3.49 \(\int \frac {(d+e x^n)^2}{(a+c x^{2 n})^2} \, dx\)

Optimal. Leaf size=203 \[ -\frac {(1-2 n) x \left (c d^2-a e^2\right ) \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n}-\frac {d e (1-n) x^{n+1} \, _2F_1\left (1,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a^2 n (n+1)}+\frac {x \left (-a e^2+c d^2+2 c d e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {e^2 x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a c} \]

[Out]

1/2*x*(c*d^2-a*e^2+2*c*d*e*x^n)/a/c/n/(a+c*x^(2*n))+e^2*x*hypergeom([1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/a/c-1/2
*(-a*e^2+c*d^2)*(1-2*n)*x*hypergeom([1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/a^2/c/n-d*e*(1-n)*x^(1+n)*hypergeom([1,
1/2*(1+n)/n],[3/2+1/2/n],-c*x^(2*n)/a)/a^2/n/(1+n)

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Rubi [A]  time = 0.17, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1437, 1431, 1418, 245, 364} \[ -\frac {(1-2 n) x \left (c d^2-a e^2\right ) \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n}-\frac {d e (1-n) x^{n+1} \, _2F_1\left (1,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a^2 n (n+1)}+\frac {x \left (-a e^2+c d^2+2 c d e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {e^2 x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(c*d^2 - a*e^2 + 2*c*d*e*x^n))/(2*a*c*n*(a + c*x^(2*n))) + (e^2*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1)
)/2, -((c*x^(2*n))/a)])/(a*c) - ((c*d^2 - a*e^2)*(1 - 2*n)*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((
c*x^(2*n))/a)])/(2*a^2*c*n) - (d*e*(1 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*
x^(2*n))/a)])/(a^2*n*(1 + 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 364

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

Rule 1418

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Dist[d, Int[1/(a + c*x^(2*n)), x], x] + D
ist[e, Int[x^n/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &
& (PosQ[a*c] ||  !IntegerQ[n])

Rule 1431

Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> -Simp[(x*(d + e*x^n)*(a + c*x^(2*n))
^(p + 1))/(2*a*n*(p + 1)), x] + Dist[1/(2*a*n*(p + 1)), Int[(d*(2*n*p + 2*n + 1) + e*(2*n*p + 3*n + 1)*x^n)*(a
+ c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && ILtQ[p, -1]

Rule 1437

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)
^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] && ((
IntegersQ[p, q] &&  !IntegerQ[n]) || IGtQ[p, 0] || (IGtQ[q, 0] &&  !IntegerQ[n]))

Rubi steps

\begin {align*} \int \frac {\left (d+e x^n\right )^2}{\left (a+c x^{2 n}\right )^2} \, dx &=\int \left (\frac {c d^2-a e^2+2 c d e x^n}{c \left (a+c x^{2 n}\right )^2}+\frac {e^2}{c \left (a+c x^{2 n}\right )}\right ) \, dx\\ &=\frac {\int \frac {c d^2-a e^2+2 c d e x^n}{\left (a+c x^{2 n}\right )^2} \, dx}{c}+\frac {e^2 \int \frac {1}{a+c x^{2 n}} \, dx}{c}\\ &=\frac {x \left (c d^2-a e^2+2 c d e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {e^2 x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a c}-\frac {\int \frac {\left (c d^2-a e^2\right ) (1-2 n)+2 c d e (1-n) x^n}{a+c x^{2 n}} \, dx}{2 a c n}\\ &=\frac {x \left (c d^2-a e^2+2 c d e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {e^2 x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a c}-\frac {\left (\left (c d^2-a e^2\right ) (1-2 n)\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{2 a c n}-\frac {(d e (1-n)) \int \frac {x^n}{a+c x^{2 n}} \, dx}{a n}\\ &=\frac {x \left (c d^2-a e^2+2 c d e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}+\frac {e^2 x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a c}-\frac {\left (c d^2-a e^2\right ) (1-2 n) x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n}-\frac {d e (1-n) x^{1+n} \, _2F_1\left (1,\frac {1+n}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a^2 n (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 136, normalized size = 0.67 \[ \frac {x \left ((n+1) \left (c d^2-a e^2\right ) \, _2F_1\left (2,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )+2 c d e x^n \, _2F_1\left (2,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )+a e^2 (n+1) \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )\right )}{a^2 c (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a*e^2*(1 + n)*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)] + (c*d^2 - a*e^2)*(1 + n)*Hy
pergeometric2F1[2, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)] + 2*c*d*e*x^n*Hypergeometric2F1[2, (1 + n)/(2*n)
, (3 + n^(-1))/2, -((c*x^(2*n))/a)]))/(a^2*c*(1 + n))

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{2} x^{2 \, n} + 2 \, d e x^{n} + d^{2}}{c^{2} x^{4 \, n} + 2 \, a c x^{2 \, n} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{n}+d \right )^{2}}{\left (c \,x^{2 n}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2 \, c d e x x^{n} + {\left (c d^{2} - a e^{2}\right )} x}{2 \, {\left (a c^{2} n x^{2 \, n} + a^{2} c n\right )}} + \int \frac {2 \, c d e {\left (n - 1\right )} x^{n} + c d^{2} {\left (2 \, n - 1\right )} + a e^{2}}{2 \, {\left (a c^{2} n x^{2 \, n} + a^{2} c n\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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