\(\int \frac {(d+e x^2)^3 (a+b \log (c x^n))}{x^3} \, dx\) [200]

Optimal result

Integrand size = 23, antiderivative size = 131 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {b d^3 n}{4 x^2}-\frac {3}{4} b d e^2 n x^2-\frac {1}{16} b e^3 n x^4-\frac {3}{2} b d^2 e n \log ^2(x)-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-1/4*b*d^3*n/x^2-3/4*b*d*e^2*n*x^2-1/16*b*e^3*n*x^4-3/2*b*d^2*e*n*ln(x)^2-1/2*d^3*(a+b*ln(c*x^n))/x^2+3/2*d*e^
2*x^2*(a+b*ln(c*x^n))+1/4*e^3*x^4*(a+b*ln(c*x^n))+3*d^2*e*ln(x)*(a+b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {272, 45, 2372, 12, 14, 2338} \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )-\frac {b d^3 n}{4 x^2}-\frac {3}{2} b d^2 e n \log ^2(x)-\frac {3}{4} b d e^2 n x^2-\frac {1}{16} b e^3 n x^4 \]

[In]

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

[Out]

-1/4*(b*d^3*n)/x^2 - (3*b*d*e^2*n*x^2)/4 - (b*e^3*n*x^4)/16 - (3*b*d^2*e*n*Log[x]^2)/2 - (d^3*(a + b*Log[c*x^n
]))/(2*x^2) + (3*d*e^2*x^2*(a + b*Log[c*x^n]))/2 + (e^3*x^4*(a + b*Log[c*x^n]))/4 + 3*d^2*e*Log[x]*(a + b*Log[
c*x^n])

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 45

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 272

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 2338

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]

Rule 2372

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]}, Dist[a + b*Log[c*x^n], u, 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*} \text {integral}& = -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )-(b n) \int \frac {-2 d^3+6 d e^2 x^4+e^3 x^6+12 d^2 e x^2 \log (x)}{4 x^3} \, dx \\ & = -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \frac {-2 d^3+6 d e^2 x^4+e^3 x^6+12 d^2 e x^2 \log (x)}{x^3} \, dx \\ & = -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \left (\frac {-2 d^3+6 d e^2 x^4+e^3 x^6}{x^3}+\frac {12 d^2 e \log (x)}{x}\right ) \, dx \\ & = -\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \frac {-2 d^3+6 d e^2 x^4+e^3 x^6}{x^3} \, dx-\left (3 b d^2 e n\right ) \int \frac {\log (x)}{x} \, dx \\ & = -\frac {3}{2} b d^2 e n \log ^2(x)-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} (b n) \int \left (-\frac {2 d^3}{x^3}+6 d e^2 x+e^3 x^3\right ) \, dx \\ & = -\frac {b d^3 n}{4 x^2}-\frac {3}{4} b d e^2 n x^2-\frac {1}{16} b e^3 n x^4-\frac {3}{2} b d^2 e n \log ^2(x)-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} e^3 x^4 \left (a+b \log \left (c x^n\right )\right )+3 d^2 e \log (x) \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.88 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {1}{16} \left (-\frac {4 b d^3 n}{x^2}-12 b d e^2 n x^2-b e^3 n x^4-\frac {8 d^3 \left (a+b \log \left (c x^n\right )\right )}{x^2}+24 d e^2 x^2 \left (a+b \log \left (c x^n\right )\right )+4 e^3 x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {24 d^2 e \left (a+b \log \left (c x^n\right )\right )^2}{b n}\right ) \]

[In]

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

[Out]

((-4*b*d^3*n)/x^2 - 12*b*d*e^2*n*x^2 - b*e^3*n*x^4 - (8*d^3*(a + b*Log[c*x^n]))/x^2 + 24*d*e^2*x^2*(a + b*Log[
c*x^n]) + 4*e^3*x^4*(a + b*Log[c*x^n]) + (24*d^2*e*(a + b*Log[c*x^n])^2)/(b*n))/16

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.14

method	result	size
parallelrisch	\(\frac {4 x^{6} \ln \left (c \,x^{n}\right ) b \,e^{3} n -x^{6} b \,e^{3} n^{2}+4 x^{6} a \,e^{3} n +24 x^{4} \ln \left (c \,x^{n}\right ) b d \,e^{2} n -12 x^{4} b d \,e^{2} n^{2}+24 x^{4} a d \,e^{2} n +48 \ln \left (x \right ) x^{2} a \,d^{2} e n +24 e \,d^{2} b \ln \left (c \,x^{n}\right )^{2} x^{2}-8 \ln \left (c \,x^{n}\right ) b \,d^{3} n -4 b \,d^{3} n^{2}-8 a \,d^{3} n}{16 x^{2} n}\)	\(149\)
risch	\(\text {Expression too large to display}\)	\(4039\)

[In]

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

[Out]

1/16/x^2*(4*x^6*ln(c*x^n)*b*e^3*n-x^6*b*e^3*n^2+4*x^6*a*e^3*n+24*x^4*ln(c*x^n)*b*d*e^2*n-12*x^4*b*d*e^2*n^2+24
*x^4*a*d*e^2*n+48*ln(x)*x^2*a*d^2*e*n+24*e*d^2*b*ln(c*x^n)^2*x^2-8*ln(c*x^n)*b*d^3*n-4*b*d^3*n^2-8*a*d^3*n)/n

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.18 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {24 \, b d^{2} e n x^{2} \log \left (x\right )^{2} - {\left (b e^{3} n - 4 \, a e^{3}\right )} x^{6} - 4 \, b d^{3} n - 12 \, {\left (b d e^{2} n - 2 \, a d e^{2}\right )} x^{4} - 8 \, a d^{3} + 4 \, {\left (b e^{3} x^{6} + 6 \, b d e^{2} x^{4} - 2 \, b d^{3}\right )} \log \left (c\right ) + 4 \, {\left (b e^{3} n x^{6} + 6 \, b d e^{2} n x^{4} + 12 \, b d^{2} e x^{2} \log \left (c\right ) + 12 \, a d^{2} e x^{2} - 2 \, b d^{3} n\right )} \log \left (x\right )}{16 \, x^{2}} \]

[In]

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

[Out]

1/16*(24*b*d^2*e*n*x^2*log(x)^2 - (b*e^3*n - 4*a*e^3)*x^6 - 4*b*d^3*n - 12*(b*d*e^2*n - 2*a*d*e^2)*x^4 - 8*a*d
^3 + 4*(b*e^3*x^6 + 6*b*d*e^2*x^4 - 2*b*d^3)*log(c) + 4*(b*e^3*n*x^6 + 6*b*d*e^2*n*x^4 + 12*b*d^2*e*x^2*log(c)
+ 12*a*d^2*e*x^2 - 2*b*d^3*n)*log(x))/x^2

Sympy [A] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.60 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\begin {cases} - \frac {a d^{3}}{2 x^{2}} + \frac {3 a d^{2} e \log {\left (c x^{n} \right )}}{n} + \frac {3 a d e^{2} x^{2}}{2} + \frac {a e^{3} x^{4}}{4} - \frac {b d^{3} n}{4 x^{2}} - \frac {b d^{3} \log {\left (c x^{n} \right )}}{2 x^{2}} + \frac {3 b d^{2} e \log {\left (c x^{n} \right )}^{2}}{2 n} - \frac {3 b d e^{2} n x^{2}}{4} + \frac {3 b d e^{2} x^{2} \log {\left (c x^{n} \right )}}{2} - \frac {b e^{3} n x^{4}}{16} + \frac {b e^{3} x^{4} \log {\left (c x^{n} \right )}}{4} & \text {for}\: n \neq 0 \\\left (a + b \log {\left (c \right )}\right ) \left (- \frac {d^{3}}{2 x^{2}} + 3 d^{2} e \log {\left (x \right )} + \frac {3 d e^{2} x^{2}}{2} + \frac {e^{3} x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a*d**3/(2*x**2) + 3*a*d**2*e*log(c*x**n)/n + 3*a*d*e**2*x**2/2 + a*e**3*x**4/4 - b*d**3*n/(4*x**2)
- b*d**3*log(c*x**n)/(2*x**2) + 3*b*d**2*e*log(c*x**n)**2/(2*n) - 3*b*d*e**2*n*x**2/4 + 3*b*d*e**2*x**2*log(c
*x**n)/2 - b*e**3*n*x**4/16 + b*e**3*x**4*log(c*x**n)/4, Ne(n, 0)), ((a + b*log(c))*(-d**3/(2*x**2) + 3*d**2*e
*log(x) + 3*d*e**2*x**2/2 + e**3*x**4/4), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.02 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {1}{16} \, b e^{3} n x^{4} + \frac {1}{4} \, b e^{3} x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a e^{3} x^{4} - \frac {3}{4} \, b d e^{2} n x^{2} + \frac {3}{2} \, b d e^{2} x^{2} \log \left (c x^{n}\right ) + \frac {3}{2} \, a d e^{2} x^{2} + \frac {3 \, b d^{2} e \log \left (c x^{n}\right )^{2}}{2 \, n} + 3 \, a d^{2} e \log \left (x\right ) - \frac {b d^{3} n}{4 \, x^{2}} - \frac {b d^{3} \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {a d^{3}}{2 \, x^{2}} \]

[In]

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

[Out]

-1/16*b*e^3*n*x^4 + 1/4*b*e^3*x^4*log(c*x^n) + 1/4*a*e^3*x^4 - 3/4*b*d*e^2*n*x^2 + 3/2*b*d*e^2*x^2*log(c*x^n)
+ 3/2*a*d*e^2*x^2 + 3/2*b*d^2*e*log(c*x^n)^2/n + 3*a*d^2*e*log(x) - 1/4*b*d^3*n/x^2 - 1/2*b*d^3*log(c*x^n)/x^2
- 1/2*a*d^3/x^2

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {1}{4} \, b e^{3} x^{4} \log \left (c\right ) + \frac {1}{4} \, a e^{3} x^{4} + \frac {3}{2} \, b d e^{2} x^{2} \log \left (c\right ) + \frac {3}{2} \, b d^{2} e n \log \left (x\right )^{2} + \frac {3}{4} \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} b d e^{2} n + \frac {1}{16} \, {\left (4 \, x^{4} \log \left (x\right ) - x^{4}\right )} b e^{3} n + \frac {3}{2} \, a d e^{2} x^{2} - \frac {1}{4} \, b d^{3} n {\left (\frac {2 \, \log \left (x\right )}{x^{2}} + \frac {1}{x^{2}}\right )} + 3 \, b d^{2} e \log \left (c\right ) \log \left ({\left | x \right |}\right ) + 3 \, a d^{2} e \log \left ({\left | x \right |}\right ) - \frac {b d^{3} \log \left (c\right )}{2 \, x^{2}} - \frac {a d^{3}}{2 \, x^{2}} \]

[In]

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

[Out]

1/4*b*e^3*x^4*log(c) + 1/4*a*e^3*x^4 + 3/2*b*d*e^2*x^2*log(c) + 3/2*b*d^2*e*n*log(x)^2 + 3/4*(2*x^2*log(x) - x
^2)*b*d*e^2*n + 1/16*(4*x^4*log(x) - x^4)*b*e^3*n + 3/2*a*d*e^2*x^2 - 1/4*b*d^3*n*(2*log(x)/x^2 + 1/x^2) + 3*b
*d^2*e*log(c)*log(abs(x)) + 3*a*d^2*e*log(abs(x)) - 1/2*b*d^3*log(c)/x^2 - 1/2*a*d^3/x^2

Mupad [B] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.24 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\ln \left (c\,x^n\right )\,\left (\frac {\frac {3\,b\,e^3\,x^6}{4}+3\,b\,d\,e^2\,x^4}{x^2}-\frac {\frac {b\,d^3}{2}+\frac {3\,b\,d^2\,e\,x^2}{2}+\frac {3\,b\,d\,e^2\,x^4}{2}+\frac {b\,e^3\,x^6}{2}}{x^2}\right )-\frac {\frac {a\,d^3}{2}+\frac {b\,d^3\,n}{4}}{x^2}+\ln \left (x\right )\,\left (3\,a\,d^2\,e+\frac {3\,b\,d^2\,e\,n}{2}\right )+\frac {e^3\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {3\,d\,e^2\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {3\,b\,d^2\,e\,{\ln \left (c\,x^n\right )}^2}{2\,n} \]

[In]

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

[Out]

log(c*x^n)*(((3*b*e^3*x^6)/4 + 3*b*d*e^2*x^4)/x^2 - ((b*d^3)/2 + (b*e^3*x^6)/2 + (3*b*d^2*e*x^2)/2 + (3*b*d*e^
2*x^4)/2)/x^2) - ((a*d^3)/2 + (b*d^3*n)/4)/x^2 + log(x)*(3*a*d^2*e + (3*b*d^2*e*n)/2) + (e^3*x^4*(4*a - b*n))/
16 + (3*d*e^2*x^2*(2*a - b*n))/4 + (3*b*d^2*e*log(c*x^n)^2)/(2*n)