3.26 \(\int \frac{(d+e x)^3 (a+b \log (c x^n))}{x^4} \, dx\)

Optimal. Leaf size=126 \[ -\frac{3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac{d^3 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac{3 d e^2 \left (a+b \log \left (c x^n\right )\right )}{x}+e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{3 b d^2 e n}{4 x^2}-\frac{b d^3 n}{9 x^3}-\frac{3 b d e^2 n}{x}-\frac{1}{2} b e^3 n \log ^2(x) \]

[Out]

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

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Rubi [A]  time = 0.107818, antiderivative size = 98, normalized size of antiderivative = 0.78, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {43, 2334, 14, 2301} \[ -\frac{1}{6} \left (\frac{9 d^2 e}{x^2}+\frac{2 d^3}{x^3}+\frac{18 d e^2}{x}-6 e^3 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{3 b d^2 e n}{4 x^2}-\frac{b d^3 n}{9 x^3}-\frac{3 b d e^2 n}{x}-\frac{1}{2} b e^3 n \log ^2(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0794775, size = 122, normalized size = 0.97 \[ -\frac{3 d^2 e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac{d^3 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac{3 d e^2 \left (a+b \log \left (c x^n\right )\right )}{x}+\frac{e^3 \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}-\frac{3 b d^2 e n}{4 x^2}-\frac{b d^3 n}{9 x^3}-\frac{3 b d e^2 n}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.166, size = 589, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(a+b*ln(c*x^n))/x^4,x)

[Out]

-1/6*b*(-6*e^3*ln(x)*x^3+18*d*e^2*x^2+9*d^2*e*x+2*d^3)/x^3*ln(x^n)-1/36*(12*a*d^3+54*I*Pi*b*d*e^2*x^2*csgn(I*c
*x^n)^2*csgn(I*c)-18*I*ln(x)*Pi*b*e^3*csgn(I*x^n)*csgn(I*c*x^n)^2*x^3-18*I*ln(x)*Pi*b*e^3*csgn(I*c*x^n)^2*csgn
(I*c)*x^3+27*I*Pi*b*d^2*e*x*csgn(I*x^n)*csgn(I*c*x^n)^2+27*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^2*csgn(I*c)-36*ln(x)*a
*e^3*x^3+108*a*d*e^2*x^2+54*a*d^2*e*x+12*ln(c)*b*d^3+54*I*Pi*b*d*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-54*I*Pi*b
*d*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+108*ln(c)*b*d*e^2*x^2+54*ln(c)*b*d^2*e*x+4*b*d^3*n-36*ln(x)*ln(
c)*b*e^3*x^3+18*b*e^3*n*ln(x)^2*x^3-27*I*Pi*b*d^2*e*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+18*I*ln(x)*Pi*b*e^3*
csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^3+27*b*d^2*e*n*x+108*b*d*e^2*n*x^2-6*I*Pi*b*d^3*csgn(I*c*x^n)^3-6*I*Pi*b
*d^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-54*I*Pi*b*d*e^2*x^2*csgn(I*c*x^n)^3-27*I*Pi*b*d^2*e*x*csgn(I*c*x^n)^3
+18*I*ln(x)*Pi*b*e^3*csgn(I*c*x^n)^3*x^3+6*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2+6*I*Pi*b*d^3*csgn(I*c*x^n)^2
*csgn(I*c))/x^3

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Maxima [A]  time = 1.08524, size = 180, normalized size = 1.43 \begin{align*} \frac{b e^{3} \log \left (c x^{n}\right )^{2}}{2 \, n} + a e^{3} \log \left (x\right ) - \frac{3 \, b d e^{2} n}{x} - \frac{3 \, b d e^{2} \log \left (c x^{n}\right )}{x} - \frac{3 \, b d^{2} e n}{4 \, x^{2}} - \frac{3 \, a d e^{2}}{x} - \frac{3 \, b d^{2} e \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac{b d^{3} n}{9 \, x^{3}} - \frac{3 \, a d^{2} e}{2 \, x^{2}} - \frac{b d^{3} \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac{a d^{3}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.04344, size = 360, normalized size = 2.86 \begin{align*} \frac{18 \, b e^{3} n x^{3} \log \left (x\right )^{2} - 4 \, b d^{3} n - 12 \, a d^{3} - 108 \,{\left (b d e^{2} n + a d e^{2}\right )} x^{2} - 27 \,{\left (b d^{2} e n + 2 \, a d^{2} e\right )} x - 6 \,{\left (18 \, b d e^{2} x^{2} + 9 \, b d^{2} e x + 2 \, b d^{3}\right )} \log \left (c\right ) + 6 \,{\left (6 \, b e^{3} x^{3} \log \left (c\right ) - 18 \, b d e^{2} n x^{2} + 6 \, a e^{3} x^{3} - 9 \, b d^{2} e n x - 2 \, b d^{3} n\right )} \log \left (x\right )}{36 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^4,x, algorithm="fricas")

[Out]

1/36*(18*b*e^3*n*x^3*log(x)^2 - 4*b*d^3*n - 12*a*d^3 - 108*(b*d*e^2*n + a*d*e^2)*x^2 - 27*(b*d^2*e*n + 2*a*d^2
*e)*x - 6*(18*b*d*e^2*x^2 + 9*b*d^2*e*x + 2*b*d^3)*log(c) + 6*(6*b*e^3*x^3*log(c) - 18*b*d*e^2*n*x^2 + 6*a*e^3
*x^3 - 9*b*d^2*e*n*x - 2*b*d^3*n)*log(x))/x^3

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Sympy [A]  time = 9.13817, size = 144, normalized size = 1.14 \begin{align*} - \frac{a d^{3}}{3 x^{3}} - \frac{3 a d^{2} e}{2 x^{2}} - \frac{3 a d e^{2}}{x} + a e^{3} \log{\left (x \right )} + b d^{3} \left (- \frac{n}{9 x^{3}} - \frac{\log{\left (c x^{n} \right )}}{3 x^{3}}\right ) + 3 b d^{2} e \left (- \frac{n}{4 x^{2}} - \frac{\log{\left (c x^{n} \right )}}{2 x^{2}}\right ) + 3 b d e^{2} \left (- \frac{n}{x} - \frac{\log{\left (c x^{n} \right )}}{x}\right ) - b e^{3} \left (\begin{cases} - \log{\left (c \right )} \log{\left (x \right )} & \text{for}\: n = 0 \\- \frac{\log{\left (c x^{n} \right )}^{2}}{2 n} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(a+b*ln(c*x**n))/x**4,x)

[Out]

-a*d**3/(3*x**3) - 3*a*d**2*e/(2*x**2) - 3*a*d*e**2/x + a*e**3*log(x) + b*d**3*(-n/(9*x**3) - log(c*x**n)/(3*x
**3)) + 3*b*d**2*e*(-n/(4*x**2) - log(c*x**n)/(2*x**2)) + 3*b*d*e**2*(-n/x - log(c*x**n)/x) - b*e**3*Piecewise
((-log(c)*log(x), Eq(n, 0)), (-log(c*x**n)**2/(2*n), True))

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Giac [A]  time = 1.36309, size = 209, normalized size = 1.66 \begin{align*} \frac{18 \, b n x^{3} e^{3} \log \left (x\right )^{2} - 108 \, b d n x^{2} e^{2} \log \left (x\right ) - 54 \, b d^{2} n x e \log \left (x\right ) + 36 \, b x^{3} e^{3} \log \left (c\right ) \log \left (x\right ) - 108 \, b d n x^{2} e^{2} - 27 \, b d^{2} n x e - 108 \, b d x^{2} e^{2} \log \left (c\right ) - 54 \, b d^{2} x e \log \left (c\right ) - 12 \, b d^{3} n \log \left (x\right ) + 36 \, a x^{3} e^{3} \log \left (x\right ) - 4 \, b d^{3} n - 108 \, a d x^{2} e^{2} - 54 \, a d^{2} x e - 12 \, b d^{3} \log \left (c\right ) - 12 \, a d^{3}}{36 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(a+b*log(c*x^n))/x^4,x, algorithm="giac")

[Out]

1/36*(18*b*n*x^3*e^3*log(x)^2 - 108*b*d*n*x^2*e^2*log(x) - 54*b*d^2*n*x*e*log(x) + 36*b*x^3*e^3*log(c)*log(x)
- 108*b*d*n*x^2*e^2 - 27*b*d^2*n*x*e - 108*b*d*x^2*e^2*log(c) - 54*b*d^2*x*e*log(c) - 12*b*d^3*n*log(x) + 36*a
*x^3*e^3*log(x) - 4*b*d^3*n - 108*a*d*x^2*e^2 - 54*a*d^2*x*e - 12*b*d^3*log(c) - 12*a*d^3)/x^3