\(\int \frac {d+e x}{(a+b x^n+c x^{2 n})^2} \, dx\) [11]

Optimal result

Integrand size = 22, antiderivative size = 589 \[ \int \frac {d+e x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n}-\frac {c e \left (8 a c (1-n)-b^2 (2-n)-b \sqrt {b^2-4 a c} (2-n)\right ) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c e \left (8 a c (1-n)-b^2 (2-n)+b \sqrt {b^2-4 a c} (2-n)\right ) x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n} \] Output:

d*x*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))+e*x^2*(b^2-2* 
a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))-c*d*(4*a*c*(1-2*n)-b^2*( 
1-n)-b*(-4*a*c+b^2)^(1/2)*(1-n))*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b- 
(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/n-c*d 
*(4*a*c*(1-2*n)-b^2*(1-n)+b*(-4*a*c+b^2)^(1/2)*(1-n))*x*hypergeom([1, 1/n] 
,[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)/(b*(-4*a*c+b^2)^( 
1/2)-4*a*c+b^2)/n-1/2*c*e*(8*a*c*(1-n)-b^2*(2-n)-b*(-4*a*c+b^2)^(1/2)*(2-n 
))*x^2*hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4 
*a*c+b^2)/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))/n-1/2*c*e*(8*a*c*(1-n)-b^2*(2-n 
)+b*(-4*a*c+b^2)^(1/2)*(2-n))*x^2*hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b 
+(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)/(b*(-4*a*c+b^2)^(1/2)-4*a*c+b^2)/n
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(4162\) vs. \(2(589)=1178\).

Time = 6.69 (sec) , antiderivative size = 4162, normalized size of antiderivative = 7.07 \[ \int \frac {d+e x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \] Input:

Integrate[(d + e*x)/(a + b*x^n + c*x^(2*n))^2,x]
 

Output:

(x*(d + e*x)*(-b^2 + 2*a*c - b*c*x^n))/(a*(-b^2 + 4*a*c)*n*(a + b*x^n + c* 
x^(2*n))) - (b*c*e*x^(2 + n)*(x^n)^(2/n - (2 + n)/n)*(-(Hypergeometric2F1[ 
-2/n, -2/n, (-2 + n)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[ 
b^2 - 4*a*c])/c + x^n))]/(Sqrt[b^2 - 4*a*c]*(x^n/(-1/2*(-b - Sqrt[b^2 - 4* 
a*c])/c + x^n))^(2/n))) + Hypergeometric2F1[-2/n, -2/n, (-2 + n)/n, -1/2*( 
-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))]/(Sqrt 
[b^2 - 4*a*c]*(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(2/n))))/(2*a* 
(-b^2 + 4*a*c)) + (b*c*e*x^(2 + n)*(x^n)^(2/n - (2 + n)/n)*(-(Hypergeometr 
ic2F1[-2/n, -2/n, (-2 + n)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - 
 Sqrt[b^2 - 4*a*c])/c + x^n))]/(Sqrt[b^2 - 4*a*c]*(x^n/(-1/2*(-b - Sqrt[b^ 
2 - 4*a*c])/c + x^n))^(2/n))) + Hypergeometric2F1[-2/n, -2/n, (-2 + n)/n, 
-1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))] 
/(Sqrt[b^2 - 4*a*c]*(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(2/n)))) 
/(a*(-b^2 + 4*a*c)*n) + (b^2*e*x^2*((1 - Hypergeometric2F1[-2/n, -2/n, (-2 
 + n)/n, -1/2*(-b - Sqrt[b^2 - 4*a*c])/(c*(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c 
 + x^n))]/(x^n/(-1/2*(-b - Sqrt[b^2 - 4*a*c])/c + x^n))^(2/n))/((b*(-b - S 
qrt[b^2 - 4*a*c]))/(2*c) + (-b - Sqrt[b^2 - 4*a*c])^2/(2*c)) + (1 - Hyperg 
eometric2F1[-2/n, -2/n, (-2 + n)/n, -1/2*(-b + Sqrt[b^2 - 4*a*c])/(c*(-1/2 
*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))]/(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c 
 + x^n))^(2/n))/((b*(-b + Sqrt[b^2 - 4*a*c]))/(2*c) + (-b + Sqrt[b^2 - ...
 

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 738, normalized size of antiderivative = 1.25, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2328, 2009}

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.

\(\displaystyle \int \frac {d+e x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\)

\(\Big \downarrow \) 2328

\(\displaystyle \int \left (\frac {d}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {e x}{\left (a+b x^n+c x^{2 n}\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b c^2 e (2-n) x^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a n (n+2) \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right )}+\frac {2 b c^2 e (2-n) x^{n+2} \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{n},2 \left (1+\frac {1}{n}\right ),-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a n (n+2) \left (b^2-4 a c\right )^{3/2} \left (\sqrt {b^2-4 a c}+b\right )}-\frac {c d x \left (-b (1-n) \sqrt {b^2-4 a c}+4 a c (1-2 n)-\left (b^2 (1-n)\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {c d x \left (b (1-n) \sqrt {b^2-4 a c}+4 a c (1-2 n)-\left (b^2 (1-n)\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )}+\frac {d x \left (-2 a c+b^2+b c x^n\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac {c e x^2 \left (4 a c (1-n)-b^2 (2-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {c e x^2 \left (4 a c (1-n)-b^2 (2-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )}+\frac {e x^2 \left (-2 a c+b^2+b c x^n\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\)

Input:

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

Output:

(d*x*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) 
+ (e*x^2*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n 
))) - (c*d*(4*a*c*(1 - 2*n) - b^2*(1 - n) - b*Sqrt[b^2 - 4*a*c]*(1 - n))*x 
*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c 
])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (c*d*(4*a*c 
*(1 - 2*n) - b^2*(1 - n) + b*Sqrt[b^2 - 4*a*c]*(1 - n))*x*Hypergeometric2F 
1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4* 
a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n) - (c*e*(4*a*c*(1 - n) - b^2*(2 
 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 
 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (c*e* 
(4*a*c*(1 - n) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2 
*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b 
^2 - 4*a*c])*n) - (2*b*c^2*e*(2 - n)*x^(2 + n)*Hypergeometric2F1[1, (2 + n 
)/n, 2*(1 + n^(-1)), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c) 
^(3/2)*(b - Sqrt[b^2 - 4*a*c])*n*(2 + n)) + (2*b*c^2*e*(2 - n)*x^(2 + n)*H 
ypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), (-2*c*x^n)/(b + Sqrt[b^2 - 
4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b + Sqrt[b^2 - 4*a*c])*n*(2 + n))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2328

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> 
Int[ExpandIntegrand[Pq*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, 
 n}, x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && ILtQ[p, -1]
 

Maple [F]

\[\int \frac {e x +d}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]

Input:

int((e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)
 

Output:

int((e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)
 

Fricas [F]

\[ \int \frac {d+e x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:

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

Output:

integral((e*x + d)/(c^2*x^(4*n) + b^2*x^(2*n) + 2*a*b*x^n + a^2 + 2*(b*c*x 
^n + a*c)*x^(2*n)), x)
 

Sympy [F(-1)]

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

integrate((e*x+d)/(a+b*x**n+c*x**(2*n))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {d+e x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:

integrate((e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")
 

Output:

((b^2*e - 2*a*c*e)*x^2 + (b*c*e*x^2 + b*c*d*x)*x^n + (b^2*d - 2*a*c*d)*x)/ 
(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) + (a*b^3*n - 4* 
a^2*b*c*n)*x^n) - integrate((2*a*c*d*(2*n - 1) - b^2*d*(n - 1) - (b*c*e*(n 
 - 2)*x + b*c*d*(n - 1))*x^n + (4*a*c*e*(n - 1) - b^2*e*(n - 2))*x)/(a^2*b 
^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) + (a*b^3*n - 4*a^2*b* 
c*n)*x^n), x)
 

Giac [F]

\[ \int \frac {d+e x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {e x + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \] Input:

integrate((e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")
 

Output:

integrate((e*x + d)/(c*x^(2*n) + b*x^n + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {d+e x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {d+e\,x}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \] Input:

int((d + e*x)/(a + b*x^n + c*x^(2*n))^2,x)
 

Output:

int((d + e*x)/(a + b*x^n + c*x^(2*n))^2, x)
 

Reduce [F]

\[ \int \frac {d+e x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\left (\int \frac {x}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \right ) e +\left (\int \frac {1}{x^{4 n} c^{2}+2 x^{3 n} b c +2 x^{2 n} a c +x^{2 n} b^{2}+2 x^{n} a b +a^{2}}d x \right ) d \] Input:

int((e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)
 

Output:

int(x/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**(2*n)*a*c + x**(2*n)*b**2 + 2 
*x**n*a*b + a**2),x)*e + int(1/(x**(4*n)*c**2 + 2*x**(3*n)*b*c + 2*x**(2*n 
)*a*c + x**(2*n)*b**2 + 2*x**n*a*b + a**2),x)*d