3.17 \(\int \frac {A+B x^n+C x^{2 n}+D x^{3 n}}{(a+b x^n+c x^{2 n})^2} \, dx\)

Optimal. Leaf size=494 \[ \frac {x \left (x^n \left (b c (a C+A c)-a b^2 D-2 a c (B c-a D)\right )+A c \left (b^2-2 a c\right )-a (a b D-2 a c C+b B c)\right )}{a c n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}+\frac {x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) \left (\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (-2 b c (a D (n+2)+B c n)+4 a c^2 C+b^3 D-b^2 c C (1-n)\right )}{\sqrt {b^2-4 a c}}-b c (1-n) (a C+A c)+a b^2 D+2 a c (B c (1-n)-a D (n+1))\right )}{a c n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) \left (-\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (-2 b c (a D (n+2)+B c n)+4 a c^2 C+b^3 D-b^2 c C (1-n)\right )}{\sqrt {b^2-4 a c}}-b c (1-n) (a C+A c)+a b^2 D+2 a c (B c (1-n)-a D (n+1))\right )}{a c n \left (b^2-4 a c\right ) \left (\sqrt {b^2-4 a c}+b\right )} \]

[Out]

x*(A*c*(-2*a*c+b^2)-a*(B*b*c-2*C*a*c+D*a*b)+(b*c*(A*c+C*a)-a*b^2*D-2*a*c*(B*c-D*a))*x^n)/a/c/(-4*a*c+b^2)/n/(a
+b*x^n+c*x^(2*n))+x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))*(a*b^2*D-b*c*(A*c+C*a)*(1-n)+2
*a*c*(B*c*(1-n)-a*D*(1+n))+(A*c^2*(4*a*c*(1-2*n)-b^2*(1-n))-a*(4*a*c^2*C+b^3*D-b^2*c*C*(1-n)-2*b*c*(B*c*n+a*D*
(2+n))))/(-4*a*c+b^2)^(1/2))/a/c/(-4*a*c+b^2)/n/(b-(-4*a*c+b^2)^(1/2))+x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(
b+(-4*a*c+b^2)^(1/2)))*(a*b^2*D-b*c*(A*c+C*a)*(1-n)+2*a*c*(B*c*(1-n)-a*D*(1+n))+(-A*c^2*(4*a*c*(1-2*n)-b^2*(1-
n))+a*(4*a*c^2*C+b^3*D-b^2*c*C*(1-n)-2*b*c*(B*c*n+a*D*(2+n))))/(-4*a*c+b^2)^(1/2))/a/c/(-4*a*c+b^2)/n/(b+(-4*a
*c+b^2)^(1/2))

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Rubi [A]  time = 1.58, antiderivative size = 494, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {1794, 1422, 245} \[ \frac {x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) \left (\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (-2 b c (a D (n+2)+B c n)+4 a c^2 C-b^2 c C (1-n)+b^3 D\right )}{\sqrt {b^2-4 a c}}-b c (1-n) (a C+A c)+a b^2 D+2 a c (B c (1-n)-a D (n+1))\right )}{a c n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) \left (-\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (-2 b c (a D (n+2)+B c n)+4 a c^2 C-b^2 c C (1-n)+b^3 D\right )}{\sqrt {b^2-4 a c}}-b c (1-n) (a C+A c)+a b^2 D+2 a c (B c (1-n)-a D (n+1))\right )}{a c n \left (b^2-4 a c\right ) \left (\sqrt {b^2-4 a c}+b\right )}+\frac {x \left (x^n \left (b c (a C+A c)-a b^2 D-2 a c (B c-a D)\right )+A c \left (b^2-2 a c\right )-a (a b D-2 a c C+b B c)\right )}{a c n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^n + C*x^(2*n) + D*x^(3*n))/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

(x*(A*c*(b^2 - 2*a*c) - a*(b*B*c - 2*a*c*C + a*b*D) + (b*c*(A*c + a*C) - a*b^2*D - 2*a*c*(B*c - a*D))*x^n))/(a
*c*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + ((a*b^2*D - b*c*(A*c + a*C)*(1 - n) + 2*a*c*(B*c*(1 - n) - a*D*(
1 + n)) + (A*c^2*(4*a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c^2*C + b^3*D - b^2*c*C*(1 - n) - 2*b*c*(B*c*n + a*D
*(2 + n))))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])
/(a*c*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*n) + ((a*b^2*D - b*c*(A*c + a*C)*(1 - n) + 2*a*c*(B*c*(1 - n) - a*
D*(1 + n)) - (A*c^2*(4*a*c*(1 - 2*n) - b^2*(1 - n)) - a*(4*a*c^2*C + b^3*D - b^2*c*C*(1 - n) - 2*b*c*(B*c*n +
a*D*(2 + n))))/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]
)])/(a*c*(b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*n)

Rule 245

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, -((b*x^n)/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rule 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 0])

Rule 1794

Int[(P3_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> With[{d = Coeff[P3, x^n, 0], e = Coef
f[P3, x^n, 1], f = Coeff[P3, x^n, 2], g = Coeff[P3, x^n, 3]}, -Simp[(x*(b^2*c*d - 2*a*c*(c*d - a*f) - a*b*(c*e
+ a*g) + (b*c*(c*d + a*f) - a*b^2*g - 2*a*c*(c*e - a*g))*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1))/(a*c*n*(p + 1)
*(b^2 - 4*a*c)), x] - Dist[1/(a*c*n*(p + 1)*(b^2 - 4*a*c)), Int[(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[a*b*(c*e
+ a*g) - b^2*c*d*(n + n*p + 1) - 2*a*c*(a*f - c*d*(2*n*(p + 1) + 1)) + (a*b^2*g*(n*(p + 2) + 1) - b*c*(c*d + a
*f)*(n*(2*p + 3) + 1) - 2*a*c*(a*g*(n + 1) - c*e*(n*(2*p + 3) + 1)))*x^n, x], x], x]] /; FreeQ[{a, b, c, n}, x
] && EqQ[n2, 2*n] && PolyQ[P3, x^n, 3] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {A+B x^n+C x^{2 n}+D x^{3 n}}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx &=\frac {x \left (A c \left (b^2-2 a c\right )-a (b B c-2 a c C+a b D)+\left (b c (A c+a C)-a b^2 D-2 a c (B c-a D)\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {\int \frac {a b (B c+a D)-2 a c (a C-A c (1-2 n))-A b^2 c (1-n)+\left (a b^2 D-b c (A c+a C) (1-n)+2 a c (B c (1-n)-a D (1+n))\right ) x^n}{a+b x^n+c x^{2 n}} \, dx}{a c \left (b^2-4 a c\right ) n}\\ &=\frac {x \left (A c \left (b^2-2 a c\right )-a (b B c-2 a c C+a b D)+\left (b c (A c+a C)-a b^2 D-2 a c (B c-a D)\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {\left (a b^2 D-b c (A c+a C) (1-n)+2 a c (B c (1-n)-a D (1+n))-\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (4 a c^2 C+b^3 D-b^2 c C (1-n)-2 b c (B c n+a D (2+n))\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 a c \left (b^2-4 a c\right ) n}+\frac {\left (a b^2 D-b c (A c+a C) (1-n)+2 a c (B c (1-n)-a D (1+n))+\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (4 a c^2 C+b^3 D-b^2 c C (1-n)-2 b c (B c n+a D (2+n))\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 a c \left (b^2-4 a c\right ) n}\\ &=\frac {x \left (A c \left (b^2-2 a c\right )-a (b B c-2 a c C+a b D)+\left (b c (A c+a C)-a b^2 D-2 a c (B c-a D)\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {\left (a b^2 D-b c (A c+a C) (1-n)+2 a c (B c (1-n)-a D (1+n))+\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (4 a c^2 C+b^3 D-b^2 c C (1-n)-2 b c (B c n+a D (2+n))\right )}{\sqrt {b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a c \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) n}+\frac {\left (a b^2 D-b c (A c+a C) (1-n)+2 a c (B c (1-n)-a D (1+n))-\frac {A c^2 \left (4 a c (1-2 n)-b^2 (1-n)\right )-a \left (4 a c^2 C+b^3 D-b^2 c C (1-n)-2 b c (B c n+a D (2+n))\right )}{\sqrt {b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a c \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) n}\\ \end {align*}

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Mathematica [B]  time = 6.89, size = 5439, normalized size = 11.01 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^n + C*x^(2*n) + D*x^(3*n))/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

Result too large to show

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {D x^{3 \, n} + C x^{2 \, n} + B x^{n} + A}{c^{2} x^{4 \, n} + b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2} + 2 \, {\left (b c x^{n} + a c\right )} x^{2 \, n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*x^n+C*x^(2*n)+D*x^(3*n))/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")

[Out]

integral((D*x^(3*n) + C*x^(2*n) + B*x^n + A)/(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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {D x^{3 \, n} + C x^{2 \, n} + B x^{n} + A}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*x^n+C*x^(2*n)+D*x^(3*n))/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")

[Out]

integrate((D*x^(3*n) + C*x^(2*n) + B*x^n + A)/(c*x^(2*n) + b*x^n + a)^2, x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {B \,x^{n}+C \,x^{2 n}+D x^{3 n}+A}{\left (b \,x^{n}+c \,x^{2 n}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*x^n+C*x^(2*n)+D*x^(3*n))/(b*x^n+c*x^(2*n)+a)^2,x)

[Out]

int((A+B*x^n+C*x^(2*n)+D*x^(3*n))/(b*x^n+c*x^(2*n)+a)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {{\left (C a b c - 2 \, B a c^{2} + A b c^{2} - {\left (a b^{2} - 2 \, a^{2} c\right )} D\right )} x x^{n} - {\left (D a^{2} b - 2 \, C a^{2} c + B a b c - {\left (b^{2} c - 2 \, a c^{2}\right )} A\right )} x}{a^{2} b^{2} c n - 4 \, a^{3} c^{2} n + {\left (a b^{2} c^{2} n - 4 \, a^{2} c^{3} n\right )} x^{2 \, n} + {\left (a b^{3} c n - 4 \, a^{2} b c^{2} n\right )} x^{n}} - \int -\frac {D a^{2} b - 2 \, C a^{2} c + B a b c - {\left (2 \, a c^{2} {\left (2 \, n - 1\right )} - b^{2} c {\left (n - 1\right )}\right )} A + {\left (C a b c {\left (n - 1\right )} - 2 \, B a c^{2} {\left (n - 1\right )} + A b c^{2} {\left (n - 1\right )} - {\left (2 \, a^{2} c {\left (n + 1\right )} - a b^{2}\right )} D\right )} x^{n}}{a^{2} b^{2} c n - 4 \, a^{3} c^{2} n + {\left (a b^{2} c^{2} n - 4 \, a^{2} c^{3} n\right )} x^{2 \, n} + {\left (a b^{3} c n - 4 \, a^{2} b c^{2} n\right )} x^{n}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*x^n+C*x^(2*n)+D*x^(3*n))/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")

[Out]

((C*a*b*c - 2*B*a*c^2 + A*b*c^2 - (a*b^2 - 2*a^2*c)*D)*x*x^n - (D*a^2*b - 2*C*a^2*c + B*a*b*c - (b^2*c - 2*a*c
^2)*A)*x)/(a^2*b^2*c*n - 4*a^3*c^2*n + (a*b^2*c^2*n - 4*a^2*c^3*n)*x^(2*n) + (a*b^3*c*n - 4*a^2*b*c^2*n)*x^n)
- integrate(-(D*a^2*b - 2*C*a^2*c + B*a*b*c - (2*a*c^2*(2*n - 1) - b^2*c*(n - 1))*A + (C*a*b*c*(n - 1) - 2*B*a
*c^2*(n - 1) + A*b*c^2*(n - 1) - (2*a^2*c*(n + 1) - a*b^2)*D)*x^n)/(a^2*b^2*c*n - 4*a^3*c^2*n + (a*b^2*c^2*n -
4*a^2*c^3*n)*x^(2*n) + (a*b^3*c*n - 4*a^2*b*c^2*n)*x^n), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+C\,x^{2\,n}+x^{3\,n}\,D+B\,x^n}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*x^(2*n) + x^(3*n)*D + B*x^n)/(a + b*x^n + c*x^(2*n))^2,x)

[Out]

int((A + C*x^(2*n) + x^(3*n)*D + B*x^n)/(a + b*x^n + c*x^(2*n))^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*x**n+C*x**(2*n)+D*x**(3*n))/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Timed out

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