3.4.35 \(\int \frac {x^m}{(a+b x^2)^2 (c+d x^2)} \, dx\) [335]

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

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 156, normalized size of antiderivative = 1.00, 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 = {483, 598, 371} \begin {gather*} \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)} \end {gather*}

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 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 483

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 598

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]

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*}

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Mathematica [A]
time = 0.17, size = 127, normalized size = 0.81 \begin {gather*} \frac {x^{1+m} \left (-a b c d \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )+a^2 d^2 \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+b c (b c-a d) \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )\right )}{a^2 c (b c-a d)^2 (1+m)} \end {gather*}

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 = 0.06, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{2} \left (d \,x^{2}+c \right )}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^m}{{\left (b\,x^2+a\right )}^2\,\left (d\,x^2+c\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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