3.4.30 \(\int \frac {x^m (a+b x^2)^2}{(c+d x^2)^3} \, dx\) [330]

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

[Out]

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

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Rubi [A]
time = 0.11, 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 = {474, 468, 371} \begin {gather*} \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} \end {gather*}

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 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 468

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 474

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 )^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*}

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Mathematica [A]
time = 0.40, size = 124, normalized size = 0.73 \begin {gather*} \frac {x^{1+m} \left (b^2 c^2 \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )-(b c-a d) \left (2 b c \, _2F_1\left (2,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )+(-b c+a d) \, _2F_1\left (3,\frac {1+m}{2};\frac {3+m}{2};-\frac {d x^2}{c}\right )\right )\right )}{c^3 d^2 (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(b^2*c^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] - (b*c - a*d)*(2*b*c*Hypergeometr
ic2F1[2, (1 + m)/2, (3 + m)/2, -((d*x^2)/c)] + (-(b*c) + a*d)*Hypergeometric2F1[3, (1 + m)/2, (3 + m)/2, -((d*
x^2)/c)])))/(c^3*d^2*(1 + m))

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \left (b \,x^{2}+a \right )^{2}}{\left (d \,x^{2}+c \right )^{3}}\, dx\]

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 = 0.00, 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)^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 = 0.00, 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)^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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{m} \left (a + b x^{2}\right )^{2}}{\left (c + d x^{2}\right )^{3}}\, dx \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)**3,x)

[Out]

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

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Giac [F]
time = 0.00, 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)^3,x, algorithm="giac")

[Out]

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

[Out]

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

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