3.1.27 \(\int \frac {(d+e x)^3 (a+b \log (c x^n))}{x^5} \, dx\) [27]

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

[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

________________________________________________________________________________________

Rubi [A]
time = 0.05, 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, 2372, 12, 45} \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(b*d^3*n)/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 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 2372

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]}, Dist[a + b*Log[c*x^n], u, 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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 109, normalized size = 1.21 \begin {gather*} -\frac {12 a \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )+b n \left (3 d^3+16 d^2 e x+36 d e^2 x^2+48 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 )}{48 x^4} \end {gather*}

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.13, size = 569, normalized size = 6.32

method	result	size
risch	\(-\frac {b \left (4 e^{3} x^{3}+6 d \,e^{2} x^{2}+4 d^{2} e x +d^{3}\right ) \ln \left (x^{n}\right )}{4 x^{4}}-\frac {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 \,d^{2} e x \,\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 c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+48 \ln \left (c \right ) b \,e^{3} x^{3}-24 i \pi b \,e^{3} x^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-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 )-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 )+12 a \,d^{3}+72 a d \,e^{2} x^{2}+48 a \,d^{2} e x +48 a \,e^{3} x^{3}+3 b \,d^{3} n +12 d^{3} b \ln \left (c \right )+72 \ln \left (c \right ) b d \,e^{2} x^{2}+48 \ln \left (c \right ) b \,d^{2} e x +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}-24 i \pi b \,d^{2} e x \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}+16 b \,d^{2} e n x +36 b d \,e^{2} n \,x^{2}-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 )+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}-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 )-6 i \pi b \,d^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+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}}{48 x^{4}}\)	\(569\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

________________________________________________________________________________________

Maxima [A]
time = 0.28, size = 140, normalized size = 1.56 \begin {gather*} -\frac {b n e^{3}}{x} - \frac {3 \, b d n e^{2}}{4 \, x^{2}} - \frac {b d^{2} n e}{3 \, x^{3}} - \frac {b e^{3} \log \left (c x^{n}\right )}{x} - \frac {3 \, b d e^{2} \log \left (c x^{n}\right )}{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 e^{3}}{x} - \frac {3 \, a d e^{2}}{2 \, x^{2}} - \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}} \end {gather*}

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

________________________________________________________________________________________

Fricas [A]
time = 0.44, size = 141, normalized size = 1.57 \begin {gather*} -\frac {3 \, b d^{3} n + 48 \, {\left (b n + a\right )} x^{3} e^{3} + 12 \, a d^{3} + 36 \, {\left (b d n + 2 \, a d\right )} x^{2} e^{2} + 16 \, {\left (b d^{2} n + 3 \, a d^{2}\right )} x e + 12 \, {\left (4 \, b x^{3} e^{3} + 6 \, b d x^{2} e^{2} + 4 \, b d^{2} x e + b d^{3}\right )} \log \left (c\right ) + 12 \, {\left (4 \, b n x^{3} e^{3} + 6 \, b d n x^{2} e^{2} + 4 \, b d^{2} n x e + b d^{3} n\right )} \log \left (x\right )}{48 \, x^{4}} \end {gather*}

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

________________________________________________________________________________________

Sympy [A]
time = 0.53, size = 158, normalized size = 1.76 \begin {gather*} - \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}{16 x^{4}} - \frac {b d^{3} \log {\left (c x^{n} \right )}}{4 x^{4}} - \frac {b d^{2} e n}{3 x^{3}} - \frac {b d^{2} e \log {\left (c x^{n} \right )}}{x^{3}} - \frac {3 b d e^{2} n}{4 x^{2}} - \frac {3 b d e^{2} \log {\left (c x^{n} \right )}}{2 x^{2}} - \frac {b e^{3} n}{x} - \frac {b e^{3} \log {\left (c x^{n} \right )}}{x} \end {gather*}

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

________________________________________________________________________________________

Giac [A]
time = 2.30, size = 158, normalized size = 1.76 \begin {gather*} -\frac {48 \, b n x^{3} e^{3} \log \left (x\right ) + 72 \, b d n x^{2} e^{2} \log \left (x\right ) + 48 \, b d^{2} n x e \log \left (x\right ) + 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 \left (c\right ) + 72 \, b d x^{2} e^{2} \log \left (c\right ) + 48 \, b d^{2} x e \log \left (c\right ) + 12 \, b d^{3} n \log \left (x\right ) + 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 \left (c\right ) + 12 \, a d^{3}}{48 \, x^{4}} \end {gather*}

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

________________________________________________________________________________________

Mupad [B]
time = 3.71, size = 118, normalized size = 1.31 \begin {gather*} -\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} \end {gather*}

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

________________________________________________________________________________________