3.4.51 \(\int (f x)^{-1+m} (d+e x^m)^3 (a+b \log (c x^n)) \, dx\) [351]

Optimal. Leaf size=171 \[ -\frac {b d^3 n x (f x)^{-1+m}}{m^2}-\frac {3 b d^2 e n x^{1+m} (f x)^{-1+m}}{4 m^2}-\frac {b d e^2 n x^{1+2 m} (f x)^{-1+m}}{3 m^2}-\frac {b e^3 n x^{1+3 m} (f x)^{-1+m}}{16 m^2}-\frac {b d^4 n x^{1-m} (f x)^{-1+m} \log (x)}{4 e m}+\frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^4 \left (a+b \log \left (c x^n\right )\right )}{4 e m} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2377, 2376, 272, 45} \begin {gather*} \frac {x^{1-m} (f x)^{m-1} \left (d+e x^m\right )^4 \left (a+b \log \left (c x^n\right )\right )}{4 e m}-\frac {b d^4 n x^{1-m} \log (x) (f x)^{m-1}}{4 e m}-\frac {b d^3 n x (f x)^{m-1}}{m^2}-\frac {3 b d^2 e n x^{m+1} (f x)^{m-1}}{4 m^2}-\frac {b d e^2 n x^{2 m+1} (f x)^{m-1}}{3 m^2}-\frac {b e^3 n x^{3 m+1} (f x)^{m-1}}{16 m^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(f*x)^(-1 + m)*(d + e*x^m)^3*(a + b*Log[c*x^n]),x]

[Out]

-((b*d^3*n*x*(f*x)^(-1 + m))/m^2) - (3*b*d^2*e*n*x^(1 + m)*(f*x)^(-1 + m))/(4*m^2) - (b*d*e^2*n*x^(1 + 2*m)*(f
*x)^(-1 + m))/(3*m^2) - (b*e^3*n*x^(1 + 3*m)*(f*x)^(-1 + m))/(16*m^2) - (b*d^4*n*x^(1 - m)*(f*x)^(-1 + m)*Log[
x])/(4*e*m) + (x^(1 - m)*(f*x)^(-1 + m)*(d + e*x^m)^4*(a + b*Log[c*x^n]))/(4*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*} \int (f x)^{-1+m} \left (d+e x^m\right )^3 \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 )^3 \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 )^4 \left (a+b \log \left (c x^n\right )\right )}{4 e m}-\frac {\left (b n x^{1-m} (f x)^{-1+m}\right ) \int \frac {\left (d+e x^m\right )^4}{x} \, dx}{4 e m}\\ &=\frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^4 \left (a+b \log \left (c x^n\right )\right )}{4 e m}-\frac {\left (b n x^{1-m} (f x)^{-1+m}\right ) \text {Subst}\left (\int \frac {(d+e x)^4}{x} \, dx,x,x^m\right )}{4 e m^2}\\ &=\frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^4 \left (a+b \log \left (c x^n\right )\right )}{4 e m}-\frac {\left (b n x^{1-m} (f x)^{-1+m}\right ) \text {Subst}\left (\int \left (4 d^3 e+\frac {d^4}{x}+6 d^2 e^2 x+4 d e^3 x^2+e^4 x^3\right ) \, dx,x,x^m\right )}{4 e m^2}\\ &=-\frac {b d^3 n x (f x)^{-1+m}}{m^2}-\frac {3 b d^2 e n x^{1+m} (f x)^{-1+m}}{4 m^2}-\frac {b d e^2 n x^{1+2 m} (f x)^{-1+m}}{3 m^2}-\frac {b e^3 n x^{1+3 m} (f x)^{-1+m}}{16 m^2}-\frac {b d^4 n x^{1-m} (f x)^{-1+m} \log (x)}{4 e m}+\frac {x^{1-m} (f x)^{-1+m} \left (d+e x^m\right )^4 \left (a+b \log \left (c x^n\right )\right )}{4 e m}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 140, normalized size = 0.82 \begin {gather*} \frac {(f x)^m \left (12 a m \left (4 d^3+6 d^2 e x^m+4 d e^2 x^{2 m}+e^3 x^{3 m}\right )-b n \left (48 d^3+36 d^2 e x^m+16 d e^2 x^{2 m}+3 e^3 x^{3 m}\right )+12 b m \left (4 d^3+6 d^2 e x^m+4 d e^2 x^{2 m}+e^3 x^{3 m}\right ) \log \left (c x^n\right )\right )}{48 f m^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^(-1 + m)*(d + e*x^m)^3*(a + b*Log[c*x^n]),x]

[Out]

((f*x)^m*(12*a*m*(4*d^3 + 6*d^2*e*x^m + 4*d*e^2*x^(2*m) + e^3*x^(3*m)) - b*n*(48*d^3 + 36*d^2*e*x^m + 16*d*e^2
*x^(2*m) + 3*e^3*x^(3*m)) + 12*b*m*(4*d^3 + 6*d^2*e*x^m + 4*d*e^2*x^(2*m) + e^3*x^(3*m))*Log[c*x^n]))/(48*f*m^
2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.20, size = 806, normalized size = 4.71

method	result	size
risch	\(\frac {b \left (e^{3} x^{3 m}+4 d \,e^{2} x^{2 m}+6 d^{2} e \,x^{m}+4 d^{3}\right ) x \,{\mathrm e}^{\frac {\left (-1+m \right ) \left (-i \pi \mathrm {csgn}\left (i f x \right )^{3}+i \pi \mathrm {csgn}\left (i f x \right )^{2} \mathrm {csgn}\left (i f \right )+i \pi \mathrm {csgn}\left (i f x \right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i f x \right ) \mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left (f \right )\right )}{2}} \ln \left (x^{n}\right )}{4 m}+\frac {\left (72 a \,d^{2} e \,x^{m} m -36 b \,d^{2} e n \,x^{m}+48 \ln \left (c \right ) b d \,e^{2} x^{2 m} m +48 a \,d^{3} m -3 b \,e^{3} n \,x^{3 m}+12 a \,e^{3} x^{3 m} m +48 \ln \left (c \right ) b \,d^{3} m -48 b \,d^{3} n +12 \ln \left (c \right ) b \,e^{3} x^{3 m} m +48 a d \,e^{2} x^{2 m} m -16 b d \,e^{2} n \,x^{2 m}-24 i \pi b \,d^{3} m \mathrm {csgn}\left (i c \,x^{n}\right )^{3}+72 \ln \left (c \right ) b \,d^{2} e \,x^{m} m +36 i \pi b \,d^{2} e \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x^{m} m +36 i \pi b \,d^{2} e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x^{m} m +24 i \pi b \,d^{3} m \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+24 i \pi b d \,e^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x^{2 m} m +24 i \pi b d \,e^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x^{2 m} m -6 i \pi b \,e^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) x^{3 m} m +6 i \pi b \,e^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x^{3 m} m -24 i \pi b d \,e^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} x^{2 m} m +6 i \pi b \,e^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} x^{3 m} m -24 i \pi b \,d^{3} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) m -36 i \pi b \,d^{2} e \mathrm {csgn}\left (i c \,x^{n}\right )^{3} x^{m} m -36 i \pi b \,d^{2} e \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) x^{m} m -24 i \pi b d \,e^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) x^{2 m} m +24 i \pi b \,d^{3} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} m -6 i \pi b \,e^{3} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} x^{3 m} m \right ) x \,{\mathrm e}^{\frac {\left (-1+m \right ) \left (-i \pi \mathrm {csgn}\left (i f x \right )^{3}+i \pi \mathrm {csgn}\left (i f x \right )^{2} \mathrm {csgn}\left (i f \right )+i \pi \mathrm {csgn}\left (i f x \right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \,\mathrm {csgn}\left (i f x \right ) \mathrm {csgn}\left (i f \right ) \mathrm {csgn}\left (i x \right )+2 \ln \left (x \right )+2 \ln \left (f \right )\right )}{2}}}{48 m^{2}}\)	\(806\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(-1+m)*(d+e*x^m)^3*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)

[Out]

1/4*b*(e^3*(x^m)^3+4*d*e^2*(x^m)^2+6*d^2*e*x^m+4*d^3)*x/m*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(x)+2*ln(f)))*ln(x^n)+1/48*
(24*I*Pi*b*d^3*csgn(I*c)*csgn(I*c*x^n)^2*m+24*I*Pi*b*d^3*csgn(I*x^n)*csgn(I*c*x^n)^2*m+12*ln(c)*b*e^3*(x^m)^3*
m+48*a*d*e^2*(x^m)^2*m+72*a*d^2*e*x^m*m-16*b*d*e^2*n*(x^m)^2-36*b*d^2*e*n*x^m+48*a*d^3*m+48*ln(c)*b*d^3*m-48*b
*d^3*n+24*I*Pi*b*d*e^2*csgn(I*c)*csgn(I*c*x^n)^2*(x^m)^2*m-3*b*e^3*n*(x^m)^3+12*a*e^3*(x^m)^3*m-36*I*Pi*b*d^2*
e*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*x^m*m-24*I*Pi*b*d^3*csgn(I*c*x^n)^3*m+6*I*Pi*b*e^3*csgn(I*x^n)*csgn(I*c*
x^n)^2*(x^m)^3*m-24*I*Pi*b*d*e^2*csgn(I*c*x^n)^3*(x^m)^2*m-24*I*Pi*b*d*e^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)
*(x^m)^2*m+24*I*Pi*b*d*e^2*csgn(I*x^n)*csgn(I*c*x^n)^2*(x^m)^2*m+36*I*Pi*b*d^2*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x
^m*m-6*I*Pi*b*e^3*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*(x^m)^3*m+36*I*Pi*b*d^2*e*csgn(I*c)*csgn(I*c*x^n)^2*x^m*
m+48*ln(c)*b*d*e^2*(x^m)^2*m+72*ln(c)*b*d^2*e*x^m*m-24*I*Pi*b*d^3*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*m+6*I*Pi
*b*e^3*csgn(I*c)*csgn(I*c*x^n)^2*(x^m)^3*m-6*I*Pi*b*e^3*csgn(I*c*x^n)^3*(x^m)^3*m-36*I*Pi*b*d^2*e*csgn(I*c*x^n
)^3*x^m*m)*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(x)+2*ln(f)))

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Maxima [A]
time = 0.30, size = 259, normalized size = 1.51 \begin {gather*} -\frac {b d^{3} f^{m - 1} n x^{m}}{m^{2}} + \frac {3 \, b d^{2} f^{m - 1} e^{\left (2 \, m \log \left (x\right ) + 1\right )} \log \left (c x^{n}\right )}{2 \, m} + \frac {3 \, a d^{2} f^{m - 1} e^{\left (2 \, m \log \left (x\right ) + 1\right )}}{2 \, m} - \frac {3 \, b d^{2} f^{m - 1} n e^{\left (2 \, m \log \left (x\right ) + 1\right )}}{4 \, m^{2}} + \frac {\left (f x\right )^{m} b d^{3} \log \left (c x^{n}\right )}{f m} + \frac {b d f^{m - 1} e^{\left (3 \, m \log \left (x\right ) + 2\right )} \log \left (c x^{n}\right )}{m} + \frac {\left (f x\right )^{m} a d^{3}}{f m} + \frac {a d f^{m - 1} e^{\left (3 \, m \log \left (x\right ) + 2\right )}}{m} - \frac {b d f^{m - 1} n e^{\left (3 \, m \log \left (x\right ) + 2\right )}}{3 \, m^{2}} + \frac {b f^{m - 1} e^{\left (4 \, m \log \left (x\right ) + 3\right )} \log \left (c x^{n}\right )}{4 \, m} + \frac {a f^{m - 1} e^{\left (4 \, m \log \left (x\right ) + 3\right )}}{4 \, m} - \frac {b f^{m - 1} n e^{\left (4 \, m \log \left (x\right ) + 3\right )}}{16 \, m^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(d+e*x^m)^3*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

-b*d^3*f^(m - 1)*n*x^m/m^2 + 3/2*b*d^2*f^(m - 1)*e^(2*m*log(x) + 1)*log(c*x^n)/m + 3/2*a*d^2*f^(m - 1)*e^(2*m*
log(x) + 1)/m - 3/4*b*d^2*f^(m - 1)*n*e^(2*m*log(x) + 1)/m^2 + (f*x)^m*b*d^3*log(c*x^n)/(f*m) + b*d*f^(m - 1)*
e^(3*m*log(x) + 2)*log(c*x^n)/m + (f*x)^m*a*d^3/(f*m) + a*d*f^(m - 1)*e^(3*m*log(x) + 2)/m - 1/3*b*d*f^(m - 1)
*n*e^(3*m*log(x) + 2)/m^2 + 1/4*b*f^(m - 1)*e^(4*m*log(x) + 3)*log(c*x^n)/m + 1/4*a*f^(m - 1)*e^(4*m*log(x) +
3)/m - 1/16*b*f^(m - 1)*n*e^(4*m*log(x) + 3)/m^2

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Fricas [A]
time = 0.36, size = 189, normalized size = 1.11 \begin {gather*} \frac {3 \, {\left (4 \, b m n e^{3} \log \left (x\right ) + 4 \, b m e^{3} \log \left (c\right ) + {\left (4 \, a m - b n\right )} e^{3}\right )} f^{m - 1} x^{4 \, m} + 16 \, {\left (3 \, b d m n e^{2} \log \left (x\right ) + 3 \, b d m e^{2} \log \left (c\right ) + {\left (3 \, a d m - b d n\right )} e^{2}\right )} f^{m - 1} x^{3 \, m} + 36 \, {\left (2 \, b d^{2} m n e \log \left (x\right ) + 2 \, b d^{2} m e \log \left (c\right ) + {\left (2 \, a d^{2} m - b d^{2} n\right )} e\right )} f^{m - 1} x^{2 \, m} + 48 \, {\left (b d^{3} m n \log \left (x\right ) + b d^{3} m \log \left (c\right ) + a d^{3} m - b d^{3} n\right )} f^{m - 1} x^{m}}{48 \, m^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(d+e*x^m)^3*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

1/48*(3*(4*b*m*n*e^3*log(x) + 4*b*m*e^3*log(c) + (4*a*m - b*n)*e^3)*f^(m - 1)*x^(4*m) + 16*(3*b*d*m*n*e^2*log(
x) + 3*b*d*m*e^2*log(c) + (3*a*d*m - b*d*n)*e^2)*f^(m - 1)*x^(3*m) + 36*(2*b*d^2*m*n*e*log(x) + 2*b*d^2*m*e*lo
g(c) + (2*a*d^2*m - b*d^2*n)*e)*f^(m - 1)*x^(2*m) + 48*(b*d^3*m*n*log(x) + b*d^3*m*log(c) + a*d^3*m - b*d^3*n)
*f^(m - 1)*x^m)/m^2

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**(-1+m)*(d+e*x**m)**3*(a+b*ln(c*x**n)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3004 deep

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (159) = 318\).
time = 3.95, size = 335, normalized size = 1.96 \begin {gather*} \frac {b d^{3} f^{m} n x^{m} \log \left (x\right )}{f m} + \frac {3 \, b d^{2} f^{m} n x^{2 \, m} e \log \left (x\right )}{2 \, f m} + \frac {b d^{3} f^{m} x^{m} \log \left (c\right )}{f m} + \frac {3 \, b d^{2} f^{m} x^{2 \, m} e \log \left (c\right )}{2 \, f m} + \frac {b d f^{m} n x^{3 \, m} e^{2} \log \left (x\right )}{f m} + \frac {a d^{3} f^{m} x^{m}}{f m} - \frac {b d^{3} f^{m} n x^{m}}{f m^{2}} + \frac {3 \, a d^{2} f^{m} x^{2 \, m} e}{2 \, f m} - \frac {3 \, b d^{2} f^{m} n x^{2 \, m} e}{4 \, f m^{2}} + \frac {b d f^{m} x^{3 \, m} e^{2} \log \left (c\right )}{f m} + \frac {b f^{m} n x^{4 \, m} e^{3} \log \left (x\right )}{4 \, f m} + \frac {a d f^{m} x^{3 \, m} e^{2}}{f m} - \frac {b d f^{m} n x^{3 \, m} e^{2}}{3 \, f m^{2}} + \frac {b f^{m} x^{4 \, m} e^{3} \log \left (c\right )}{4 \, f m} + \frac {a f^{m} x^{4 \, m} e^{3}}{4 \, f m} - \frac {b f^{m} n x^{4 \, m} e^{3}}{16 \, f m^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(d+e*x^m)^3*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

b*d^3*f^m*n*x^m*log(x)/(f*m) + 3/2*b*d^2*f^m*n*x^(2*m)*e*log(x)/(f*m) + b*d^3*f^m*x^m*log(c)/(f*m) + 3/2*b*d^2
*f^m*x^(2*m)*e*log(c)/(f*m) + b*d*f^m*n*x^(3*m)*e^2*log(x)/(f*m) + a*d^3*f^m*x^m/(f*m) - b*d^3*f^m*n*x^m/(f*m^
2) + 3/2*a*d^2*f^m*x^(2*m)*e/(f*m) - 3/4*b*d^2*f^m*n*x^(2*m)*e/(f*m^2) + b*d*f^m*x^(3*m)*e^2*log(c)/(f*m) + 1/
4*b*f^m*n*x^(4*m)*e^3*log(x)/(f*m) + a*d*f^m*x^(3*m)*e^2/(f*m) - 1/3*b*d*f^m*n*x^(3*m)*e^2/(f*m^2) + 1/4*b*f^m
*x^(4*m)*e^3*log(c)/(f*m) + 1/4*a*f^m*x^(4*m)*e^3/(f*m) - 1/16*b*f^m*n*x^(4*m)*e^3/(f*m^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (f\,x\right )}^{m-1}\,{\left (d+e\,x^m\right )}^3\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^(m - 1)*(d + e*x^m)^3*(a + b*log(c*x^n)),x)

[Out]

int((f*x)^(m - 1)*(d + e*x^m)^3*(a + b*log(c*x^n)), x)

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