\(\int x^3 (d+e x)^3 (a+b \log (c x^n)) \, dx\) [19]

Optimal result

Integrand size = 21, antiderivative size = 100 \[ \int x^3 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} b d^3 n x^4-\frac {3}{25} b d^2 e n x^5-\frac {1}{12} b d e^2 n x^6-\frac {1}{49} b e^3 n x^7+\frac {1}{140} \left (35 d^3 x^4+84 d^2 e x^5+70 d e^2 x^6+20 e^3 x^7\right ) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-1/16*b*d^3*n*x^4-3/25*b*d^2*e*n*x^5-1/12*b*d*e^2*n*x^6-1/49*b*e^3*n*x^7+1/140*(20*e^3*x^7+70*d*e^2*x^6+84*d^2
*e*x^5+35*d^3*x^4)*(a+b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {45, 2371, 12, 14} \[ \int x^3 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{140} \left (35 d^3 x^4+84 d^2 e x^5+70 d e^2 x^6+20 e^3 x^7\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{16} b d^3 n x^4-\frac {3}{25} b d^2 e n x^5-\frac {1}{12} b d e^2 n x^6-\frac {1}{49} b e^3 n x^7 \]

[In]

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

[Out]

-1/16*(b*d^3*n*x^4) - (3*b*d^2*e*n*x^5)/25 - (b*d*e^2*n*x^6)/12 - (b*e^3*n*x^7)/49 + ((35*d^3*x^4 + 84*d^2*e*x
^5 + 70*d*e^2*x^6 + 20*e^3*x^7)*(a + b*Log[c*x^n]))/140

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 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}{140} \left (35 d^3 x^4+84 d^2 e x^5+70 d e^2 x^6+20 e^3 x^7\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {1}{140} x^3 \left (35 d^3+84 d^2 e x+70 d e^2 x^2+20 e^3 x^3\right ) \, dx \\ & = \frac {1}{140} \left (35 d^3 x^4+84 d^2 e x^5+70 d e^2 x^6+20 e^3 x^7\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{140} (b n) \int x^3 \left (35 d^3+84 d^2 e x+70 d e^2 x^2+20 e^3 x^3\right ) \, dx \\ & = \frac {1}{140} \left (35 d^3 x^4+84 d^2 e x^5+70 d e^2 x^6+20 e^3 x^7\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{140} (b n) \int \left (35 d^3 x^3+84 d^2 e x^4+70 d e^2 x^5+20 e^3 x^6\right ) \, dx \\ & = -\frac {1}{16} b d^3 n x^4-\frac {3}{25} b d^2 e n x^5-\frac {1}{12} b d e^2 n x^6-\frac {1}{49} b e^3 n x^7+\frac {1}{140} \left (35 d^3 x^4+84 d^2 e x^5+70 d e^2 x^6+20 e^3 x^7\right ) \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.33 \[ \int x^3 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} b d^3 n x^4-\frac {3}{25} b d^2 e n x^5-\frac {1}{12} b d e^2 n x^6-\frac {1}{49} b e^3 n x^7+\frac {1}{4} d^3 x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {3}{5} d^2 e x^5 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} d e^2 x^6 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{7} e^3 x^7 \left (a+b \log \left (c x^n\right )\right ) \]

[In]

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

[Out]

-1/16*(b*d^3*n*x^4) - (3*b*d^2*e*n*x^5)/25 - (b*d*e^2*n*x^6)/12 - (b*e^3*n*x^7)/49 + (d^3*x^4*(a + b*Log[c*x^n
]))/4 + (3*d^2*e*x^5*(a + b*Log[c*x^n]))/5 + (d*e^2*x^6*(a + b*Log[c*x^n]))/2 + (e^3*x^7*(a + b*Log[c*x^n]))/7

Maple [A] (verified)

Time = 11.00 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.44

method	result	size
parallelrisch	\(\frac {x^{7} \ln \left (c \,x^{n}\right ) b \,e^{3}}{7}-\frac {b \,e^{3} n \,x^{7}}{49}+\frac {a \,e^{3} x^{7}}{7}+\frac {x^{6} \ln \left (c \,x^{n}\right ) b d \,e^{2}}{2}-\frac {b d \,e^{2} n \,x^{6}}{12}+\frac {a d \,e^{2} x^{6}}{2}+\frac {3 x^{5} \ln \left (c \,x^{n}\right ) b \,d^{2} e}{5}-\frac {3 b \,d^{2} e n \,x^{5}}{25}+\frac {3 a \,d^{2} e \,x^{5}}{5}+\frac {x^{4} \ln \left (c \,x^{n}\right ) b \,d^{3}}{4}-\frac {b \,d^{3} n \,x^{4}}{16}+\frac {a \,d^{3} x^{4}}{4}\)	\(144\)
risch	\(\frac {a \,e^{3} x^{7}}{7}+\frac {a \,d^{3} x^{4}}{4}+\frac {3 a \,d^{2} e \,x^{5}}{5}+\frac {a d \,e^{2} x^{6}}{2}+\frac {i \pi b \,e^{3} x^{7} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{14}+\frac {i \pi b \,e^{3} x^{7} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{14}-\frac {i \pi b d \,e^{2} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {b \,x^{4} \left (20 e^{3} x^{3}+70 d \,e^{2} x^{2}+84 d^{2} e x +35 d^{3}\right ) \ln \left (x^{n}\right )}{140}+\frac {\ln \left (c \right ) b \,e^{3} x^{7}}{7}+\frac {\ln \left (c \right ) b \,d^{3} x^{4}}{4}-\frac {i \pi b \,d^{3} x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}-\frac {i \pi b \,e^{3} x^{7} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{14}-\frac {3 i \pi b \,d^{2} e \,x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{10}+\frac {i \pi b \,d^{3} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}+\frac {i \pi b \,d^{3} x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b d \,e^{2} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{4}-\frac {3 i \pi b \,d^{2} 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 {3 b \,d^{2} e n \,x^{5}}{25}-\frac {b d \,e^{2} n \,x^{6}}{12}-\frac {b \,d^{3} n \,x^{4}}{16}-\frac {b \,e^{3} n \,x^{7}}{49}+\frac {3 \ln \left (c \right ) b \,d^{2} e \,x^{5}}{5}+\frac {\ln \left (c \right ) b d \,e^{2} x^{6}}{2}+\frac {i \pi b d \,e^{2} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}-\frac {i \pi b \,e^{3} x^{7} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{14}-\frac {i \pi b \,d^{3} 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 {3 i \pi b \,d^{2} e \,x^{5} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}+\frac {i \pi b d \,e^{2} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {3 i \pi b \,d^{2} e \,x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}\)	\(600\)

[In]

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

[Out]

1/7*x^7*ln(c*x^n)*b*e^3-1/49*b*e^3*n*x^7+1/7*a*e^3*x^7+1/2*x^6*ln(c*x^n)*b*d*e^2-1/12*b*d*e^2*n*x^6+1/2*a*d*e^
2*x^6+3/5*x^5*ln(c*x^n)*b*d^2*e-3/25*b*d^2*e*n*x^5+3/5*a*d^2*e*x^5+1/4*x^4*ln(c*x^n)*b*d^3-1/16*b*d^3*n*x^4+1/
4*a*d^3*x^4

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.67 \[ \int x^3 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{49} \, {\left (b e^{3} n - 7 \, a e^{3}\right )} x^{7} - \frac {1}{12} \, {\left (b d e^{2} n - 6 \, a d e^{2}\right )} x^{6} - \frac {3}{25} \, {\left (b d^{2} e n - 5 \, a d^{2} e\right )} x^{5} - \frac {1}{16} \, {\left (b d^{3} n - 4 \, a d^{3}\right )} x^{4} + \frac {1}{140} \, {\left (20 \, b e^{3} x^{7} + 70 \, b d e^{2} x^{6} + 84 \, b d^{2} e x^{5} + 35 \, b d^{3} x^{4}\right )} \log \left (c\right ) + \frac {1}{140} \, {\left (20 \, b e^{3} n x^{7} + 70 \, b d e^{2} n x^{6} + 84 \, b d^{2} e n x^{5} + 35 \, b d^{3} n x^{4}\right )} \log \left (x\right ) \]

[In]

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

[Out]

-1/49*(b*e^3*n - 7*a*e^3)*x^7 - 1/12*(b*d*e^2*n - 6*a*d*e^2)*x^6 - 3/25*(b*d^2*e*n - 5*a*d^2*e)*x^5 - 1/16*(b*
d^3*n - 4*a*d^3)*x^4 + 1/140*(20*b*e^3*x^7 + 70*b*d*e^2*x^6 + 84*b*d^2*e*x^5 + 35*b*d^3*x^4)*log(c) + 1/140*(2
0*b*e^3*n*x^7 + 70*b*d*e^2*n*x^6 + 84*b*d^2*e*n*x^5 + 35*b*d^3*n*x^4)*log(x)

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.70 \[ \int x^3 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d^{3} x^{4}}{4} + \frac {3 a d^{2} e x^{5}}{5} + \frac {a d e^{2} x^{6}}{2} + \frac {a e^{3} x^{7}}{7} - \frac {b d^{3} n x^{4}}{16} + \frac {b d^{3} x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {3 b d^{2} e n x^{5}}{25} + \frac {3 b d^{2} e x^{5} \log {\left (c x^{n} \right )}}{5} - \frac {b d e^{2} n x^{6}}{12} + \frac {b d e^{2} x^{6} \log {\left (c x^{n} \right )}}{2} - \frac {b e^{3} n x^{7}}{49} + \frac {b e^{3} x^{7} \log {\left (c x^{n} \right )}}{7} \]

[In]

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

[Out]

a*d**3*x**4/4 + 3*a*d**2*e*x**5/5 + a*d*e**2*x**6/2 + a*e**3*x**7/7 - b*d**3*n*x**4/16 + b*d**3*x**4*log(c*x**
n)/4 - 3*b*d**2*e*n*x**5/25 + 3*b*d**2*e*x**5*log(c*x**n)/5 - b*d*e**2*n*x**6/12 + b*d*e**2*x**6*log(c*x**n)/2
- b*e**3*n*x**7/49 + b*e**3*x**7*log(c*x**n)/7

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.43 \[ \int x^3 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{49} \, b e^{3} n x^{7} + \frac {1}{7} \, b e^{3} x^{7} \log \left (c x^{n}\right ) - \frac {1}{12} \, b d e^{2} n x^{6} + \frac {1}{7} \, a e^{3} x^{7} + \frac {1}{2} \, b d e^{2} x^{6} \log \left (c x^{n}\right ) - \frac {3}{25} \, b d^{2} e n x^{5} + \frac {1}{2} \, a d e^{2} x^{6} + \frac {3}{5} \, b d^{2} e x^{5} \log \left (c x^{n}\right ) - \frac {1}{16} \, b d^{3} n x^{4} + \frac {3}{5} \, a d^{2} e x^{5} + \frac {1}{4} \, b d^{3} x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d^{3} x^{4} \]

[In]

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

[Out]

-1/49*b*e^3*n*x^7 + 1/7*b*e^3*x^7*log(c*x^n) - 1/12*b*d*e^2*n*x^6 + 1/7*a*e^3*x^7 + 1/2*b*d*e^2*x^6*log(c*x^n)
- 3/25*b*d^2*e*n*x^5 + 1/2*a*d*e^2*x^6 + 3/5*b*d^2*e*x^5*log(c*x^n) - 1/16*b*d^3*n*x^4 + 3/5*a*d^2*e*x^5 + 1/
4*b*d^3*x^4*log(c*x^n) + 1/4*a*d^3*x^4

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.77 \[ \int x^3 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{7} \, b e^{3} n x^{7} \log \left (x\right ) - \frac {1}{49} \, b e^{3} n x^{7} + \frac {1}{7} \, b e^{3} x^{7} \log \left (c\right ) + \frac {1}{2} \, b d e^{2} n x^{6} \log \left (x\right ) - \frac {1}{12} \, b d e^{2} n x^{6} + \frac {1}{7} \, a e^{3} x^{7} + \frac {1}{2} \, b d e^{2} x^{6} \log \left (c\right ) + \frac {3}{5} \, b d^{2} e n x^{5} \log \left (x\right ) - \frac {3}{25} \, b d^{2} e n x^{5} + \frac {1}{2} \, a d e^{2} x^{6} + \frac {3}{5} \, b d^{2} e x^{5} \log \left (c\right ) + \frac {1}{4} \, b d^{3} n x^{4} \log \left (x\right ) - \frac {1}{16} \, b d^{3} n x^{4} + \frac {3}{5} \, a d^{2} e x^{5} + \frac {1}{4} \, b d^{3} x^{4} \log \left (c\right ) + \frac {1}{4} \, a d^{3} x^{4} \]

[In]

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

[Out]

1/7*b*e^3*n*x^7*log(x) - 1/49*b*e^3*n*x^7 + 1/7*b*e^3*x^7*log(c) + 1/2*b*d*e^2*n*x^6*log(x) - 1/12*b*d*e^2*n*x
^6 + 1/7*a*e^3*x^7 + 1/2*b*d*e^2*x^6*log(c) + 3/5*b*d^2*e*n*x^5*log(x) - 3/25*b*d^2*e*n*x^5 + 1/2*a*d*e^2*x^6
+ 3/5*b*d^2*e*x^5*log(c) + 1/4*b*d^3*n*x^4*log(x) - 1/16*b*d^3*n*x^4 + 3/5*a*d^2*e*x^5 + 1/4*b*d^3*x^4*log(c)
+ 1/4*a*d^3*x^4

Mupad [B] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.13 \[ \int x^3 (d+e x)^3 \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3\,x^4}{4}+\frac {3\,b\,d^2\,e\,x^5}{5}+\frac {b\,d\,e^2\,x^6}{2}+\frac {b\,e^3\,x^7}{7}\right )+\frac {d^3\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {e^3\,x^7\,\left (7\,a-b\,n\right )}{49}+\frac {3\,d^2\,e\,x^5\,\left (5\,a-b\,n\right )}{25}+\frac {d\,e^2\,x^6\,\left (6\,a-b\,n\right )}{12} \]

[In]

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

[Out]

log(c*x^n)*((b*d^3*x^4)/4 + (b*e^3*x^7)/7 + (3*b*d^2*e*x^5)/5 + (b*d*e^2*x^6)/2) + (d^3*x^4*(4*a - b*n))/16 +
(e^3*x^7*(7*a - b*n))/49 + (3*d^2*e*x^5*(5*a - b*n))/25 + (d*e^2*x^6*(6*a - b*n))/12