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

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

[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)

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Rubi [A]  time = 0.21477, antiderivative size = 171, normalized size of antiderivative = 1., 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 = {2339, 2338, 266, 43} \[ \frac{x^{1-m} (f x)^{m-1} \left (d+e x^m\right )^4 \left (a+b \log \left (c x^n\right )\right )}{4 e m}-\frac{3 b d^2 e n x^{m+1} (f x)^{m-1}}{4 m^2}-\frac{b d^4 n x^{1-m} \log (x) (f x)^{m-1}}{4 e m}-\frac{b d^3 n x (f x)^{m-1}}{m^2}-\frac{b d e^2 n x^{2 m+1} (f x)^{m-1}}{3 m^2}-\frac{b e^3 n x^{3 m+1} (f x)^{m-1}}{16 m^2} \]

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 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 )^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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.155689, size = 140, normalized size = 0.82 \[ \frac{(f x)^m \left (12 a m \left (6 d^2 e x^m+4 d^3+4 d e^2 x^{2 m}+e^3 x^{3 m}\right )+12 b m \log \left (c x^n\right ) \left (6 d^2 e x^m+4 d^3+4 d e^2 x^{2 m}+e^3 x^{3 m}\right )-b n \left (36 d^2 e x^m+48 d^3+16 d e^2 x^{2 m}+3 e^3 x^{3 m}\right )\right )}{48 f m^2} \]

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]  time = 0.242, size = 806, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

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)

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

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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)^3*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.38314, size = 479, normalized size = 2.8 \begin{align*} \frac{3 \,{\left (4 \, b e^{3} m n \log \left (x\right ) + 4 \, b e^{3} m \log \left (c\right ) + 4 \, a e^{3} m - b e^{3} n\right )} f^{m - 1} x^{4 \, m} + 16 \,{\left (3 \, b d e^{2} m n \log \left (x\right ) + 3 \, b d e^{2} m \log \left (c\right ) + 3 \, a d e^{2} m - b d e^{2} n\right )} f^{m - 1} x^{3 \, m} + 36 \,{\left (2 \, b d^{2} e m n \log \left (x\right ) + 2 \, b d^{2} e m \log \left (c\right ) + 2 \, a d^{2} e m - b d^{2} e n\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{align*}

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

[Out]

Timed out

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Giac [B]  time = 1.47814, size = 452, normalized size = 2.64 \begin{align*} \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{align*}

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)