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

Optimal. Leaf size=120 \[ -\frac {x^{m+1} (b c-a d) (a d (1-m)+b c (m+3)) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{2 c^2 d^2 (m+1)}+\frac {x^{m+1} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {b^2 x^{m+1}}{d^2 (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {463, 459, 364} \[ -\frac {x^{m+1} (b c-a d) (a d (1-m)+b c (m+3)) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )}{2 c^2 d^2 (m+1)}+\frac {x^{m+1} (b c-a d)^2}{2 c d^2 \left (c+d x^2\right )}+\frac {b^2 x^{m+1}}{d^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x^(1 + m))/(d^2*(1 + m)) + ((b*c - a*d)^2*x^(1 + m))/(2*c*d^2*(c + d*x^2)) - ((b*c - a*d)*(a*d*(1 - m) +
b*c*(3 + m))*x^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(2*c^2*d^2*(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 459

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

Rule 463

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 98, normalized size = 0.82 \[ \frac {x^{m+1} \left (-2 b c (b c-a d) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )+(b c-a d)^2 \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};-\frac {d x^2}{c}\right )+b^2 c^2\right )}{c^2 d^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} x^{m}}{d^{2} x^{4} + 2 \, c d x^{2} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \,x^{2}+a \right )^{2} x^{m}}{\left (d \,x^{2}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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