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

Optimal. Leaf size=80 \[ d^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+2 d e x \left (a+b \log \left (c x^n\right )\right )+\frac{1}{2} e^2 x^2 \left (a+b \log \left (c x^n\right )\right )-\frac{1}{2} b d^2 n \log ^2(x)-\frac{1}{4} b n (4 d+e x)^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0714581, antiderivative size = 63, normalized size of antiderivative = 0.79, 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} \[ \frac{1}{2} \left (2 d^2 \log (x)+4 d e x+e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{2} b d^2 n \log ^2(x)-\frac{1}{4} b n (4 d+e x)^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0457333, size = 83, normalized size = 1.04 \[ \frac{d^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+\frac{1}{2} e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+2 a d e x+2 b d e x \log \left (c x^n\right )-2 b d e n x-\frac{1}{4} b e^2 n x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.239, size = 410, normalized size = 5.1 \begin{align*} \left ({\frac{b{e}^{2}{x}^{2}}{2}}+2\,bdex+b{d}^{2}\ln \left ( x \right ) \right ) \ln \left ({x}^{n} \right ) -{\frac{b{d}^{2}n \left ( \ln \left ( x \right ) \right ) ^{2}}{2}}-{\frac{i}{4}}\pi \,b{e}^{2}{x}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}-{\frac{i}{2}}\ln \left ( x \right ) \pi \,b{d}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}-i\pi \,bdex \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}-{\frac{i}{2}}\ln \left ( x \right ) \pi \,b{d}^{2}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{2}}\ln \left ( x \right ) \pi \,b{d}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +i\pi \,bdex{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-i\pi \,bdex{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) +i\pi \,bdex \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{\ln \left ( c \right ) b{e}^{2}{x}^{2}}{2}}-{\frac{b{e}^{2}n{x}^{2}}{4}}+2\,\ln \left ( c \right ) bdex+{\frac{a{e}^{2}{x}^{2}}{2}}-2\,bdenx+2\,adex+{\frac{i}{2}}\ln \left ( x \right ) \pi \,b{d}^{2}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-{\frac{i}{4}}\pi \,b{e}^{2}{x}^{2}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) +{\frac{i}{4}}\pi \,b{e}^{2}{x}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{i}{4}}\pi \,b{e}^{2}{x}^{2}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}+\ln \left ( x \right ) \ln \left ( c \right ) b{d}^{2}+\ln \left ( x \right ) a{d}^{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,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.07129, size = 113, normalized size = 1.41 \begin{align*} -\frac{1}{4} \, b e^{2} n x^{2} + \frac{1}{2} \, b e^{2} x^{2} \log \left (c x^{n}\right ) - 2 \, b d e n x + \frac{1}{2} \, a e^{2} x^{2} + 2 \, b d e x \log \left (c x^{n}\right ) + 2 \, a d e x + \frac{b d^{2} \log \left (c x^{n}\right )^{2}}{2 \, n} + a d^{2} \log \left (x\right ) \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,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.994326, size = 244, normalized size = 3.05 \begin{align*} \frac{1}{2} \, b d^{2} n \log \left (x\right )^{2} - \frac{1}{4} \,{\left (b e^{2} n - 2 \, a e^{2}\right )} x^{2} - 2 \,{\left (b d e n - a d e\right )} x + \frac{1}{2} \,{\left (b e^{2} x^{2} + 4 \, b d e x\right )} \log \left (c\right ) + \frac{1}{2} \,{\left (b e^{2} n x^{2} + 4 \, b d e n x + 2 \, b d^{2} \log \left (c\right ) + 2 \, a d^{2}\right )} \log \left (x\right ) \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,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.38067, size = 128, normalized size = 1.6 \begin{align*} a d^{2} \log{\left (x \right )} + 2 a d e x + \frac{a e^{2} x^{2}}{2} + \frac{b d^{2} n \log{\left (x \right )}^{2}}{2} + b d^{2} \log{\left (c \right )} \log{\left (x \right )} + 2 b d e n x \log{\left (x \right )} - 2 b d e n x + 2 b d e x \log{\left (c \right )} + \frac{b e^{2} n x^{2} \log{\left (x \right )}}{2} - \frac{b e^{2} n x^{2}}{4} + \frac{b e^{2} x^{2} \log{\left (c \right )}}{2} \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,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.32801, size = 135, normalized size = 1.69 \begin{align*} \frac{1}{2} \, b n x^{2} e^{2} \log \left (x\right ) + 2 \, b d n x e \log \left (x\right ) + \frac{1}{2} \, b d^{2} n \log \left (x\right )^{2} - \frac{1}{4} \, b n x^{2} e^{2} - 2 \, b d n x e + \frac{1}{2} \, b x^{2} e^{2} \log \left (c\right ) + 2 \, b d x e \log \left (c\right ) + b d^{2} \log \left (c\right ) \log \left (x\right ) + \frac{1}{2} \, a x^{2} e^{2} + 2 \, a d x e + a d^{2} \log \left (x\right ) \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,x, algorithm="giac")

[Out]

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