3.1033 \(\int \frac{x^2}{(a+b x^n)^2 (c+d x^n)} \, dx\)

Optimal. Leaf size=142 \[ \frac{b x^3 (a d (3-2 n)-b c (3-n)) \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{b x^n}{a}\right )}{3 a^2 n (b c-a d)^2}+\frac{d^2 x^3 \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{d x^n}{c}\right )}{3 c (b c-a d)^2}+\frac{b x^3}{a n (b c-a d) \left (a+b x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.224274, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {504, 597, 364} \[ \frac{b x^3 (a d (3-2 n)-b c (3-n)) \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{b x^n}{a}\right )}{3 a^2 n (b c-a d)^2}+\frac{d^2 x^3 \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{d x^n}{c}\right )}{3 c (b c-a d)^2}+\frac{b x^3}{a n (b c-a d) \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 504

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*(e*x
)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*e*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d)*(
p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n*(
p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && In
tBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 597

Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n_)))/((c_) + (d_.)*(x_)^(n_)), x_Sy
mbol] :> Int[ExpandIntegrand[((g*x)^m*(a + b*x^n)^p*(e + f*x^n))/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
f, g, m, n, p}, x]

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

Mathematica [A]  time = 0.159323, size = 135, normalized size = 0.95 \[ \frac{x^3 \left (a \left (a d^2 n \left (a+b x^n\right ) \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{d x^n}{c}\right )+3 b c (b c-a d)\right )+b c \left (a+b x^n\right ) (a d (3-2 n)+b c (n-3)) \, _2F_1\left (1,\frac{3}{n};\frac{n+3}{n};-\frac{b x^n}{a}\right )\right )}{3 a^2 c n (b c-a d)^2 \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.148, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{2}}{ \left ( a+b{x}^{n} \right ) ^{2} \left ( c+d{x}^{n} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{b x^{3}}{a^{2} b c n - a^{3} d n +{\left (a b^{2} c n - a^{2} b d n\right )} x^{n}} + d^{2} \int \frac{x^{2}}{b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} +{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\right )} x^{n}}\,{d x} -{\left (a b d{\left (2 \, n - 3\right )} - b^{2} c{\left (n - 3\right )}\right )} \int \frac{x^{2}}{a^{2} b^{2} c^{2} n - 2 \, a^{3} b c d n + a^{4} d^{2} n +{\left (a b^{3} c^{2} n - 2 \, a^{2} b^{2} c d n + a^{3} b d^{2} n\right )} x^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2}}{{\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^2/(a+b*x^n)^2/(c+d*x^n),x, algorithm="giac")

[Out]

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