)*csgn(I*x^n)^2*csgn(I*c*x^n)^3-16*Pi^2*b^2*d^2*
csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)^4-4*Pi^2*b^2*d^2*csgn(I*c)^2*csgn(I*c*x^n)^4+8*Pi^2*b^2*d^2*csgn(I*x^n)*cs
gn(I*c*x^n)^5-8*Pi^2*b^2*d*e*x^2*csgn(I*c)^2*csgn(I*c*x^n)^4+16*Pi^2*b^2*d*e*x^2*csgn(I*c)*csgn(I*c*x^n)^5-24*
ln(c)*b^2*d*e*n*x^2+64*ln(c)*a*b*d*e*x^2+8*Pi^2*b^2*d^2*csgn(I*c)*csgn(I*c*x^n)^5+12*I*Pi*b^2*d*e*n*x^2*csgn(I
*c)*csgn(I*x^n)*csgn(I*c*x^n)-8*ln(c)*b^2*e^2*n*x^4+32*ln(c)*a*b*e^2*x^4-8*I*Pi*b^2*d^2*n*csgn(I*c)*csgn(I*c*x
^n)^2-8*I*Pi*b^2*d^2*n*csgn(I*x^n)*csgn(I*c*x^n)^2)/e/(-2*I*b*Pi*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+2*I*b*Pi*
csgn(I*c)*csgn(I*c*x^n)^2+2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2-2*I*b*Pi*csgn(I*c*x^n)^3+4*b*ln(c)-b*n+4*a)
________________________________________________________________________________________
Maxima [A]
time = 0.28, size = 60, normalized size = 1.28 \begin {gather*} -\frac {1}{16} \, b n x^{4} e + \frac {1}{4} \, b x^{4} e \log \left (c x^{n}\right ) + \frac {1}{4} \, a x^{4} e - \frac {1}{4} \, b d n x^{2} + \frac {1}{2} \, b d x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(e*x^2+d)*(a+b*log(c*x^n)),x, algorithm="maxima")
[Out]
-1/16*b*n*x^4*e + 1/4*b*x^4*e*log(c*x^n) + 1/4*a*x^4*e - 1/4*b*d*n*x^2 + 1/2*b*d*x^2*log(c*x^n) + 1/2*a*d*x^2
________________________________________________________________________________________
Fricas [A]
time = 0.41, size = 69, normalized size = 1.47 \begin {gather*} -\frac {1}{16} \, {\left (b n - 4 \, a\right )} x^{4} e - \frac {1}{4} \, {\left (b d n - 2 \, a d\right )} x^{2} + \frac {1}{4} \, {\left (b x^{4} e + 2 \, b d x^{2}\right )} \log \left (c\right ) + \frac {1}{4} \, {\left (b n x^{4} e + 2 \, b d n x^{2}\right )} \log \left (x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(e*x^2+d)*(a+b*log(c*x^n)),x, algorithm="fricas")
[Out]
-1/16*(b*n - 4*a)*x^4*e - 1/4*(b*d*n - 2*a*d)*x^2 + 1/4*(b*x^4*e + 2*b*d*x^2)*log(c) + 1/4*(b*n*x^4*e + 2*b*d*
n*x^2)*log(x)
________________________________________________________________________________________
Sympy [A]
time = 0.29, size = 66, normalized size = 1.40 \begin {gather*} \frac {a d x^{2}}{2} + \frac {a e x^{4}}{4} - \frac {b d n x^{2}}{4} + \frac {b d x^{2} \log {\left (c x^{n} \right )}}{2} - \frac {b e n x^{4}}{16} + \frac {b e x^{4} \log {\left (c x^{n} \right )}}{4} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(e*x**2+d)*(a+b*ln(c*x**n)),x)
[Out]
a*d*x**2/2 + a*e*x**4/4 - b*d*n*x**2/4 + b*d*x**2*log(c*x**n)/2 - b*e*n*x**4/16 + b*e*x**4*log(c*x**n)/4
________________________________________________________________________________________
Giac [A]
time = 2.34, size = 73, normalized size = 1.55 \begin {gather*} \frac {1}{4} \, b n x^{4} e \log \left (x\right ) - \frac {1}{16} \, b n x^{4} e + \frac {1}{4} \, b x^{4} e \log \left (c\right ) + \frac {1}{4} \, a x^{4} e + \frac {1}{2} \, b d n x^{2} \log \left (x\right ) - \frac {1}{4} \, b d n x^{2} + \frac {1}{2} \, b d x^{2} \log \left (c\right ) + \frac {1}{2} \, a d x^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x*(e*x^2+d)*(a+b*log(c*x^n)),x, algorithm="giac")
[Out]
1/4*b*n*x^4*e*log(x) - 1/16*b*n*x^4*e + 1/4*b*x^4*e*log(c) + 1/4*a*x^4*e + 1/2*b*d*n*x^2*log(x) - 1/4*b*d*n*x^
2 + 1/2*b*d*x^2*log(c) + 1/2*a*d*x^2
________________________________________________________________________________________
Mupad [B]
time = 3.39, size = 51, normalized size = 1.09 \begin {gather*} \ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^4}{4}+\frac {b\,d\,x^2}{2}\right )+\frac {d\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {e\,x^4\,\left (4\,a-b\,n\right )}{16} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x*(d + e*x^2)*(a + b*log(c*x^n)),x)
[Out]
log(c*x^n)*((b*d*x^2)/2 + (b*e*x^4)/4) + (d*x^2*(2*a - b*n))/4 + (e*x^4*(4*a - b*n))/16
________________________________________________________________________________________