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

Optimal. Leaf size=84 \[ -\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac{2 d e \left (a+b \log \left (c x^n\right )\right )}{x}+e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{b n (d+4 e x)^2}{4 x^2}-\frac{1}{2} b e^2 n \log ^2(x) \]

[Out]

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

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Rubi [A]  time = 0.0788719, antiderivative size = 67, normalized size of antiderivative = 0.8, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {43, 2334, 37, 2301} \[ -\frac{1}{2} \left (\frac{d^2}{x^2}+\frac{4 d e}{x}-2 e^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{b n (d+4 e x)^2}{4 x^2}-\frac{1}{2} b e^2 n \log ^2(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0547216, size = 84, normalized size = 1. \[ -\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac{2 d e \left (a+b \log \left (c x^n\right )\right )}{x}+\frac{e^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}-\frac{b d^2 n}{4 x^2}-\frac{2 b d e n}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.149, size = 418, normalized size = 5. \begin{align*} -{\frac{b \left ( -2\,{e}^{2}\ln \left ( x \right ){x}^{2}+4\,dex+{d}^{2} \right ) \ln \left ({x}^{n} \right ) }{2\,{x}^{2}}}-{\frac{-2\,i\ln \left ( x \right ) \pi \,b{e}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ){x}^{2}-i\pi \,b{d}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}-4\,i\pi \,bdex \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+i\pi \,b{d}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +4\,i\pi \,bdex{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}+4\,i\pi \,bdex \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -2\,i\ln \left ( x \right ) \pi \,b{e}^{2}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{x}^{2}+2\,i\ln \left ( x \right ) \pi \,b{e}^{2}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ){x}^{2}+2\,i\ln \left ( x \right ) \pi \,b{e}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}{x}^{2}+i\pi \,b{d}^{2}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-i\pi \,b{d}^{2}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -4\,i\pi \,bdex{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) +2\,b{e}^{2}n \left ( \ln \left ( x \right ) \right ) ^{2}{x}^{2}-4\,\ln \left ( x \right ) \ln \left ( c \right ) b{e}^{2}{x}^{2}-4\,\ln \left ( x \right ) a{e}^{2}{x}^{2}+8\,\ln \left ( c \right ) bdex+8\,bdenx+2\,\ln \left ( c \right ) b{d}^{2}+8\,adex+b{d}^{2}n+2\,a{d}^{2}}{4\,{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*b*(-2*e^2*ln(x)*x^2+4*d*e*x+d^2)/x^2*ln(x^n)-1/4*(-2*I*ln(x)*Pi*b*e^2*csgn(I*c*x^n)^2*csgn(I*c)*x^2-I*Pi*
b*d^2*csgn(I*c*x^n)^3-4*I*Pi*b*d*e*x*csgn(I*c*x^n)^3+I*Pi*b*d^2*csgn(I*c*x^n)^2*csgn(I*c)+4*I*Pi*b*d*e*x*csgn(
I*x^n)*csgn(I*c*x^n)^2+4*I*Pi*b*d*e*x*csgn(I*c*x^n)^2*csgn(I*c)-2*I*ln(x)*Pi*b*e^2*csgn(I*x^n)*csgn(I*c*x^n)^2
*x^2+2*I*ln(x)*Pi*b*e^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^2+2*I*ln(x)*Pi*b*e^2*csgn(I*c*x^n)^3*x^2+I*Pi*b*
d^2*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-4*I*Pi*b*d*e*x*csgn(I*x^n)*csgn
(I*c*x^n)*csgn(I*c)+2*b*e^2*n*ln(x)^2*x^2-4*ln(x)*ln(c)*b*e^2*x^2-4*ln(x)*a*e^2*x^2+8*ln(c)*b*d*e*x+8*b*d*e*n*
x+2*ln(c)*b*d^2+8*a*d*e*x+b*d^2*n+2*a*d^2)/x^2

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Maxima [A]  time = 1.1419, size = 122, normalized size = 1.45 \begin{align*} \frac{b e^{2} \log \left (c x^{n}\right )^{2}}{2 \, n} + a e^{2} \log \left (x\right ) - \frac{2 \, b d e n}{x} - \frac{2 \, b d e \log \left (c x^{n}\right )}{x} - \frac{b d^{2} n}{4 \, x^{2}} - \frac{2 \, a d e}{x} - \frac{b d^{2} \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac{a d^{2}}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.06644, size = 242, normalized size = 2.88 \begin{align*} \frac{2 \, b e^{2} n x^{2} \log \left (x\right )^{2} - b d^{2} n - 2 \, a d^{2} - 8 \,{\left (b d e n + a d e\right )} x - 2 \,{\left (4 \, b d e x + b d^{2}\right )} \log \left (c\right ) + 2 \,{\left (2 \, b e^{2} x^{2} \log \left (c\right ) - 4 \, b d e n x + 2 \, a e^{2} x^{2} - b d^{2} n\right )} \log \left (x\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b*e^2*n*x^2*log(x)^2 - b*d^2*n - 2*a*d^2 - 8*(b*d*e*n + a*d*e)*x - 2*(4*b*d*e*x + b*d^2)*log(c) + 2*(2*
b*e^2*x^2*log(c) - 4*b*d*e*n*x + 2*a*e^2*x^2 - b*d^2*n)*log(x))/x^2

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Sympy [A]  time = 20.7117, size = 99, normalized size = 1.18 \begin{align*} - \frac{a d^{2}}{2 x^{2}} - \frac{2 a d e}{x} + a e^{2} \log{\left (x \right )} + b d^{2} \left (- \frac{n}{4 x^{2}} - \frac{\log{\left (c x^{n} \right )}}{2 x^{2}}\right ) + 2 b d e \left (- \frac{n}{x} - \frac{\log{\left (c x^{n} \right )}}{x}\right ) - b e^{2} \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)**2*(a+b*ln(c*x**n))/x**3,x)

[Out]

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

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Giac [A]  time = 1.34803, size = 142, normalized size = 1.69 \begin{align*} \frac{2 \, b n x^{2} e^{2} \log \left (x\right )^{2} - 8 \, b d n x e \log \left (x\right ) + 4 \, b x^{2} e^{2} \log \left (c\right ) \log \left (x\right ) - 8 \, b d n x e - 8 \, b d x e \log \left (c\right ) - 2 \, b d^{2} n \log \left (x\right ) + 4 \, a x^{2} e^{2} \log \left (x\right ) - b d^{2} n - 8 \, a d x e - 2 \, b d^{2} \log \left (c\right ) - 2 \, a d^{2}}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*b*n*x^2*e^2*log(x)^2 - 8*b*d*n*x*e*log(x) + 4*b*x^2*e^2*log(c)*log(x) - 8*b*d*n*x*e - 8*b*d*x*e*log(c)
- 2*b*d^2*n*log(x) + 4*a*x^2*e^2*log(x) - b*d^2*n - 8*a*d*x*e - 2*b*d^2*log(c) - 2*a*d^2)/x^2