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

Optimal. Leaf size=153 \[ \frac{d^2 (f x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{f (m+1)}+\frac{2 d e (f x)^{m+2} \left (a+b \log \left (c x^n\right )\right )}{f^2 (m+2)}+\frac{e^2 (f x)^{m+3} \left (a+b \log \left (c x^n\right )\right )}{f^3 (m+3)}-\frac{b d^2 n (f x)^{m+1}}{f (m+1)^2}-\frac{2 b d e n (f x)^{m+2}}{f^2 (m+2)^2}-\frac{b e^2 n (f x)^{m+3}}{f^3 (m+3)^2} \]

[Out]

-((b*d^2*n*(f*x)^(1 + m))/(f*(1 + m)^2)) - (2*b*d*e*n*(f*x)^(2 + m))/(f^2*(2 + m)^2) - (b*e^2*n*(f*x)^(3 + m))
/(f^3*(3 + m)^2) + (d^2*(f*x)^(1 + m)*(a + b*Log[c*x^n]))/(f*(1 + m)) + (2*d*e*(f*x)^(2 + m)*(a + b*Log[c*x^n]
))/(f^2*(2 + m)) + (e^2*(f*x)^(3 + m)*(a + b*Log[c*x^n]))/(f^3*(3 + m))

________________________________________________________________________________________

Rubi [A]  time = 0.17256, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {43, 2350, 12, 14} \[ \frac{d^2 (f x)^{m+1} \left (a+b \log \left (c x^n\right )\right )}{f (m+1)}+\frac{2 d e (f x)^{m+2} \left (a+b \log \left (c x^n\right )\right )}{f^2 (m+2)}+\frac{e^2 (f x)^{m+3} \left (a+b \log \left (c x^n\right )\right )}{f^3 (m+3)}-\frac{b d^2 n (f x)^{m+1}}{f (m+1)^2}-\frac{2 b d e n (f x)^{m+2}}{f^2 (m+2)^2}-\frac{b e^2 n (f x)^{m+3}}{f^3 (m+3)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*d^2*n*(f*x)^(1 + m))/(f*(1 + m)^2)) - (2*b*d*e*n*(f*x)^(2 + m))/(f^2*(2 + m)^2) - (b*e^2*n*(f*x)^(3 + m))
/(f^3*(3 + m)^2) + (d^2*(f*x)^(1 + m)*(a + b*Log[c*x^n]))/(f*(1 + m)) + (2*d*e*(f*x)^(2 + m)*(a + b*Log[c*x^n]
))/(f^2*(2 + m)) + (e^2*(f*x)^(3 + m)*(a + b*Log[c*x^n]))/(f^3*(3 + m))

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])

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 (d+e x)^2 \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{d^2 (f x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{f (1+m)}+\frac{2 d e (f x)^{2+m} \left (a+b \log \left (c x^n\right )\right )}{f^2 (2+m)}+\frac{e^2 (f x)^{3+m} \left (a+b \log \left (c x^n\right )\right )}{f^3 (3+m)}-(b n) \int \frac{(f x)^m \left (d^2 (2+m) (3+m)+2 d e (1+m) (3+m) x+e^2 (1+m) (2+m) x^2\right )}{(1+m) (2+m) (3+m)} \, dx\\ &=\frac{d^2 (f x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{f (1+m)}+\frac{2 d e (f x)^{2+m} \left (a+b \log \left (c x^n\right )\right )}{f^2 (2+m)}+\frac{e^2 (f x)^{3+m} \left (a+b \log \left (c x^n\right )\right )}{f^3 (3+m)}-\frac{(b n) \int (f x)^m \left (d^2 (2+m) (3+m)+2 d e (1+m) (3+m) x+e^2 (1+m) (2+m) x^2\right ) \, dx}{6+11 m+6 m^2+m^3}\\ &=\frac{d^2 (f x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{f (1+m)}+\frac{2 d e (f x)^{2+m} \left (a+b \log \left (c x^n\right )\right )}{f^2 (2+m)}+\frac{e^2 (f x)^{3+m} \left (a+b \log \left (c x^n\right )\right )}{f^3 (3+m)}-\frac{(b n) \int \left (d^2 (2+m) (3+m) (f x)^m+\frac{2 d e (1+m) (3+m) (f x)^{1+m}}{f}+\frac{e^2 (1+m) (2+m) (f x)^{2+m}}{f^2}\right ) \, dx}{6+11 m+6 m^2+m^3}\\ &=-\frac{b d^2 n (f x)^{1+m}}{f (1+m)^2}-\frac{2 b d e n (f x)^{2+m}}{f^2 (2+m)^2}-\frac{b e^2 n (f x)^{3+m}}{f^3 (3+m)^2}+\frac{d^2 (f x)^{1+m} \left (a+b \log \left (c x^n\right )\right )}{f (1+m)}+\frac{2 d e (f x)^{2+m} \left (a+b \log \left (c x^n\right )\right )}{f^2 (2+m)}+\frac{e^2 (f x)^{3+m} \left (a+b \log \left (c x^n\right )\right )}{f^3 (3+m)}\\ \end{align*}

Mathematica [A]  time = 0.129306, size = 108, normalized size = 0.71 \[ x (f x)^m \left (\frac{d^2 \left (a+b \log \left (c x^n\right )\right )}{m+1}+\frac{2 d e x \left (a+b \log \left (c x^n\right )\right )}{m+2}+\frac{e^2 x^2 \left (a+b \log \left (c x^n\right )\right )}{m+3}-\frac{b d^2 n}{(m+1)^2}-\frac{2 b d e n x}{(m+2)^2}-\frac{b e^2 n x^2}{(m+3)^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(f*x)^m*(d + e*x)^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)/(2 + m)^2 - (b*e^2*n*x^2)/(3 + m)^2 + (d^2*(a + b*Log[c*x^n]
))/(1 + m) + (2*d*e*x*(a + b*Log[c*x^n]))/(2 + m) + (e^2*x^2*(a + b*Log[c*x^n]))/(3 + m))

________________________________________________________________________________________

Maple [C]  time = 0.276, size = 2702, normalized size = 17.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x*(e^2*m^2*x^2+2*d*e*m^2*x+3*e^2*m*x^2+d^2*m^2+8*d*e*m*x+2*e^2*x^2+5*d^2*m+6*d*e*x+6*d^2)/(1+m)/(2+m)/(3+m)*
exp(1/2*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)+1/2*x*(72*ln(c)*b*d^2+12*I*Pi*b*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+
12*I*Pi*b*e^2*x^2*csgn(I*c*x^n)^2*csgn(I*c)+96*I*Pi*b*d^2*m*csgn(I*x^n)*csgn(I*c*x^n)^2+72*a*d^2+72*a*d*e*x+2*
a*e^2*m^5*x^2-2*b*d^2*m^4*n+228*a*d*e*m*x+4*a*d*e*m^5*x-2*b*e^2*m^4*n*x^2+62*ln(c)*b*e^2*m^3*x^2+102*ln(c)*b*e
^2*m^2*x^2+80*ln(c)*b*e^2*m*x^2+2*ln(c)*b*e^2*m^5*x^2+18*ln(c)*b*e^2*m^4*x^2+24*a*e^2*x^2-12*b*e^2*m^3*n*x^2+4
0*a*d*e*m^4*x-74*b*d^2*m^2*n-120*b*d^2*m*n+18*a*e^2*m^4*x^2-20*b*d^2*m^3*n-72*b*d^2*n+2*ln(c)*b*d^2*m^5+22*ln(
c)*b*d^2*m^4+94*ln(c)*b*d^2*m^3+194*ln(c)*b*d^2*m^2+192*ln(c)*b*d^2*m-51*I*Pi*b*e^2*m^2*x^2*csgn(I*c*x^n)^3+47
*I*Pi*b*d^2*m^3*csgn(I*x^n)*csgn(I*c*x^n)^2+47*I*Pi*b*d^2*m^3*csgn(I*c*x^n)^2*csgn(I*c)-36*b*d*e*n*x-76*I*Pi*b
*d*e*m^3*x*csgn(I*c*x^n)^3+51*I*Pi*b*e^2*m^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+51*I*Pi*b*e^2*m^2*x^2*csgn(I*c*x^
n)^2*csgn(I*c)-2*I*Pi*b*d*e*m^5*x*csgn(I*c*x^n)^3+22*a*d^2*m^4-I*Pi*b*d^2*m^5*csgn(I*c*x^n)^3+9*I*Pi*b*e^2*m^4
*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+9*I*Pi*b*e^2*m^4*x^2*csgn(I*c*x^n)^2*csgn(I*c)+76*I*Pi*b*d*e*m^3*x*csgn(I*x^n
)*csgn(I*c*x^n)^2+76*I*Pi*b*d*e*m^3*x*csgn(I*c*x^n)^2*csgn(I*c)+94*a*d^2*m^3+194*a*d^2*m^2+192*a*d^2*m-32*b*d*
e*m^3*n*x-4*b*d*e*m^4*n*x+72*ln(c)*b*d*e*x-36*I*Pi*b*d^2*csgn(I*c*x^n)^3+24*ln(c)*b*e^2*x^2-40*I*Pi*b*e^2*m*x^
2*csgn(I*c*x^n)^3+97*I*Pi*b*d^2*m^2*csgn(I*x^n)*csgn(I*c*x^n)^2+11*I*Pi*b*d^2*m^4*csgn(I*x^n)*csgn(I*c*x^n)^2+
11*I*Pi*b*d^2*m^4*csgn(I*c*x^n)^2*csgn(I*c)-26*b*e^2*m^2*n*x^2-88*b*d*e*m^2*n*x-24*b*e^2*m*n*x^2-96*b*d*e*m*n*
x-12*I*Pi*b*e^2*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-96*I*Pi*b*d^2*m*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+I*
Pi*b*e^2*m^5*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+I*Pi*b*e^2*m^5*x^2*csgn(I*c*x^n)^2*csgn(I*c)-97*I*Pi*b*d^2*m^2*cs
gn(I*c*x^n)^3-31*I*Pi*b*e^2*m^3*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+2*a*d^2*m^5-51*I*Pi*b*e^2*m^2*x^2*csgn
(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+136*I*Pi*b*d*e*m^2*x*csgn(I*x^n)*csgn(I*c*x^n)^2+136*I*Pi*b*d*e*m^2*x*csgn(I*c
*x^n)^2*csgn(I*c)-36*I*Pi*b*d*e*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+272*a*d*e*m^2*x-I*Pi*b*d^2*m^5*csgn(I*x^
n)*csgn(I*c*x^n)*csgn(I*c)-20*I*Pi*b*d*e*m^4*x*csgn(I*c*x^n)^3-I*Pi*b*e^2*m^5*x^2*csgn(I*c*x^n)^3-9*I*Pi*b*e^2
*m^4*x^2*csgn(I*c*x^n)^3+I*Pi*b*d^2*m^5*csgn(I*x^n)*csgn(I*c*x^n)^2+I*Pi*b*d^2*m^5*csgn(I*c*x^n)^2*csgn(I*c)+9
7*I*Pi*b*d^2*m^2*csgn(I*c*x^n)^2*csgn(I*c)-114*I*Pi*b*d*e*m*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-2*I*Pi*b*d*e
*m^5*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-20*I*Pi*b*d*e*m^4*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+80*a*e^2*m*
x^2+152*a*d*e*m^3*x+36*I*Pi*b*d*e*x*csgn(I*x^n)*csgn(I*c*x^n)^2-47*I*Pi*b*d^2*m^3*csgn(I*x^n)*csgn(I*c*x^n)*cs
gn(I*c)-136*I*Pi*b*d*e*m^2*x*csgn(I*c*x^n)^3-9*I*Pi*b*e^2*m^4*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+20*I*Pi*
b*d*e*m^4*x*csgn(I*x^n)*csgn(I*c*x^n)^2+20*I*Pi*b*d*e*m^4*x*csgn(I*c*x^n)^2*csgn(I*c)-40*I*Pi*b*e^2*m*x^2*csgn
(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+114*I*Pi*b*d*e*m*x*csgn(I*x^n)*csgn(I*c*x^n)^2+114*I*Pi*b*d*e*m*x*csgn(I*c*x^n
)^2*csgn(I*c)+62*a*e^2*m^3*x^2+102*a*e^2*m^2*x^2-11*I*Pi*b*d^2*m^4*csgn(I*c*x^n)^3+152*ln(c)*b*d*e*m^3*x+272*l
n(c)*b*d*e*m^2*x+228*ln(c)*b*d*e*m*x+40*ln(c)*b*d*e*m^4*x+4*ln(c)*b*d*e*m^5*x-97*I*Pi*b*d^2*m^2*csgn(I*x^n)*cs
gn(I*c*x^n)*csgn(I*c)-114*I*Pi*b*d*e*m*x*csgn(I*c*x^n)^3-I*Pi*b*e^2*m^5*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c
)+2*I*Pi*b*d*e*m^5*x*csgn(I*x^n)*csgn(I*c*x^n)^2+2*I*Pi*b*d*e*m^5*x*csgn(I*c*x^n)^2*csgn(I*c)-76*I*Pi*b*d*e*m^
3*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-136*I*Pi*b*d*e*m^2*x*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-8*b*e^2*n*x^2
+96*I*Pi*b*d^2*m*csgn(I*c*x^n)^2*csgn(I*c)-36*I*Pi*b*d*e*x*csgn(I*c*x^n)^3-36*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*
x^n)*csgn(I*c)-31*I*Pi*b*e^2*m^3*x^2*csgn(I*c*x^n)^3-12*I*Pi*b*e^2*x^2*csgn(I*c*x^n)^3-96*I*Pi*b*d^2*m*csgn(I*
c*x^n)^3+36*I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c*x^n)^2+36*I*Pi*b*d^2*csgn(I*c*x^n)^2*csgn(I*c)-47*I*Pi*b*d^2*m^3*c
sgn(I*c*x^n)^3-11*I*Pi*b*d^2*m^4*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+36*I*Pi*b*d*e*x*csgn(I*c*x^n)^2*csgn(I*c)
+40*I*Pi*b*e^2*m*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+40*I*Pi*b*e^2*m*x^2*csgn(I*c*x^n)^2*csgn(I*c)+31*I*Pi*b*e^2*m
^3*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2+31*I*Pi*b*e^2*m^3*x^2*csgn(I*c*x^n)^2*csgn(I*c))/(3+m)^2/(1+m)^2/(2+m)^2*ex
p(1/2*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)^m*(e*x+d)^2*(a+b*log(c*x^n)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.33504, size = 1474, normalized size = 9.63 \begin{align*} \frac{{\left ({\left (a e^{2} m^{5} + 9 \, a e^{2} m^{4} + 31 \, a e^{2} m^{3} + 51 \, a e^{2} m^{2} + 40 \, a e^{2} m + 12 \, a e^{2} -{\left (b e^{2} m^{4} + 6 \, b e^{2} m^{3} + 13 \, b e^{2} m^{2} + 12 \, b e^{2} m + 4 \, b e^{2}\right )} n\right )} x^{3} + 2 \,{\left (a d e m^{5} + 10 \, a d e m^{4} + 38 \, a d e m^{3} + 68 \, a d e m^{2} + 57 \, a d e m + 18 \, a d e -{\left (b d e m^{4} + 8 \, b d e m^{3} + 22 \, b d e m^{2} + 24 \, b d e m + 9 \, b d e\right )} n\right )} x^{2} +{\left (a d^{2} m^{5} + 11 \, a d^{2} m^{4} + 47 \, a d^{2} m^{3} + 97 \, a d^{2} m^{2} + 96 \, a d^{2} m + 36 \, a d^{2} -{\left (b d^{2} m^{4} + 10 \, b d^{2} m^{3} + 37 \, b d^{2} m^{2} + 60 \, b d^{2} m + 36 \, b d^{2}\right )} n\right )} x +{\left ({\left (b e^{2} m^{5} + 9 \, b e^{2} m^{4} + 31 \, b e^{2} m^{3} + 51 \, b e^{2} m^{2} + 40 \, b e^{2} m + 12 \, b e^{2}\right )} x^{3} + 2 \,{\left (b d e m^{5} + 10 \, b d e m^{4} + 38 \, b d e m^{3} + 68 \, b d e m^{2} + 57 \, b d e m + 18 \, b d e\right )} x^{2} +{\left (b d^{2} m^{5} + 11 \, b d^{2} m^{4} + 47 \, b d^{2} m^{3} + 97 \, b d^{2} m^{2} + 96 \, b d^{2} m + 36 \, b d^{2}\right )} x\right )} \log \left (c\right ) +{\left ({\left (b e^{2} m^{5} + 9 \, b e^{2} m^{4} + 31 \, b e^{2} m^{3} + 51 \, b e^{2} m^{2} + 40 \, b e^{2} m + 12 \, b e^{2}\right )} n x^{3} + 2 \,{\left (b d e m^{5} + 10 \, b d e m^{4} + 38 \, b d e m^{3} + 68 \, b d e m^{2} + 57 \, b d e m + 18 \, b d e\right )} n x^{2} +{\left (b d^{2} m^{5} + 11 \, b d^{2} m^{4} + 47 \, b d^{2} m^{3} + 97 \, b d^{2} m^{2} + 96 \, b d^{2} m + 36 \, b d^{2}\right )} n x\right )} \log \left (x\right )\right )} e^{\left (m \log \left (f\right ) + m \log \left (x\right )\right )}}{m^{6} + 12 \, m^{5} + 58 \, m^{4} + 144 \, m^{3} + 193 \, m^{2} + 132 \, m + 36} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*e^2*m^5 + 9*a*e^2*m^4 + 31*a*e^2*m^3 + 51*a*e^2*m^2 + 40*a*e^2*m + 12*a*e^2 - (b*e^2*m^4 + 6*b*e^2*m^3 + 1
3*b*e^2*m^2 + 12*b*e^2*m + 4*b*e^2)*n)*x^3 + 2*(a*d*e*m^5 + 10*a*d*e*m^4 + 38*a*d*e*m^3 + 68*a*d*e*m^2 + 57*a*
d*e*m + 18*a*d*e - (b*d*e*m^4 + 8*b*d*e*m^3 + 22*b*d*e*m^2 + 24*b*d*e*m + 9*b*d*e)*n)*x^2 + (a*d^2*m^5 + 11*a*
d^2*m^4 + 47*a*d^2*m^3 + 97*a*d^2*m^2 + 96*a*d^2*m + 36*a*d^2 - (b*d^2*m^4 + 10*b*d^2*m^3 + 37*b*d^2*m^2 + 60*
b*d^2*m + 36*b*d^2)*n)*x + ((b*e^2*m^5 + 9*b*e^2*m^4 + 31*b*e^2*m^3 + 51*b*e^2*m^2 + 40*b*e^2*m + 12*b*e^2)*x^
3 + 2*(b*d*e*m^5 + 10*b*d*e*m^4 + 38*b*d*e*m^3 + 68*b*d*e*m^2 + 57*b*d*e*m + 18*b*d*e)*x^2 + (b*d^2*m^5 + 11*b
*d^2*m^4 + 47*b*d^2*m^3 + 97*b*d^2*m^2 + 96*b*d^2*m + 36*b*d^2)*x)*log(c) + ((b*e^2*m^5 + 9*b*e^2*m^4 + 31*b*e
^2*m^3 + 51*b*e^2*m^2 + 40*b*e^2*m + 12*b*e^2)*n*x^3 + 2*(b*d*e*m^5 + 10*b*d*e*m^4 + 38*b*d*e*m^3 + 68*b*d*e*m
^2 + 57*b*d*e*m + 18*b*d*e)*n*x^2 + (b*d^2*m^5 + 11*b*d^2*m^4 + 47*b*d^2*m^3 + 97*b*d^2*m^2 + 96*b*d^2*m + 36*
b*d^2)*n*x)*log(x))*e^(m*log(f) + m*log(x))/(m^6 + 12*m^5 + 58*m^4 + 144*m^3 + 193*m^2 + 132*m + 36)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

________________________________________________________________________________________

Giac [B]  time = 1.34403, size = 505, normalized size = 3.3 \begin{align*} \frac{b f^{2} f^{m} x^{3} x^{m} e^{2} \log \left (c\right )}{f^{2} m + 3 \, f^{2}} + \frac{b f^{m} m n x^{3} x^{m} e^{2} \log \left (x\right )}{m^{2} + 6 \, m + 9} + \frac{2 \, b d f^{m} m n x^{2} x^{m} e \log \left (x\right )}{m^{2} + 4 \, m + 4} + \frac{a f^{2} f^{m} x^{3} x^{m} e^{2}}{f^{2} m + 3 \, f^{2}} + \frac{b d^{2} f^{m} m n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} + \frac{3 \, b f^{m} n x^{3} x^{m} e^{2} \log \left (x\right )}{m^{2} + 6 \, m + 9} + \frac{4 \, b d f^{m} n x^{2} x^{m} e \log \left (x\right )}{m^{2} + 4 \, m + 4} - \frac{b f^{m} n x^{3} x^{m} e^{2}}{m^{2} + 6 \, m + 9} - \frac{2 \, b d f^{m} n x^{2} x^{m} e}{m^{2} + 4 \, m + 4} + \frac{2 \, b d f^{m} x^{2} x^{m} e \log \left (c\right )}{m + 2} + \frac{b d^{2} f^{m} n x x^{m} \log \left (x\right )}{m^{2} + 2 \, m + 1} - \frac{b d^{2} f^{m} n x x^{m}}{m^{2} + 2 \, m + 1} + \frac{2 \, a d f^{m} x^{2} x^{m} e}{m + 2} + \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*(e*x+d)^2*(a+b*log(c*x^n)),x, algorithm="giac")

[Out]

b*f^2*f^m*x^3*x^m*e^2*log(c)/(f^2*m + 3*f^2) + b*f^m*m*n*x^3*x^m*e^2*log(x)/(m^2 + 6*m + 9) + 2*b*d*f^m*m*n*x^
2*x^m*e*log(x)/(m^2 + 4*m + 4) + a*f^2*f^m*x^3*x^m*e^2/(f^2*m + 3*f^2) + b*d^2*f^m*m*n*x*x^m*log(x)/(m^2 + 2*m
+ 1) + 3*b*f^m*n*x^3*x^m*e^2*log(x)/(m^2 + 6*m + 9) + 4*b*d*f^m*n*x^2*x^m*e*log(x)/(m^2 + 4*m + 4) - b*f^m*n*
x^3*x^m*e^2/(m^2 + 6*m + 9) - 2*b*d*f^m*n*x^2*x^m*e/(m^2 + 4*m + 4) + 2*b*d*f^m*x^2*x^m*e*log(c)/(m + 2) + b*d
^2*f^m*n*x*x^m*log(x)/(m^2 + 2*m + 1) - b*d^2*f^m*n*x*x^m/(m^2 + 2*m + 1) + 2*a*d*f^m*x^2*x^m*e/(m + 2) + (f*x
)^m*b*d^2*x*log(c)/(m + 1) + (f*x)^m*a*d^2*x/(m + 1)