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

Optimal. Leaf size=156 \[ \frac{b x^{m+1} (b c (1-m)-a d (3-m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^2}+\frac{d^2 x^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c (m+1) (b c-a d)^2}+\frac{b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.160639, antiderivative size = 156, 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 = {472, 584, 364} \[ \frac{b x^{m+1} (b c (1-m)-a d (3-m)) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )}{2 a^2 (m+1) (b c-a d)^2}+\frac{d^2 x^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{c (m+1) (b c-a d)^2}+\frac{b x^{m+1}}{2 a \left (a+b x^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 472

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, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p
, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 584

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

Mathematica [A]  time = 0.0721027, size = 127, normalized size = 0.81 \[ \frac{x^{m+1} \left (a^2 d^2 \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )-a b c d \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )+b c (b c-a d) \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{b x^2}{a}\right )\right )}{a^2 c (m+1) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m}}{b^{2} d x^{6} +{\left (b^{2} c + 2 \, a b d\right )} x^{4} + a^{2} c +{\left (2 \, a b c + a^{2} d\right )} x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m/(b^2*d*x^6 + (b^2*c + 2*a*b*d)*x^4 + a^2*c + (2*a*b*c + a^2*d)*x^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(x**m/(b*x**2+a)**2/(d*x**2+c),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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