\(\int \frac {(d+e x^2) (a+b \log (c x^n))}{x} \, dx\) [174]

Optimal result

Integrand size = 21, antiderivative size = 52 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {1}{4} b e n x^2+\frac {1}{2} e x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {14, 2393, 2338, 2341} \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+\frac {1}{2} e x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b e n x^2 \]

[In]

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

[Out]

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

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 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 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2393

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.10 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {1}{2} a e x^2-\frac {1}{4} b e n x^2+a d \log (x)+\frac {1}{2} b e x^2 \log \left (c x^n\right )+\frac {b d \log ^2\left (c x^n\right )}{2 n} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12

method	result	size
parallelrisch	\(\frac {2 x^{2} \ln \left (c \,x^{n}\right ) b e n -x^{2} b e \,n^{2}+2 x^{2} a e n +4 \ln \left (x \right ) a d n +2 b d \ln \left (c \,x^{n}\right )^{2}}{4 n}\)	\(58\)
risch	\(\left (\frac {b e \,x^{2}}{2}+b d \ln \left (x \right )\right ) \ln \left (x^{n}\right )-\frac {b d n \ln \left (x \right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) b e \,x^{2}}{4}+\frac {i \pi \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} b e \,x^{2}}{4}+\frac {i \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} b e \,x^{2}}{4}-\frac {i \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3} b e \,x^{2}}{4}+\frac {\ln \left (c \right ) b e \,x^{2}}{2}-\frac {b e n \,x^{2}}{4}+\frac {a e \,x^{2}}{2}-\frac {i \ln \left (x \right ) \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )}{2}+\frac {i \ln \left (x \right ) \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}+\frac {i \ln \left (x \right ) \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i \ln \left (x \right ) \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\ln \left (x \right ) \ln \left (c \right ) b d +\ln \left (x \right ) a d\)	\(257\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {1}{2} \, b e x^{2} \log \left (c\right ) + \frac {1}{2} \, b d n \log \left (x\right )^{2} - \frac {1}{4} \, {\left (b e n - 2 \, a e\right )} x^{2} + \frac {1}{2} \, {\left (b e n x^{2} + 2 \, b d \log \left (c\right ) + 2 \, a d\right )} \log \left (x\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.50 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\begin {cases} \frac {a d \log {\left (c x^{n} \right )}}{n} + \frac {a e x^{2}}{2} + \frac {b d \log {\left (c x^{n} \right )}^{2}}{2 n} - \frac {b e n x^{2}}{4} + \frac {b e x^{2} \log {\left (c x^{n} \right )}}{2} & \text {for}\: n \neq 0 \\\left (a + b \log {\left (c \right )}\right ) \left (d \log {\left (x \right )} + \frac {e x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {1}{4} \, b e n x^{2} + \frac {1}{2} \, b e x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a e x^{2} + \frac {b d \log \left (c x^{n}\right )^{2}}{2 \, n} + a d \log \left (x\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {1}{2} \, b e n x^{2} \log \left (x\right ) + \frac {1}{2} \, b d n \log \left (x\right )^{2} - \frac {1}{4} \, {\left (b e n - 2 \, b e \log \left (c\right ) - 2 \, a e\right )} x^{2} + {\left (b d \log \left (c\right ) + a d\right )} \log \left (x\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.92 \[ \int \frac {\left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=a\,d\,\ln \left (x\right )+\frac {e\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {b\,e\,x^2\,\ln \left (c\,x^n\right )}{2}+\frac {b\,d\,{\ln \left (c\,x^n\right )}^2}{2\,n} \]

[In]

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

[Out]

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