3.136 \(\int \frac{x^2 (c+d x^2+e x^4+f x^6)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=167 \[ \frac{x^3 \left (c-\frac{a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac{x \left (-7 a^2 b e+11 a^3 f+3 a b^2 d+b^3 c\right )}{8 a b^4 \left (a+b x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (-15 a^2 b e+35 a^3 f+3 a b^2 d+b^3 c\right )}{8 a^{3/2} b^{9/2}}+\frac{x (b e-3 a f)}{b^4}+\frac{f x^3}{3 b^3} \]

[Out]

((b*e - 3*a*f)*x)/b^4 + (f*x^3)/(3*b^3) + ((c - (a*(b^2*d - a*b*e + a^2*f))/b^3)*x^3)/(4*a*(a + b*x^2)^2) - ((
b^3*c + 3*a*b^2*d - 7*a^2*b*e + 11*a^3*f)*x)/(8*a*b^4*(a + b*x^2)) + ((b^3*c + 3*a*b^2*d - 15*a^2*b*e + 35*a^3
*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(9/2))

________________________________________________________________________________________

Rubi [A]  time = 0.262115, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1804, 1585, 1257, 1153, 205} \[ \frac{x^3 \left (c-\frac{a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}-\frac{x \left (-7 a^2 b e+11 a^3 f+3 a b^2 d+b^3 c\right )}{8 a b^4 \left (a+b x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (-15 a^2 b e+35 a^3 f+3 a b^2 d+b^3 c\right )}{8 a^{3/2} b^{9/2}}+\frac{x (b e-3 a f)}{b^4}+\frac{f x^3}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^3,x]

[Out]

((b*e - 3*a*f)*x)/b^4 + (f*x^3)/(3*b^3) + ((c - (a*(b^2*d - a*b*e + a^2*f))/b^3)*x^3)/(4*a*(a + b*x^2)^2) - ((
b^3*c + 3*a*b^2*d - 7*a^2*b*e + 11*a^3*f)*x)/(8*a*b^4*(a + b*x^2)) + ((b^3*c + 3*a*b^2*d - 15*a^2*b*e + 35*a^3
*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(9/2))

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]

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 1257

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

Rule 1153

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

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 )^3} \, dx &=\frac{\left (c-\frac{a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\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-4 a \left (e-\frac{a f}{b}\right ) x^3-4 a f x^5\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac{\left (c-\frac{a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac{\int \frac{x^2 \left (-b c-3 a d+\frac{3 a^2 e}{b}-\frac{3 a^3 f}{b^2}-4 a \left (e-\frac{a f}{b}\right ) x^2-4 a f x^4\right )}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac{\left (c-\frac{a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac{\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac{\int \frac{b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f+8 a b (b e-2 a f) x^2+8 a b^2 f x^4}{a+b x^2} \, dx}{8 a b^4}\\ &=\frac{\left (c-\frac{a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac{\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac{\int \left (8 a (b e-3 a f)+8 a b f x^2+\frac{b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f}{a+b x^2}\right ) \, dx}{8 a b^4}\\ &=\frac{(b e-3 a f) x}{b^4}+\frac{f x^3}{3 b^3}+\frac{\left (c-\frac{a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac{\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac{\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \int \frac{1}{a+b x^2} \, dx}{8 a b^4}\\ &=\frac{(b e-3 a f) x}{b^4}+\frac{f x^3}{3 b^3}+\frac{\left (c-\frac{a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^3}{4 a \left (a+b x^2\right )^2}-\frac{\left (b^3 c+3 a b^2 d-7 a^2 b e+11 a^3 f\right ) x}{8 a b^4 \left (a+b x^2\right )}+\frac{\left (b^3 c+3 a b^2 d-15 a^2 b e+35 a^3 f\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.120743, size = 156, normalized size = 0.93 \[ \frac{x \left (a^2 b^2 \left (-9 d+75 e x^2-56 f x^4\right )+5 a^3 b \left (9 e-35 f x^2\right )-105 a^4 f+a b^3 \left (-3 c-15 d x^2+24 e x^4+8 f x^6\right )+3 b^4 c x^2\right )}{24 a b^4 \left (a+b x^2\right )^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right ) \left (-15 a^2 b e+35 a^3 f+3 a b^2 d+b^3 c\right )}{8 a^{3/2} b^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(c + d*x^2 + e*x^4 + f*x^6))/(a + b*x^2)^3,x]

[Out]

(x*(-105*a^4*f + 3*b^4*c*x^2 + 5*a^3*b*(9*e - 35*f*x^2) + a^2*b^2*(-9*d + 75*e*x^2 - 56*f*x^4) + a*b^3*(-3*c -
15*d*x^2 + 24*e*x^4 + 8*f*x^6)))/(24*a*b^4*(a + b*x^2)^2) + ((b^3*c + 3*a*b^2*d - 15*a^2*b*e + 35*a^3*f)*ArcT
an[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(3/2)*b^(9/2))

________________________________________________________________________________________

Maple [A]  time = 0.011, size = 259, normalized size = 1.6 \begin{align*}{\frac{f{x}^{3}}{3\,{b}^{3}}}-3\,{\frac{afx}{{b}^{4}}}+{\frac{ex}{{b}^{3}}}-{\frac{13\,{x}^{3}{a}^{2}f}{8\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{9\,a{x}^{3}e}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{5\,{x}^{3}d}{8\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{{x}^{3}c}{8\, \left ( b{x}^{2}+a \right ) ^{2}a}}-{\frac{11\,{a}^{3}fx}{8\,{b}^{4} \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{7\,{a}^{2}ex}{8\,{b}^{3} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{3\,adx}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}}-{\frac{cx}{8\,b \left ( b{x}^{2}+a \right ) ^{2}}}+{\frac{35\,{a}^{2}f}{8\,{b}^{4}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{15\,ae}{8\,{b}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,d}{8\,{b}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{c}{8\,ab}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

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)^3,x)

[Out]

1/3*f*x^3/b^3-3/b^4*a*f*x+1/b^3*x*e-13/8/b^3/(b*x^2+a)^2*x^3*a^2*f+9/8/b^2/(b*x^2+a)^2*x^3*a*e-5/8/b/(b*x^2+a)
^2*x^3*d+1/8/(b*x^2+a)^2/a*x^3*c-11/8/b^4/(b*x^2+a)^2*a^3*f*x+7/8/b^3/(b*x^2+a)^2*a^2*e*x-3/8/b^2/(b*x^2+a)^2*
a*d*x-1/8/b/(b*x^2+a)^2*c*x+35/8/b^4*a^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*f-15/8/b^3*a/(a*b)^(1/2)*arctan(b
*x/(a*b)^(1/2))*e+3/8/b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*d+1/8/b/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.5005, size = 1195, normalized size = 7.16 \begin{align*} \left [\frac{16 \, a^{2} b^{4} f x^{7} + 16 \,{\left (3 \, a^{2} b^{4} e - 7 \, a^{3} b^{3} f\right )} x^{5} + 2 \,{\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 75 \, a^{3} b^{3} e - 175 \, a^{4} b^{2} f\right )} x^{3} - 3 \,{\left (a^{2} b^{3} c + 3 \, a^{3} b^{2} d - 15 \, a^{4} b e + 35 \, a^{5} f +{\left (b^{5} c + 3 \, a b^{4} d - 15 \, a^{2} b^{3} e + 35 \, a^{3} b^{2} f\right )} x^{4} + 2 \,{\left (a b^{4} c + 3 \, a^{2} b^{3} d - 15 \, a^{3} b^{2} e + 35 \, a^{4} 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 ) - 6 \,{\left (a^{2} b^{4} c + 3 \, a^{3} b^{3} d - 15 \, a^{4} b^{2} e + 35 \, a^{5} b f\right )} x}{48 \,{\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}, \frac{8 \, a^{2} b^{4} f x^{7} + 8 \,{\left (3 \, a^{2} b^{4} e - 7 \, a^{3} b^{3} f\right )} x^{5} +{\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 75 \, a^{3} b^{3} e - 175 \, a^{4} b^{2} f\right )} x^{3} + 3 \,{\left (a^{2} b^{3} c + 3 \, a^{3} b^{2} d - 15 \, a^{4} b e + 35 \, a^{5} f +{\left (b^{5} c + 3 \, a b^{4} d - 15 \, a^{2} b^{3} e + 35 \, a^{3} b^{2} f\right )} x^{4} + 2 \,{\left (a b^{4} c + 3 \, a^{2} b^{3} d - 15 \, a^{3} b^{2} e + 35 \, a^{4} b f\right )} x^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right ) - 3 \,{\left (a^{2} b^{4} c + 3 \, a^{3} b^{3} d - 15 \, a^{4} b^{2} e + 35 \, a^{5} b f\right )} x}{24 \,{\left (a^{2} b^{7} x^{4} + 2 \, a^{3} b^{6} x^{2} + a^{4} b^{5}\right )}}\right ] \end{align*}

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)^3,x, algorithm="fricas")

[Out]

[1/48*(16*a^2*b^4*f*x^7 + 16*(3*a^2*b^4*e - 7*a^3*b^3*f)*x^5 + 2*(3*a*b^5*c - 15*a^2*b^4*d + 75*a^3*b^3*e - 17
5*a^4*b^2*f)*x^3 - 3*(a^2*b^3*c + 3*a^3*b^2*d - 15*a^4*b*e + 35*a^5*f + (b^5*c + 3*a*b^4*d - 15*a^2*b^3*e + 35
*a^3*b^2*f)*x^4 + 2*(a*b^4*c + 3*a^2*b^3*d - 15*a^3*b^2*e + 35*a^4*b*f)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a
*b)*x - a)/(b*x^2 + a)) - 6*(a^2*b^4*c + 3*a^3*b^3*d - 15*a^4*b^2*e + 35*a^5*b*f)*x)/(a^2*b^7*x^4 + 2*a^3*b^6*
x^2 + a^4*b^5), 1/24*(8*a^2*b^4*f*x^7 + 8*(3*a^2*b^4*e - 7*a^3*b^3*f)*x^5 + (3*a*b^5*c - 15*a^2*b^4*d + 75*a^3
*b^3*e - 175*a^4*b^2*f)*x^3 + 3*(a^2*b^3*c + 3*a^3*b^2*d - 15*a^4*b*e + 35*a^5*f + (b^5*c + 3*a*b^4*d - 15*a^2
*b^3*e + 35*a^3*b^2*f)*x^4 + 2*(a*b^4*c + 3*a^2*b^3*d - 15*a^3*b^2*e + 35*a^4*b*f)*x^2)*sqrt(a*b)*arctan(sqrt(
a*b)*x/a) - 3*(a^2*b^4*c + 3*a^3*b^3*d - 15*a^4*b^2*e + 35*a^5*b*f)*x)/(a^2*b^7*x^4 + 2*a^3*b^6*x^2 + a^4*b^5)
]

________________________________________________________________________________________

Sympy [A]  time = 12.9172, size = 258, normalized size = 1.54 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{3} b^{9}}} \left (35 a^{3} f - 15 a^{2} b e + 3 a b^{2} d + b^{3} c\right ) \log{\left (- a^{2} b^{4} \sqrt{- \frac{1}{a^{3} b^{9}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{3} b^{9}}} \left (35 a^{3} f - 15 a^{2} b e + 3 a b^{2} d + b^{3} c\right ) \log{\left (a^{2} b^{4} \sqrt{- \frac{1}{a^{3} b^{9}}} + x \right )}}{16} - \frac{x^{3} \left (13 a^{3} b f - 9 a^{2} b^{2} e + 5 a b^{3} d - b^{4} c\right ) + x \left (11 a^{4} f - 7 a^{3} b e + 3 a^{2} b^{2} d + a b^{3} c\right )}{8 a^{3} b^{4} + 16 a^{2} b^{5} x^{2} + 8 a b^{6} x^{4}} + \frac{f x^{3}}{3 b^{3}} - \frac{x \left (3 a f - b e\right )}{b^{4}} \end{align*}

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)**3,x)

[Out]

-sqrt(-1/(a**3*b**9))*(35*a**3*f - 15*a**2*b*e + 3*a*b**2*d + b**3*c)*log(-a**2*b**4*sqrt(-1/(a**3*b**9)) + x)
/16 + sqrt(-1/(a**3*b**9))*(35*a**3*f - 15*a**2*b*e + 3*a*b**2*d + b**3*c)*log(a**2*b**4*sqrt(-1/(a**3*b**9))
+ x)/16 - (x**3*(13*a**3*b*f - 9*a**2*b**2*e + 5*a*b**3*d - b**4*c) + x*(11*a**4*f - 7*a**3*b*e + 3*a**2*b**2*
d + a*b**3*c))/(8*a**3*b**4 + 16*a**2*b**5*x**2 + 8*a*b**6*x**4) + f*x**3/(3*b**3) - x*(3*a*f - b*e)/b**4

________________________________________________________________________________________

Giac [A]  time = 1.22776, size = 234, normalized size = 1.4 \begin{align*} \frac{{\left (b^{3} c + 3 \, a b^{2} d + 35 \, a^{3} f - 15 \, a^{2} b e\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a b^{4}} + \frac{b^{4} c x^{3} - 5 \, a b^{3} d x^{3} - 13 \, a^{3} b f x^{3} + 9 \, a^{2} b^{2} x^{3} e - a b^{3} c x - 3 \, a^{2} b^{2} d x - 11 \, a^{4} f x + 7 \, a^{3} b x e}{8 \,{\left (b x^{2} + a\right )}^{2} a b^{4}} + \frac{b^{6} f x^{3} - 9 \, a b^{5} f x + 3 \, b^{6} x e}{3 \, b^{9}} \end{align*}

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)^3,x, algorithm="giac")

[Out]

1/8*(b^3*c + 3*a*b^2*d + 35*a^3*f - 15*a^2*b*e)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^4) + 1/8*(b^4*c*x^3 - 5*a
*b^3*d*x^3 - 13*a^3*b*f*x^3 + 9*a^2*b^2*x^3*e - a*b^3*c*x - 3*a^2*b^2*d*x - 11*a^4*f*x + 7*a^3*b*x*e)/((b*x^2
+ a)^2*a*b^4) + 1/3*(b^6*f*x^3 - 9*a*b^5*f*x + 3*b^6*x*e)/b^9