3.353 \(\int (f x)^{-1+m} (d+e x^m) (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=90 \[ \frac{d (f x)^m \left (a+b \log \left (c x^n\right )\right )}{f m}+\frac{e x^m (f x)^m \left (a+b \log \left (c x^n\right )\right )}{2 f m}-\frac{b d n (f x)^m}{f m^2}-\frac{b e n x^m (f x)^m}{4 f m^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.116969, antiderivative size = 113, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2339, 2338, 266, 43} \[ \frac{x^{1-m} (f x)^{m-1} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{2 e m}-\frac{b d^2 n x^{1-m} \log (x) (f x)^{m-1}}{2 e m}-\frac{b d n x (f x)^{m-1}}{m^2}-\frac{b e n x^{m+1} (f x)^{m-1}}{4 m^2} \]

Antiderivative was successfully verified.

[In]

Int[(f*x)^(-1 + m)*(d + e*x^m)*(a + b*Log[c*x^n]),x]

[Out]

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

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :>
Dist[(f*x)^m/x^m, Int[x^m*(d + e*x^r)^q*(a + b*Log[c*x^n])^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r},
x] && EqQ[m, r - 1] && IGtQ[p, 0] &&  !(IntegerQ[m] || GtQ[f, 0])

Rule 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :
> Simp[(f^m*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^p)/(e*r*(q + 1)), x] - Dist[(b*f^m*n*p)/(e*r*(q + 1)), Int[
((d + e*x^r)^(q + 1)*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[
m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.063729, size = 61, normalized size = 0.68 \[ \frac{(f x)^m \left (2 a m \left (2 d+e x^m\right )+2 b m \log \left (c x^n\right ) \left (2 d+e x^m\right )-b n \left (4 d+e x^m\right )\right )}{4 f m^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^(-1 + m)*(d + e*x^m)*(a + b*Log[c*x^n]),x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.165, size = 426, normalized size = 4.7 \begin{align*}{\frac{b \left ( e{x}^{m}+2\,d \right ) x\ln \left ({x}^{n} \right ) }{2\,m}{{\rm e}^{{\frac{ \left ( -1+m \right ) \left ( -i\pi \, \left ({\it csgn} \left ( ifx \right ) \right ) ^{3}+i\pi \, \left ({\it csgn} \left ( ifx \right ) \right ) ^{2}{\it csgn} \left ( if \right ) +i\pi \, \left ({\it csgn} \left ( ifx \right ) \right ) ^{2}{\it csgn} \left ( ix \right ) -i\pi \,{\it csgn} \left ( ifx \right ){\it csgn} \left ( if \right ){\it csgn} \left ( ix \right ) +2\,\ln \left ( f \right ) +2\,\ln \left ( x \right ) \right ) }{2}}}}}+{\frac{ \left ( i\pi \,be{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{x}^{m}m-i\pi \,be{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ){x}^{m}m-i\pi \,be \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}{x}^{m}m+i\pi \,be \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ){x}^{m}m+2\,i\pi \,bdm{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-2\,i\pi \,bdm{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -2\,i\pi \,bdm \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+2\,i\pi \,bdm \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +2\,\ln \left ( c \right ) be{x}^{m}m+4\,\ln \left ( c \right ) bdm+2\,{x}^{m}aem-{x}^{m}ben+4\,adm-4\,bdn \right ) x}{4\,{m}^{2}}{{\rm e}^{{\frac{ \left ( -1+m \right ) \left ( -i\pi \, \left ({\it csgn} \left ( ifx \right ) \right ) ^{3}+i\pi \, \left ({\it csgn} \left ( ifx \right ) \right ) ^{2}{\it csgn} \left ( if \right ) +i\pi \, \left ({\it csgn} \left ( ifx \right ) \right ) ^{2}{\it csgn} \left ( ix \right ) -i\pi \,{\it csgn} \left ( ifx \right ){\it csgn} \left ( if \right ){\it csgn} \left ( ix \right ) +2\,\ln \left ( f \right ) +2\,\ln \left ( x \right ) \right ) }{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(-1+m)*(d+e*x^m)*(a+b*ln(c*x^n)),x)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(d+e*x^m)*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.32975, size = 201, normalized size = 2.23 \begin{align*} \frac{{\left (2 \, b e m n \log \left (x\right ) + 2 \, b e m \log \left (c\right ) + 2 \, a e m - b e n\right )} f^{m - 1} x^{2 \, m} + 4 \,{\left (b d m n \log \left (x\right ) + b d m \log \left (c\right ) + a d m - b d n\right )} f^{m - 1} x^{m}}{4 \, m^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(d+e*x^m)*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**(-1+m)*(d+e*x**m)*(a+b*ln(c*x**n)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.33761, size = 203, normalized size = 2.26 \begin{align*} \frac{b d f^{m} n x^{m} \log \left (x\right )}{f m} + \frac{b f^{m} n x^{2 \, m} e \log \left (x\right )}{2 \, f m} + \frac{b d f^{m} x^{m} \log \left (c\right )}{f m} + \frac{b f^{m} x^{2 \, m} e \log \left (c\right )}{2 \, f m} + \frac{a d f^{m} x^{m}}{f m} - \frac{b d f^{m} n x^{m}}{f m^{2}} + \frac{a f^{m} x^{2 \, m} e}{2 \, f m} - \frac{b f^{m} n x^{2 \, m} e}{4 \, f m^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(d+e*x^m)*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

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