3.370 \(\int \frac {(d+e x^r) (a+b \log (c x^n))}{x} \, dx\)

Optimal. Leaf size=53 \[ \frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+\frac {e x^r \left (a+b \log \left (c x^n\right )\right )}{r}-\frac {b e n x^r}{r^2} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {14, 2351, 2301, 2304} \[ \frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+\frac {e x^r \left (a+b \log \left (c x^n\right )\right )}{r}-\frac {b e n x^r}{r^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*e*n*x^r)/r^2) + (e*x^r*(a + b*Log[c*x^n]))/r + (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 2301

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 2304

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 2351

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

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Mathematica [A]  time = 0.09, size = 54, normalized size = 1.02 \[ \frac {e x^r (a r-b n)}{r^2}+a d \log (x)+\frac {b d \log ^2\left (c x^n\right )}{2 n}+\frac {b e x^r \log \left (c x^n\right )}{r} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 64, normalized size = 1.21 \[ \frac {b d n r^{2} \log \relax (x)^{2} + 2 \, {\left (b e n r \log \relax (x) + b e r \log \relax (c) - b e n + a e r\right )} x^{r} + 2 \, {\left (b d r^{2} \log \relax (c) + a d r^{2}\right )} \log \relax (x)}{2 \, r^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 69, normalized size = 1.30 \[ \frac {1}{2} \, b d n \log \relax (x)^{2} + \frac {b n x^{r} e \log \relax (x)}{r} + b d \log \relax (c) \log \relax (x) + \frac {b x^{r} e \log \relax (c)}{r} + a d \log \relax (x) - \frac {b n x^{r} e}{r^{2}} + \frac {a x^{r} e}{r} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.26, size = 278, normalized size = 5.25 \[ -\frac {i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \relax (x )}{2}+\frac {i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2}+\frac {i \pi b d \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \relax (x )}{2}-\frac {i \pi b d \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \relax (x )}{2}-\frac {b d n \ln \relax (x )^{2}}{2}-\frac {i \pi b e \,x^{r} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{2 r}+\frac {i \pi b e \,x^{r} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 r}+\frac {i \pi b e \,x^{r} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{2 r}-\frac {i \pi b e \,x^{r} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{2 r}+b d \ln \relax (c ) \ln \relax (x )+a d \ln \relax (x )+\frac {b e \,x^{r} \ln \relax (c )}{r}+\frac {a e \,x^{r}}{r}-\frac {b e n \,x^{r}}{r^{2}}+\frac {\left (d r \ln \relax (x )+e \,x^{r}\right ) b \ln \left (x^{n}\right )}{r} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+e*x^r)*(b*ln(c*x^n)+a)/x,x)

[Out]

b*(d*r*ln(x)+e*x^r)/r*ln(x^n)+1/2*I*Pi*ln(x)*b*d*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*csgn(I*c)*csgn(I*c*x^n)*csg
n(I*x^n)*d*b*ln(x)*Pi-1/2*I*csgn(I*c*x^n)^3*d*b*ln(x)*Pi+1/2*I*Pi*ln(x)*b*d*csgn(I*c*x^n)^2*csgn(I*c)+1/2*I/r*
Pi*b*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x^r-1/2*I/r*Pi*b*e*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^r-1/2*I/r*Pi*b*e*c
sgn(I*c*x^n)^3*x^r+1/2*I/r*Pi*b*e*csgn(I*c*x^n)^2*csgn(I*c)*x^r-1/2*b*d*n*ln(x)^2+b*d*ln(c)*ln(x)+1/r*b*e*x^r*
ln(c)+a*d*ln(x)+1/r*x^r*a*e-b*e*n*x^r/r^2

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maxima [A]  time = 1.02, size = 56, normalized size = 1.06 \[ \frac {b e x^{r} \log \left (c x^{n}\right )}{r} + \frac {b d \log \left (c x^{n}\right )^{2}}{2 \, n} + a d \log \relax (x) - \frac {b e n x^{r}}{r^{2}} + \frac {a e x^{r}}{r} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\left (d+e\,x^r\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 9.86, size = 112, normalized size = 2.11 \[ \begin {cases} a d \log {\relax (x )} + \frac {a e x^{r}}{r} + \frac {b d n \log {\relax (x )}^{2}}{2} + b d \log {\relax (c )} \log {\relax (x )} + \frac {b e n x^{r} \log {\relax (x )}}{r} - \frac {b e n x^{r}}{r^{2}} + \frac {b e x^{r} \log {\relax (c )}}{r} & \text {for}\: r \neq 0 \\\left (d + e\right ) \left (\begin {cases} a \log {\relax (x )} & \text {for}\: b = 0 \\- \left (- a - b \log {\relax (c )}\right ) \log {\relax (x )} & \text {for}\: n = 0 \\\frac {\left (- a - b \log {\left (c x^{n} \right )}\right )^{2}}{2 b n} & \text {otherwise} \end {cases}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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