3.2.22 \(\int \frac {x^2 (c+d x^2+e x^4+f x^6)}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=163 \[ \frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{2 a \left (a+b x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-7 a^3 f+5 a^2 b e-3 a b^2 d+b^3 c\right )}{2 \sqrt {a} b^{9/2}}-\frac {x \left (-7 a^3 f+5 a^2 b e-3 a b^2 d+b^3 c\right )}{2 a b^4}+\frac {x^3 (b e-2 a f)}{3 b^3}+\frac {f x^5}{5 b^2} \]

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Rubi [A]  time = 0.23, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1804, 1585, 1261, 205} \begin {gather*} \frac {x^3 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{2 a \left (a+b x^2\right )}-\frac {x \left (5 a^2 b e-7 a^3 f-3 a b^2 d+b^3 c\right )}{2 a b^4}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (5 a^2 b e-7 a^3 f-3 a b^2 d+b^3 c\right )}{2 \sqrt {a} b^{9/2}}+\frac {x^3 (b e-2 a f)}{3 b^3}+\frac {f x^5}{5 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^2,x]

[Out]

-((b^3*c - 3*a*b^2*d + 5*a^2*b*e - 7*a^3*f)*x)/(2*a*b^4) + ((b*e - 2*a*f)*x^3)/(3*b^3) + (f*x^5)/(5*b^2) + ((c
- (a*(b^2*d - a*b*e + a^2*f))/b^3)*x^3)/(2*a*(a + b*x^2)) + ((b^3*c - 3*a*b^2*d + 5*a^2*b*e - 7*a^3*f)*ArcTan
[(Sqrt[b]*x)/Sqrt[a]])/(2*Sqrt[a]*b^(9/2))

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rule 1804

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[((c*x)^m*(a + b*x^2)^(p + 1)*(a*g - b*f*x))/(2*a*b*(p + 1)), x] + Dist[c/(2*a*b*(p + 1)), Int[
(c*x)^(m - 1)*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*x*Q - a*g*m + b*f*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && GtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^2} \, dx &=\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{2 a \left (a+b x^2\right )}-\frac {\int \frac {x \left (\left (b c-3 a d+\frac {3 a^2 e}{b}-\frac {3 a^3 f}{b^2}\right ) x-2 a \left (e-\frac {a f}{b}\right ) x^3-2 a f x^5\right )}{a+b x^2} \, dx}{2 a b}\\ &=\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{2 a \left (a+b x^2\right )}-\frac {\int \frac {x^2 \left (b c-3 a d+\frac {3 a^2 e}{b}-\frac {3 a^3 f}{b^2}-2 a \left (e-\frac {a f}{b}\right ) x^2-2 a f x^4\right )}{a+b x^2} \, dx}{2 a b}\\ &=\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{2 a \left (a+b x^2\right )}-\frac {\int \left (c-\frac {a \left (3 b^2 d-5 a b e+7 a^2 f\right )}{b^3}-\frac {2 a (b e-2 a f) x^2}{b^2}-\frac {2 a f x^4}{b}+\frac {-a b^3 c+3 a^2 b^2 d-5 a^3 b e+7 a^4 f}{b^3 \left (a+b x^2\right )}\right ) \, dx}{2 a b}\\ &=-\frac {\left (b^3 c-3 a b^2 d+5 a^2 b e-7 a^3 f\right ) x}{2 a b^4}+\frac {(b e-2 a f) x^3}{3 b^3}+\frac {f x^5}{5 b^2}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{2 a \left (a+b x^2\right )}+\frac {\left (b^3 c-3 a b^2 d+5 a^2 b e-7 a^3 f\right ) \int \frac {1}{a+b x^2} \, dx}{2 b^4}\\ &=-\frac {\left (b^3 c-3 a b^2 d+5 a^2 b e-7 a^3 f\right ) x}{2 a b^4}+\frac {(b e-2 a f) x^3}{3 b^3}+\frac {f x^5}{5 b^2}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{2 a \left (a+b x^2\right )}+\frac {\left (b^3 c-3 a b^2 d+5 a^2 b e-7 a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 148, normalized size = 0.91 \begin {gather*} \frac {x \left (3 a^2 f-2 a b e+b^2 d\right )}{b^4}-\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (7 a^3 f-5 a^2 b e+3 a b^2 d-b^3 c\right )}{2 \sqrt {a} b^{9/2}}-\frac {x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{2 b^4 \left (a+b x^2\right )}+\frac {x^3 (b e-2 a f)}{3 b^3}+\frac {f x^5}{5 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^2,x]

[Out]

((b^2*d - 2*a*b*e + 3*a^2*f)*x)/b^4 + ((b*e - 2*a*f)*x^3)/(3*b^3) + (f*x^5)/(5*b^2) - ((b^3*c - a*b^2*d + a^2*
b*e - a^3*f)*x)/(2*b^4*(a + b*x^2)) - ((-(b^3*c) + 3*a*b^2*d - 5*a^2*b*e + 7*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]
])/(2*Sqrt[a]*b^(9/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^2 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^2,x]

[Out]

IntegrateAlgebraic[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^2, x]

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fricas [A]  time = 1.17, size = 418, normalized size = 2.56 \begin {gather*} \left [\frac {12 \, a b^{4} f x^{7} + 4 \, {\left (5 \, a b^{4} e - 7 \, a^{2} b^{3} f\right )} x^{5} + 20 \, {\left (3 \, a b^{4} d - 5 \, a^{2} b^{3} e + 7 \, a^{3} b^{2} f\right )} x^{3} + 15 \, {\left (a b^{3} c - 3 \, a^{2} b^{2} d + 5 \, a^{3} b e - 7 \, a^{4} f + {\left (b^{4} c - 3 \, a b^{3} d + 5 \, a^{2} b^{2} e - 7 \, a^{3} b f\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 30 \, {\left (a b^{4} c - 3 \, a^{2} b^{3} d + 5 \, a^{3} b^{2} e - 7 \, a^{4} b f\right )} x}{60 \, {\left (a b^{6} x^{2} + a^{2} b^{5}\right )}}, \frac {6 \, a b^{4} f x^{7} + 2 \, {\left (5 \, a b^{4} e - 7 \, a^{2} b^{3} f\right )} x^{5} + 10 \, {\left (3 \, a b^{4} d - 5 \, a^{2} b^{3} e + 7 \, a^{3} b^{2} f\right )} x^{3} + 15 \, {\left (a b^{3} c - 3 \, a^{2} b^{2} d + 5 \, a^{3} b e - 7 \, a^{4} f + {\left (b^{4} c - 3 \, a b^{3} d + 5 \, a^{2} b^{2} e - 7 \, a^{3} b f\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 15 \, {\left (a b^{4} c - 3 \, a^{2} b^{3} d + 5 \, a^{3} b^{2} e - 7 \, a^{4} b f\right )} x}{30 \, {\left (a b^{6} x^{2} + a^{2} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/60*(12*a*b^4*f*x^7 + 4*(5*a*b^4*e - 7*a^2*b^3*f)*x^5 + 20*(3*a*b^4*d - 5*a^2*b^3*e + 7*a^3*b^2*f)*x^3 + 15*
(a*b^3*c - 3*a^2*b^2*d + 5*a^3*b*e - 7*a^4*f + (b^4*c - 3*a*b^3*d + 5*a^2*b^2*e - 7*a^3*b*f)*x^2)*sqrt(-a*b)*l
og((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 30*(a*b^4*c - 3*a^2*b^3*d + 5*a^3*b^2*e - 7*a^4*b*f)*x)/(a*b^6*
x^2 + a^2*b^5), 1/30*(6*a*b^4*f*x^7 + 2*(5*a*b^4*e - 7*a^2*b^3*f)*x^5 + 10*(3*a*b^4*d - 5*a^2*b^3*e + 7*a^3*b^
2*f)*x^3 + 15*(a*b^3*c - 3*a^2*b^2*d + 5*a^3*b*e - 7*a^4*f + (b^4*c - 3*a*b^3*d + 5*a^2*b^2*e - 7*a^3*b*f)*x^2
)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - 15*(a*b^4*c - 3*a^2*b^3*d + 5*a^3*b^2*e - 7*a^4*b*f)*x)/(a*b^6*x^2 + a^2*b
^5)]

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giac [A]  time = 0.37, size = 152, normalized size = 0.93 \begin {gather*} \frac {{\left (b^{3} c - 3 \, a b^{2} d - 7 \, a^{3} f + 5 \, a^{2} b e\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{4}} - \frac {b^{3} c x - a b^{2} d x - a^{3} f x + a^{2} b x e}{2 \, {\left (b x^{2} + a\right )} b^{4}} + \frac {3 \, b^{8} f x^{5} - 10 \, a b^{7} f x^{3} + 5 \, b^{8} x^{3} e + 15 \, b^{8} d x + 45 \, a^{2} b^{6} f x - 30 \, a b^{7} x e}{15 \, b^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

1/2*(b^3*c - 3*a*b^2*d - 7*a^3*f + 5*a^2*b*e)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) - 1/2*(b^3*c*x - a*b^2*d*x
- a^3*f*x + a^2*b*x*e)/((b*x^2 + a)*b^4) + 1/15*(3*b^8*f*x^5 - 10*a*b^7*f*x^3 + 5*b^8*x^3*e + 15*b^8*d*x + 45
*a^2*b^6*f*x - 30*a*b^7*x*e)/b^10

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maple [A]  time = 0.01, size = 212, normalized size = 1.30 \begin {gather*} \frac {f \,x^{5}}{5 b^{2}}-\frac {2 a f \,x^{3}}{3 b^{3}}+\frac {e \,x^{3}}{3 b^{2}}+\frac {a^{3} f x}{2 \left (b \,x^{2}+a \right ) b^{4}}-\frac {7 a^{3} f \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{4}}-\frac {a^{2} e x}{2 \left (b \,x^{2}+a \right ) b^{3}}+\frac {5 a^{2} e \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{3}}+\frac {a d x}{2 \left (b \,x^{2}+a \right ) b^{2}}-\frac {3 a d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b^{2}}-\frac {c x}{2 \left (b \,x^{2}+a \right ) b}+\frac {c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, b}+\frac {3 a^{2} f x}{b^{4}}-\frac {2 a e x}{b^{3}}+\frac {d x}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*f*x^5/b^2-2/3/b^3*x^3*a*f+1/3/b^2*x^3*e+3/b^4*a^2*f*x-2/b^3*a*e*x+1/b^2*d*x+1/2/b^4*x/(b*x^2+a)*a^3*f-1/2/
b^3*x/(b*x^2+a)*a^2*e+1/2/b^2*x/(b*x^2+a)*a*d-1/2/(b*x^2+a)/b*c*x-7/2/b^4/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x
)*a^3*f+5/2/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*a^2*e-3/2/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*a*d+
1/2/(a*b)^(1/2)/b*c*arctan(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 2.96, size = 140, normalized size = 0.86 \begin {gather*} -\frac {{\left (b^{3} c - a b^{2} d + a^{2} b e - a^{3} f\right )} x}{2 \, {\left (b^{5} x^{2} + a b^{4}\right )}} + \frac {{\left (b^{3} c - 3 \, a b^{2} d + 5 \, a^{2} b e - 7 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{4}} + \frac {3 \, b^{2} f x^{5} + 5 \, {\left (b^{2} e - 2 \, a b f\right )} x^{3} + 15 \, {\left (b^{2} d - 2 \, a b e + 3 \, a^{2} f\right )} x}{15 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

-1/2*(b^3*c - a*b^2*d + a^2*b*e - a^3*f)*x/(b^5*x^2 + a*b^4) + 1/2*(b^3*c - 3*a*b^2*d + 5*a^2*b*e - 7*a^3*f)*a
rctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + 1/15*(3*b^2*f*x^5 + 5*(b^2*e - 2*a*b*f)*x^3 + 15*(b^2*d - 2*a*b*e + 3*a
^2*f)*x)/b^4

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mupad [B]  time = 1.00, size = 153, normalized size = 0.94 \begin {gather*} x^3\,\left (\frac {e}{3\,b^2}-\frac {2\,a\,f}{3\,b^3}\right )-x\,\left (\frac {a^2\,f}{b^4}-\frac {d}{b^2}+\frac {2\,a\,\left (\frac {e}{b^2}-\frac {2\,a\,f}{b^3}\right )}{b}\right )-\frac {x\,\left (-\frac {f\,a^3}{2}+\frac {e\,a^2\,b}{2}-\frac {d\,a\,b^2}{2}+\frac {c\,b^3}{2}\right )}{b^5\,x^2+a\,b^4}+\frac {f\,x^5}{5\,b^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (-7\,f\,a^3+5\,e\,a^2\,b-3\,d\,a\,b^2+c\,b^3\right )}{2\,\sqrt {a}\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^3*(e/(3*b^2) - (2*a*f)/(3*b^3)) - x*((a^2*f)/b^4 - d/b^2 + (2*a*(e/b^2 - (2*a*f)/b^3))/b) - (x*((b^3*c)/2 -
(a^3*f)/2 - (a*b^2*d)/2 + (a^2*b*e)/2))/(a*b^4 + b^5*x^2) + (f*x^5)/(5*b^2) + (atan((b^(1/2)*x)/a^(1/2))*(b^3*
c - 7*a^3*f - 3*a*b^2*d + 5*a^2*b*e))/(2*a^(1/2)*b^(9/2))

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sympy [A]  time = 3.04, size = 221, normalized size = 1.36 \begin {gather*} x^{3} \left (- \frac {2 a f}{3 b^{3}} + \frac {e}{3 b^{2}}\right ) + x \left (\frac {3 a^{2} f}{b^{4}} - \frac {2 a e}{b^{3}} + \frac {d}{b^{2}}\right ) + \frac {x \left (a^{3} f - a^{2} b e + a b^{2} d - b^{3} c\right )}{2 a b^{4} + 2 b^{5} x^{2}} + \frac {\sqrt {- \frac {1}{a b^{9}}} \left (7 a^{3} f - 5 a^{2} b e + 3 a b^{2} d - b^{3} c\right ) \log {\left (- a b^{4} \sqrt {- \frac {1}{a b^{9}}} + x \right )}}{4} - \frac {\sqrt {- \frac {1}{a b^{9}}} \left (7 a^{3} f - 5 a^{2} b e + 3 a b^{2} d - b^{3} c\right ) \log {\left (a b^{4} \sqrt {- \frac {1}{a b^{9}}} + x \right )}}{4} + \frac {f x^{5}}{5 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(f*x**6+e*x**4+d*x**2+c)/(b*x**2+a)**2,x)

[Out]

x**3*(-2*a*f/(3*b**3) + e/(3*b**2)) + x*(3*a**2*f/b**4 - 2*a*e/b**3 + d/b**2) + x*(a**3*f - a**2*b*e + a*b**2*
d - b**3*c)/(2*a*b**4 + 2*b**5*x**2) + sqrt(-1/(a*b**9))*(7*a**3*f - 5*a**2*b*e + 3*a*b**2*d - b**3*c)*log(-a*
b**4*sqrt(-1/(a*b**9)) + x)/4 - sqrt(-1/(a*b**9))*(7*a**3*f - 5*a**2*b*e + 3*a*b**2*d - b**3*c)*log(a*b**4*sqr
t(-1/(a*b**9)) + x)/4 + f*x**5/(5*b**2)

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