3.24 \(\int \frac {(d+e x)^3 (a+b \log (c x^n))}{x^2} \, dx\)

Optimal. Leaf size=119 \[ -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )+3 d e^2 x \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b d^3 n}{x}-\frac {3}{2} b d^2 e n \log ^2(x)-3 b d e^2 n x-\frac {1}{4} b e^3 n x^2 \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 92, normalized size of antiderivative = 0.77, number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {43, 2334, 2301} \[ -\frac {1}{2} \left (-6 d^2 e \log (x)+\frac {2 d^3}{x}-6 d e^2 x-e^3 x^2\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {3}{2} b d^2 e n \log ^2(x)-\frac {b d^3 n}{x}-3 b d e^2 n x-\frac {1}{4} b e^3 n x^2 \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(a + b*Log[c*x^n]))/x^2,x]

[Out]

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

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

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Mathematica [A]  time = 0.08, size = 118, normalized size = 0.99 \[ -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {3 d^2 e \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+\frac {1}{2} e^3 x^2 \left (a+b \log \left (c x^n\right )\right )+3 a d e^2 x+3 b d e^2 x \log \left (c x^n\right )-\frac {b d^3 n}{x}-3 b d e^2 n x-\frac {1}{4} b e^3 n x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(a + b*Log[c*x^n]))/x^2,x]

[Out]

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

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fricas [A]  time = 0.57, size = 149, normalized size = 1.25 \[ \frac {6 \, b d^{2} e n x \log \relax (x)^{2} - 4 \, b d^{3} n - 4 \, a d^{3} - {\left (b e^{3} n - 2 \, a e^{3}\right )} x^{3} - 12 \, {\left (b d e^{2} n - a d e^{2}\right )} x^{2} + 2 \, {\left (b e^{3} x^{3} + 6 \, b d e^{2} x^{2} - 2 \, b d^{3}\right )} \log \relax (c) + 2 \, {\left (b e^{3} n x^{3} + 6 \, b d e^{2} n x^{2} + 6 \, b d^{2} e x \log \relax (c) - 2 \, b d^{3} n + 6 \, a d^{2} e x\right )} \log \relax (x)}{4 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 154, normalized size = 1.29 \[ \frac {6 \, b d^{2} n x e \log \relax (x)^{2} + 2 \, b n x^{3} e^{3} \log \relax (x) + 12 \, b d n x^{2} e^{2} \log \relax (x) + 12 \, b d^{2} x e \log \relax (c) \log \relax (x) - b n x^{3} e^{3} - 12 \, b d n x^{2} e^{2} + 2 \, b x^{3} e^{3} \log \relax (c) + 12 \, b d x^{2} e^{2} \log \relax (c) - 4 \, b d^{3} n \log \relax (x) + 12 \, a d^{2} x e \log \relax (x) - 4 \, b d^{3} n + 2 \, a x^{3} e^{3} + 12 \, a d x^{2} e^{2} - 4 \, b d^{3} \log \relax (c) - 4 \, a d^{3}}{4 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^2,x, algorithm="giac")

[Out]

1/4*(6*b*d^2*n*x*e*log(x)^2 + 2*b*n*x^3*e^3*log(x) + 12*b*d*n*x^2*e^2*log(x) + 12*b*d^2*x*e*log(c)*log(x) - b*
n*x^3*e^3 - 12*b*d*n*x^2*e^2 + 2*b*x^3*e^3*log(c) + 12*b*d*x^2*e^2*log(c) - 4*b*d^3*n*log(x) + 12*a*d^2*x*e*lo
g(x) - 4*b*d^3*n + 2*a*x^3*e^3 + 12*a*d*x^2*e^2 - 4*b*d^3*log(c) - 4*a*d^3)/x

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maple [C]  time = 0.33, size = 588, normalized size = 4.94 \[ -\frac {\left (-e^{3} x^{3}-6 d^{2} e x \ln \relax (x )-6 d \,e^{2} x^{2}+2 d^{3}\right ) b \ln \left (x^{n}\right )}{2 x}-\frac {-12 b d \,e^{2} x^{2} \ln \relax (c )-12 a d \,e^{2} x^{2}-12 a \,d^{2} e x \ln \relax (x )+4 a \,d^{3}-2 a \,e^{3} x^{3}+4 b \,d^{3} n +4 b \,d^{3} \ln \relax (c )-2 b \,e^{3} x^{3} \ln \relax (c )+6 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 )+6 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 ) \ln \relax (x )-2 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 )+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 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-6 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+b \,e^{3} n \,x^{3}-2 i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-12 b \,d^{2} e x \ln \relax (c ) \ln \relax (x )+6 b \,d^{2} e n x \ln \relax (x )^{2}+i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+6 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+2 i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+2 i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-6 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )-6 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )+6 i \pi b \,d^{2} e x \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )+12 b d \,e^{2} n \,x^{2}}{4 x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(b*ln(c*x^n)+a)/x^2,x)

[Out]

-1/2*b*(-e^3*x^3-6*d^2*e*ln(x)*x-6*d*e^2*x^2+2*d^3)/x*ln(x^n)-1/4*(-12*b*d*e^2*x^2*ln(c)+6*I*ln(x)*Pi*b*d^2*e*
csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x-12*a*d*e^2*x^2-12*ln(x)*a*d^2*e*x+4*a*d^3-6*I*ln(x)*Pi*b*d^2*e*csgn(I*c*
x^n)^2*csgn(I*c)*x-2*a*e^3*x^3+4*b*d^3*n+4*ln(c)*b*d^3-2*b*e^3*x^3*ln(c)+6*I*Pi*b*d*e^2*x^2*csgn(I*x^n)*csgn(I
*c*x^n)*csgn(I*c)-6*I*ln(x)*Pi*b*d^2*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x+I*Pi*b*e^3*x^3*csgn(I*x^n)*csgn(I*c*x^n)*
csgn(I*c)-6*I*Pi*b*d*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-6*I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^2*csgn(I*c)-2*I*Pi*b
*d^3*csgn(I*c*x^n)^3+b*e^3*n*x^3+6*I*ln(x)*Pi*b*d^2*e*csgn(I*c*x^n)^3*x-12*ln(x)*ln(c)*b*d^2*e*x+6*b*d^2*e*n*l
n(x)^2*x+I*Pi*b*e^3*x^3*csgn(I*c*x^n)^3+2*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2+2*I*Pi*b*d^3*csgn(I*c*x^n)^2*
csgn(I*c)-I*Pi*b*e^3*x^3*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*e^3*x^3*csgn(I*c*x^n)^2*csgn(I*c)-2*I*Pi*b*d^3*csg
n(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+6*I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^3+12*b*d*e^2*n*x^2)/x

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maxima [A]  time = 0.57, size = 127, normalized size = 1.07 \[ -\frac {1}{4} \, b e^{3} n x^{2} + \frac {1}{2} \, b e^{3} x^{2} \log \left (c x^{n}\right ) - 3 \, b d e^{2} n x + \frac {1}{2} \, a e^{3} x^{2} + 3 \, b d e^{2} x \log \left (c x^{n}\right ) + 3 \, a d e^{2} x + \frac {3 \, b d^{2} e \log \left (c x^{n}\right )^{2}}{2 \, n} + 3 \, a d^{2} e \log \relax (x) - \frac {b d^{3} n}{x} - \frac {b d^{3} \log \left (c x^{n}\right )}{x} - \frac {a d^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.65, size = 154, normalized size = 1.29 \[ \ln \relax (x)\,\left (3\,a\,d^2\,e+3\,b\,d^2\,e\,n\right )-\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3+3\,b\,d^2\,e\,x+3\,b\,d\,e^2\,x^2+b\,e^3\,x^3}{x}-\frac {\frac {3\,b\,e^3\,x^3}{2}+6\,b\,d\,e^2\,x^2}{x}\right )-\frac {a\,d^3+b\,d^3\,n}{x}+\frac {e^3\,x^2\,\left (2\,a-b\,n\right )}{4}+3\,d\,e^2\,x\,\left (a-b\,n\right )+\frac {3\,b\,d^2\,e\,{\ln \left (c\,x^n\right )}^2}{2\,n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*log(c*x^n))*(d + e*x)^3)/x^2,x)

[Out]

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

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sympy [A]  time = 1.87, size = 182, normalized size = 1.53 \[ - \frac {a d^{3}}{x} + 3 a d^{2} e \log {\relax (x )} + 3 a d e^{2} x + \frac {a e^{3} x^{2}}{2} - \frac {b d^{3} n \log {\relax (x )}}{x} - \frac {b d^{3} n}{x} - \frac {b d^{3} \log {\relax (c )}}{x} + \frac {3 b d^{2} e n \log {\relax (x )}^{2}}{2} + 3 b d^{2} e \log {\relax (c )} \log {\relax (x )} + 3 b d e^{2} n x \log {\relax (x )} - 3 b d e^{2} n x + 3 b d e^{2} x \log {\relax (c )} + \frac {b e^{3} n x^{2} \log {\relax (x )}}{2} - \frac {b e^{3} n x^{2}}{4} + \frac {b e^{3} x^{2} \log {\relax (c )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a+b*ln(c*x**n))/x**2,x)

[Out]

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

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