3.1043 \(\int \frac{x^{-1+2 n}}{(a+b x^n)^2 (c+d x^n)} \, dx\)

Optimal. Leaf size=75 \[ \frac{a}{b n (b c-a d) \left (a+b x^n\right )}+\frac{c \log \left (a+b x^n\right )}{n (b c-a d)^2}-\frac{c \log \left (c+d x^n\right )}{n (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.0624375, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {446, 77} \[ \frac{a}{b n (b c-a d) \left (a+b x^n\right )}+\frac{c \log \left (a+b x^n\right )}{n (b c-a d)^2}-\frac{c \log \left (c+d x^n\right )}{n (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0974601, size = 58, normalized size = 0.77 \[ \frac{\frac{a (b c-a d)}{b \left (a+b x^n\right )}+c \log \left (a+b x^n\right )-c \log \left (c+d x^n\right )}{n (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 109, normalized size = 1.5 \begin{align*}{\frac{{{\rm e}^{n\ln \left ( x \right ) }}}{ \left ( ad-bc \right ) n \left ( a+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) }}+{\frac{c\ln \left ( a+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) }{n \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) }}-{\frac{c\ln \left ( c+d{{\rm e}^{n\ln \left ( x \right ) }} \right ) }{n \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(a*d-b*c)/n*exp(n*ln(x))/(a+b*exp(n*ln(x)))+c/n/(a^2*d^2-2*a*b*c*d+b^2*c^2)*ln(a+b*exp(n*ln(x)))-c/n/(a^2*d^
2-2*a*b*c*d+b^2*c^2)*ln(c+d*exp(n*ln(x)))

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Maxima [A]  time = 0.95208, size = 163, normalized size = 2.17 \begin{align*} \frac{c \log \left (\frac{b x^{n} + a}{b}\right )}{b^{2} c^{2} n - 2 \, a b c d n + a^{2} d^{2} n} - \frac{c \log \left (\frac{d x^{n} + c}{d}\right )}{b^{2} c^{2} n - 2 \, a b c d n + a^{2} d^{2} n} + \frac{a}{a b^{2} c n - a^{2} b d n +{\left (b^{3} c n - a b^{2} d n\right )} x^{n}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*log((b*x^n + a)/b)/(b^2*c^2*n - 2*a*b*c*d*n + a^2*d^2*n) - c*log((d*x^n + c)/d)/(b^2*c^2*n - 2*a*b*c*d*n + a
^2*d^2*n) + a/(a*b^2*c*n - a^2*b*d*n + (b^3*c*n - a*b^2*d*n)*x^n)

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Fricas [A]  time = 1.07292, size = 244, normalized size = 3.25 \begin{align*} \frac{a b c - a^{2} d +{\left (b^{2} c x^{n} + a b c\right )} \log \left (b x^{n} + a\right ) -{\left (b^{2} c x^{n} + a b c\right )} \log \left (d x^{n} + c\right )}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} n x^{n} +{\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*b*c - a^2*d + (b^2*c*x^n + a*b*c)*log(b*x^n + a) - (b^2*c*x^n + a*b*c)*log(d*x^n + c))/((b^4*c^2 - 2*a*b^3*
c*d + a^2*b^2*d^2)*n*x^n + (a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*n)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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