\(\int \frac {(d+e x^2) (a+b \log (c x^n))}{x^4} \, dx\) [181]

Optimal result

Integrand size = 21, antiderivative size = 53 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b d n}{9 x^3}-\frac {b e n}{x}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{x} \]

[Out]

-1/9*b*d*n/x^3-b*e*n/x-1/3*d*(a+b*ln(c*x^n))/x^3-e*(a+b*ln(c*x^n))/x

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {14, 2372, 12} \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {b d n}{9 x^3}-\frac {b e n}{x} \]

[In]

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

[Out]

-1/9*(b*d*n)/x^3 - (b*e*n)/x - (d*(a + b*Log[c*x^n]))/(3*x^3) - (e*(a + b*Log[c*x^n]))/x

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 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 \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{x}-(b n) \int \frac {-d-3 e x^2}{3 x^4} \, dx \\ & = -\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {1}{3} (b n) \int \frac {-d-3 e x^2}{x^4} \, dx \\ & = -\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {1}{3} (b n) \int \left (-\frac {d}{x^4}-\frac {3 e}{x^2}\right ) \, dx \\ & = -\frac {b d n}{9 x^3}-\frac {b e n}{x}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.19 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {a d}{3 x^3}-\frac {b d n}{9 x^3}-\frac {a e}{x}-\frac {b e n}{x}-\frac {b d \log \left (c x^n\right )}{3 x^3}-\frac {b e \log \left (c x^n\right )}{x} \]

[In]

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

[Out]

-1/3*(a*d)/x^3 - (b*d*n)/(9*x^3) - (a*e)/x - (b*e*n)/x - (b*d*Log[c*x^n])/(3*x^3) - (b*e*Log[c*x^n])/x

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(-\frac {9 b e \,x^{2} \ln \left (c \,x^{n}\right )+9 b e n \,x^{2}+9 a e \,x^{2}+3 b \ln \left (c \,x^{n}\right ) d +b d n +3 a d}{9 x^{3}}\)	\(53\)
risch	\(-\frac {b \left (3 e \,x^{2}+d \right ) \ln \left (x^{n}\right )}{3 x^{3}}-\frac {-9 i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) b e \,x^{2}+9 i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} b e \,x^{2}+9 i \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} b e \,x^{2}-9 i \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3} b e \,x^{2}+18 \ln \left (c \right ) b e \,x^{2}+18 b e n \,x^{2}+18 a e \,x^{2}-3 i \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )+3 i \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+3 i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-3 i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+6 d b \ln \left (c \right )+2 b d n +6 a d}{18 x^{3}}\)	\(249\)

[In]

int((e*x^2+d)*(a+b*ln(c*x^n))/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/9/x^3*(9*b*e*x^2*ln(c*x^n)+9*b*e*n*x^2+9*a*e*x^2+3*b*ln(c*x^n)*d+b*d*n+3*a*d)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b d n + 9 \, {\left (b e n + a e\right )} x^{2} + 3 \, a d + 3 \, {\left (3 \, b e x^{2} + b d\right )} \log \left (c\right ) + 3 \, {\left (3 \, b e n x^{2} + b d n\right )} \log \left (x\right )}{9 \, x^{3}} \]

[In]

integrate((e*x^2+d)*(a+b*log(c*x^n))/x^4,x, algorithm="fricas")

[Out]

-1/9*(b*d*n + 9*(b*e*n + a*e)*x^2 + 3*a*d + 3*(3*b*e*x^2 + b*d)*log(c) + 3*(3*b*e*n*x^2 + b*d*n)*log(x))/x^3

Sympy [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=- \frac {a d}{3 x^{3}} - \frac {a e}{x} - \frac {b d n}{9 x^{3}} - \frac {b d \log {\left (c x^{n} \right )}}{3 x^{3}} - \frac {b e n}{x} - \frac {b e \log {\left (c x^{n} \right )}}{x} \]

[In]

integrate((e*x**2+d)*(a+b*ln(c*x**n))/x**4,x)

[Out]

-a*d/(3*x**3) - a*e/x - b*d*n/(9*x**3) - b*d*log(c*x**n)/(3*x**3) - b*e*n/x - b*e*log(c*x**n)/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b e n}{x} - \frac {b e \log \left (c x^{n}\right )}{x} - \frac {a e}{x} - \frac {b d n}{9 \, x^{3}} - \frac {b d \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {a d}{3 \, x^{3}} \]

[In]

integrate((e*x^2+d)*(a+b*log(c*x^n))/x^4,x, algorithm="maxima")

[Out]

-b*e*n/x - b*e*log(c*x^n)/x - a*e/x - 1/9*b*d*n/x^3 - 1/3*b*d*log(c*x^n)/x^3 - 1/3*a*d/x^3

Giac [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.23 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {{\left (3 \, b e n x^{2} + b d n\right )} \log \left (x\right )}{3 \, x^{3}} - \frac {9 \, b e n x^{2} + 9 \, b e x^{2} \log \left (c\right ) + 9 \, a e x^{2} + b d n + 3 \, b d \log \left (c\right ) + 3 \, a d}{9 \, x^{3}} \]

[In]

integrate((e*x^2+d)*(a+b*log(c*x^n))/x^4,x, algorithm="giac")

[Out]

-1/3*(3*b*e*n*x^2 + b*d*n)*log(x)/x^3 - 1/9*(9*b*e*n*x^2 + 9*b*e*x^2*log(c) + 9*a*e*x^2 + b*d*n + 3*b*d*log(c)
+ 3*a*d)/x^3

Mupad [B] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {\left (3\,a\,e+3\,b\,e\,n\right )\,x^2+a\,d+\frac {b\,d\,n}{3}}{3\,x^3}-\frac {\ln \left (c\,x^n\right )\,\left (b\,e\,x^2+\frac {b\,d}{3}\right )}{x^3} \]

[In]

int(((d + e*x^2)*(a + b*log(c*x^n)))/x^4,x)

[Out]

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