\(\int x (d+e x^2) (a+b \log (c x^n)) \, dx\) [173]

Optimal result

Integrand size = 19, antiderivative size = 47 \[ \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{4} b d n x^2-\frac {1}{16} b e n x^4+\frac {1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[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] (verified)

Time = 0.02 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {14, 2371, 12} \[ \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} \left (2 d x^2+e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b d n x^2-\frac {1}{16} b e n x^4 \]

[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*} \text {integral}& = \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] (verified)

Time = 0.00 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.47 \[ \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{2} a d x^2-\frac {1}{4} b d n x^2+\frac {1}{4} a e x^4-\frac {1}{16} b e n x^4+\frac {1}{2} b d x^2 \log \left (c x^n\right )+\frac {1}{4} b e x^4 \log \left (c x^n\right ) \]

[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 [A] (verified)

Time = 0.17 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.23

method	result	size
parallelrisch	\(\frac {x^{4} b e \ln \left (c \,x^{n}\right )}{4}-\frac {b e n \,x^{4}}{16}+\frac {x^{4} a e}{4}+\frac {x^{2} \ln \left (c \,x^{n}\right ) b d}{2}-\frac {b d n \,x^{2}}{4}+\frac {a d \,x^{2}}{2}\)	\(58\)
risch	\(\text {Expression too large to display}\)	\(2346\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.43 \[ \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} \, {\left (b e n - 4 \, a e\right )} x^{4} - \frac {1}{4} \, {\left (b d n - 2 \, a d\right )} x^{2} + \frac {1}{4} \, {\left (b e x^{4} + 2 \, b d x^{2}\right )} \log \left (c\right ) + \frac {1}{4} \, {\left (b e n x^{4} + 2 \, b d n x^{2}\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/16*(b*e*n - 4*a*e)*x^4 - 1/4*(b*d*n - 2*a*d)*x^2 + 1/4*(b*e*x^4 + 2*b*d*x^2)*log(c) + 1/4*(b*e*n*x^4 + 2*b*
d*n*x^2)*log(x)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.40 \[ \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d x^{2}}{2} + \frac {a e x^{4}}{4} - \frac {b d n x^{2}}{4} + \frac {b d x^{2} \log {\left (c x^{n} \right )}}{2} - \frac {b e n x^{4}}{16} + \frac {b e x^{4} \log {\left (c x^{n} \right )}}{4} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.21 \[ \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} \, b e n x^{4} + \frac {1}{4} \, b e x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a e x^{4} - \frac {1}{4} \, b d n x^{2} + \frac {1}{2} \, b d x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d x^{2} \]

[In]

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

[Out]

-1/16*b*e*n*x^4 + 1/4*b*e*x^4*log(c*x^n) + 1/4*a*e*x^4 - 1/4*b*d*n*x^2 + 1/2*b*d*x^2*log(c*x^n) + 1/2*a*d*x^2

Giac [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.47 \[ \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} \, b e n x^{4} \log \left (x\right ) - \frac {1}{16} \, b e n x^{4} + \frac {1}{4} \, b e x^{4} \log \left (c\right ) + \frac {1}{4} \, a e x^{4} + \frac {1}{2} \, b d n x^{2} \log \left (x\right ) - \frac {1}{4} \, b d n x^{2} + \frac {1}{2} \, b d x^{2} \log \left (c\right ) + \frac {1}{2} \, a d x^{2} \]

[In]

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

[Out]

1/4*b*e*n*x^4*log(x) - 1/16*b*e*n*x^4 + 1/4*b*e*x^4*log(c) + 1/4*a*e*x^4 + 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] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.09 \[ \int x \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^4}{4}+\frac {b\,d\,x^2}{2}\right )+\frac {d\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {e\,x^4\,\left (4\,a-b\,n\right )}{16} \]

[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