3.23 \(\int \frac {(d+e x)^3 (a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=122 \[ d^3 \log (x) \left (a+b \log \left (c x^n\right )\right )+3 d^2 e x \left (a+b \log \left (c x^n\right )\right )+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{3} e^3 x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} b d^3 n \log ^2(x)-3 b d^2 e n x-\frac {3}{4} b d e^2 n x^2-\frac {1}{9} b e^3 n x^3 \]

[Out]

-3*b*d^2*e*n*x-3/4*b*d*e^2*n*x^2-1/9*b*e^3*n*x^3-1/2*b*d^3*n*ln(x)^2+3*d^2*e*x*(a+b*ln(c*x^n))+3/2*d*e^2*x^2*(
a+b*ln(c*x^n))+1/3*e^3*x^3*(a+b*ln(c*x^n))+d^3*ln(x)*(a+b*ln(c*x^n))

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Rubi [A]  time = 0.09, antiderivative size = 94, normalized size of antiderivative = 0.77, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {43, 2334, 2301} \[ \frac {1}{6} \left (18 d^2 e x+6 d^3 \log (x)+9 d e^2 x^2+2 e^3 x^3\right ) \left (a+b \log \left (c x^n\right )\right )-3 b d^2 e n x-\frac {1}{2} b d^3 n \log ^2(x)-\frac {3}{4} b d e^2 n x^2-\frac {1}{9} b e^3 n x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3*b*d^2*e*n*x - (3*b*d*e^2*n*x^2)/4 - (b*e^3*n*x^3)/9 - (b*d^3*n*Log[x]^2)/2 + ((18*d^2*e*x + 9*d*e^2*x^2 + 2
*e^3*x^3 + 6*d^3*Log[x])*(a + b*Log[c*x^n]))/6

Rule 43

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2334

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] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {align*} \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x} \, dx &=\frac {1}{6} \left (18 d^2 e x+9 d e^2 x^2+2 e^3 x^3+6 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (\frac {1}{6} e \left (18 d^2+9 d e x+2 e^2 x^2\right )+\frac {d^3 \log (x)}{x}\right ) \, dx\\ &=\frac {1}{6} \left (18 d^2 e x+9 d e^2 x^2+2 e^3 x^3+6 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\left (b d^3 n\right ) \int \frac {\log (x)}{x} \, dx-\frac {1}{6} (b e n) \int \left (18 d^2+9 d e x+2 e^2 x^2\right ) \, dx\\ &=-3 b d^2 e n x-\frac {3}{4} b d e^2 n x^2-\frac {1}{9} b e^3 n x^3-\frac {1}{2} b d^3 n \log ^2(x)+\frac {1}{6} \left (18 d^2 e x+9 d e^2 x^2+2 e^3 x^3+6 d^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 123, normalized size = 1.01 \[ \frac {d^3 \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{3} e^3 x^3 \left (a+b \log \left (c x^n\right )\right )+3 a d^2 e x+3 b d^2 e x \log \left (c x^n\right )-3 b d^2 e n x-\frac {3}{4} b d e^2 n x^2-\frac {1}{9} b e^3 n x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

3*a*d^2*e*x - 3*b*d^2*e*n*x - (3*b*d*e^2*n*x^2)/4 - (b*e^3*n*x^3)/9 + 3*b*d^2*e*x*Log[c*x^n] + (3*d*e^2*x^2*(a
+ b*Log[c*x^n]))/2 + (e^3*x^3*(a + b*Log[c*x^n]))/3 + (d^3*(a + b*Log[c*x^n])^2)/(2*b*n)

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fricas [A]  time = 0.54, size = 149, normalized size = 1.22 \[ \frac {1}{2} \, b d^{3} n \log \relax (x)^{2} - \frac {1}{9} \, {\left (b e^{3} n - 3 \, a e^{3}\right )} x^{3} - \frac {3}{4} \, {\left (b d e^{2} n - 2 \, a d e^{2}\right )} x^{2} - 3 \, {\left (b d^{2} e n - a d^{2} e\right )} x + \frac {1}{6} \, {\left (2 \, b e^{3} x^{3} + 9 \, b d e^{2} x^{2} + 18 \, b d^{2} e x\right )} \log \relax (c) + \frac {1}{6} \, {\left (2 \, b e^{3} n x^{3} + 9 \, b d e^{2} n x^{2} + 18 \, b d^{2} e n x + 6 \, b d^{3} \log \relax (c) + 6 \, a d^{3}\right )} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*d^3*n*log(x)^2 - 1/9*(b*e^3*n - 3*a*e^3)*x^3 - 3/4*(b*d*e^2*n - 2*a*d*e^2)*x^2 - 3*(b*d^2*e*n - a*d^2*e)
*x + 1/6*(2*b*e^3*x^3 + 9*b*d*e^2*x^2 + 18*b*d^2*e*x)*log(c) + 1/6*(2*b*e^3*n*x^3 + 9*b*d*e^2*n*x^2 + 18*b*d^2
*e*n*x + 6*b*d^3*log(c) + 6*a*d^3)*log(x)

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giac [A]  time = 0.30, size = 150, normalized size = 1.23 \[ \frac {1}{3} \, b n x^{3} e^{3} \log \relax (x) + \frac {3}{2} \, b d n x^{2} e^{2} \log \relax (x) + 3 \, b d^{2} n x e \log \relax (x) + \frac {1}{2} \, b d^{3} n \log \relax (x)^{2} - \frac {1}{9} \, b n x^{3} e^{3} - \frac {3}{4} \, b d n x^{2} e^{2} - 3 \, b d^{2} n x e + \frac {1}{3} \, b x^{3} e^{3} \log \relax (c) + \frac {3}{2} \, b d x^{2} e^{2} \log \relax (c) + 3 \, b d^{2} x e \log \relax (c) + b d^{3} \log \relax (c) \log \relax (x) + \frac {1}{3} \, a x^{3} e^{3} + \frac {3}{2} \, a d x^{2} e^{2} + 3 \, a d^{2} x e + a d^{3} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*n*x^3*e^3*log(x) + 3/2*b*d*n*x^2*e^2*log(x) + 3*b*d^2*n*x*e*log(x) + 1/2*b*d^3*n*log(x)^2 - 1/9*b*n*x^3*
e^3 - 3/4*b*d*n*x^2*e^2 - 3*b*d^2*n*x*e + 1/3*b*x^3*e^3*log(c) + 3/2*b*d*x^2*e^2*log(c) + 3*b*d^2*x*e*log(c) +
b*d^3*log(c)*log(x) + 1/3*a*x^3*e^3 + 3/2*a*d*x^2*e^2 + 3*a*d^2*x*e + a*d^3*log(x)

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maple [C]  time = 0.30, size = 579, normalized size = 4.75 \[ \frac {3 b d \,e^{2} x^{2} \ln \relax (c )}{2}+3 b \,d^{2} e x \ln \relax (c )-\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )}{2}+\frac {3 a d \,e^{2} x^{2}}{2}+3 a \,d^{2} e x +\frac {a \,e^{3} x^{3}}{3}+\frac {b \,e^{3} x^{3} \ln \relax (c )}{3}+\left (\frac {b \,e^{3} x^{3}}{3}+\frac {3 b d \,e^{2} x^{2}}{2}+b \,d^{3} \ln \relax (x )+3 b \,d^{2} e x \right ) \ln \left (x^{n}\right )+b \,d^{3} \ln \relax (c ) \ln \relax (x )+a \,d^{3} \ln \relax (x )-\frac {3 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2}-\frac {3 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4}-\frac {i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{6}+\frac {3 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}+\frac {3 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4}-\frac {b \,e^{3} n \,x^{3}}{9}+\frac {3 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {3 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{6}-\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2}+\frac {i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6}+\frac {i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {3 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2}+\frac {i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2}-\frac {3 i \pi b \,d^{2} e x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2}-\frac {b \,d^{3} n \ln \relax (x )^{2}}{2}-3 b \,d^{2} e n x -\frac {3 b d \,e^{2} n \,x^{2}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/2*ln(c)*b*d*e^2*x^2+3*ln(c)*b*d^2*e*x+3/2*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^2*csgn(I*c)-1/2*I*ln(x)*Pi*b*d^3*csgn
(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+3/2*a*d*e^2*x^2+3*a*d^2*e*x+3/4*I*Pi*b*d*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+3
/4*I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^2*csgn(I*c)+3/2*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2*e*x-1/6*I*Pi*b*e^3*x^
3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/3*a*e^3*x^3+1/3*ln(c)*b*e^3*x^3+(1/3*b*e^3*x^3+3/2*b*d*e^2*x^2+3*b*d^2
*e*x+b*d^3*ln(x))*ln(x^n)+ln(x)*ln(c)*b*d^3+ln(x)*a*d^3-3/4*I*Pi*b*d*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*
c)-3/2*I*Pi*b*d^2*e*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/9*b*e^3*n*x^3-1/6*I*Pi*b*e^3*x^3*csgn(I*c*x^n)^3-1
/2*I*ln(x)*Pi*b*d^3*csgn(I*c*x^n)^3-3/2*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^3+1/6*I*Pi*b*e^3*x^3*csgn(I*x^n)*csgn(I*c
*x^n)^2-3/4*I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^3+1/2*I*ln(x)*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2+1/2*I*ln(x)*Pi*b
*d^3*csgn(I*c*x^n)^2*csgn(I*c)+1/6*I*Pi*b*e^3*x^3*csgn(I*c*x^n)^2*csgn(I*c)-1/2*b*d^3*n*ln(x)^2-3*b*d^2*e*n*x-
3/4*b*d*e^2*n*x^2

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maxima [A]  time = 0.63, size = 127, normalized size = 1.04 \[ -\frac {1}{9} \, b e^{3} n x^{3} + \frac {1}{3} \, b e^{3} x^{3} \log \left (c x^{n}\right ) - \frac {3}{4} \, b d e^{2} n x^{2} + \frac {1}{3} \, a e^{3} x^{3} + \frac {3}{2} \, b d e^{2} x^{2} \log \left (c x^{n}\right ) - 3 \, b d^{2} e n x + \frac {3}{2} \, a d e^{2} x^{2} + 3 \, b d^{2} e x \log \left (c x^{n}\right ) + 3 \, a d^{2} e x + \frac {b d^{3} \log \left (c x^{n}\right )^{2}}{2 \, n} + a d^{3} \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*b*e^3*n*x^3 + 1/3*b*e^3*x^3*log(c*x^n) - 3/4*b*d*e^2*n*x^2 + 1/3*a*e^3*x^3 + 3/2*b*d*e^2*x^2*log(c*x^n) -
3*b*d^2*e*n*x + 3/2*a*d*e^2*x^2 + 3*b*d^2*e*x*log(c*x^n) + 3*a*d^2*e*x + 1/2*b*d^3*log(c*x^n)^2/n + a*d^3*log
(x)

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mupad [B]  time = 3.64, size = 106, normalized size = 0.87 \[ \ln \left (c\,x^n\right )\,\left (3\,b\,d^2\,e\,x+\frac {3\,b\,d\,e^2\,x^2}{2}+\frac {b\,e^3\,x^3}{3}\right )+\frac {e^3\,x^3\,\left (3\,a-b\,n\right )}{9}+a\,d^3\,\ln \relax (x)+\frac {b\,d^3\,{\ln \left (c\,x^n\right )}^2}{2\,n}+\frac {3\,d\,e^2\,x^2\,\left (2\,a-b\,n\right )}{4}+3\,d^2\,e\,x\,\left (a-b\,n\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(c*x^n)*((b*e^3*x^3)/3 + 3*b*d^2*e*x + (3*b*d*e^2*x^2)/2) + (e^3*x^3*(3*a - b*n))/9 + a*d^3*log(x) + (b*d^3
*log(c*x^n)^2)/(2*n) + (3*d*e^2*x^2*(2*a - b*n))/4 + 3*d^2*e*x*(a - b*n)

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sympy [A]  time = 1.86, size = 199, normalized size = 1.63 \[ a d^{3} \log {\relax (x )} + 3 a d^{2} e x + \frac {3 a d e^{2} x^{2}}{2} + \frac {a e^{3} x^{3}}{3} + \frac {b d^{3} n \log {\relax (x )}^{2}}{2} + b d^{3} \log {\relax (c )} \log {\relax (x )} + 3 b d^{2} e n x \log {\relax (x )} - 3 b d^{2} e n x + 3 b d^{2} e x \log {\relax (c )} + \frac {3 b d e^{2} n x^{2} \log {\relax (x )}}{2} - \frac {3 b d e^{2} n x^{2}}{4} + \frac {3 b d e^{2} x^{2} \log {\relax (c )}}{2} + \frac {b e^{3} n x^{3} \log {\relax (x )}}{3} - \frac {b e^{3} n x^{3}}{9} + \frac {b e^{3} x^{3} \log {\relax (c )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*d**3*log(x) + 3*a*d**2*e*x + 3*a*d*e**2*x**2/2 + a*e**3*x**3/3 + b*d**3*n*log(x)**2/2 + b*d**3*log(c)*log(x)
+ 3*b*d**2*e*n*x*log(x) - 3*b*d**2*e*n*x + 3*b*d**2*e*x*log(c) + 3*b*d*e**2*n*x**2*log(x)/2 - 3*b*d*e**2*n*x*
*2/4 + 3*b*d*e**2*x**2*log(c)/2 + b*e**3*n*x**3*log(x)/3 - b*e**3*n*x**3/9 + b*e**3*x**3*log(c)/3

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