3.1049 \(\int \frac{x^{-1+3 n}}{(a+b x^n)^2 (c+d x^n)} \, dx\)

Optimal. Leaf size=95 \[ -\frac{a^2}{b^2 n (b c-a d) \left (a+b x^n\right )}-\frac{a (2 b c-a d) \log \left (a+b x^n\right )}{b^2 n (b c-a d)^2}+\frac{c^2 \log \left (c+d x^n\right )}{d n (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.0903976, antiderivative size = 95, 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, 88} \[ -\frac{a^2}{b^2 n (b c-a d) \left (a+b x^n\right )}-\frac{a (2 b c-a d) \log \left (a+b x^n\right )}{b^2 n (b c-a d)^2}+\frac{c^2 \log \left (c+d x^n\right )}{d n (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2/(b^2*(b*c - a*d)*n*(a + b*x^n))) - (a*(2*b*c - a*d)*Log[a + b*x^n])/(b^2*(b*c - a*d)^2*n) + (c^2*Log[c +
d*x^n])/(d*(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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

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

Mathematica [A]  time = 0.096695, size = 90, normalized size = 0.95 \[ \frac{-\frac{a^2}{b^2 (b c-a d) \left (a+b x^n\right )}-\frac{a (2 b c-a d) \log \left (a+b x^n\right )}{b^2 (b c-a d)^2}+\frac{c^2 \log \left (c+d x^n\right )}{d (b c-a d)^2}}{n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 163, normalized size = 1.7 \begin{align*}{\frac{{a}^{2}}{ \left ( ad-bc \right ){b}^{2}n \left ( a+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) }}+{\frac{{c}^{2}\ln \left ( c+d{{\rm e}^{n\ln \left ( x \right ) }} \right ) }{dn \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) }}+{\frac{{a}^{2}\ln \left ( a+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) d}{{b}^{2} \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) n}}-2\,{\frac{a\ln \left ( a+b{{\rm e}^{n\ln \left ( x \right ) }} \right ) c}{ \left ({a}^{2}{d}^{2}-2\,abcd+{b}^{2}{c}^{2} \right ) bn}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.949438, size = 198, normalized size = 2.08 \begin{align*} \frac{c^{2} \log \left (\frac{d x^{n} + c}{d}\right )}{b^{2} c^{2} d n - 2 \, a b c d^{2} n + a^{2} d^{3} n} - \frac{a^{2}}{a b^{3} c n - a^{2} b^{2} d n +{\left (b^{4} c n - a b^{3} d n\right )} x^{n}} - \frac{{\left (2 \, a b c - a^{2} d\right )} \log \left (\frac{b x^{n} + a}{b}\right )}{b^{4} c^{2} n - 2 \, a b^{3} c d n + a^{2} b^{2} d^{2} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.10814, size = 324, normalized size = 3.41 \begin{align*} -\frac{a^{2} b c d - a^{3} d^{2} +{\left (2 \, a^{2} b c d - a^{3} d^{2} +{\left (2 \, a b^{2} c d - a^{2} b d^{2}\right )} x^{n}\right )} \log \left (b x^{n} + a\right ) -{\left (b^{3} c^{2} x^{n} + a b^{2} c^{2}\right )} \log \left (d x^{n} + c\right )}{{\left (b^{5} c^{2} d - 2 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} n x^{n} +{\left (a b^{4} c^{2} d - 2 \, a^{2} b^{3} c d^{2} + a^{3} b^{2} d^{3}\right )} n} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a^2*b*c*d - a^3*d^2 + (2*a^2*b*c*d - a^3*d^2 + (2*a*b^2*c*d - a^2*b*d^2)*x^n)*log(b*x^n + a) - (b^3*c^2*x^n
+ a*b^2*c^2)*log(d*x^n + c))/((b^5*c^2*d - 2*a*b^4*c*d^2 + a^2*b^3*d^3)*n*x^n + (a*b^4*c^2*d - 2*a^2*b^3*c*d^2
+ a^3*b^2*d^3)*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+3*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^{3 \, 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+3*n)/(a+b*x^n)^2/(c+d*x^n),x, algorithm="giac")

[Out]

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