\(\int \frac {(d+e x^n)^3}{(a+c x^{2 n})^3} \, dx\) [53]

Optimal result

Integrand size = 21, antiderivative size = 424 \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}+\frac {e^2 x \left (3 d+e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}-\frac {x \left (d \left (c d^2-3 a e^2\right ) (1-4 n)+e \left (3 c d^2-a e^2\right ) (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {d \left (c d^2-3 a e^2\right ) (1-4 n) (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2}-\frac {3 d e^2 (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n}+\frac {e \left (3 c d^2-a e^2\right ) (1-3 n) (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2 (1+n)}-\frac {e^3 (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n (1+n)} \]

[Out]

1/4*x*(d*(-3*a*e^2+c*d^2)+e*(-a*e^2+3*c*d^2)*x^n)/a/c/n/(a+c*x^(2*n))^2+1/2*e^2*x*(3*d+e*x^n)/a/c/n/(a+c*x^(2*
n))-1/8*x*(d*(-3*a*e^2+c*d^2)*(1-4*n)+e*(-a*e^2+3*c*d^2)*(1-3*n)*x^n)/a^2/c/n^2/(a+c*x^(2*n))+1/8*d*(-3*a*e^2+
c*d^2)*(1-4*n)*(1-2*n)*x*hypergeom([1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/a^3/c/n^2-3/2*d*e^2*(1-2*n)*x*hypergeom(
[1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/a^2/c/n+1/8*e*(-a*e^2+3*c*d^2)*(1-3*n)*(1-n)*x^(1+n)*hypergeom([1, 1/2*(1+n
)/n],[3/2+1/2/n],-c*x^(2*n)/a)/a^3/c/n^2/(1+n)-1/2*e^3*(1-n)*x^(1+n)*hypergeom([1, 1/2*(1+n)/n],[3/2+1/2/n],-c
*x^(2*n)/a)/a^2/c/n/(1+n)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1451, 1445, 1432, 251, 371} \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {e (1-3 n) (1-n) x^{n+1} \left (3 c d^2-a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2 (n+1)}+\frac {d (1-4 n) (1-2 n) x \left (c d^2-3 a e^2\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{8 a^3 c n^2}-\frac {x \left (e (1-3 n) x^n \left (3 c d^2-a e^2\right )+d (1-4 n) \left (c d^2-3 a e^2\right )\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}-\frac {3 d e^2 (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n}-\frac {e^3 (1-n) x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 c n (n+1)}+\frac {x \left (e x^n \left (3 c d^2-a e^2\right )+d \left (c d^2-3 a e^2\right )\right )}{4 a c n \left (a+c x^{2 n}\right )^2}+\frac {e^2 x \left (3 d+e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )} \]

[In]

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

[Out]

(x*(d*(c*d^2 - 3*a*e^2) + e*(3*c*d^2 - a*e^2)*x^n))/(4*a*c*n*(a + c*x^(2*n))^2) + (e^2*x*(3*d + e*x^n))/(2*a*c
*n*(a + c*x^(2*n))) - (x*(d*(c*d^2 - 3*a*e^2)*(1 - 4*n) + e*(3*c*d^2 - a*e^2)*(1 - 3*n)*x^n))/(8*a^2*c*n^2*(a
+ c*x^(2*n))) + (d*(c*d^2 - 3*a*e^2)*(1 - 4*n)*(1 - 2*n)*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*
x^(2*n))/a)])/(8*a^3*c*n^2) - (3*d*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) + (e*(3*c*d^2 - a*e^2)*(1 - 3*n)*(1 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 +
n^(-1))/2, -((c*x^(2*n))/a)])/(8*a^3*c*n^2*(1 + n)) - (e^3*(1 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n
), (3 + n^(-1))/2, -((c*x^(2*n))/a)])/(2*a^2*c*n*(1 + n))

Rule 251

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1432

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 1445

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 1451

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*} \text {integral}& = \int \left (\frac {c d^3-3 a d e^2+\left (3 c d^2 e-a e^3\right ) x^n}{c \left (a+c x^{2 n}\right )^3}+\frac {e^2 \left (3 d+e x^n\right )}{c \left (a+c x^{2 n}\right )^2}\right ) \, dx \\ & = \frac {\int \frac {c d^3-3 a d e^2+\left (3 c d^2 e-a e^3\right ) x^n}{\left (a+c x^{2 n}\right )^3} \, dx}{c}+\frac {e^2 \int \frac {3 d+e x^n}{\left (a+c x^{2 n}\right )^2} \, dx}{c} \\ & = \frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}+\frac {e^2 x \left (3 d+e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}-\frac {\int \frac {\left (c d^3-3 a d e^2\right ) (1-4 n)+\left (3 c d^2 e-a e^3\right ) (1-3 n) x^n}{\left (a+c x^{2 n}\right )^2} \, dx}{4 a c n}-\frac {e^2 \int \frac {3 d (1-2 n)+e (1-n) x^n}{a+c x^{2 n}} \, dx}{2 a c n} \\ & = \frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}+\frac {e^2 x \left (3 d+e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}-\frac {x \left (d \left (c d^2-3 a e^2\right ) (1-4 n)+e \left (3 c d^2-a e^2\right ) (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {\int \frac {\left (c d^3-3 a d e^2\right ) (1-4 n) (1-2 n)+\left (3 c d^2 e-a e^3\right ) (1-3 n) (1-n) x^n}{a+c x^{2 n}} \, dx}{8 a^2 c n^2}-\frac {\left (3 d e^2 (1-2 n)\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{2 a c n}-\frac {\left (e^3 (1-n)\right ) \int \frac {x^n}{a+c x^{2 n}} \, dx}{2 a c n} \\ & = \frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}+\frac {e^2 x \left (3 d+e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}-\frac {x \left (d \left (c d^2-3 a e^2\right ) (1-4 n)+e \left (3 c d^2-a e^2\right ) (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}-\frac {3 d e^2 (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 {e^3 (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 )}{2 a^2 c n (1+n)}+\frac {\left (d \left (c d^2-3 a e^2\right ) (1-4 n) (1-2 n)\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{8 a^2 c n^2}+\frac {\left (e \left (3 c d^2-a e^2\right ) (1-3 n) (1-n)\right ) \int \frac {x^n}{a+c x^{2 n}} \, dx}{8 a^2 c n^2} \\ & = \frac {x \left (d \left (c d^2-3 a e^2\right )+e \left (3 c d^2-a e^2\right ) x^n\right )}{4 a c n \left (a+c x^{2 n}\right )^2}+\frac {e^2 x \left (3 d+e x^n\right )}{2 a c n \left (a+c x^{2 n}\right )}-\frac {x \left (d \left (c d^2-3 a e^2\right ) (1-4 n)+e \left (3 c d^2-a e^2\right ) (1-3 n) x^n\right )}{8 a^2 c n^2 \left (a+c x^{2 n}\right )}+\frac {d \left (c d^2-3 a e^2\right ) (1-4 n) (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 )}{8 a^3 c n^2}-\frac {3 d e^2 (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 {e \left (3 c d^2-a e^2\right ) (1-3 n) (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 )}{8 a^3 c n^2 (1+n)}-\frac {e^3 (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 )}{2 a^2 c n (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.74 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.44 \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^3} \, dx=\frac {x \left (3 a d e^2 \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\frac {a e^3 x^n \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{1+n}+d \left (c d^2-3 a e^2\right ) \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\frac {e \left (3 c d^2-a e^2\right ) x^n \operatorname {Hypergeometric2F1}\left (3,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{1+n}\right )}{a^3 c} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {\left (d +e \,x^{n}\right )^{3}}{\left (a +c \,x^{2 n}\right )^{3}}d x\]

[In]

int((d+e*x^n)^3/(a+c*x^(2*n))^3,x)

[Out]

int((d+e*x^n)^3/(a+c*x^(2*n))^3,x)

Fricas [F]

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

[In]

integrate((d+e*x^n)^3/(a+c*x^(2*n))^3,x, algorithm="fricas")

[Out]

integral((e^3*x^(3*n) + 3*d*e^2*x^(2*n) + 3*d^2*e*x^n + d^3)/(c^3*x^(6*n) + 3*a*c^2*x^(4*n) + 3*a^2*c*x^(2*n)
+ a^3), x)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^3} \, dx=\text {Timed out} \]

[In]

integrate((d+e*x**n)**3/(a+c*x**(2*n))**3,x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate((d+e*x^n)^3/(a+c*x^(2*n))^3,x, algorithm="maxima")

[Out]

1/8*((3*c^2*d^2*e*(3*n - 1) + a*c*e^3*(n + 1))*x*x^(3*n) + (c^2*d^3*(4*n - 1) + 3*a*c*d*e^2)*x*x^(2*n) + (3*a*
c*d^2*e*(5*n - 1) - a^2*e^3*(n - 1))*x*x^n + (a*c*d^3*(6*n - 1) - 3*a^2*d*e^2*(2*n - 1))*x)/(a^2*c^3*n^2*x^(4*
n) + 2*a^3*c^2*n^2*x^(2*n) + a^4*c*n^2) + integrate(1/8*((8*n^2 - 6*n + 1)*c*d^3 + 3*a*d*e^2*(2*n - 1) + (3*(3
*n^2 - 4*n + 1)*c*d^2*e + (n^2 - 1)*a*e^3)*x^n)/(a^2*c^2*n^2*x^(2*n) + a^3*c*n^2), x)

Giac [F]

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

[In]

integrate((d+e*x^n)^3/(a+c*x^(2*n))^3,x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+c x^{2 n}\right )^3} \, dx=\int \frac {{\left (d+e\,x^n\right )}^3}{{\left (a+c\,x^{2\,n}\right )}^3} \,d x \]

[In]

int((d + e*x^n)^3/(a + c*x^(2*n))^3,x)

[Out]

int((d + e*x^n)^3/(a + c*x^(2*n))^3, x)