\(\int (f x)^m (d+e x^n)^q (c d^2-b d e-b e^2 x^n-c e^2 x^{2 n})^p \, dx\) [128]

Optimal result

Integrand size = 48, antiderivative size = 150 \[ \int (f x)^m \left (d+e x^n\right )^q \left (c d^2-b d e-b e^2 x^n-c e^2 x^{2 n}\right )^p \, dx=\frac {(f x)^{1+m} \left (d+e x^n\right )^q \left (1+\frac {e x^n}{d}\right )^{-p-q} \left (1-\frac {c e x^n}{c d-b e}\right )^{-p} \left (d (c d-b e)-b e^2 x^n-c e^2 x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {1+m}{n},-p-q,-p,\frac {1+m+n}{n},-\frac {e x^n}{d},\frac {c e x^n}{c d-b e}\right )}{f (1+m)} \] Output:

(f*x)^(1+m)*(d+e*x^n)^q*(1+e*x^n/d)^(-p-q)*(d*(-b*e+c*d)-b*e^2*x^n-c*e^2*x 
^(2*n))^p*AppellF1((1+m)/n,-p,-p-q,(1+m+n)/n,c*e*x^n/(-b*e+c*d),-e*x^n/d)/ 
f/(1+m)/((1-c*e*x^n/(-b*e+c*d))^p)
 

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.97 \[ \int (f x)^m \left (d+e x^n\right )^q \left (c d^2-b d e-b e^2 x^n-c e^2 x^{2 n}\right )^p \, dx=\frac {x (f x)^m \left (d+e x^n\right )^q \left (\frac {c d-b e-c e x^n}{c d-b e}\right )^{-p} \left (1+\frac {e x^n}{d}\right )^{-p-q} \left (\left (d+e x^n\right ) \left (-b e+c \left (d-e x^n\right )\right )\right )^p \operatorname {AppellF1}\left (\frac {1+m}{n},-p-q,-p,\frac {1+m+n}{n},-\frac {e x^n}{d},\frac {c e x^n}{c d-b e}\right )}{1+m} \] Input:

Integrate[(f*x)^m*(d + e*x^n)^q*(c*d^2 - b*d*e - b*e^2*x^n - c*e^2*x^(2*n) 
)^p,x]
 

Output:

(x*(f*x)^m*(d + e*x^n)^q*(1 + (e*x^n)/d)^(-p - q)*((d + e*x^n)*(-(b*e) + c 
*(d - e*x^n)))^p*AppellF1[(1 + m)/n, -p - q, -p, (1 + m + n)/n, -((e*x^n)/ 
d), (c*e*x^n)/(c*d - b*e)])/((1 + m)*((c*d - b*e - c*e*x^n)/(c*d - b*e))^p 
)
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1395, 1013, 1013, 1012}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (f x)^m \left (d+e x^n\right )^q \left (-b d e-b e^2 x^n+c d^2-c e^2 x^{2 n}\right )^p \, dx\)

\(\Big \downarrow \) 1395

\(\displaystyle \left (d+e x^n\right )^{-p} \left (-b e+c d-c e x^n\right )^{-p} \left (d (c d-b e)-b e^2 x^n-c e^2 x^{2 n}\right )^p \int (f x)^m \left (e x^n+d\right )^{p+q} \left (-c e x^n+c d-b e\right )^pdx\)

\(\Big \downarrow \) 1013

\(\displaystyle \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-p-q} \left (-b e+c d-c e x^n\right )^{-p} \left (d (c d-b e)-b e^2 x^n-c e^2 x^{2 n}\right )^p \int (f x)^m \left (-c e x^n+c d-b e\right )^p \left (\frac {e x^n}{d}+1\right )^{p+q}dx\)

\(\Big \downarrow \) 1013

\(\displaystyle \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-p-q} \left (1-\frac {c e x^n}{c d-b e}\right )^{-p} \left (d (c d-b e)-b e^2 x^n-c e^2 x^{2 n}\right )^p \int (f x)^m \left (\frac {e x^n}{d}+1\right )^{p+q} \left (1-\frac {c e x^n}{c d-b e}\right )^pdx\)

\(\Big \downarrow \) 1012

\(\displaystyle \frac {(f x)^{m+1} \left (d+e x^n\right )^q \left (\frac {e x^n}{d}+1\right )^{-p-q} \left (1-\frac {c e x^n}{c d-b e}\right )^{-p} \left (d (c d-b e)-b e^2 x^n-c e^2 x^{2 n}\right )^p \operatorname {AppellF1}\left (\frac {m+1}{n},-p-q,-p,\frac {m+n+1}{n},-\frac {e x^n}{d},\frac {c e x^n}{c d-b e}\right )}{f (m+1)}\)

Input:

Int[(f*x)^m*(d + e*x^n)^q*(c*d^2 - b*d*e - b*e^2*x^n - c*e^2*x^(2*n))^p,x]
 

Output:

((f*x)^(1 + m)*(d + e*x^n)^q*(1 + (e*x^n)/d)^(-p - q)*(d*(c*d - b*e) - b*e 
^2*x^n - c*e^2*x^(2*n))^p*AppellF1[(1 + m)/n, -p - q, -p, (1 + m + n)/n, - 
((e*x^n)/d), (c*e*x^n)/(c*d - b*e)])/(f*(1 + m)*(1 - (c*e*x^n)/(c*d - b*e) 
)^p)
 

Defintions of rubi rules used

rule 1012

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m 
+ 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, 
 b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n 
 - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1013

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ 
n/a))^FracPart[p])   Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; 
 FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & 
& NeQ[m, n - 1] &&  !(IntegerQ[p] || GtQ[a, 0])
 

rule 1395

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( 
x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d 
+ e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p 
+ q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E 
qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 
1] && EqQ[n, 2])
 

Maple [F]

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

Input:

int((f*x)^m*(d+e*x^n)^q*(c*d^2-b*d*e-b*e^2*x^n-c*e^2*x^(2*n))^p,x)
 

Output:

int((f*x)^m*(d+e*x^n)^q*(c*d^2-b*d*e-b*e^2*x^n-c*e^2*x^(2*n))^p,x)
 

Fricas [F]

\[ \int (f x)^m \left (d+e x^n\right )^q \left (c d^2-b d e-b e^2 x^n-c e^2 x^{2 n}\right )^p \, dx=\int { {\left (-c e^{2} x^{2 \, n} - b e^{2} x^{n} + c d^{2} - b d e\right )}^{p} {\left (e x^{n} + d\right )}^{q} \left (f x\right )^{m} \,d x } \] Input:

integrate((f*x)^m*(d+e*x^n)^q*(c*d^2-b*d*e-b*e^2*x^n-c*e^2*x^(2*n))^p,x, a 
lgorithm="fricas")
 

Output:

integral((-c*e^2*x^(2*n) - b*e^2*x^n + c*d^2 - b*d*e)^p*(e*x^n + d)^q*(f*x 
)^m, x)
 

Sympy [F(-1)]

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

integrate((f*x)**m*(d+e*x**n)**q*(c*d**2-b*d*e-b*e**2*x**n-c*e**2*x**(2*n) 
)**p,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (f x)^m \left (d+e x^n\right )^q \left (c d^2-b d e-b e^2 x^n-c e^2 x^{2 n}\right )^p \, dx=\int { {\left (-c e^{2} x^{2 \, n} - b e^{2} x^{n} + c d^{2} - b d e\right )}^{p} {\left (e x^{n} + d\right )}^{q} \left (f x\right )^{m} \,d x } \] Input:

integrate((f*x)^m*(d+e*x^n)^q*(c*d^2-b*d*e-b*e^2*x^n-c*e^2*x^(2*n))^p,x, a 
lgorithm="maxima")
 

Output:

integrate((-c*e^2*x^(2*n) - b*e^2*x^n + c*d^2 - b*d*e)^p*(e*x^n + d)^q*(f* 
x)^m, x)
 

Giac [F]

\[ \int (f x)^m \left (d+e x^n\right )^q \left (c d^2-b d e-b e^2 x^n-c e^2 x^{2 n}\right )^p \, dx=\int { {\left (-c e^{2} x^{2 \, n} - b e^{2} x^{n} + c d^{2} - b d e\right )}^{p} {\left (e x^{n} + d\right )}^{q} \left (f x\right )^{m} \,d x } \] Input:

integrate((f*x)^m*(d+e*x^n)^q*(c*d^2-b*d*e-b*e^2*x^n-c*e^2*x^(2*n))^p,x, a 
lgorithm="giac")
 

Output:

integrate((-c*e^2*x^(2*n) - b*e^2*x^n + c*d^2 - b*d*e)^p*(e*x^n + d)^q*(f* 
x)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (f x)^m \left (d+e x^n\right )^q \left (c d^2-b d e-b e^2 x^n-c e^2 x^{2 n}\right )^p \, dx=\int {\left (f\,x\right )}^m\,{\left (d+e\,x^n\right )}^q\,{\left (c\,d^2-c\,e^2\,x^{2\,n}-b\,d\,e-b\,e^2\,x^n\right )}^p \,d x \] Input:

int((f*x)^m*(d + e*x^n)^q*(c*d^2 - c*e^2*x^(2*n) - b*d*e - b*e^2*x^n)^p,x)
 

Output:

int((f*x)^m*(d + e*x^n)^q*(c*d^2 - c*e^2*x^(2*n) - b*d*e - b*e^2*x^n)^p, x 
)
 

Reduce [F]

\[ \int (f x)^m \left (d+e x^n\right )^q \left (c d^2-b d e-b e^2 x^n-c e^2 x^{2 n}\right )^p \, dx=\int \left (f x \right )^{m} \left (x^{n} e +d \right )^{q} \left (c \,d^{2}-b d e -b \,e^{2} x^{n}-x^{2 n} c \,e^{2}\right )^{p}d x \] Input:

int((f*x)^m*(d+e*x^n)^q*(c*d^2-b*d*e-b*e^2*x^n-c*e^2*x^(2*n))^p,x)
 

Output:

int((f*x)^m*(d+e*x^n)^q*(c*d^2-b*d*e-b*e^2*x^n-c*e^2*x^(2*n))^p,x)