\(\int \frac {(d+e x^2)^3 (a+b \log (c x^n))}{x^2} \, dx\) [205]

Optimal result

Integrand size = 23, antiderivative size = 118 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {b d^3 n}{x}-3 b d^2 e n x-\frac {1}{3} b d e^2 n x^3-\frac {1}{25} b e^3 n x^5-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d^2 e x \left (a+b \log \left (c x^n\right )\right )+d e^2 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^3 x^5 \left (a+b \log \left (c x^n\right )\right ) \]

[Out]

-b*d^3*n/x-3*b*d^2*e*n*x-1/3*b*d*e^2*n*x^3-1/25*b*e^3*n*x^5-d^3*(a+b*ln(c*x^n))/x+3*d^2*e*x*(a+b*ln(c*x^n))+d*
e^2*x^3*(a+b*ln(c*x^n))+1/5*e^3*x^5*(a+b*ln(c*x^n))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {276, 2372} \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d^2 e x \left (a+b \log \left (c x^n\right )\right )+d e^2 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^3 x^5 \left (a+b \log \left (c x^n\right )\right )-\frac {b d^3 n}{x}-3 b d^2 e n x-\frac {1}{3} b d e^2 n x^3-\frac {1}{25} b e^3 n x^5 \]

[In]

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

[Out]

-((b*d^3*n)/x) - 3*b*d^2*e*n*x - (b*d*e^2*n*x^3)/3 - (b*e^3*n*x^5)/25 - (d^3*(a + b*Log[c*x^n]))/x + 3*d^2*e*x
*(a + b*Log[c*x^n]) + d*e^2*x^3*(a + b*Log[c*x^n]) + (e^3*x^5*(a + b*Log[c*x^n]))/5

Rule 276

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 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 )}{x}+3 d^2 e x \left (a+b \log \left (c x^n\right )\right )+d e^2 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^3 x^5 \left (a+b \log \left (c x^n\right )\right )-(b n) \int \left (3 d^2 e-\frac {d^3}{x^2}+d e^2 x^2+\frac {e^3 x^4}{5}\right ) \, dx \\ & = -\frac {b d^3 n}{x}-3 b d^2 e n x-\frac {1}{3} b d e^2 n x^3-\frac {1}{25} b e^3 n x^5-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+3 d^2 e x \left (a+b \log \left (c x^n\right )\right )+d e^2 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^3 x^5 \left (a+b \log \left (c x^n\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.04 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {b d^3 n}{x}+3 a d^2 e x-3 b d^2 e n x-\frac {1}{3} b d e^2 n x^3-\frac {1}{25} b e^3 n x^5+3 b d^2 e x \log \left (c x^n\right )-\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{x}+d e^2 x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {1}{5} e^3 x^5 \left (a+b \log \left (c x^n\right )\right ) \]

[In]

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

[Out]

-((b*d^3*n)/x) + 3*a*d^2*e*x - 3*b*d^2*e*n*x - (b*d*e^2*n*x^3)/3 - (b*e^3*n*x^5)/25 + 3*b*d^2*e*x*Log[c*x^n] -
(d^3*(a + b*Log[c*x^n]))/x + d*e^2*x^3*(a + b*Log[c*x^n]) + (e^3*x^5*(a + b*Log[c*x^n]))/5

Maple [A] (verified)

Time = 0.68 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.19

method	result	size
parallelrisch	\(-\frac {-15 x^{6} b \ln \left (c \,x^{n}\right ) e^{3}+3 b \,e^{3} n \,x^{6}-15 x^{6} a \,e^{3}-75 x^{4} b \ln \left (c \,x^{n}\right ) d \,e^{2}+25 b d \,e^{2} n \,x^{4}-75 x^{4} a d \,e^{2}-225 b \ln \left (c \,x^{n}\right ) d^{2} e \,x^{2}+225 b \,d^{2} e n \,x^{2}-225 a \,d^{2} e \,x^{2}+75 b \ln \left (c \,x^{n}\right ) d^{3}+75 b \,d^{3} n +75 a \,d^{3}}{75 x}\)	\(140\)
risch	\(-\frac {b \left (-e^{3} x^{6}-5 e^{2} d \,x^{4}-15 d^{2} e \,x^{2}+5 d^{3}\right ) \ln \left (x^{n}\right )}{5 x}-\frac {-30 x^{6} a \,e^{3}-150 \ln \left (c \right ) b d \,e^{2} x^{4}+75 i \pi b d \,e^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+225 i e \pi b \,d^{2} x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-75 i \pi b \,d^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-150 x^{4} a d \,e^{2}-450 a \,d^{2} e \,x^{2}+150 a \,d^{3}+150 d^{3} b \ln \left (c \right )-30 \ln \left (c \right ) b \,e^{3} x^{6}-450 e \ln \left (c \right ) b \,d^{2} x^{2}+150 b \,d^{3} n -15 i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+75 i \pi b d \,e^{2} x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-75 i \pi b \,d^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+15 i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+225 i e \pi b \,d^{2} x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-15 i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-225 i e \pi b \,d^{2} x^{2} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-225 i e \pi b \,d^{2} x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+15 i \pi b \,e^{3} x^{6} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )-75 i \pi b d \,e^{2} x^{4} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-75 i \pi b d \,e^{2} x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+50 b d \,e^{2} n \,x^{4}+450 b \,d^{2} e n \,x^{2}+6 b \,e^{3} n \,x^{6}+75 i \pi b \,d^{3} \operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+75 i \pi b \,d^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{150 x}\)	\(587\)

[In]

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

[Out]

-1/75/x*(-15*x^6*b*ln(c*x^n)*e^3+3*b*e^3*n*x^6-15*x^6*a*e^3-75*x^4*b*ln(c*x^n)*d*e^2+25*b*d*e^2*n*x^4-75*x^4*a
*d*e^2-225*b*ln(c*x^n)*d^2*e*x^2+225*b*d^2*e*n*x^2-225*a*d^2*e*x^2+75*b*ln(c*x^n)*d^3+75*b*d^3*n+75*a*d^3)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.35 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {3 \, {\left (b e^{3} n - 5 \, a e^{3}\right )} x^{6} + 75 \, b d^{3} n + 25 \, {\left (b d e^{2} n - 3 \, a d e^{2}\right )} x^{4} + 75 \, a d^{3} + 225 \, {\left (b d^{2} e n - a d^{2} e\right )} x^{2} - 15 \, {\left (b e^{3} x^{6} + 5 \, b d e^{2} x^{4} + 15 \, b d^{2} e x^{2} - 5 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b e^{3} n x^{6} + 5 \, b d e^{2} n x^{4} + 15 \, b d^{2} e n x^{2} - 5 \, b d^{3} n\right )} \log \left (x\right )}{75 \, x} \]

[In]

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

[Out]

-1/75*(3*(b*e^3*n - 5*a*e^3)*x^6 + 75*b*d^3*n + 25*(b*d*e^2*n - 3*a*d*e^2)*x^4 + 75*a*d^3 + 225*(b*d^2*e*n - a
*d^2*e)*x^2 - 15*(b*e^3*x^6 + 5*b*d*e^2*x^4 + 15*b*d^2*e*x^2 - 5*b*d^3)*log(c) - 15*(b*e^3*n*x^6 + 5*b*d*e^2*n
*x^4 + 15*b*d^2*e*n*x^2 - 5*b*d^3*n)*log(x))/x

Sympy [A] (verification not implemented)

Time = 0.63 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.24 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=- \frac {a d^{3}}{x} + 3 a d^{2} e x + a d e^{2} x^{3} + \frac {a e^{3} x^{5}}{5} - \frac {b d^{3} n}{x} - \frac {b d^{3} \log {\left (c x^{n} \right )}}{x} - 3 b d^{2} e n x + 3 b d^{2} e x \log {\left (c x^{n} \right )} - \frac {b d e^{2} n x^{3}}{3} + b d e^{2} x^{3} \log {\left (c x^{n} \right )} - \frac {b e^{3} n x^{5}}{25} + \frac {b e^{3} x^{5} \log {\left (c x^{n} \right )}}{5} \]

[In]

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

[Out]

-a*d**3/x + 3*a*d**2*e*x + a*d*e**2*x**3 + a*e**3*x**5/5 - b*d**3*n/x - b*d**3*log(c*x**n)/x - 3*b*d**2*e*n*x
+ 3*b*d**2*e*x*log(c*x**n) - b*d*e**2*n*x**3/3 + b*d*e**2*x**3*log(c*x**n) - b*e**3*n*x**5/25 + b*e**3*x**5*lo
g(c*x**n)/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.14 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {1}{25} \, b e^{3} n x^{5} + \frac {1}{5} \, b e^{3} x^{5} \log \left (c x^{n}\right ) + \frac {1}{5} \, a e^{3} x^{5} - \frac {1}{3} \, b d e^{2} n x^{3} + b d e^{2} x^{3} \log \left (c x^{n}\right ) + a d e^{2} x^{3} - 3 \, b d^{2} e n x + 3 \, b d^{2} e x \log \left (c x^{n}\right ) + 3 \, a d^{2} e x - \frac {b d^{3} n}{x} - \frac {b d^{3} \log \left (c x^{n}\right )}{x} - \frac {a d^{3}}{x} \]

[In]

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

[Out]

-1/25*b*e^3*n*x^5 + 1/5*b*e^3*x^5*log(c*x^n) + 1/5*a*e^3*x^5 - 1/3*b*d*e^2*n*x^3 + b*d*e^2*x^3*log(c*x^n) + a*
d*e^2*x^3 - 3*b*d^2*e*n*x + 3*b*d^2*e*x*log(c*x^n) + 3*a*d^2*e*x - b*d^3*n/x - b*d^3*log(c*x^n)/x - a*d^3/x

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.28 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {1}{25} \, {\left (b e^{3} n - 5 \, b e^{3} \log \left (c\right ) - 5 \, a e^{3}\right )} x^{5} - \frac {1}{3} \, {\left (b d e^{2} n - 3 \, b d e^{2} \log \left (c\right ) - 3 \, a d e^{2}\right )} x^{3} - 3 \, {\left (b d^{2} e n - b d^{2} e \log \left (c\right ) - a d^{2} e\right )} x + \frac {1}{5} \, {\left (b e^{3} n x^{5} + 5 \, b d e^{2} n x^{3} + 15 \, b d^{2} e n x - \frac {5 \, b d^{3} n}{x}\right )} \log \left (x\right ) - \frac {b d^{3} n + b d^{3} \log \left (c\right ) + a d^{3}}{x} \]

[In]

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

[Out]

-1/25*(b*e^3*n - 5*b*e^3*log(c) - 5*a*e^3)*x^5 - 1/3*(b*d*e^2*n - 3*b*d*e^2*log(c) - 3*a*d*e^2)*x^3 - 3*(b*d^2
*e*n - b*d^2*e*log(c) - a*d^2*e)*x + 1/5*(b*e^3*n*x^5 + 5*b*d*e^2*n*x^3 + 15*b*d^2*e*n*x - 5*b*d^3*n/x)*log(x)
- (b*d^3*n + b*d^3*log(c) + a*d^3)/x

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.23 \[ \int \frac {\left (d+e x^2\right )^3 \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\ln \left (c\,x^n\right )\,\left (\frac {6\,b\,d^2\,e\,x^2+4\,b\,d\,e^2\,x^4+\frac {6\,b\,e^3\,x^6}{5}}{x}-\frac {b\,d^3+3\,b\,d^2\,e\,x^2+3\,b\,d\,e^2\,x^4+b\,e^3\,x^6}{x}\right )-\frac {a\,d^3+b\,d^3\,n}{x}+\frac {e^3\,x^5\,\left (5\,a-b\,n\right )}{25}+\frac {d\,e^2\,x^3\,\left (3\,a-b\,n\right )}{3}+3\,d^2\,e\,x\,\left (a-b\,n\right ) \]

[In]

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

[Out]

log(c*x^n)*(((6*b*e^3*x^6)/5 + 6*b*d^2*e*x^2 + 4*b*d*e^2*x^4)/x - (b*d^3 + b*e^3*x^6 + 3*b*d^2*e*x^2 + 3*b*d*e
^2*x^4)/x) - (a*d^3 + b*d^3*n)/x + (e^3*x^5*(5*a - b*n))/25 + (d*e^2*x^3*(3*a - b*n))/3 + 3*d^2*e*x*(a - b*n)