3.582 \(\int \frac {c+d x^{-1+n}}{(a+b x^n)^2} \, dx\)

Optimal. Leaf size=44 \[ \frac {c x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2}-\frac {d}{b n \left (a+b x^n\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {1891, 245, 261} \[ \frac {c x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2}-\frac {d}{b n \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^(-1 + n))/(a + b*x^n)^2,x]

[Out]

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

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 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1891

Int[((A_) + (B_.)*(x_)^(m_.))*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[A, Int[(a + b*x^n)^p, x], x] +
Dist[B, Int[x^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, A, B, m, n, p}, x] && EqQ[m - n + 1, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 44, normalized size = 1.00 \[ \frac {c x \, _2F_1\left (2,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2}-\frac {d}{a b n+b^2 n x^n} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^(-1 + n))/(a + b*x^n)^2,x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d x^{n - 1} + c}{b^{2} x^{2 \, n} + 2 \, a b x^{n} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 1.03, size = 0, normalized size = 0.00 \[ \int \frac {d \,x^{n -1}+c}{\left (b \,x^{n}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 5.35, size = 49, normalized size = 1.11 \[ \frac {c\,x\,{{}}_2{\mathrm {F}}_1\left (2,\frac {1}{n};\ \frac {1}{n}+1;\ -\frac {b\,x^n}{a}\right )}{a^2}-\frac {a\,d}{b\,\left (a^2\,n+a\,b\,n\,x^n\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [C]  time = 51.15, size = 299, normalized size = 6.80 \[ c \left (\frac {n x \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, \frac {1}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{a \left (a n^{3} \Gamma \left (1 + \frac {1}{n}\right ) + b n^{3} x^{n} \Gamma \left (1 + \frac {1}{n}\right )\right )} + \frac {n x \Gamma \left (\frac {1}{n}\right )}{a \left (a n^{3} \Gamma \left (1 + \frac {1}{n}\right ) + b n^{3} x^{n} \Gamma \left (1 + \frac {1}{n}\right )\right )} - \frac {x \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, \frac {1}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{a \left (a n^{3} \Gamma \left (1 + \frac {1}{n}\right ) + b n^{3} x^{n} \Gamma \left (1 + \frac {1}{n}\right )\right )} + \frac {b n x x^{n} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, \frac {1}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{a^{2} \left (a n^{3} \Gamma \left (1 + \frac {1}{n}\right ) + b n^{3} x^{n} \Gamma \left (1 + \frac {1}{n}\right )\right )} - \frac {b x x^{n} \Phi \left (\frac {b x^{n} e^{i \pi }}{a}, 1, \frac {1}{n}\right ) \Gamma \left (\frac {1}{n}\right )}{a^{2} \left (a n^{3} \Gamma \left (1 + \frac {1}{n}\right ) + b n^{3} x^{n} \Gamma \left (1 + \frac {1}{n}\right )\right )}\right ) + d \left (\begin {cases} \frac {\log {\relax (x )}}{a^{2}} & \text {for}\: b = 0 \wedge n = 0 \\\frac {x^{n}}{a^{2} n} & \text {for}\: b = 0 \\\frac {\log {\relax (x )}}{\left (a + b\right )^{2}} & \text {for}\: n = 0 \\- \frac {1}{a b n + b^{2} n x^{n}} & \text {otherwise} \end {cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(n*x*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, 1/n)*gamma(1/n)/(a*(a*n**3*gamma(1 + 1/n) + b*n**3*x**n*gamma(1 +
1/n))) + n*x*gamma(1/n)/(a*(a*n**3*gamma(1 + 1/n) + b*n**3*x**n*gamma(1 + 1/n))) - x*lerchphi(b*x**n*exp_pola
r(I*pi)/a, 1, 1/n)*gamma(1/n)/(a*(a*n**3*gamma(1 + 1/n) + b*n**3*x**n*gamma(1 + 1/n))) + b*n*x*x**n*lerchphi(b
*x**n*exp_polar(I*pi)/a, 1, 1/n)*gamma(1/n)/(a**2*(a*n**3*gamma(1 + 1/n) + b*n**3*x**n*gamma(1 + 1/n))) - b*x*
x**n*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, 1/n)*gamma(1/n)/(a**2*(a*n**3*gamma(1 + 1/n) + b*n**3*x**n*gamma(1
+ 1/n)))) + d*Piecewise((log(x)/a**2, Eq(b, 0) & Eq(n, 0)), (x**n/(a**2*n), Eq(b, 0)), (log(x)/(a + b)**2, Eq(
n, 0)), (-1/(a*b*n + b**2*n*x**n), True))

________________________________________________________________________________________