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]
-1/25*b*d^2*n/x^5-2*b*d*e*n*x^(-5+r)/(5-r)^2-b*e^2*n*x^(-5+2*r)/(5-2*r)^2-1/5*d^2*(a+b*ln(c*x^n))/x^5-2*d*e*x^
(-5+r)*(a+b*ln(c*x^n))/(5-r)-e^2*x^(-5+2*r)*(a+b*ln(c*x^n))/(5-2*r)
________________________________________________________________________________________
Rubi [A] time = 0.17, 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 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 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])
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 = 0.31, size = 127, normalized size = 1.00 \[ \frac {a \left (-5 d^2+\frac {50 d e x^r}{r-5}+\frac {25 e^2 x^{2 r}}{2 r-5}\right )+5 b \log \left (c x^n\right ) \left (-d^2+\frac {10 d e x^r}{r-5}+\frac {5 e^2 x^{2 r}}{2 r-5}\right )+b n \left (-d^2-\frac {50 d e x^r}{(r-5)^2}-\frac {25 e^2 x^{2 r}}{(5-2 r)^2}\right )}{25 x^5} \]
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)
________________________________________________________________________________________
fricas [B] time = 0.48, size = 466, normalized size = 3.67 \[ -\frac {4 \, {\left (b d^{2} n + 5 \, a d^{2}\right )} r^{4} + 625 \, b d^{2} n - 60 \, {\left (b d^{2} n + 5 \, a d^{2}\right )} r^{3} + 3125 \, a d^{2} + 325 \, {\left (b d^{2} n + 5 \, a d^{2}\right )} r^{2} - 750 \, {\left (b d^{2} n + 5 \, a d^{2}\right )} r - 25 \, {\left (2 \, a e^{2} r^{3} - 25 \, b e^{2} n - 125 \, a e^{2} - {\left (b e^{2} n + 25 \, a e^{2}\right )} r^{2} + 10 \, {\left (b e^{2} n + 10 \, a e^{2}\right )} r + {\left (2 \, b e^{2} r^{3} - 25 \, b e^{2} r^{2} + 100 \, b e^{2} r - 125 \, b e^{2}\right )} \log \relax (c) + {\left (2 \, b e^{2} n r^{3} - 25 \, b e^{2} n r^{2} + 100 \, b e^{2} n r - 125 \, b e^{2} n\right )} \log \relax (x)\right )} x^{2 \, r} - 50 \, {\left (4 \, a d e r^{3} - 25 \, b d e n - 125 \, a d e - 4 \, {\left (b d e n + 10 \, a d e\right )} r^{2} + 5 \, {\left (4 \, b d e n + 25 \, a d e\right )} r + {\left (4 \, b d e r^{3} - 40 \, b d e r^{2} + 125 \, b d e r - 125 \, b d e\right )} \log \relax (c) + {\left (4 \, b d e n r^{3} - 40 \, b d e n r^{2} + 125 \, b d e n r - 125 \, b d e n\right )} \log \relax (x)\right )} x^{r} + 5 \, {\left (4 \, b d^{2} r^{4} - 60 \, b d^{2} r^{3} + 325 \, b d^{2} r^{2} - 750 \, b d^{2} r + 625 \, b d^{2}\right )} \log \relax (c) + 5 \, {\left (4 \, b d^{2} n r^{4} - 60 \, b d^{2} n r^{3} + 325 \, b d^{2} n r^{2} - 750 \, b d^{2} n r + 625 \, b d^{2} n\right )} \log \relax (x)}{25 \, {\left (4 \, r^{4} - 60 \, r^{3} + 325 \, r^{2} - 750 \, r + 625\right )} x^{5}} \]
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)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{r} + d\right )}^{2} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{6}}\,{d x} \]
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)
________________________________________________________________________________________
maple [C] time = 0.32, size = 1930, normalized size = 15.20 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d+e*x^r)^2*(b*ln(c*x^n)+a)/x^6,x)
[Out]
-1/5*b*(-5*e^2*(x^r)^2*r+2*d^2*r^2-20*d*e*r*x^r+25*(x^r)^2*e^2-15*d^2*r+50*d*e*x^r+25*d^2)/x^5/(-5+2*r)/(r-5)*
ln(x^n)-1/50*(6250*I*Pi*b*d*e*r*x^r*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+6250*ln(c)*b*e^2*(x^r)^2-100*a*e^2*r^3
*(x^r)^2+12500*a*d*e*x^r+1250*a*e^2*r^2*(x^r)^2-5000*a*e^2*r*(x^r)^2+1250*b*e^2*n*(x^r)^2+40*b*d^2*r^4*ln(c)-6
00*b*d^2*r^3*ln(c)+3250*b*d^2*r^2*ln(c)-7500*b*d^2*r*ln(c)+6250*a*d^2+8*b*d^2*n*r^4-120*b*d^2*n*r^3+1250*b*d^2
*n+6250*a*e^2*(x^r)^2+6250*b*d^2*ln(c)+40*a*d^2*r^4+650*b*d^2*n*r^2-1500*b*d^2*n*r+3250*a*d^2*r^2-7500*a*d^2*r
-600*a*d^2*r^3-625*I*Pi*b*e^2*r^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*(x^r)^2-3125*I*Pi*b*d^2*csgn(I*x^n)*csgn
(I*c*x^n)*csgn(I*c)-200*I*Pi*b*d*e*r^3*x^r*csgn(I*c)*csgn(I*c*x^n)^2+50*I*Pi*b*e^2*r^3*csgn(I*x^n)*csgn(I*c*x^
n)*csgn(I*c)*(x^r)^2+3750*I*Pi*b*d^2*r*csgn(I*c*x^n)^3+50*b*e^2*n*r^2*(x^r)^2-400*a*d*e*r^3*x^r+625*I*Pi*b*e^2
*r^2*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2-20*I*Pi*b*d^2*r^4*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)^2-50*I*Pi*b*e^2*r^3*csgn(I*c*x^n)^2*csgn(I*c)*(x^r)^2+6250*I*Pi*b*d
*e*r*x^r*csgn(I*c*x^n)^3+4000*a*d*e*r^2*x^r-12500*a*d*e*r*x^r-500*b*e^2*n*r*(x^r)^2+2500*b*d*e*n*x^r+1250*ln(c
)*b*e^2*r^2*(x^r)^2-5000*ln(c)*b*e^2*r*(x^r)^2+12500*b*d*e*x^r*ln(c)-100*ln(c)*b*e^2*r^3*(x^r)^2-2000*I*Pi*b*d
*e*r^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^r-2000*b*d*e*n*r*x^r+400*b*d*e*n*r^2*x^r+4000*b*d*e*r^2*x^r*ln(c)
-12500*b*d*e*r*x^r*ln(c)-400*b*d*e*r^3*x^r*ln(c)+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+2000*I*Pi*b*d*e*r^2*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r-3125*I*Pi*b*e
^2*csgn(I*c*x^n)^3*(x^r)^2+200*I*Pi*b*d*e*r^3*x^r*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)+1625*I*Pi*b*d^2*r^2*csgn
(I*x^n)*csgn(I*c*x^n)^2-50*I*Pi*b*e^2*r^3*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2+200*I*Pi*b*d*e*r^3*x^r*csgn(I*c*
x^n)^3+2500*I*Pi*b*e^2*r*csgn(I*c*x^n)^3*(x^r)^2-3125*I*Pi*b*d^2*csgn(I*c*x^n)^3-6250*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+3750*I*Pi*b*d^2*r*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-6250*I
*Pi*b*d*e*r*x^r*csgn(I*x^n)*csgn(I*c*x^n)^2-2500*I*Pi*b*e^2*r*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2-6250*I*Pi*b*
d*e*r*x^r*csgn(I*c)*csgn(I*c*x^n)^2-300*I*Pi*b*d^2*r^3*csgn(I*x^n)*csgn(I*c*x^n)^2+50*I*Pi*b*e^2*r^3*csgn(I*c*
x^n)^3*(x^r)^2+300*I*Pi*b*d^2*r^3*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-2500*I*Pi*b*e^2*r*csgn(I*c*x^n)^2*csgn(I
*c)*(x^r)^2-200*I*Pi*b*d*e*r^3*x^r*csgn(I*x^n)*csgn(I*c*x^n)^2-1625*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+3125*I*Pi*b*d^2*csgn(I*c*x^n)^2*csgn(I*c)+3125*I*Pi*b*e^2*csgn(I*c*x^n)^2*csg
n(I*c)*(x^r)^2-300*I*Pi*b*d^2*r^3*csgn(I*c)*csgn(I*c*x^n)^2+6250*I*Pi*b*d*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r+25
00*I*Pi*b*e^2*r*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*(x^r)^2+300*I*Pi*b*d^2*r^3*csgn(I*c*x^n)^3+1625*I*Pi*b*d^2
*r^2*csgn(I*c*x^n)^2*csgn(I*c)+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*csgn(I*c*x^n)^2
*csgn(I*c)+3125*I*Pi*b*e^2*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^r)^2-20*I*Pi*b*d^2*r^4*csgn(I*c*x^n)^3-3750*I*Pi*b*d
^2*r*csgn(I*c)*csgn(I*c*x^n)^2-3750*I*Pi*b*d^2*r*csgn(I*x^n)*csgn(I*c*x^n)^2+625*I*Pi*b*e^2*r^2*csgn(I*c*x^n)^
2*csgn(I*c)*(x^r)^2-2000*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*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)/(-5+2*r)^2/x^5/(r-5)^2
________________________________________________________________________________________
maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(r-6>0)', see `assume?` for mor
e details)Is r-6 equal to -1?
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x^r\right )}^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^6} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((d + e*x^r)^2*(a + b*log(c*x^n)))/x^6,x)
[Out]
int(((d + e*x^r)^2*(a + b*log(c*x^n)))/x^6, x)
________________________________________________________________________________________
sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
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
________________________________________________________________________________________