3.588 \(\int \frac{c+d x^{n/2}+e x^n+f x^{3 n/2}}{(a+b x^n)^2} \, dx\)

Optimal. Leaf size=162 \[ \frac{x (a e-b c (1-n)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2 b n}-\frac{x^{\frac{n+2}{2}} (b d (2-n)-a f (n+2)) \, _2F_1\left (1,\frac{1}{2} \left (1+\frac{2}{n}\right );\frac{1}{2} \left (3+\frac{2}{n}\right );-\frac{b x^n}{a}\right )}{a^2 b n (n+2)}+\frac{x \left (x^{n/2} (b d-a f)-a e+b c\right )}{a b n \left (a+b x^n\right )} \]

[Out]

(x*(b*c - a*e + (b*d - a*f)*x^(n/2)))/(a*b*n*(a + b*x^n)) - ((b*d*(2 - n) - a*f*(2 + n))*x^((2 + n)/2)*Hyperge
ometric2F1[1, (1 + 2/n)/2, (3 + 2/n)/2, -((b*x^n)/a)])/(a^2*b*n*(2 + n)) + ((a*e - b*c*(1 - n))*x*Hypergeometr
ic2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^2*b*n)

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Rubi [A]  time = 0.124185, antiderivative size = 162, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {1892, 1418, 245, 364} \[ \frac{x (a e-b c (1-n)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2 b n}-\frac{x^{\frac{n+2}{2}} (b d (2-n)-a f (n+2)) \, _2F_1\left (1,\frac{1}{2} \left (1+\frac{2}{n}\right );\frac{1}{2} \left (3+\frac{2}{n}\right );-\frac{b x^n}{a}\right )}{a^2 b n (n+2)}+\frac{x \left (x^{n/2} (b d-a f)-a e+b c\right )}{a b n \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(b*c - a*e + (b*d - a*f)*x^(n/2)))/(a*b*n*(a + b*x^n)) - ((b*d*(2 - n) - a*f*(2 + n))*x^((2 + n)/2)*Hyperge
ometric2F1[1, (1 + 2/n)/2, (3 + 2/n)/2, -((b*x^n)/a)])/(a^2*b*n*(2 + n)) + ((a*e - b*c*(1 - n))*x*Hypergeometr
ic2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^2*b*n)

Rule 1892

Int[(P3_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{A = Coeff[P3, x^(n/2), 0], B = Coeff[P3, x^(n/2),
1], C = Coeff[P3, x^(n/2), 2], D = Coeff[P3, x^(n/2), 3]}, -Simp[(x*(b*A - a*C + (b*B - a*D)*x^(n/2))*(a + b*x
^n)^(p + 1))/(a*b*n*(p + 1)), x] - Dist[1/(2*a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*Simp[2*a*C - 2*b*A*(n*(p
+ 1) + 1) + (a*D*(n + 2) - b*B*(n*(2*p + 3) + 2))*x^(n/2), x], x], x]] /; FreeQ[{a, b, n}, x] && PolyQ[P3, x^(
n/2), 3] && ILtQ[p, -1]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.338448, size = 147, normalized size = 0.91 \[ \frac{x \left ((b c-a e) \, _2F_1\left (2,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )+\frac{2 x^{n/2} (b d-a f) \, _2F_1\left (2,\frac{1}{2}+\frac{1}{n};\frac{3}{2}+\frac{1}{n};-\frac{b x^n}{a}\right )}{n+2}+a e \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )+\frac{2 a f x^{n/2} \, _2F_1\left (1,\frac{1}{2}+\frac{1}{n};\frac{3}{2}+\frac{1}{n};-\frac{b x^n}{a}\right )}{n+2}\right )}{a^2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*((2*a*f*x^(n/2)*Hypergeometric2F1[1, 1/2 + n^(-1), 3/2 + n^(-1), -((b*x^n)/a)])/(2 + n) + a*e*Hypergeometri
c2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)] + (2*(b*d - a*f)*x^(n/2)*Hypergeometric2F1[2, 1/2 + n^(-1), 3/2 + n^
(-1), -((b*x^n)/a)])/(2 + n) + (b*c - a*e)*Hypergeometric2F1[2, n^(-1), 1 + n^(-1), -((b*x^n)/a)]))/(a^2*b)

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Maple [F]  time = 0.432, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+b{x}^{n} \right ) ^{2}} \left ( c+d{x}^{{\frac{n}{2}}}+e{x}^{n}+f{x}^{{\frac{3\,n}{2}}} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (b d - a f\right )} x x^{\frac{1}{2} \, n} +{\left (b c - a e\right )} x}{a b^{2} n x^{n} + a^{2} b n} + \int \frac{2 \, b c{\left (n - 1\right )} + 2 \, a e +{\left (a f{\left (n + 2\right )} + b d{\left (n - 2\right )}\right )} x^{\frac{1}{2} \, n}}{2 \,{\left (a b^{2} n x^{n} + a^{2} b n\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^(1/2*n)+e*x^n+f*x^(3/2*n))/(a+b*x^n)^2,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{f x^{\frac{3}{2} \, n} + d x^{\frac{1}{2} \, n} + e x^{n} + c}{b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^(1/2*n)+e*x^n+f*x^(3/2*n))/(a+b*x^n)^2,x, algorithm="fricas")

[Out]

integral((f*x^(3/2*n) + d*x^(1/2*n) + e*x^n + c)/(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f x^{\frac{3}{2} \, n} + d x^{\frac{1}{2} \, n} + e x^{n} + c}{{\left (b x^{n} + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*x^(1/2*n)+e*x^n+f*x^(3/2*n))/(a+b*x^n)^2,x, algorithm="giac")

[Out]

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