3.27 \(\int \frac {(d+e x)^3 (a+b \log (c x^n))}{x^5} \, dx\)

Optimal. Leaf size=90 \[ -\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 d x^4}-\frac {b d^3 n}{16 x^4}-\frac {b d^2 e n}{3 x^3}+\frac {b e^4 n \log (x)}{4 d}-\frac {3 b d e^2 n}{4 x^2}-\frac {b e^3 n}{x} \]

[Out]

-1/16*b*d^3*n/x^4-1/3*b*d^2*e*n/x^3-3/4*b*d*e^2*n/x^2-b*e^3*n/x+1/4*b*e^4*n*ln(x)/d-1/4*(e*x+d)^4*(a+b*ln(c*x^
n))/d/x^4

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Rubi [A]  time = 0.08, antiderivative size = 90, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {37, 2334, 12, 43} \[ -\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 d x^4}-\frac {b d^2 e n}{3 x^3}-\frac {b d^3 n}{16 x^4}-\frac {3 b d e^2 n}{4 x^2}+\frac {b e^4 n \log (x)}{4 d}-\frac {b e^3 n}{x} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(a + b*Log[c*x^n]))/x^5,x]

[Out]

-(b*d^3*n)/(16*x^4) - (b*d^2*e*n)/(3*x^3) - (3*b*d*e^2*n)/(4*x^2) - (b*e^3*n)/x + (b*e^4*n*Log[x])/(4*d) - ((d
+ e*x)^4*(a + b*Log[c*x^n]))/(4*d*x^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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

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 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^5} \, dx &=-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 d x^4}-(b n) \int -\frac {(d+e x)^4}{4 d x^5} \, dx\\ &=-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 d x^4}+\frac {(b n) \int \frac {(d+e x)^4}{x^5} \, dx}{4 d}\\ &=-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 d x^4}+\frac {(b n) \int \left (\frac {d^4}{x^5}+\frac {4 d^3 e}{x^4}+\frac {6 d^2 e^2}{x^3}+\frac {4 d e^3}{x^2}+\frac {e^4}{x}\right ) \, dx}{4 d}\\ &=-\frac {b d^3 n}{16 x^4}-\frac {b d^2 e n}{3 x^3}-\frac {3 b d e^2 n}{4 x^2}-\frac {b e^3 n}{x}+\frac {b e^4 n \log (x)}{4 d}-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{4 d x^4}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 109, normalized size = 1.21 \[ -\frac {12 a \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )+12 b \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right ) \log \left (c x^n\right )+b n \left (3 d^3+16 d^2 e x+36 d e^2 x^2+48 e^3 x^3\right )}{48 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(a + b*Log[c*x^n]))/x^5,x]

[Out]

-1/48*(12*a*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) + b*n*(3*d^3 + 16*d^2*e*x + 36*d*e^2*x^2 + 48*e^3*x^3)
+ 12*b*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3)*Log[c*x^n])/x^4

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fricas [A]  time = 0.42, size = 152, normalized size = 1.69 \[ -\frac {3 \, b d^{3} n + 12 \, a d^{3} + 48 \, {\left (b e^{3} n + a e^{3}\right )} x^{3} + 36 \, {\left (b d e^{2} n + 2 \, a d e^{2}\right )} x^{2} + 16 \, {\left (b d^{2} e n + 3 \, a d^{2} e\right )} x + 12 \, {\left (4 \, b e^{3} x^{3} + 6 \, b d e^{2} x^{2} + 4 \, b d^{2} e x + b d^{3}\right )} \log \relax (c) + 12 \, {\left (4 \, b e^{3} n x^{3} + 6 \, b d e^{2} n x^{2} + 4 \, b d^{2} e n x + b d^{3} n\right )} \log \relax (x)}{48 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="fricas")

[Out]

-1/48*(3*b*d^3*n + 12*a*d^3 + 48*(b*e^3*n + a*e^3)*x^3 + 36*(b*d*e^2*n + 2*a*d*e^2)*x^2 + 16*(b*d^2*e*n + 3*a*
d^2*e)*x + 12*(4*b*e^3*x^3 + 6*b*d*e^2*x^2 + 4*b*d^2*e*x + b*d^3)*log(c) + 12*(4*b*e^3*n*x^3 + 6*b*d*e^2*n*x^2
+ 4*b*d^2*e*n*x + b*d^3*n)*log(x))/x^4

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giac [A]  time = 0.32, size = 158, normalized size = 1.76 \[ -\frac {48 \, b n x^{3} e^{3} \log \relax (x) + 72 \, b d n x^{2} e^{2} \log \relax (x) + 48 \, b d^{2} n x e \log \relax (x) + 48 \, b n x^{3} e^{3} + 36 \, b d n x^{2} e^{2} + 16 \, b d^{2} n x e + 48 \, b x^{3} e^{3} \log \relax (c) + 72 \, b d x^{2} e^{2} \log \relax (c) + 48 \, b d^{2} x e \log \relax (c) + 12 \, b d^{3} n \log \relax (x) + 3 \, b d^{3} n + 48 \, a x^{3} e^{3} + 72 \, a d x^{2} e^{2} + 48 \, a d^{2} x e + 12 \, b d^{3} \log \relax (c) + 12 \, a d^{3}}{48 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="giac")

[Out]

-1/48*(48*b*n*x^3*e^3*log(x) + 72*b*d*n*x^2*e^2*log(x) + 48*b*d^2*n*x*e*log(x) + 48*b*n*x^3*e^3 + 36*b*d*n*x^2
*e^2 + 16*b*d^2*n*x*e + 48*b*x^3*e^3*log(c) + 72*b*d*x^2*e^2*log(c) + 48*b*d^2*x*e*log(c) + 12*b*d^3*n*log(x)
+ 3*b*d^3*n + 48*a*x^3*e^3 + 72*a*d*x^2*e^2 + 48*a*d^2*x*e + 12*b*d^3*log(c) + 12*a*d^3)/x^4

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maple [C]  time = 0.18, size = 569, normalized size = 6.32 \[ -\frac {\left (4 e^{3} x^{3}+6 d \,e^{2} x^{2}+4 d^{2} e x +d^{3}\right ) b \ln \left (x^{n}\right )}{4 x^{4}}-\frac {72 b d \,e^{2} x^{2} \ln \relax (c )+48 b \,d^{2} e x \ln \relax (c )+72 a d \,e^{2} x^{2}+48 a \,d^{2} e x +12 a \,d^{3}+48 a \,e^{3} x^{3}+3 b \,d^{3} n +12 b \,d^{3} \ln \relax (c )+48 b \,e^{3} x^{3} \ln \relax (c )-24 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 )-36 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^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )-24 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 )+36 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+36 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+48 b \,e^{3} n \,x^{3}-6 i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+24 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+24 i \pi b \,d^{2} e x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-24 i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+24 i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+24 i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-36 i \pi b d \,e^{2} x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-24 i \pi b \,d^{2} e x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+6 i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+6 i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+16 b \,d^{2} e n x +36 b d \,e^{2} n \,x^{2}}{48 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(b*ln(c*x^n)+a)/x^5,x)

[Out]

-1/4*b*(4*e^3*x^3+6*d*e^2*x^2+4*d^2*e*x+d^3)/x^4*ln(x^n)-1/48*(72*b*d*e^2*x^2*ln(c)+48*b*d^2*e*x*ln(c)+72*a*d*
e^2*x^2+48*a*d^2*e*x+12*a*d^3+48*a*e^3*x^3+3*b*d^3*n+24*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2*e*x-24*I*Pi*b*e
^3*x^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+12*b*d^3*ln(c)+48*b*e^3*x^3*ln(c)+24*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^2
*csgn(I*c)-36*I*Pi*b*d*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-24*I*Pi*b*d^2*e*x*csgn(I*x^n)*csgn(I*c*x^n)
*csgn(I*c)+48*b*e^3*n*x^3+36*I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^2*csgn(I*c)-6*I*Pi*b*d^3*csgn(I*c*x^n)^3+36*I*Pi*b
*d*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+24*I*Pi*b*e^3*x^3*csgn(I*c*x^n)^2*csgn(I*c)+24*I*Pi*b*e^3*x^3*csgn(I*x^
n)*csgn(I*c*x^n)^2-24*I*Pi*b*e^3*x^3*csgn(I*c*x^n)^3+6*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2+6*I*Pi*b*d^3*csg
n(I*c)*csgn(I*c*x^n)^2-36*I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^3-24*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^3+16*b*d^2*e*n*x+36
*b*d*e^2*n*x^2-6*I*Pi*b*d^3*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n))/x^4

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maxima [A]  time = 0.67, size = 143, normalized size = 1.59 \[ -\frac {b e^{3} n}{x} - \frac {b e^{3} \log \left (c x^{n}\right )}{x} - \frac {3 \, b d e^{2} n}{4 \, x^{2}} - \frac {a e^{3}}{x} - \frac {3 \, b d e^{2} \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {b d^{2} e n}{3 \, x^{3}} - \frac {3 \, a d e^{2}}{2 \, x^{2}} - \frac {b d^{2} e \log \left (c x^{n}\right )}{x^{3}} - \frac {b d^{3} n}{16 \, x^{4}} - \frac {a d^{2} e}{x^{3}} - \frac {b d^{3} \log \left (c x^{n}\right )}{4 \, x^{4}} - \frac {a d^{3}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^5,x, algorithm="maxima")

[Out]

-b*e^3*n/x - b*e^3*log(c*x^n)/x - 3/4*b*d*e^2*n/x^2 - a*e^3/x - 3/2*b*d*e^2*log(c*x^n)/x^2 - 1/3*b*d^2*e*n/x^3
- 3/2*a*d*e^2/x^2 - b*d^2*e*log(c*x^n)/x^3 - 1/16*b*d^3*n/x^4 - a*d^2*e/x^3 - 1/4*b*d^3*log(c*x^n)/x^4 - 1/4*
a*d^3/x^4

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mupad [B]  time = 3.71, size = 118, normalized size = 1.31 \[ -\frac {x^3\,\left (4\,a\,e^3+4\,b\,e^3\,n\right )+x\,\left (4\,a\,d^2\,e+\frac {4\,b\,d^2\,e\,n}{3}\right )+a\,d^3+x^2\,\left (6\,a\,d\,e^2+3\,b\,d\,e^2\,n\right )+\frac {b\,d^3\,n}{4}}{4\,x^4}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3}{4}+b\,d^2\,e\,x+\frac {3\,b\,d\,e^2\,x^2}{2}+b\,e^3\,x^3\right )}{x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*log(c*x^n))*(d + e*x)^3)/x^5,x)

[Out]

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

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sympy [B]  time = 2.99, size = 206, normalized size = 2.29 \[ - \frac {a d^{3}}{4 x^{4}} - \frac {a d^{2} e}{x^{3}} - \frac {3 a d e^{2}}{2 x^{2}} - \frac {a e^{3}}{x} - \frac {b d^{3} n \log {\relax (x )}}{4 x^{4}} - \frac {b d^{3} n}{16 x^{4}} - \frac {b d^{3} \log {\relax (c )}}{4 x^{4}} - \frac {b d^{2} e n \log {\relax (x )}}{x^{3}} - \frac {b d^{2} e n}{3 x^{3}} - \frac {b d^{2} e \log {\relax (c )}}{x^{3}} - \frac {3 b d e^{2} n \log {\relax (x )}}{2 x^{2}} - \frac {3 b d e^{2} n}{4 x^{2}} - \frac {3 b d e^{2} \log {\relax (c )}}{2 x^{2}} - \frac {b e^{3} n \log {\relax (x )}}{x} - \frac {b e^{3} n}{x} - \frac {b e^{3} \log {\relax (c )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a+b*ln(c*x**n))/x**5,x)

[Out]

-a*d**3/(4*x**4) - a*d**2*e/x**3 - 3*a*d*e**2/(2*x**2) - a*e**3/x - b*d**3*n*log(x)/(4*x**4) - b*d**3*n/(16*x*
*4) - b*d**3*log(c)/(4*x**4) - b*d**2*e*n*log(x)/x**3 - b*d**2*e*n/(3*x**3) - b*d**2*e*log(c)/x**3 - 3*b*d*e**
2*n*log(x)/(2*x**2) - 3*b*d*e**2*n/(4*x**2) - 3*b*d*e**2*log(c)/(2*x**2) - b*e**3*n*log(x)/x - b*e**3*n/x - b*
e**3*log(c)/x

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