3.390 \(\int \frac{(d+e x^r)^2 (a+b \log (c x^n))}{x^6} \, dx\)

Optimal. Leaf size=127 \[ -\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{5 x^5}-\frac{2 d e x^{r-5} \left (a+b \log \left (c x^n\right )\right )}{5-r}-\frac{e^2 x^{2 r-5} \left (a+b \log \left (c x^n\right )\right )}{5-2 r}-\frac{b d^2 n}{25 x^5}-\frac{2 b d e n x^{r-5}}{(5-r)^2}-\frac{b e^2 n x^{2 r-5}}{(5-2 r)^2} \]

[Out]

-(b*d^2*n)/(25*x^5) - (2*b*d*e*n*x^(-5 + r))/(5 - r)^2 - (b*e^2*n*x^(-5 + 2*r))/(5 - 2*r)^2 - (d^2*(a + b*Log[
c*x^n]))/(5*x^5) - (2*d*e*x^(-5 + r)*(a + b*Log[c*x^n]))/(5 - r) - (e^2*x^(-5 + 2*r)*(a + b*Log[c*x^n]))/(5 -
2*r)

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Rubi [A]  time = 0.173024, antiderivative size = 109, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {270, 2334, 12, 14} \[ -\frac{1}{5} \left (\frac{d^2}{x^5}+\frac{10 d e x^{r-5}}{5-r}+\frac{5 e^2 x^{2 r-5}}{5-2 r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{b d^2 n}{25 x^5}-\frac{2 b d e n x^{r-5}}{(5-r)^2}-\frac{b e^2 n x^{2 r-5}}{(5-2 r)^2} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^r)^2*(a + b*Log[c*x^n]))/x^6,x]

[Out]

-(b*d^2*n)/(25*x^5) - (2*b*d*e*n*x^(-5 + r))/(5 - r)^2 - (b*e^2*n*x^(-5 + 2*r))/(5 - 2*r)^2 - ((d^2/x^5 + (10*
d*e*x^(-5 + r))/(5 - r) + (5*e^2*x^(-5 + 2*r))/(5 - 2*r))*(a + b*Log[c*x^n]))/5

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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]]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx &=-\frac{1}{5} \left (\frac{d^2}{x^5}+\frac{10 d e x^{-5+r}}{5-r}+\frac{5 e^2 x^{-5+2 r}}{5-2 r}\right ) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac{-d^2+\frac{10 d e x^r}{-5+r}+\frac{5 e^2 x^{2 r}}{-5+2 r}}{5 x^6} \, dx\\ &=-\frac{1}{5} \left (\frac{d^2}{x^5}+\frac{10 d e x^{-5+r}}{5-r}+\frac{5 e^2 x^{-5+2 r}}{5-2 r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{5} (b n) \int \frac{-d^2+\frac{10 d e x^r}{-5+r}+\frac{5 e^2 x^{2 r}}{-5+2 r}}{x^6} \, dx\\ &=-\frac{1}{5} \left (\frac{d^2}{x^5}+\frac{10 d e x^{-5+r}}{5-r}+\frac{5 e^2 x^{-5+2 r}}{5-2 r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{5} (b n) \int \left (-\frac{d^2}{x^6}+\frac{10 d e x^{-6+r}}{-5+r}+\frac{5 e^2 x^{2 (-3+r)}}{-5+2 r}\right ) \, dx\\ &=-\frac{b d^2 n}{25 x^5}-\frac{2 b d e n x^{-5+r}}{(5-r)^2}-\frac{b e^2 n x^{-5+2 r}}{(5-2 r)^2}-\frac{1}{5} \left (\frac{d^2}{x^5}+\frac{10 d e x^{-5+r}}{5-r}+\frac{5 e^2 x^{-5+2 r}}{5-2 r}\right ) \left (a+b \log \left (c x^n\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.29933, size = 127, normalized size = 1. \[ \frac{a \left (-5 d^2+\frac{50 d e x^r}{r-5}+\frac{25 e^2 x^{2 r}}{2 r-5}\right )+5 b \log \left (c x^n\right ) \left (-d^2+\frac{10 d e x^r}{r-5}+\frac{5 e^2 x^{2 r}}{2 r-5}\right )+b n \left (-d^2-\frac{50 d e x^r}{(r-5)^2}-\frac{25 e^2 x^{2 r}}{(5-2 r)^2}\right )}{25 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^r)^2*(a + b*Log[c*x^n]))/x^6,x]

[Out]

(b*n*(-d^2 - (50*d*e*x^r)/(-5 + r)^2 - (25*e^2*x^(2*r))/(5 - 2*r)^2) + a*(-5*d^2 + (50*d*e*x^r)/(-5 + r) + (25
*e^2*x^(2*r))/(-5 + 2*r)) + 5*b*(-d^2 + (10*d*e*x^r)/(-5 + r) + (5*e^2*x^(2*r))/(-5 + 2*r))*Log[c*x^n])/(25*x^
5)

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Maple [C]  time = 0.237, size = 1930, normalized size = 15.2 \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^6,x)

[Out]

-1/5*b*(-5*e^2*(x^r)^2*r+2*d^2*r^2-20*d*e*x^r*r+25*e^2*(x^r)^2-15*d^2*r+50*d*e*x^r+25*d^2)/x^5/(-5+2*r)/(-5+r)
*ln(x^n)-1/50*(8*b*d^2*n*r^4-120*b*d^2*n*r^3+6250*ln(c)*b*d^2+200*I*Pi*b*d*e*r^3*csgn(I*x^n)*csgn(I*c*x^n)*csg
n(I*c)*x^r+650*b*d^2*n*r^2-1500*b*d^2*n*r-100*a*e^2*r^3*(x^r)^2+1250*a*e^2*r^2*(x^r)^2+6250*a*d^2+12500*a*d*e*
x^r+6250*I*Pi*b*d*e*r*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^r-3750*I*Pi*b*d^2*r*csgn(I*c*x^n)^2*csgn(I*c)-2500
*I*Pi*b*e^2*r*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2-2500*I*Pi*b*e^2*r*csgn(I*c*x^n)^2*csgn(I*c)*(x^r)^2+6250*I*P
i*b*d*e*r*csgn(I*c*x^n)^3*x^r+300*I*Pi*b*d^2*r^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-5000*a*e^2*r*(x^r)^2+3250
*a*d^2*r^2-7500*a*d^2*r+40*a*d^2*r^4-600*a*d^2*r^3+3750*I*Pi*b*d^2*r*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-50*I*
Pi*b*e^2*r^3*csgn(I*c*x^n)^2*csgn(I*c)*(x^r)^2+1250*b*d^2*n+2500*b*d*e*n*x^r-100*ln(c)*b*e^2*r^3*(x^r)^2+12500
*ln(c)*b*d*e*x^r+1250*ln(c)*b*e^2*r^2*(x^r)^2-5000*ln(c)*b*e^2*r*(x^r)^2-50*I*Pi*b*e^2*r^3*csgn(I*x^n)*csgn(I*
c*x^n)^2*(x^r)^2+6250*ln(c)*b*e^2*(x^r)^2+1250*b*e^2*n*(x^r)^2+3250*ln(c)*b*d^2*r^2-7500*ln(c)*b*d^2*r+6250*a*
e^2*(x^r)^2+40*ln(c)*b*d^2*r^4-600*ln(c)*b*d^2*r^3-1625*I*Pi*b*d^2*r^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+625
0*I*Pi*b*d*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r+2500*I*Pi*b*e^2*r*csgn(I*c*x^n)^3*(x^r)^2+200*I*Pi*b*d*e*r^3*csgn
(I*c*x^n)^3*x^r+300*I*Pi*b*d^2*r^3*csgn(I*c*x^n)^3+50*b*e^2*n*r^2*(x^r)^2-400*a*d*e*r^3*x^r+4000*a*d*e*r^2*x^r
-12500*a*d*e*r*x^r-500*b*e^2*n*r*(x^r)^2+2000*I*Pi*b*d*e*r^2*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r-300*I*Pi*b*d^2*r^
3*csgn(I*x^n)*csgn(I*c*x^n)^2-300*I*Pi*b*d^2*r^3*csgn(I*c*x^n)^2*csgn(I*c)-3125*I*Pi*b*d^2*csgn(I*c*x^n)^3-312
5*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-3125*I*Pi*b*e^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*(x^r)^2-2
0*I*Pi*b*d^2*r^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+6250*I*Pi*b*d*e*csgn(I*c*x^n)^2*csgn(I*c)*x^r-2000*I*Pi*b
*d*e*r^2*csgn(I*c*x^n)^3*x^r-1625*I*Pi*b*d^2*r^2*csgn(I*c*x^n)^3+3125*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2+6
25*I*Pi*b*e^2*r^2*csgn(I*c*x^n)^2*csgn(I*c)*(x^r)^2+625*I*Pi*b*e^2*r^2*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2+162
5*I*Pi*b*d^2*r^2*csgn(I*x^n)*csgn(I*c*x^n)^2+20*I*Pi*b*d^2*r^4*csgn(I*x^n)*csgn(I*c*x^n)^2+20*I*Pi*b*d^2*r^4*c
sgn(I*c*x^n)^2*csgn(I*c)-3750*I*Pi*b*d^2*r*csgn(I*x^n)*csgn(I*c*x^n)^2+3125*I*Pi*b*d^2*csgn(I*c*x^n)^2*csgn(I*
c)-20*I*Pi*b*d^2*r^4*csgn(I*c*x^n)^3-3125*I*Pi*b*e^2*csgn(I*c*x^n)^3*(x^r)^2-6250*I*Pi*b*d*e*r*csgn(I*x^n)*csg
n(I*c*x^n)^2*x^r-6250*I*Pi*b*d*e*r*csgn(I*c*x^n)^2*csgn(I*c)*x^r-200*I*Pi*b*d*e*r^3*csgn(I*c*x^n)^2*csgn(I*c)*
x^r+50*I*Pi*b*e^2*r^3*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*(x^r)^2+2000*I*Pi*b*d*e*r^2*csgn(I*c*x^n)^2*csgn(I*c
)*x^r-6250*I*Pi*b*d*e*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^r-625*I*Pi*b*e^2*r^2*csgn(I*x^n)*csgn(I*c*x^n)*csg
n(I*c)*(x^r)^2+3750*I*Pi*b*d^2*r*csgn(I*c*x^n)^3+1625*I*Pi*b*d^2*r^2*csgn(I*c*x^n)^2*csgn(I*c)+4000*ln(c)*b*d*
e*r^2*x^r-12500*ln(c)*b*d*e*r*x^r-400*ln(c)*b*d*e*r^3*x^r+3125*I*Pi*b*e^2*csgn(I*c*x^n)^2*csgn(I*c)*(x^r)^2-62
50*I*Pi*b*d*e*csgn(I*c*x^n)^3*x^r-625*I*Pi*b*e^2*r^2*csgn(I*c*x^n)^3*(x^r)^2+3125*I*Pi*b*e^2*csgn(I*x^n)*csgn(
I*c*x^n)^2*(x^r)^2+2500*I*Pi*b*e^2*r*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*(x^r)^2-2000*b*d*e*n*r*x^r+400*b*d*e*
n*r^2*x^r+50*I*Pi*b*e^2*r^3*csgn(I*c*x^n)^3*(x^r)^2-200*I*Pi*b*d*e*r^3*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r-2000*I*
Pi*b*d*e*r^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^r)/(-5+2*r)^2/x^5/(-5+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^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.43625, size = 1154, normalized size = 9.09 \begin{align*} -\frac{4 \,{\left (b d^{2} n + 5 \, a d^{2}\right )} r^{4} + 625 \, b d^{2} n - 60 \,{\left (b d^{2} n + 5 \, a d^{2}\right )} r^{3} + 3125 \, a d^{2} + 325 \,{\left (b d^{2} n + 5 \, a d^{2}\right )} r^{2} - 750 \,{\left (b d^{2} n + 5 \, a d^{2}\right )} r - 25 \,{\left (2 \, a e^{2} r^{3} - 25 \, b e^{2} n - 125 \, a e^{2} -{\left (b e^{2} n + 25 \, a e^{2}\right )} r^{2} + 10 \,{\left (b e^{2} n + 10 \, a e^{2}\right )} r +{\left (2 \, b e^{2} r^{3} - 25 \, b e^{2} r^{2} + 100 \, b e^{2} r - 125 \, b e^{2}\right )} \log \left (c\right ) +{\left (2 \, b e^{2} n r^{3} - 25 \, b e^{2} n r^{2} + 100 \, b e^{2} n r - 125 \, b e^{2} n\right )} \log \left (x\right )\right )} x^{2 \, r} - 50 \,{\left (4 \, a d e r^{3} - 25 \, b d e n - 125 \, a d e - 4 \,{\left (b d e n + 10 \, a d e\right )} r^{2} + 5 \,{\left (4 \, b d e n + 25 \, a d e\right )} r +{\left (4 \, b d e r^{3} - 40 \, b d e r^{2} + 125 \, b d e r - 125 \, b d e\right )} \log \left (c\right ) +{\left (4 \, b d e n r^{3} - 40 \, b d e n r^{2} + 125 \, b d e n r - 125 \, b d e n\right )} \log \left (x\right )\right )} x^{r} + 5 \,{\left (4 \, b d^{2} r^{4} - 60 \, b d^{2} r^{3} + 325 \, b d^{2} r^{2} - 750 \, b d^{2} r + 625 \, b d^{2}\right )} \log \left (c\right ) + 5 \,{\left (4 \, b d^{2} n r^{4} - 60 \, b d^{2} n r^{3} + 325 \, b d^{2} n r^{2} - 750 \, b d^{2} n r + 625 \, b d^{2} n\right )} \log \left (x\right )}{25 \,{\left (4 \, r^{4} - 60 \, r^{3} + 325 \, r^{2} - 750 \, r + 625\right )} x^{5}} \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^6,x, algorithm="fricas")

[Out]

-1/25*(4*(b*d^2*n + 5*a*d^2)*r^4 + 625*b*d^2*n - 60*(b*d^2*n + 5*a*d^2)*r^3 + 3125*a*d^2 + 325*(b*d^2*n + 5*a*
d^2)*r^2 - 750*(b*d^2*n + 5*a*d^2)*r - 25*(2*a*e^2*r^3 - 25*b*e^2*n - 125*a*e^2 - (b*e^2*n + 25*a*e^2)*r^2 + 1
0*(b*e^2*n + 10*a*e^2)*r + (2*b*e^2*r^3 - 25*b*e^2*r^2 + 100*b*e^2*r - 125*b*e^2)*log(c) + (2*b*e^2*n*r^3 - 25
*b*e^2*n*r^2 + 100*b*e^2*n*r - 125*b*e^2*n)*log(x))*x^(2*r) - 50*(4*a*d*e*r^3 - 25*b*d*e*n - 125*a*d*e - 4*(b*
d*e*n + 10*a*d*e)*r^2 + 5*(4*b*d*e*n + 25*a*d*e)*r + (4*b*d*e*r^3 - 40*b*d*e*r^2 + 125*b*d*e*r - 125*b*d*e)*lo
g(c) + (4*b*d*e*n*r^3 - 40*b*d*e*n*r^2 + 125*b*d*e*n*r - 125*b*d*e*n)*log(x))*x^r + 5*(4*b*d^2*r^4 - 60*b*d^2*
r^3 + 325*b*d^2*r^2 - 750*b*d^2*r + 625*b*d^2)*log(c) + 5*(4*b*d^2*n*r^4 - 60*b*d^2*n*r^3 + 325*b*d^2*n*r^2 -
750*b*d^2*n*r + 625*b*d^2*n)*log(x))/((4*r^4 - 60*r^3 + 325*r^2 - 750*r + 625)*x^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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**6,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{r} + d\right )}^{2}{\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{6}}\,{d x} \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^6,x, algorithm="giac")

[Out]

integrate((e*x^r + d)^2*(b*log(c*x^n) + a)/x^6, x)