3.441 \(\int (f x)^m (d+e x^r)^2 (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=165 \[ \frac{d^2 (f x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{f (m+1)}+\frac{2 d e x^{r+1} (f x)^m \left (a+b \log \left (c x^n\right )\right )}{m+r+1}+\frac{e^2 x^{2 r+1} (f x)^m \left (a+b \log \left (c x^n\right )\right )}{m+2 r+1}-\frac{b d^2 n (f x)^{m+1}}{f (m+1)^2}-\frac{2 b d e n x^{r+1} (f x)^m}{(m+r+1)^2}-\frac{b e^2 n x^{2 r+1} (f x)^m}{(m+2 r+1)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.185399, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {270, 20, 30, 2350, 14} \[ \frac{d^2 (f x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{f (m+1)}+\frac{2 d e x^{r+1} (f x)^m \left (a+b \log \left (c x^n\right )\right )}{m+r+1}+\frac{e^2 x^{2 r+1} (f x)^m \left (a+b \log \left (c x^n\right )\right )}{m+2 r+1}-\frac{b d^2 n (f x)^{m+1}}{f (m+1)^2}-\frac{2 b d e n x^{r+1} (f x)^m}{(m+r+1)^2}-\frac{b e^2 n x^{2 r+1} (f x)^m}{(m+2 r+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 20

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(b^IntPart[n]*(b*v)^FracPart[n])/(a^IntPart[n
]*(a*v)^FracPart[n]), Int[u*(a*v)^(m + n), x], x] /; FreeQ[{a, b, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]
&&  !IntegerQ[m + n]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

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]]

Rubi steps

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

Mathematica [A]  time = 0.244075, size = 124, normalized size = 0.75 \[ x (f x)^m \left (\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{m+1}+\frac{2 d e x^r \left (a+b \log \left (c x^n\right )\right )}{m+r+1}+\frac{e^2 x^{2 r} \left (a+b \log \left (c x^n\right )\right )}{m+2 r+1}-\frac{b d^2 n}{(m+1)^2}-\frac{2 b d e n x^r}{(m+r+1)^2}-\frac{b e^2 n x^{2 r}}{(m+2 r+1)^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^m*(d + e*x^r)^2*(a + b*Log[c*x^n]),x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.727, size = 8737, normalized size = 53. \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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)^m*(d+e*x^r)^2*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.58715, size = 4199, normalized size = 25.45 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(d+e*x^r)^2*(a+b*log(c*x^n)),x, algorithm="fricas")

[Out]

(((b*e^2*m^5 + 5*b*e^2*m^4 + 10*b*e^2*m^3 + 10*b*e^2*m^2 + 5*b*e^2*m + 2*(b*e^2*m^2 + 2*b*e^2*m + b*e^2)*r^3 +
b*e^2 + 5*(b*e^2*m^3 + 3*b*e^2*m^2 + 3*b*e^2*m + b*e^2)*r^2 + 4*(b*e^2*m^4 + 4*b*e^2*m^3 + 6*b*e^2*m^2 + 4*b*
e^2*m + b*e^2)*r)*x*log(c) + (2*(b*e^2*m^2 + 2*b*e^2*m + b*e^2)*n*r^3 + 5*(b*e^2*m^3 + 3*b*e^2*m^2 + 3*b*e^2*m
+ b*e^2)*n*r^2 + 4*(b*e^2*m^4 + 4*b*e^2*m^3 + 6*b*e^2*m^2 + 4*b*e^2*m + b*e^2)*n*r + (b*e^2*m^5 + 5*b*e^2*m^4
+ 10*b*e^2*m^3 + 10*b*e^2*m^2 + 5*b*e^2*m + b*e^2)*n)*x*log(x) + (a*e^2*m^5 + 5*a*e^2*m^4 + 10*a*e^2*m^3 + 10
*a*e^2*m^2 + 5*a*e^2*m + 2*(a*e^2*m^2 + 2*a*e^2*m + a*e^2)*r^3 + a*e^2 + (5*a*e^2*m^3 + 15*a*e^2*m^2 + 15*a*e^
2*m + 5*a*e^2 - (b*e^2*m^2 + 2*b*e^2*m + b*e^2)*n)*r^2 - (b*e^2*m^4 + 4*b*e^2*m^3 + 6*b*e^2*m^2 + 4*b*e^2*m +
b*e^2)*n + 2*(2*a*e^2*m^4 + 8*a*e^2*m^3 + 12*a*e^2*m^2 + 8*a*e^2*m + 2*a*e^2 - (b*e^2*m^3 + 3*b*e^2*m^2 + 3*b*
e^2*m + b*e^2)*n)*r)*x)*x^(2*r)*e^(m*log(f) + m*log(x)) + 2*((b*d*e*m^5 + 5*b*d*e*m^4 + 10*b*d*e*m^3 + 10*b*d*
e*m^2 + 5*b*d*e*m + 4*(b*d*e*m^2 + 2*b*d*e*m + b*d*e)*r^3 + b*d*e + 8*(b*d*e*m^3 + 3*b*d*e*m^2 + 3*b*d*e*m + b
*d*e)*r^2 + 5*(b*d*e*m^4 + 4*b*d*e*m^3 + 6*b*d*e*m^2 + 4*b*d*e*m + b*d*e)*r)*x*log(c) + (4*(b*d*e*m^2 + 2*b*d*
e*m + b*d*e)*n*r^3 + 8*(b*d*e*m^3 + 3*b*d*e*m^2 + 3*b*d*e*m + b*d*e)*n*r^2 + 5*(b*d*e*m^4 + 4*b*d*e*m^3 + 6*b*
d*e*m^2 + 4*b*d*e*m + b*d*e)*n*r + (b*d*e*m^5 + 5*b*d*e*m^4 + 10*b*d*e*m^3 + 10*b*d*e*m^2 + 5*b*d*e*m + b*d*e)
*n)*x*log(x) + (a*d*e*m^5 + 5*a*d*e*m^4 + 10*a*d*e*m^3 + 10*a*d*e*m^2 + 5*a*d*e*m + 4*(a*d*e*m^2 + 2*a*d*e*m +
a*d*e)*r^3 + a*d*e + 4*(2*a*d*e*m^3 + 6*a*d*e*m^2 + 6*a*d*e*m + 2*a*d*e - (b*d*e*m^2 + 2*b*d*e*m + b*d*e)*n)*
r^2 - (b*d*e*m^4 + 4*b*d*e*m^3 + 6*b*d*e*m^2 + 4*b*d*e*m + b*d*e)*n + (5*a*d*e*m^4 + 20*a*d*e*m^3 + 30*a*d*e*m
^2 + 20*a*d*e*m + 5*a*d*e - 4*(b*d*e*m^3 + 3*b*d*e*m^2 + 3*b*d*e*m + b*d*e)*n)*r)*x)*x^r*e^(m*log(f) + m*log(x
)) + ((b*d^2*m^5 + 5*b*d^2*m^4 + 10*b*d^2*m^3 + 10*b*d^2*m^2 + 4*(b*d^2*m + b*d^2)*r^4 + 5*b*d^2*m + 12*(b*d^2
*m^2 + 2*b*d^2*m + b*d^2)*r^3 + b*d^2 + 13*(b*d^2*m^3 + 3*b*d^2*m^2 + 3*b*d^2*m + b*d^2)*r^2 + 6*(b*d^2*m^4 +
4*b*d^2*m^3 + 6*b*d^2*m^2 + 4*b*d^2*m + b*d^2)*r)*x*log(c) + (4*(b*d^2*m + b*d^2)*n*r^4 + 12*(b*d^2*m^2 + 2*b*
d^2*m + b*d^2)*n*r^3 + 13*(b*d^2*m^3 + 3*b*d^2*m^2 + 3*b*d^2*m + b*d^2)*n*r^2 + 6*(b*d^2*m^4 + 4*b*d^2*m^3 + 6
*b*d^2*m^2 + 4*b*d^2*m + b*d^2)*n*r + (b*d^2*m^5 + 5*b*d^2*m^4 + 10*b*d^2*m^3 + 10*b*d^2*m^2 + 5*b*d^2*m + b*d
^2)*n)*x*log(x) + (a*d^2*m^5 + 5*a*d^2*m^4 + 10*a*d^2*m^3 + 10*a*d^2*m^2 + 4*(a*d^2*m - b*d^2*n + a*d^2)*r^4 +
5*a*d^2*m + 12*(a*d^2*m^2 + 2*a*d^2*m + a*d^2 - (b*d^2*m + b*d^2)*n)*r^3 + a*d^2 + 13*(a*d^2*m^3 + 3*a*d^2*m^
2 + 3*a*d^2*m + a*d^2 - (b*d^2*m^2 + 2*b*d^2*m + b*d^2)*n)*r^2 - (b*d^2*m^4 + 4*b*d^2*m^3 + 6*b*d^2*m^2 + 4*b*
d^2*m + b*d^2)*n + 6*(a*d^2*m^4 + 4*a*d^2*m^3 + 6*a*d^2*m^2 + 4*a*d^2*m + a*d^2 - (b*d^2*m^3 + 3*b*d^2*m^2 + 3
*b*d^2*m + b*d^2)*n)*r)*x)*e^(m*log(f) + m*log(x)))/(m^6 + 6*m^5 + 4*(m^2 + 2*m + 1)*r^4 + 15*m^4 + 12*(m^3 +
3*m^2 + 3*m + 1)*r^3 + 20*m^3 + 13*(m^4 + 4*m^3 + 6*m^2 + 4*m + 1)*r^2 + 15*m^2 + 6*(m^5 + 5*m^4 + 10*m^3 + 10
*m^2 + 5*m + 1)*r + 6*m + 1)

________________________________________________________________________________________

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)**m*(d+e*x**r)**2*(a+b*ln(c*x**n)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.37214, size = 713, normalized size = 4.32 \begin{align*} \frac{2 \, b d f^{m} m n x x^{m} x^{r} e \log \left (x\right )}{m^{2} + 2 \, m r + r^{2} + 2 \, m + 2 \, r + 1} + \frac{2 \, b d f^{m} n r x x^{m} x^{r} e \log \left (x\right )}{m^{2} + 2 \, m r + r^{2} + 2 \, m + 2 \, r + 1} + \frac{b d^{2} f^{m} m n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} + \frac{b f^{m} m n x x^{m} x^{2 \, r} e^{2} \log \left (x\right )}{m^{2} + 4 \, m r + 4 \, r^{2} + 2 \, m + 4 \, r + 1} + \frac{2 \, b f^{m} n r x x^{m} x^{2 \, r} e^{2} \log \left (x\right )}{m^{2} + 4 \, m r + 4 \, r^{2} + 2 \, m + 4 \, r + 1} + \frac{2 \, b d f^{m} n x x^{m} x^{r} e \log \left (x\right )}{m^{2} + 2 \, m r + r^{2} + 2 \, m + 2 \, r + 1} - \frac{2 \, b d f^{m} n x x^{m} x^{r} e}{m^{2} + 2 \, m r + r^{2} + 2 \, m + 2 \, r + 1} + \frac{2 \, b d f^{m} x x^{m} x^{r} e \log \left (c\right )}{m + r + 1} + \frac{b d^{2} f^{m} n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} + \frac{b f^{m} n x x^{m} x^{2 \, r} e^{2} \log \left (x\right )}{m^{2} + 4 \, m r + 4 \, r^{2} + 2 \, m + 4 \, r + 1} - \frac{b d^{2} f^{m} n x x^{m}}{m^{2} + 2 \, m + 1} - \frac{b f^{m} n x x^{m} x^{2 \, r} e^{2}}{m^{2} + 4 \, m r + 4 \, r^{2} + 2 \, m + 4 \, r + 1} + \frac{2 \, a d f^{m} x x^{m} x^{r} e}{m + r + 1} + \frac{b f^{m} x x^{m} x^{2 \, r} e^{2} \log \left (c\right )}{m + 2 \, r + 1} + \frac{a f^{m} x x^{m} x^{2 \, r} e^{2}}{m + 2 \, r + 1} + \frac{\left (f x\right )^{m} b d^{2} x \log \left (c\right )}{m + 1} + \frac{\left (f x\right )^{m} a d^{2} x}{m + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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