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

Optimal. Leaf size=104 \[ d^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{2 d e x^r \left (a+b \log \left (c x^n\right )\right )}{r}+\frac{e^2 x^{2 r} \left (a+b \log \left (c x^n\right )\right )}{2 r}-\frac{1}{2} b d^2 n \log ^2(x)-\frac{2 b d e n x^r}{r^2}-\frac{b e^2 n x^{2 r}}{4 r^2} \]

[Out]

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

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Rubi [A]  time = 0.133859, antiderivative size = 87, normalized size of antiderivative = 0.84, number of steps used = 5, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {266, 43, 2334, 12, 14, 2301} \[ \frac{1}{2} \left (2 d^2 \log (x)+\frac{4 d e x^r}{r}+\frac{e^2 x^{2 r}}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{2} b d^2 n \log ^2(x)-\frac{2 b d e n x^r}{r^2}-\frac{b e^2 n x^{2 r}}{4 r^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.209372, size = 90, normalized size = 0.87 \[ \frac{1}{4} \left (\frac{e x^r \left (2 a r \left (4 d+e x^r\right )-b n \left (8 d+e x^r\right )\right )}{r^2}+4 a d^2 \log (x)+\frac{2 b d^2 \log ^2\left (c x^n\right )}{n}+\frac{2 b e x^r \log \left (c x^n\right ) \left (4 d+e x^r\right )}{r}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e*x^r*(2*a*r*(4*d + e*x^r) - b*n*(8*d + e*x^r)))/r^2 + 4*a*d^2*Log[x] + (2*b*e*x^r*(4*d + e*x^r)*Log[c*x^n])
/r + (2*b*d^2*Log[c*x^n]^2)/n)/4

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Maple [C]  time = 0.139, size = 487, 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((d+e*x^r)^2*(a+b*ln(c*x^n))/x,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.37679, size = 286, normalized size = 2.75 \begin{align*} \frac{2 \, b d^{2} n r^{2} \log \left (x\right )^{2} +{\left (2 \, b e^{2} n r \log \left (x\right ) + 2 \, b e^{2} r \log \left (c\right ) - b e^{2} n + 2 \, a e^{2} r\right )} x^{2 \, r} + 8 \,{\left (b d e n r \log \left (x\right ) + b d e r \log \left (c\right ) - b d e n + a d e r\right )} x^{r} + 4 \,{\left (b d^{2} r^{2} \log \left (c\right ) + a d^{2} r^{2}\right )} \log \left (x\right )}{4 \, r^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+e*x**r)**2*(a+b*ln(c*x**n))/x,x)

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.31555, size = 189, normalized size = 1.82 \begin{align*} \frac{1}{2} \, b d^{2} n \log \left (x\right )^{2} + \frac{2 \, b d n x^{r} e \log \left (x\right )}{r} + b d^{2} \log \left (c\right ) \log \left (x\right ) + \frac{2 \, b d x^{r} e \log \left (c\right )}{r} + a d^{2} \log \left (x\right ) + \frac{b n x^{2 \, r} e^{2} \log \left (x\right )}{2 \, r} - \frac{2 \, b d n x^{r} e}{r^{2}} + \frac{2 \, a d x^{r} e}{r} + \frac{b x^{2 \, r} e^{2} \log \left (c\right )}{2 \, r} - \frac{b n x^{2 \, r} e^{2}}{4 \, r^{2}} + \frac{a x^{2 \, r} e^{2}}{2 \, r} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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