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)) + ((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))

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Rubi [A]  time = 0.103223, antiderivative size = 120, normalized size of antiderivative = 1., 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 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]

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 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 )^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*}

Mathematica [A]  time = 0.108257, size = 118, normalized size = 0.98 \[ \frac{x^{m+1} \left (\frac{a^2 \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{m+1}+b x^2 \left (\frac{2 a \, _2F_1\left (2,\frac{m+3}{2};\frac{m+5}{2};-\frac{d x^2}{c}\right )}{m+3}+\frac{b x^2 \, _2F_1\left (2,\frac{m+5}{2};\frac{m+7}{2};-\frac{d x^2}{c}\right )}{m+5}\right )\right )}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*((a^2*Hypergeometric2F1[2, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(1 + m) + b*x^2*((2*a*Hypergeometri
c2F1[2, (3 + m)/2, (5 + m)/2, -((d*x^2)/c)])/(3 + m) + (b*x^2*Hypergeometric2F1[2, (5 + m)/2, (7 + m)/2, -((d*
x^2)/c)])/(5 + m))))/c^2

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Maple [F]  time = 0.048, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{2}{x}^{m}}{ \left ( d{x}^{2}+c \right ) ^{2}}}\, 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)^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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{2} x^{m}}{{\left (d x^{2} + c\right )}^{2}}\,{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)^2,x, algorithm="maxima")

[Out]

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

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

[Out]

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

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

[Out]

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