\(\int (f x)^{-1+m} (d+e x^m) (a+b \log (c x^n)) \, dx\) [353]

Optimal result

Integrand size = 25, antiderivative size = 90 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b d n (f x)^m}{f m^2}-\frac {b e n x^m (f x)^m}{4 f m^2}+\frac {d (f x)^m \left (a+b \log \left (c x^n\right )\right )}{f m}+\frac {e x^m (f x)^m \left (a+b \log \left (c x^n\right )\right )}{2 f m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.26, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2377, 2376, 272, 45} \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{1-m} (f x)^{m-1} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{2 e m}-\frac {b d^2 n x^{1-m} \log (x) (f x)^{m-1}}{2 e m}-\frac {b d n x (f x)^{m-1}}{m^2}-\frac {b e n x^{m+1} (f x)^{m-1}}{4 m^2} \]

[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 45

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])

Rule 272

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 2376

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 2377

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])

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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] (verified)

Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.68 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {(f x)^m \left (2 a m \left (2 d+e x^m\right )-b n \left (4 d+e x^m\right )+2 b m \left (2 d+e x^m\right ) \log \left (c x^n\right )\right )}{4 f m^2} \]

[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 [A] (verified)

Time = 1.74 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.17

method	result	size
parallelrisch	\(-\frac {-2 x \,x^{m} \ln \left (c \,x^{n}\right ) \left (f x \right )^{m -1} b e m -2 x \,x^{m} \left (f x \right )^{m -1} a e m +x \,x^{m} \left (f x \right )^{m -1} b e n -4 x \ln \left (c \,x^{n}\right ) \left (f x \right )^{m -1} b d m -4 x \left (f x \right )^{m -1} a d m +4 x \left (f x \right )^{m -1} b d n}{4 m^{2}}\)	\(105\)
risch	\(\frac {b \left (e \,x^{m}+2 d \right ) x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \operatorname {csgn}\left (i f x \right )^{3}+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right )+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left (f \right )\right )}{2}} \ln \left (x^{n}\right )}{2 m}-\frac {\left (i \pi b e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) x^{m} m -i \pi b e \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{m} m -i \pi b e \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x^{m} m +i \pi b e \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x^{m} m +2 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 ) m -2 i \pi b d \,\operatorname {csgn}\left (i c \right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} m -2 i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} m +2 i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3} m -2 \ln \left (c \right ) b e \,x^{m} m -4 \ln \left (c \right ) b d m -2 x^{m} a e m +x^{m} b e n -4 a d m +4 b d n \right ) x \,{\mathrm e}^{\frac {\left (m -1\right ) \left (-i \pi \operatorname {csgn}\left (i f x \right )^{3}+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i f \right )+i \pi \operatorname {csgn}\left (i f x \right )^{2} \operatorname {csgn}\left (i x \right )-i \pi \,\operatorname {csgn}\left (i f x \right ) \operatorname {csgn}\left (i f \right ) \operatorname {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left (f \right )\right )}{2}}}{4 m^{2}}\)	\(425\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {{\left (2 \, b e m n \log \left (x\right ) + 2 \, b e m \log \left (c\right ) + 2 \, a e m - b e n\right )} f^{m - 1} x^{2 \, m} + 4 \, {\left (b d m n \log \left (x\right ) + b d m \log \left (c\right ) + a d m - b d n\right )} f^{m - 1} x^{m}}{4 \, m^{2}} \]

[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 [A] (verification not implemented)

Time = 3.01 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.73 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\begin {cases} \frac {a d x \left (f x\right )^{m - 1}}{m} + \frac {a e x x^{m} \left (f x\right )^{m - 1}}{2 m} + \frac {b d x \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{m} - \frac {b d n x \left (f x\right )^{m - 1}}{m^{2}} + \frac {b e x x^{m} \left (f x\right )^{m - 1} \log {\left (c x^{n} \right )}}{2 m} - \frac {b e n x x^{m} \left (f x\right )^{m - 1}}{4 m^{2}} & \text {for}\: m \neq 0 \\\frac {\left (d + e\right ) \left (\begin {cases} a \log {\left (x \right )} & \text {for}\: b = 0 \\- \left (- a - b \log {\left (c \right )}\right ) \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {\left (- a - b \log {\left (c x^{n} \right )}\right )^{2}}{2 b n} & \text {otherwise} \end {cases}\right )}{f} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((a*d*x*(f*x)**(m - 1)/m + a*e*x*x**m*(f*x)**(m - 1)/(2*m) + b*d*x*(f*x)**(m - 1)*log(c*x**n)/m - b*d
*n*x*(f*x)**(m - 1)/m**2 + b*e*x*x**m*(f*x)**(m - 1)*log(c*x**n)/(2*m) - b*e*n*x*x**m*(f*x)**(m - 1)/(4*m**2),
Ne(m, 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))/f, True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.21 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b e f^{m - 1} x^{2 \, m} \log \left (c x^{n}\right )}{2 \, m} + \frac {a e f^{m - 1} x^{2 \, m}}{2 \, m} - \frac {b e f^{m - 1} n x^{2 \, m}}{4 \, m^{2}} - \frac {b d f^{m - 1} n x^{m}}{m^{2}} + \frac {\left (f x\right )^{m} b d \log \left (c x^{n}\right )}{f m} + \frac {\left (f x\right )^{m} a d}{f m} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.62 \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {b e f^{m} n x^{2 \, m} \log \left (x\right )}{2 \, f m} + \frac {b d f^{m} n x^{m} \log \left (x\right )}{f m} + \frac {b e f^{m} x^{2 \, m} \log \left (c\right )}{2 \, f m} + \frac {b d f^{m} x^{m} \log \left (c\right )}{f m} + \frac {a e f^{m} x^{2 \, m}}{2 \, f m} - \frac {b e f^{m} n x^{2 \, m}}{4 \, f m^{2}} + \frac {a d f^{m} x^{m}}{f m} - \frac {b d f^{m} n x^{m}}{f m^{2}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right ) \, dx=\int {\left (f\,x\right )}^{m-1}\,\left (d+e\,x^m\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

[In]

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

[Out]

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