\(\int (f x)^m (d+e x^r)^2 (a+b \log (c x^n))^p \, dx\) [449]

Optimal result

Integrand size = 27, antiderivative size = 350 \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\frac {d^2 e^{-\frac {a (1+m)}{b n}} (f x)^{1+m} \left (c x^n\right )^{-\frac {1+m}{n}} \Gamma \left (1+p,-\frac {(1+m) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(1+m) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{f (1+m)}+\frac {2 d e e^{-\frac {a (1+m+r)}{b n}} x^{1+r} (f x)^m \left (c x^n\right )^{-\frac {1+m+r}{n}} \Gamma \left (1+p,-\frac {(1+m+r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(1+m+r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m+r}+\frac {e^2 e^{-\frac {a (1+m+2 r)}{b n}} x^{1+2 r} (f x)^m \left (c x^n\right )^{-\frac {1+m+2 r}{n}} \Gamma \left (1+p,-\frac {(1+m+2 r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(1+m+2 r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m+2 r} \]

[Out]

d^2*(f*x)^(1+m)*GAMMA(p+1,-(1+m)*(a+b*ln(c*x^n))/b/n)*(a+b*ln(c*x^n))^p/exp(a*(1+m)/b/n)/f/(1+m)/((c*x^n)^((1+
m)/n))/((-(1+m)*(a+b*ln(c*x^n))/b/n)^p)+2*d*e*x^(1+r)*(f*x)^m*GAMMA(p+1,-(1+m+r)*(a+b*ln(c*x^n))/b/n)*(a+b*ln(
c*x^n))^p/exp(a*(1+m+r)/b/n)/(1+m+r)/((c*x^n)^((1+m+r)/n))/((-(1+m+r)*(a+b*ln(c*x^n))/b/n)^p)+e^2*x^(1+2*r)*(f
*x)^m*GAMMA(p+1,-(1+m+2*r)*(a+b*ln(c*x^n))/b/n)*(a+b*ln(c*x^n))^p/exp(a*(1+m+2*r)/b/n)/(1+m+2*r)/((c*x^n)^((1+
m+2*r)/n))/((-(1+m+2*r)*(a+b*ln(c*x^n))/b/n)^p)

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2395, 2347, 2212, 20} \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\frac {d^2 (f x)^{m+1} e^{-\frac {a (m+1)}{b n}} \left (c x^n\right )^{-\frac {m+1}{n}} \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(m+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p} \Gamma \left (p+1,-\frac {(m+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{f (m+1)}+\frac {2 d e x^{r+1} (f x)^m e^{-\frac {a (m+r+1)}{b n}} \left (c x^n\right )^{-\frac {m+r+1}{n}} \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(m+r+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p} \Gamma \left (p+1,-\frac {(m+r+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{m+r+1}+\frac {e^2 x^{2 r+1} (f x)^m e^{-\frac {a (m+2 r+1)}{b n}} \left (c x^n\right )^{-\frac {m+2 r+1}{n}} \left (a+b \log \left (c x^n\right )\right )^p \left (-\frac {(m+2 r+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p} \Gamma \left (p+1,-\frac {(m+2 r+1) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{m+2 r+1} \]

[In]

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

[Out]

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

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 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 2347

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)/n)*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rule 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

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

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 304, normalized size of antiderivative = 0.87 \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=x^{-m} (f x)^m \left (a+b \log \left (c x^n\right )\right )^p \left (\frac {d^2 e^{-\frac {(1+m) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} \Gamma \left (1+p,-\frac {(1+m) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (-\frac {(1+m) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m}+e \left (\frac {2 d e^{-\frac {(1+m+r) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} \Gamma \left (1+p,-\frac {(1+m+r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (-\frac {(1+m+r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m+r}+\frac {e e^{-\frac {(1+m+2 r) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{b n}} \Gamma \left (1+p,-\frac {(1+m+2 r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right ) \left (-\frac {(1+m+2 r) \left (a+b \log \left (c x^n\right )\right )}{b n}\right )^{-p}}{1+m+2 r}\right )\right ) \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (f x \right )^{m} \left (d +e \,x^{r}\right )^{2} {\left (a +b \ln \left (c \,x^{n}\right )\right )}^{p}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\int { {\left (e x^{r} + d\right )}^{2} \left (f x\right )^{m} {\left (b \log \left (c x^{n}\right ) + a\right )}^{p} \,d x } \]

[In]

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

[Out]

integral((e^2*x^(2*r) + 2*d*e*x^r + d^2)*(f*x)^m*(b*log(c*x^n) + a)^p, x)

Sympy [F(-1)]

Timed out. \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

Giac [F]

\[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\int { {\left (e x^{r} + d\right )}^{2} \left (f x\right )^{m} {\left (b \log \left (c x^{n}\right ) + a\right )}^{p} \,d x } \]

[In]

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

[Out]

integrate((e*x^r + d)^2*(f*x)^m*(b*log(c*x^n) + a)^p, x)

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \left (d+e x^r\right )^2 \left (a+b \log \left (c x^n\right )\right )^p \, dx=\int {\left (f\,x\right )}^m\,{\left (d+e\,x^r\right )}^2\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^p \,d x \]

[In]

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

[Out]

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