3.336 \(\int \frac {x^m}{(a+b x^2)^3 (c+d x^2)} \, dx\)

Optimal. Leaf size=234 \[ \frac {b x^{m+1} (b c (3-m)-a d (7-m))}{8 a^2 \left (a+b x^2\right ) (b c-a d)^2}+\frac {b x^{m+1} \left (a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (m^2-6 m+5\right )+b^2 c^2 \left (m^2-4 m+3\right )\right ) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{8 a^3 (m+1) (b c-a d)^3}-\frac {d^3 x^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)^3}+\frac {b x^{m+1}}{4 a \left (a+b x^2\right )^2 (b c-a d)} \]

[Out]

1/4*b*x^(1+m)/a/(-a*d+b*c)/(b*x^2+a)^2+1/8*b*(b*c*(3-m)-a*d*(7-m))*x^(1+m)/a^2/(-a*d+b*c)^2/(b*x^2+a)+1/8*b*(a
^2*d^2*(m^2-8*m+15)-2*a*b*c*d*(m^2-6*m+5)+b^2*c^2*(m^2-4*m+3))*x^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-b
*x^2/a)/a^3/(-a*d+b*c)^3/(1+m)-d^3*x^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-d*x^2/c)/c/(-a*d+b*c)^3/(1+m)

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Rubi [A]  time = 0.37, antiderivative size = 234, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {472, 579, 584, 364} \[ \frac {b x^{m+1} \left (a^2 d^2 \left (m^2-8 m+15\right )-2 a b c d \left (m^2-6 m+5\right )+b^2 c^2 \left (m^2-4 m+3\right )\right ) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )}{8 a^3 (m+1) (b c-a d)^3}+\frac {b x^{m+1} (b c (3-m)-a d (7-m))}{8 a^2 \left (a+b x^2\right ) (b c-a d)^2}-\frac {d^3 x^{m+1} \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{c (m+1) (b c-a d)^3}+\frac {b x^{m+1}}{4 a \left (a+b x^2\right )^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^(1 + m))/(4*a*(b*c - a*d)*(a + b*x^2)^2) + (b*(b*c*(3 - m) - a*d*(7 - m))*x^(1 + m))/(8*a^2*(b*c - a*d)^2
*(a + b*x^2)) + (b*(a^2*d^2*(15 - 8*m + m^2) - 2*a*b*c*d*(5 - 6*m + m^2) + b^2*c^2*(3 - 4*m + m^2))*x^(1 + m)*
Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(8*a^3*(b*c - a*d)^3*(1 + m)) - (d^3*x^(1 + m)*Hyper
geometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(c*(b*c - a*d)^3*(1 + m))

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])

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 579

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

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]

Rubi steps

\begin {align*} \int \frac {x^m}{\left (a+b x^2\right )^3 \left (c+d x^2\right )} \, dx &=\frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}-\frac {\int \frac {x^m \left (4 a d-b c (3-m)-b d (3-m) x^2\right )}{\left (a+b x^2\right )^2 \left (c+d x^2\right )} \, dx}{4 a (b c-a d)}\\ &=\frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {\int \frac {x^m \left (8 a^2 d^2-a b c d \left (7-8 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )+b d (b c (3-m)-a d (7-m)) (1-m) x^2\right )}{\left (a+b x^2\right ) \left (c+d x^2\right )} \, dx}{8 a^2 (b c-a d)^2}\\ &=\frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {\int \left (\frac {b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right ) x^m}{(b c-a d) \left (a+b x^2\right )}+\frac {8 a^2 d^3 x^m}{(-b c+a d) \left (c+d x^2\right )}\right ) \, dx}{8 a^2 (b c-a d)^2}\\ &=\frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}-\frac {d^3 \int \frac {x^m}{c+d x^2} \, dx}{(b c-a d)^3}+\frac {\left (b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right )\right ) \int \frac {x^m}{a+b x^2} \, dx}{8 a^2 (b c-a d)^3}\\ &=\frac {b x^{1+m}}{4 a (b c-a d) \left (a+b x^2\right )^2}+\frac {b (b c (3-m)-a d (7-m)) x^{1+m}}{8 a^2 (b c-a d)^2 \left (a+b x^2\right )}+\frac {b \left (a^2 d^2 \left (15-8 m+m^2\right )-2 a b c d \left (5-6 m+m^2\right )+b^2 c^2 \left (3-4 m+m^2\right )\right ) x^{1+m} \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {b x^2}{a}\right )}{8 a^3 (b c-a d)^3 (1+m)}-\frac {d^3 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)^3 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 170, normalized size = 0.73 \[ \frac {x^{m+1} \left (a^3 d^3 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )-a^2 b c d^2 \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )-b c (a d-b c) \left ((a d-b c) \, _2F_1\left (3,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )+a d \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {b x^2}{a}\right )\right )\right )}{a^3 c (m+1) (a d-b c)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(-(a^2*b*c*d^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)]) + a^3*d^3*Hypergeometric2F
1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] - b*c*(-(b*c) + a*d)*(a*d*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2,
-((b*x^2)/a)] + (-(b*c) + a*d)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])))/(a^3*c*(-(b*c) + a*
d)^3*(1 + m))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{m}}{b^{3} d x^{8} + {\left (b^{3} c + 3 \, a b^{2} d\right )} x^{6} + 3 \, {\left (a b^{2} c + a^{2} b d\right )} x^{4} + a^{3} c + {\left (3 \, a^{2} b c + a^{3} d\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^m/(b^3*d*x^8 + (b^3*c + 3*a*b^2*d)*x^6 + 3*(a*b^2*c + a^2*b*d)*x^4 + a^3*c + (3*a^2*b*c + a^3*d)*x^
2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x^2+a)^3/(d*x^2+c),x, algorithm="giac")

[Out]

integrate(x^m/((b*x^2 + a)^3*(d*x^2 + c)), x)

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{\left (b \,x^{2}+a \right )^{3} \left (d \,x^{2}+c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(b*x^2+a)^3/(d*x^2+c),x)

[Out]

int(x^m/(b*x^2+a)^3/(d*x^2+c),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{m}}{{\left (b x^{2} + a\right )}^{3} {\left (d x^{2} + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/((b*x^2 + a)^3*(d*x^2 + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^m}{{\left (b\,x^2+a\right )}^3\,\left (d\,x^2+c\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/((a + b*x^2)^3*(c + d*x^2)),x)

[Out]

int(x^m/((a + b*x^2)^3*(c + d*x^2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m/(b*x**2+a)**3/(d*x**2+c),x)

[Out]

Timed out

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