3.361 \(\int (f x)^{-1+m} (d+e x^m) (a+b \log (c x^n))^2 \, dx\)

Optimal. Leaf size=226 \[ -\frac{b d^2 n x^{1-m} \log (x) (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{e m}+\frac{x^{1-m} (f x)^{m-1} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e m}-\frac{2 b d n x (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{m^2}-\frac{b e n x^{m+1} (f x)^{m-1} \left (a+b \log \left (c x^n\right )\right )}{2 m^2}+\frac{b^2 d^2 n^2 x^{1-m} \log ^2(x) (f x)^{m-1}}{2 e m}+\frac{2 b^2 d n^2 x (f x)^{m-1}}{m^3}+\frac{b^2 e n^2 x^{m+1} (f x)^{m-1}}{4 m^3} \]

[Out]

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

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Rubi [A]  time = 0.302762, antiderivative size = 195, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2339, 2338, 266, 43, 2334, 12, 14, 2301} \[ -\frac{b n x^{1-m} (f x)^{m-1} \left (2 d^2 \log (x)+\frac{4 d e x^m}{m}+\frac{e^2 x^{2 m}}{m}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e m}+\frac{x^{1-m} (f x)^{m-1} \left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )^2}{2 e m}+\frac{b^2 d^2 n^2 x^{1-m} \log ^2(x) (f x)^{m-1}}{2 e m}+\frac{2 b^2 d n^2 x (f x)^{m-1}}{m^3}+\frac{b^2 e n^2 x^{m+1} (f x)^{m-1}}{4 m^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

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]]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps

\begin{align*} \int (f x)^{-1+m} \left (d+e x^m\right ) \left (a+b \log \left (c x^n\right )\right )^2 \, dx &=\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 )^2 \, 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}{2 e m}-\frac{\left (b n x^{1-m} (f x)^{-1+m}\right ) \int \frac{\left (d+e x^m\right )^2 \left (a+b \log \left (c x^n\right )\right )}{x} \, dx}{e m}\\ &=-\frac{b n x^{1-m} (f x)^{-1+m} \left (\frac{4 d e x^m}{m}+\frac{e^2 x^{2 m}}{m}+2 d^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )}{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}{2 e m}+\frac{\left (b^2 n^2 x^{1-m} (f x)^{-1+m}\right ) \int \frac{e x^m \left (4 d+e x^m\right )+2 d^2 m \log (x)}{2 m x} \, dx}{e m}\\ &=-\frac{b n x^{1-m} (f x)^{-1+m} \left (\frac{4 d e x^m}{m}+\frac{e^2 x^{2 m}}{m}+2 d^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )}{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}{2 e m}+\frac{\left (b^2 n^2 x^{1-m} (f x)^{-1+m}\right ) \int \frac{e x^m \left (4 d+e x^m\right )+2 d^2 m \log (x)}{x} \, dx}{2 e m^2}\\ &=-\frac{b n x^{1-m} (f x)^{-1+m} \left (\frac{4 d e x^m}{m}+\frac{e^2 x^{2 m}}{m}+2 d^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )}{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}{2 e m}+\frac{\left (b^2 n^2 x^{1-m} (f x)^{-1+m}\right ) \int \left (4 d e x^{-1+m}+e^2 x^{-1+2 m}+\frac{2 d^2 m \log (x)}{x}\right ) \, dx}{2 e m^2}\\ &=\frac{2 b^2 d n^2 x (f x)^{-1+m}}{m^3}+\frac{b^2 e n^2 x^{1+m} (f x)^{-1+m}}{4 m^3}-\frac{b n x^{1-m} (f x)^{-1+m} \left (\frac{4 d e x^m}{m}+\frac{e^2 x^{2 m}}{m}+2 d^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )}{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}{2 e m}+\frac{\left (b^2 d^2 n^2 x^{1-m} (f x)^{-1+m}\right ) \int \frac{\log (x)}{x} \, dx}{e m}\\ &=\frac{2 b^2 d n^2 x (f x)^{-1+m}}{m^3}+\frac{b^2 e n^2 x^{1+m} (f x)^{-1+m}}{4 m^3}+\frac{b^2 d^2 n^2 x^{1-m} (f x)^{-1+m} \log ^2(x)}{2 e m}-\frac{b n x^{1-m} (f x)^{-1+m} \left (\frac{4 d e x^m}{m}+\frac{e^2 x^{2 m}}{m}+2 d^2 \log (x)\right ) \left (a+b \log \left (c x^n\right )\right )}{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}{2 e m}\\ \end{align*}

Mathematica [A]  time = 0.13111, size = 125, normalized size = 0.55 \[ \frac{(f x)^m \left (2 a^2 m^2 \left (2 d+e x^m\right )-2 b m \log \left (c x^n\right ) \left (b n \left (4 d+e x^m\right )-2 a m \left (2 d+e x^m\right )\right )-2 a b m n \left (4 d+e x^m\right )+2 b^2 m^2 \log ^2\left (c x^n\right ) \left (2 d+e x^m\right )+b^2 n^2 \left (8 d+e x^m\right )\right )}{4 f m^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((f*x)^m*(2*a^2*m^2*(2*d + e*x^m) - 2*a*b*m*n*(4*d + e*x^m) + b^2*n^2*(8*d + e*x^m) - 2*b*m*(-2*a*m*(2*d + e*x
^m) + b*n*(4*d + e*x^m))*Log[c*x^n] + 2*b^2*m^2*(2*d + e*x^m)*Log[c*x^n]^2))/(4*f*m^3)

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Maple [C]  time = 0.291, size = 1920, normalized size = 8.5 \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)*(a+b*ln(c*x^n))^2,x)

[Out]

1/2*b^2*(e*x^m+2*d)*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*cs
gn(I*x)+I*Pi*csgn(I*f*x)*csgn(I*f)*csgn(I*x)-2*ln(f)-2*ln(x)))*ln(x^n)^2+1/2*b*(I*Pi*b*e*csgn(I*x^n)*csgn(I*c*
x^n)^2*x^m*m-I*Pi*b*e*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^m*m-I*Pi*b*e*csgn(I*c*x^n)^3*x^m*m+I*Pi*b*e*csgn(I
*c*x^n)^2*csgn(I*c)*x^m*m+2*I*Pi*b*d*m*csgn(I*x^n)*csgn(I*c*x^n)^2-2*I*Pi*b*d*m*csgn(I*x^n)*csgn(I*c*x^n)*csgn
(I*c)-2*I*Pi*b*d*m*csgn(I*c*x^n)^3+2*I*Pi*b*d*m*csgn(I*c*x^n)^2*csgn(I*c)+2*ln(c)*b*e*x^m*m+4*ln(c)*b*d*m+2*x^
m*a*e*m-x^m*b*e*n+4*a*d*m-4*b*d*n)*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)))*ln(x^n)+1/8*(8*a^2*d*m^2+2*I*P
i*b^2*e*m*n*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^m-4*I*Pi*ln(c)*b^2*e*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^m
*m^2-4*I*Pi*a*b*e*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*x^m*m^2+8*ln(c)^2*b^2*d*m^2+4*a^2*e*x^m*m^2+2*b^2*e*n^2*
x^m-16*ln(c)*b^2*d*m*n+16*ln(c)*a*b*d*m^2+16*b^2*d*n^2-4*a*b*e*m*n*x^m-4*ln(c)*b^2*e*m*n*x^m+8*ln(c)*a*b*e*x^m
*m^2-16*a*b*d*m*n-4*I*Pi*ln(c)*b^2*e*csgn(I*c*x^n)^3*x^m*m^2+2*Pi^2*b^2*e*csgn(I*x^n)^2*csgn(I*c*x^n)^3*csgn(I
*c)*x^m*m^2-Pi^2*b^2*e*csgn(I*x^n)^2*csgn(I*c*x^n)^2*csgn(I*c)^2*x^m*m^2-4*Pi^2*b^2*e*csgn(I*x^n)*csgn(I*c*x^n
)^4*csgn(I*c)*x^m*m^2-2*Pi^2*b^2*d*csgn(I*c*x^n)^6*m^2+4*ln(c)^2*b^2*e*x^m*m^2-Pi^2*b^2*e*csgn(I*c*x^n)^6*x^m*
m^2-2*Pi^2*b^2*d*csgn(I*x^n)^2*csgn(I*c*x^n)^4*m^2+4*Pi^2*b^2*d*csgn(I*x^n)*csgn(I*c*x^n)^5*m^2+4*Pi^2*b^2*d*c
sgn(I*c*x^n)^5*csgn(I*c)*m^2-2*Pi^2*b^2*d*csgn(I*c*x^n)^4*csgn(I*c)^2*m^2-4*I*Pi*a*b*e*csgn(I*c*x^n)^3*x^m*m^2
+4*I*Pi*ln(c)*b^2*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x^m*m^2+4*I*Pi*ln(c)*b^2*e*csgn(I*c*x^n)^2*csgn(I*c)*x^m*m^2+4
*I*Pi*a*b*e*csgn(I*x^n)*csgn(I*c*x^n)^2*x^m*m^2+2*I*Pi*b^2*e*m*n*csgn(I*c*x^n)^3*x^m+8*I*Pi*ln(c)*b^2*d*csgn(I
*x^n)*csgn(I*c*x^n)^2*m^2+8*I*Pi*ln(c)*b^2*d*csgn(I*c*x^n)^2*csgn(I*c)*m^2-8*I*Pi*a*b*d*csgn(I*x^n)*csgn(I*c*x
^n)*csgn(I*c)*m^2+8*I*Pi*b^2*d*m*n*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-8*I*Pi*ln(c)*b^2*d*csgn(I*x^n)*csgn(I*c
*x^n)*csgn(I*c)*m^2+8*I*Pi*a*b*d*csgn(I*x^n)*csgn(I*c*x^n)^2*m^2+8*I*Pi*a*b*d*csgn(I*c*x^n)^2*csgn(I*c)*m^2-8*
I*Pi*b^2*d*m*n*csgn(I*x^n)*csgn(I*c*x^n)^2-8*I*Pi*b^2*d*m*n*csgn(I*c*x^n)^2*csgn(I*c)+2*Pi^2*b^2*e*csgn(I*c*x^
n)^5*csgn(I*c)*x^m*m^2-Pi^2*b^2*e*csgn(I*c*x^n)^4*csgn(I*c)^2*x^m*m^2-Pi^2*b^2*e*csgn(I*x^n)^2*csgn(I*c*x^n)^4
*x^m*m^2+2*Pi^2*b^2*e*csgn(I*x^n)*csgn(I*c*x^n)^5*x^m*m^2-8*I*Pi*ln(c)*b^2*d*csgn(I*c*x^n)^3*m^2-8*I*Pi*a*b*d*
csgn(I*c*x^n)^3*m^2+8*I*Pi*b^2*d*m*n*csgn(I*c*x^n)^3+2*Pi^2*b^2*e*csgn(I*x^n)*csgn(I*c*x^n)^3*csgn(I*c)^2*x^m*
m^2+4*Pi^2*b^2*d*csgn(I*x^n)^2*csgn(I*c*x^n)^3*csgn(I*c)*m^2-2*Pi^2*b^2*d*csgn(I*x^n)^2*csgn(I*c*x^n)^2*csgn(I
*c)^2*m^2-8*Pi^2*b^2*d*csgn(I*x^n)*csgn(I*c*x^n)^4*csgn(I*c)*m^2+4*Pi^2*b^2*d*csgn(I*x^n)*csgn(I*c*x^n)^3*csgn
(I*c)^2*m^2+4*I*Pi*a*b*e*csgn(I*c*x^n)^2*csgn(I*c)*x^m*m^2-2*I*Pi*b^2*e*m*n*csgn(I*x^n)*csgn(I*c*x^n)^2*x^m-2*
I*Pi*b^2*e*m*n*csgn(I*c*x^n)^2*csgn(I*c)*x^m)*x/m^3*exp(-1/2*(-1+m)*(I*Pi*csgn(I*f*x)^3-I*Pi*csgn(I*f*x)^2*csg
n(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)*(a+b*log(c*x^n))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45839, size = 566, normalized size = 2.5 \begin{align*} \frac{{\left (2 \, b^{2} e m^{2} n^{2} \log \left (x\right )^{2} + 2 \, b^{2} e m^{2} \log \left (c\right )^{2} + 2 \, a^{2} e m^{2} - 2 \, a b e m n + b^{2} e n^{2} + 2 \,{\left (2 \, a b e m^{2} - b^{2} e m n\right )} \log \left (c\right ) + 2 \,{\left (2 \, b^{2} e m^{2} n \log \left (c\right ) + 2 \, a b e m^{2} n - b^{2} e m n^{2}\right )} \log \left (x\right )\right )} f^{m - 1} x^{2 \, m} + 4 \,{\left (b^{2} d m^{2} n^{2} \log \left (x\right )^{2} + b^{2} d m^{2} \log \left (c\right )^{2} + a^{2} d m^{2} - 2 \, a b d m n + 2 \, b^{2} d n^{2} + 2 \,{\left (a b d m^{2} - b^{2} d m n\right )} \log \left (c\right ) + 2 \,{\left (b^{2} d m^{2} n \log \left (c\right ) + a b d m^{2} n - b^{2} d m n^{2}\right )} \log \left (x\right )\right )} f^{m - 1} x^{m}}{4 \, m^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(d+e*x^m)*(a+b*log(c*x^n))^2,x, algorithm="fricas")

[Out]

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

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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)*(a+b*ln(c*x**n))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.42363, size = 601, normalized size = 2.66 \begin{align*} \frac{b^{2} d f^{m} n^{2} x^{m} \log \left (x\right )^{2}}{f m} + \frac{b^{2} f^{m} n^{2} x^{2 \, m} e \log \left (x\right )^{2}}{2 \, f m} + \frac{2 \, b^{2} d f^{m} n x^{m} \log \left (c\right ) \log \left (x\right )}{f m} + \frac{b^{2} f^{m} n x^{2 \, m} e \log \left (c\right ) \log \left (x\right )}{f m} + \frac{b^{2} d f^{m} x^{m} \log \left (c\right )^{2}}{f m} + \frac{b^{2} f^{m} x^{2 \, m} e \log \left (c\right )^{2}}{2 \, f m} + \frac{2 \, a b d f^{m} n x^{m} \log \left (x\right )}{f m} - \frac{2 \, b^{2} d f^{m} n^{2} x^{m} \log \left (x\right )}{f m^{2}} + \frac{a b f^{m} n x^{2 \, m} e \log \left (x\right )}{f m} - \frac{b^{2} f^{m} n^{2} x^{2 \, m} e \log \left (x\right )}{2 \, f m^{2}} + \frac{2 \, a b d f^{m} x^{m} \log \left (c\right )}{f m} - \frac{2 \, b^{2} d f^{m} n x^{m} \log \left (c\right )}{f m^{2}} + \frac{a b f^{m} x^{2 \, m} e \log \left (c\right )}{f m} - \frac{b^{2} f^{m} n x^{2 \, m} e \log \left (c\right )}{2 \, f m^{2}} + \frac{a^{2} d f^{m} x^{m}}{f m} - \frac{2 \, a b d f^{m} n x^{m}}{f m^{2}} + \frac{2 \, b^{2} d f^{m} n^{2} x^{m}}{f m^{3}} + \frac{a^{2} f^{m} x^{2 \, m} e}{2 \, f m} - \frac{a b f^{m} n x^{2 \, m} e}{2 \, f m^{2}} + \frac{b^{2} f^{m} n^{2} x^{2 \, m} e}{4 \, f m^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^(-1+m)*(d+e*x^m)*(a+b*log(c*x^n))^2,x, algorithm="giac")

[Out]

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