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

Optimal. Leaf size=52 \[ \frac{d \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+\frac{1}{2} e x^2 \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} b e n x^2 \]

[Out]

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

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Rubi [A]  time = 0.064304, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {14, 2351, 2301, 2304} \[ \frac{d \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+\frac{1}{2} e x^2 \left (a+b \log \left (c x^n\right )\right )-\frac{1}{4} b e n x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*e*n*x^2)/4 + (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 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]))

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
]

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\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 \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]  time = 0.0024418, size = 57, normalized size = 1.1 \[ a d \log (x)+\frac{1}{2} a e x^2+\frac{b d \log ^2\left (c x^n\right )}{2 n}+\frac{1}{2} b e x^2 \log \left (c x^n\right )-\frac{1}{4} b e n x^2 \]

Antiderivative was successfully verified.

[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)

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Maple [C]  time = 0.23, size = 257, normalized size = 4.9 \begin{align*} \left ({\frac{eb{x}^{2}}{2}}+bd\ln \left ( x \right ) \right ) \ln \left ({x}^{n} \right ) -{\frac{bdn \left ( \ln \left ( x \right ) \right ) ^{2}}{2}}+{\frac{i}{4}}\pi \,be{x}^{2}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-{\frac{i}{4}}\pi \,be{x}^{2}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{4}}\pi \,be{x}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+{\frac{i}{4}}\pi \,be{x}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +{\frac{\ln \left ( c \right ) be{x}^{2}}{2}}-{\frac{ben{x}^{2}}{4}}+{\frac{ae{x}^{2}}{2}}+{\frac{i}{2}}\ln \left ( x \right ) \pi \,bd{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-{\frac{i}{2}}\ln \left ( x \right ) \pi \,bd{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -{\frac{i}{2}}\ln \left ( x \right ) \pi \,bd \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+{\frac{i}{2}}\ln \left ( x \right ) \pi \,bd \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +\ln \left ( x \right ) \ln \left ( c \right ) bd+\ln \left ( x \right ) ad \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.09173, size = 66, normalized size = 1.27 \begin{align*} -\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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 1.3388, size = 155, normalized size = 2.98 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 1.18411, size = 71, normalized size = 1.37 \begin{align*} a d \log{\left (x \right )} + \frac{a e x^{2}}{2} + \frac{b d n \log{\left (x \right )}^{2}}{2} + b d \log{\left (c \right )} \log{\left (x \right )} + \frac{b e n x^{2} \log{\left (x \right )}}{2} - \frac{b e n x^{2}}{4} + \frac{b e x^{2} \log{\left (c \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20174, size = 81, normalized size = 1.56 \begin{align*} \frac{1}{2} \, b n x^{2} e \log \left (x\right ) - \frac{1}{4} \, b n x^{2} e + \frac{1}{2} \, b x^{2} e \log \left (c\right ) + \frac{1}{2} \, b d n \log \left (x\right )^{2} + \frac{1}{2} \, a x^{2} e + b d \log \left (c\right ) \log \left (x\right ) + a d \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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