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

Optimal. Leaf size=171 \[ \frac{x^{m+1} \left (a^2 d^2 \left (m^2-4 m+3\right )+2 a b c d \left (1-m^2\right )+b^2 c^2 \left (m^2+4 m+3\right )\right ) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{8 c^3 d^2 (m+1)}-\frac{x^{m+1} (b c-a d) (a d (3-m)+b c (m+5))}{8 c^2 d^2 \left (c+d x^2\right )}+\frac{x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.153598, antiderivative size = 166, normalized size of antiderivative = 0.97, 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, 457, 364} \[ \frac{x^{m+1} \left (\frac{(1-m) \left (4 a^2 d^2-(m+1) (b c-a d)^2\right )}{c^2 (m+1)}+4 b^2\right ) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{8 c d^2}-\frac{x^{m+1} (b c-a d) (a d (3-m)+b c (m+5))}{8 c^2 d^2 \left (c+d x^2\right )}+\frac{x^{m+1} (b c-a d)^2}{4 c d^2 \left (c+d x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*c - a*d)^2*x^(1 + m))/(4*c*d^2*(c + d*x^2)^2) - ((b*c - a*d)*(a*d*(3 - m) + b*c*(5 + m))*x^(1 + m))/(8*c^2
*d^2*(c + d*x^2)) + ((4*b^2 + ((1 - m)*(4*a^2*d^2 - (b*c - a*d)^2*(1 + m)))/(c^2*(1 + m)))*x^(1 + m)*Hypergeom
etric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)])/(8*c*d^2)

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 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

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

Mathematica [A]  time = 0.107786, size = 118, normalized size = 0.69 \[ \frac{x^{m+1} \left (\frac{a^2 \, _2F_1\left (3,\frac{m+1}{2};\frac{m+3}{2};-\frac{d x^2}{c}\right )}{m+1}+b x^2 \left (\frac{2 a \, _2F_1\left (3,\frac{m+3}{2};\frac{m+5}{2};-\frac{d x^2}{c}\right )}{m+3}+\frac{b x^2 \, _2F_1\left (3,\frac{m+5}{2};\frac{m+7}{2};-\frac{d x^2}{c}\right )}{m+5}\right )\right )}{c^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.064, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( b{x}^{2}+a \right ) ^{2}{x}^{m}}{ \left ( d{x}^{2}+c \right ) ^{3}}}\, 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)^3,x)

[Out]

int(x^m*(b*x^2+a)^2/(d*x^2+c)^3,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 )}^{3}}\,{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)^3,x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^2*x^m/(d*x^2 + c)^3, 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^{3} x^{6} + 3 \, c d^{2} x^{4} + 3 \, c^{2} d x^{2} + c^{3}}, 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)^3,x, algorithm="fricas")

[Out]

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

[Out]

Integral(x**m*(a + b*x**2)**2/(c + d*x**2)**3, 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 )}^{3}}\,{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)^3,x, algorithm="giac")

[Out]

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