\(\int x^3 (d+e x) (a+b \log (c x^n)) \, dx\) [1]

Optimal result

Integrand size = 19, antiderivative size = 48 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} b d n x^4-\frac {1}{25} b e n x^5+\frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-1/16*b*d*n*x^4-1/25*b*e*n*x^5+1/20*(4*e*x^5+5*d*x^4)*(a+b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 48, 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 = {45, 2371, 12} \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b d n x^4-\frac {1}{25} b e n x^5 \]

[In]

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

[Out]

-1/16*(b*d*n*x^4) - (b*e*n*x^5)/25 + ((5*d*x^4 + 4*e*x^5)*(a + b*Log[c*x^n]))/20

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {1}{20} x^3 (5 d+4 e x) \, dx \\ & = \frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{20} (b n) \int x^3 (5 d+4 e x) \, dx \\ & = \frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{20} (b n) \int \left (5 d x^3+4 e x^4\right ) \, dx \\ & = -\frac {1}{16} b d n x^4-\frac {1}{25} b e n x^5+\frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{400} x^4 \left (20 a (5 d+4 e x)-b n (25 d+16 e x)+20 b (5 d+4 e x) \log \left (c x^n\right )\right ) \]

[In]

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

[Out]

(x^4*(20*a*(5*d + 4*e*x) - b*n*(25*d + 16*e*x) + 20*b*(5*d + 4*e*x)*Log[c*x^n]))/400

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(\frac {x^{5} \ln \left (c \,x^{n}\right ) b e}{5}-\frac {b e n \,x^{5}}{25}+\frac {a e \,x^{5}}{5}+\frac {x^{4} \ln \left (c \,x^{n}\right ) b d}{4}-\frac {b d n \,x^{4}}{16}+\frac {a d \,x^{4}}{4}\)	\(58\)
risch	\(\frac {b \,x^{4} \left (4 e x +5 d \right ) \ln \left (x^{n}\right )}{20}-\frac {i \pi b e \,x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{10}+\frac {i \pi b e \,x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}+\frac {i \pi b e \,x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}-\frac {i \pi b e \,x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{10}+\frac {\ln \left (c \right ) b e \,x^{5}}{5}-\frac {b e n \,x^{5}}{25}+\frac {a e \,x^{5}}{5}-\frac {i \pi b d \,x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{8}+\frac {i \pi b d \,x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b d \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b d \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {\ln \left (c \right ) b d \,x^{4}}{4}-\frac {b d n \,x^{4}}{16}+\frac {a d \,x^{4}}{4}\)	\(264\)

[In]

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

[Out]

1/5*x^5*ln(c*x^n)*b*e-1/25*b*e*n*x^5+1/5*a*e*x^5+1/4*x^4*ln(c*x^n)*b*d-1/16*b*d*n*x^4+1/4*a*d*x^4

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{25} \, {\left (b e n - 5 \, a e\right )} x^{5} - \frac {1}{16} \, {\left (b d n - 4 \, a d\right )} x^{4} + \frac {1}{20} \, {\left (4 \, b e x^{5} + 5 \, b d x^{4}\right )} \log \left (c\right ) + \frac {1}{20} \, {\left (4 \, b e n x^{5} + 5 \, b d n x^{4}\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/25*(b*e*n - 5*a*e)*x^5 - 1/16*(b*d*n - 4*a*d)*x^4 + 1/20*(4*b*e*x^5 + 5*b*d*x^4)*log(c) + 1/20*(4*b*e*n*x^5
+ 5*b*d*n*x^4)*log(x)

Sympy [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d x^{4}}{4} + \frac {a e x^{5}}{5} - \frac {b d n x^{4}}{16} + \frac {b d x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {b e n x^{5}}{25} + \frac {b e x^{5} \log {\left (c x^{n} \right )}}{5} \]

[In]

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

[Out]

a*d*x**4/4 + a*e*x**5/5 - b*d*n*x**4/16 + b*d*x**4*log(c*x**n)/4 - b*e*n*x**5/25 + b*e*x**5*log(c*x**n)/5

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{25} \, b e n x^{5} + \frac {1}{5} \, b e x^{5} \log \left (c x^{n}\right ) - \frac {1}{16} \, b d n x^{4} + \frac {1}{5} \, a e x^{5} + \frac {1}{4} \, b d x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d x^{4} \]

[In]

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

[Out]

-1/25*b*e*n*x^5 + 1/5*b*e*x^5*log(c*x^n) - 1/16*b*d*n*x^4 + 1/5*a*e*x^5 + 1/4*b*d*x^4*log(c*x^n) + 1/4*a*d*x^4

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{5} \, b e n x^{5} \log \left (x\right ) - \frac {1}{25} \, b e n x^{5} + \frac {1}{5} \, b e x^{5} \log \left (c\right ) + \frac {1}{4} \, b d n x^{4} \log \left (x\right ) - \frac {1}{16} \, b d n x^{4} + \frac {1}{5} \, a e x^{5} + \frac {1}{4} \, b d x^{4} \log \left (c\right ) + \frac {1}{4} \, a d x^{4} \]

[In]

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

[Out]

1/5*b*e*n*x^5*log(x) - 1/25*b*e*n*x^5 + 1/5*b*e*x^5*log(c) + 1/4*b*d*n*x^4*log(x) - 1/16*b*d*n*x^4 + 1/5*a*e*x
^5 + 1/4*b*d*x^4*log(c) + 1/4*a*d*x^4

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^5}{5}+\frac {b\,d\,x^4}{4}\right )+\frac {d\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {e\,x^5\,\left (5\,a-b\,n\right )}{25} \]

[In]

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

[Out]

log(c*x^n)*((b*d*x^4)/4 + (b*e*x^5)/5) + (d*x^4*(4*a - b*n))/16 + (e*x^5*(5*a - b*n))/25