3.224 \(\int \frac {(f x)^m (d+e x^2)}{(a+b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=392 \[ \frac {c (f x)^{m+1} \left (b \left (d (1-m) \sqrt {b^2-4 a c}+4 a e\right )-2 a \left (e (1-m) \sqrt {b^2-4 a c}+2 c d (3-m)\right )+b^2 (d-d m)\right ) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{2 a f (m+1) \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (f x)^{m+1} \left (b \left (4 a e-d (1-m) \sqrt {b^2-4 a c}\right )+2 a \left (e (1-m) \sqrt {b^2-4 a c}-2 c d (3-m)\right )+b^2 d (1-m)\right ) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{2 a f (m+1) \left (b^2-4 a c\right )^{3/2} \left (\sqrt {b^2-4 a c}+b\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

[Out]

1/2*(f*x)^(1+m)*(b^2*d-2*a*c*d-a*b*e+c*(-2*a*e+b*d)*x^2)/a/(-4*a*c+b^2)/f/(c*x^4+b*x^2+a)-1/2*c*(f*x)^(1+m)*hy
pergeom([1, 1/2+1/2*m],[3/2+1/2*m],-2*c*x^2/(b+(-4*a*c+b^2)^(1/2)))*(b^2*d*(1-m)+b*(4*a*e-d*(1-m)*(-4*a*c+b^2)
^(1/2))+2*a*(-2*c*d*(3-m)+e*(1-m)*(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/f/(1+m)/(b+(-4*a*c+b^2)^(1/2))+1/2
*c*(f*x)^(1+m)*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-2*c*x^2/(b-(-4*a*c+b^2)^(1/2)))*(b^2*(-d*m+d)+b*(4*a*e+d*
(1-m)*(-4*a*c+b^2)^(1/2))-2*a*(2*c*d*(3-m)+e*(1-m)*(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/f/(1+m)/(b-(-4*a*
c+b^2)^(1/2))

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Rubi [A]  time = 2.65, antiderivative size = 358, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1277, 1285, 364} \[ \frac {c (f x)^{m+1} \left ((1-m) \sqrt {b^2-4 a c} (b d-2 a e)+4 a b e-4 a c d (3-m)+b^2 (d-d m)\right ) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}\right )}{2 a f (m+1) \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c (f x)^{m+1} \left (-(1-m) \sqrt {b^2-4 a c} (b d-2 a e)+4 a b e-4 a c d (3-m)+b^2 (d-d m)\right ) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}}\right )}{2 a f (m+1) \left (b^2-4 a c\right )^{3/2} \left (\sqrt {b^2-4 a c}+b\right )}+\frac {(f x)^{m+1} \left (c x^2 (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{2 a f \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[((f*x)^m*(d + e*x^2))/(a + b*x^2 + c*x^4)^2,x]

[Out]

((f*x)^(1 + m)*(b^2*d - 2*a*c*d - a*b*e + c*(b*d - 2*a*e)*x^2))/(2*a*(b^2 - 4*a*c)*f*(a + b*x^2 + c*x^4)) + (c
*(4*a*b*e + Sqrt[b^2 - 4*a*c]*(b*d - 2*a*e)*(1 - m) - 4*a*c*d*(3 - m) + b^2*(d - d*m))*(f*x)^(1 + m)*Hypergeom
etric2F1[1, (1 + m)/2, (3 + m)/2, (-2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])])/(2*a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2
- 4*a*c])*f*(1 + m)) - (c*(4*a*b*e - Sqrt[b^2 - 4*a*c]*(b*d - 2*a*e)*(1 - m) - 4*a*c*d*(3 - m) + b^2*(d - d*m)
)*(f*x)^(1 + m)*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])])/(2*a*(b^2 - 4*
a*c)^(3/2)*(b + Sqrt[b^2 - 4*a*c])*f*(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 1277

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> -Simp[((f
*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1)*(d*(b^2 - 2*a*c) - a*b*e + (b*d - 2*a*e)*c*x^2))/(2*a*f*(p + 1)*(b^2 -
4*a*c)), x] + Dist[1/(2*a*(p + 1)*(b^2 - 4*a*c)), Int[(f*x)^m*(a + b*x^2 + c*x^4)^(p + 1)*Simp[d*(b^2*(m + 2*
(p + 1) + 1) - 2*a*c*(m + 4*(p + 1) + 1)) - a*b*e*(m + 1) + c*(m + 2*(2*p + 3) + 1)*(b*d - 2*a*e)*x^2, x], x],
x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p] && (IntegerQ[p] |
| IntegerQ[m])

Rule 1285

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2))/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt
[b^2 - 4*a*c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[(f*x)^m/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d
- b*e)/(2*q), Int[(f*x)^m/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b^2 - 4*a*c,
0]

Rubi steps

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

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Mathematica [C]  time = 0.23, size = 160, normalized size = 0.41 \[ \frac {x (f x)^m \left (d (m+3) F_1\left (\frac {m+1}{2};2,2;\frac {m+3}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )+e (m+1) x^2 F_1\left (\frac {m+3}{2};2,2;\frac {m+5}{2};-\frac {2 c x^2}{b+\sqrt {b^2-4 a c}},\frac {2 c x^2}{\sqrt {b^2-4 a c}-b}\right )\right )}{a^2 (m+1) (m+3)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((f*x)^m*(d + e*x^2))/(a + b*x^2 + c*x^4)^2,x]

[Out]

(x*(f*x)^m*(d*(3 + m)*AppellF1[(1 + m)/2, 2, 2, (3 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^2)/(-b +
Sqrt[b^2 - 4*a*c])] + e*(1 + m)*x^2*AppellF1[(3 + m)/2, 2, 2, (5 + m)/2, (-2*c*x^2)/(b + Sqrt[b^2 - 4*a*c]),
(2*c*x^2)/(-b + Sqrt[b^2 - 4*a*c])]))/(a^2*(1 + m)*(3 + m))

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fricas [F]  time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{c^{2} x^{8} + 2 \, b c x^{6} + {\left (b^{2} + 2 \, a c\right )} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(f*x)^m/(c*x^4 + b*x^2 + a)^2, x)

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{2}+d \right ) \left (f x \right )^{m}}{\left (c \,x^{4}+b \,x^{2}+a \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^2,x)

[Out]

int((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )} \left (f x\right )^{m}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(e*x^2+d)/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")

[Out]

integrate((e*x^2 + d)*(f*x)^m/(c*x^4 + b*x^2 + a)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (f\,x\right )}^m\,\left (e\,x^2+d\right )}{{\left (c\,x^4+b\,x^2+a\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((f*x)^m*(d + e*x^2))/(a + b*x^2 + c*x^4)^2,x)

[Out]

int(((f*x)^m*(d + e*x^2))/(a + b*x^2 + c*x^4)^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((f*x)**m*(e*x**2+d)/(c*x**4+b*x**2+a)**2,x)

[Out]

Timed out

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