\(\int \frac {x^4 (c+d x^2+e x^4+f x^6)}{(a+b x^2)^3} \, dx\) [135]

Optimal result

Integrand size = 30, antiderivative size = 207 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {\left (b^3 c-5 a b^2 d+13 a^2 b e-25 a^3 f\right ) x}{4 a b^5}+\frac {(b e-3 a f) x^3}{3 b^4}+\frac {f x^5}{5 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^5}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c-5 a b^2 d+9 a^2 b e-13 a^3 f\right ) x}{8 b^5 \left (a+b x^2\right )}+\frac {\left (3 b^3 c-15 a b^2 d+35 a^2 b e-63 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{11/2}} \]

[Out]

-1/4*(-25*a^3*f+13*a^2*b*e-5*a*b^2*d+b^3*c)*x/a/b^5+1/3*(-3*a*f+b*e)*x^3/b^4+1/5*f*x^5/b^3+1/4*(c-a*(a^2*f-a*b
*e+b^2*d)/b^3)*x^5/a/(b*x^2+a)^2-1/8*(-13*a^3*f+9*a^2*b*e-5*a*b^2*d+b^3*c)*x/b^5/(b*x^2+a)+1/8*(-63*a^3*f+35*a
^2*b*e-15*a*b^2*d+3*b^3*c)*arctan(x*b^(1/2)/a^(1/2))/b^(11/2)/a^(1/2)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1818, 1599, 1271, 1824, 211} \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x^5 \left (c-\frac {a \left (a^2 f-a b e+b^2 d\right )}{b^3}\right )}{4 a \left (a+b x^2\right )^2}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-63 a^3 f+35 a^2 b e-15 a b^2 d+3 b^3 c\right )}{8 \sqrt {a} b^{11/2}}-\frac {x \left (-13 a^3 f+9 a^2 b e-5 a b^2 d+b^3 c\right )}{8 b^5 \left (a+b x^2\right )}-\frac {x \left (-25 a^3 f+13 a^2 b e-5 a b^2 d+b^3 c\right )}{4 a b^5}+\frac {x^3 (b e-3 a f)}{3 b^4}+\frac {f x^5}{5 b^3} \]

[In]

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

[Out]

-1/4*((b^3*c - 5*a*b^2*d + 13*a^2*b*e - 25*a^3*f)*x)/(a*b^5) + ((b*e - 3*a*f)*x^3)/(3*b^4) + (f*x^5)/(5*b^3) +
((c - (a*(b^2*d - a*b*e + a^2*f))/b^3)*x^5)/(4*a*(a + b*x^2)^2) - ((b^3*c - 5*a*b^2*d + 9*a^2*b*e - 13*a^3*f)
*x)/(8*b^5*(a + b*x^2)) + ((3*b^3*c - 15*a*b^2*d + 35*a^2*b*e - 63*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt
[a]*b^(11/2))

Rule 211

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

Rule 1271

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/(d + e*x^2))*(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))], 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 1599

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 1818

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 1824

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^5}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {x^3 \left (\left (b c-5 a d+\frac {5 a^2 e}{b}-\frac {5 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^5}{4 a \left (a+b x^2\right )^2}-\frac {\int \frac {x^4 \left (b c-5 a d+\frac {5 a^2 e}{b}-\frac {5 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^5}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c-5 a b^2 d+9 a^2 b e-13 a^3 f\right ) x}{8 b^5 \left (a+b x^2\right )}+\frac {\int \frac {a \left (b^3 c-5 a b^2 d+9 a^2 b e-13 a^3 f\right )-2 b \left (b^3 c-5 a b^2 d+9 a^2 b e-13 a^3 f\right ) x^2+8 a b^2 (b e-2 a f) x^4+8 a b^3 f x^6}{a+b x^2} \, dx}{8 a b^5} \\ & = \frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^5}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c-5 a b^2 d+9 a^2 b e-13 a^3 f\right ) x}{8 b^5 \left (a+b x^2\right )}+\frac {\int \left (-2 \left (b^3 c-5 a b^2 d+13 a^2 b e-25 a^3 f\right )+8 a b (b e-3 a f) x^2+8 a b^2 f x^4+\frac {3 a b^3 c-15 a^2 b^2 d+35 a^3 b e-63 a^4 f}{a+b x^2}\right ) \, dx}{8 a b^5} \\ & = -\frac {\left (b^3 c-5 a b^2 d+13 a^2 b e-25 a^3 f\right ) x}{4 a b^5}+\frac {(b e-3 a f) x^3}{3 b^4}+\frac {f x^5}{5 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^5}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c-5 a b^2 d+9 a^2 b e-13 a^3 f\right ) x}{8 b^5 \left (a+b x^2\right )}+\frac {\left (3 b^3 c-15 a b^2 d+35 a^2 b e-63 a^3 f\right ) \int \frac {1}{a+b x^2} \, dx}{8 b^5} \\ & = -\frac {\left (b^3 c-5 a b^2 d+13 a^2 b e-25 a^3 f\right ) x}{4 a b^5}+\frac {(b e-3 a f) x^3}{3 b^4}+\frac {f x^5}{5 b^3}+\frac {\left (c-\frac {a \left (b^2 d-a b e+a^2 f\right )}{b^3}\right ) x^5}{4 a \left (a+b x^2\right )^2}-\frac {\left (b^3 c-5 a b^2 d+9 a^2 b e-13 a^3 f\right ) x}{8 b^5 \left (a+b x^2\right )}+\frac {\left (3 b^3 c-15 a b^2 d+35 a^2 b e-63 a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.85 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {x \left (945 a^4 f-525 a^3 b \left (e-3 f x^2\right )+a^2 b^2 \left (225 d-875 e x^2+504 f x^4\right )-a b^3 \left (45 c-375 d x^2+280 e x^4+72 f x^6\right )+b^4 x^2 \left (-75 c+8 \left (15 d x^2+5 e x^4+3 f x^6\right )\right )\right )}{120 b^5 \left (a+b x^2\right )^2}+\frac {\left (3 b^3 c-15 a b^2 d+35 a^2 b e-63 a^3 f\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{11/2}} \]

[In]

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

[Out]

(x*(945*a^4*f - 525*a^3*b*(e - 3*f*x^2) + a^2*b^2*(225*d - 875*e*x^2 + 504*f*x^4) - a*b^3*(45*c - 375*d*x^2 +
280*e*x^4 + 72*f*x^6) + b^4*x^2*(-75*c + 8*(15*d*x^2 + 5*e*x^4 + 3*f*x^6))))/(120*b^5*(a + b*x^2)^2) + ((3*b^3
*c - 15*a*b^2*d + 35*a^2*b*e - 63*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(11/2))

Maple [A] (verified)

Time = 3.50 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {\frac {1}{5} f \,x^{5} b^{2}-a b f \,x^{3}+\frac {1}{3} b^{2} e \,x^{3}+6 a^{2} f x -3 a b e x +b^{2} d x}{b^{5}}-\frac {\frac {\left (-\frac {17}{8} a^{3} b f +\frac {13}{8} a^{2} e \,b^{2}-\frac {9}{8} a \,b^{3} d +\frac {5}{8} b^{4} c \right ) x^{3}-\frac {a \left (15 f \,a^{3}-11 a^{2} b e +7 a \,b^{2} d -3 b^{3} c \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (63 f \,a^{3}-35 a^{2} b e +15 a \,b^{2} d -3 b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{b^{5}}\)	\(177\)
risch	\(\frac {f \,x^{5}}{5 b^{3}}-\frac {a f \,x^{3}}{b^{4}}+\frac {e \,x^{3}}{3 b^{3}}+\frac {6 a^{2} f x}{b^{5}}-\frac {3 a e x}{b^{4}}+\frac {d x}{b^{3}}+\frac {\left (\frac {17}{8} a^{3} b f -\frac {13}{8} a^{2} e \,b^{2}+\frac {9}{8} a \,b^{3} d -\frac {5}{8} b^{4} c \right ) x^{3}+\frac {a \left (15 f \,a^{3}-11 a^{2} b e +7 a \,b^{2} d -3 b^{3} c \right ) x}{8}}{b^{5} \left (b \,x^{2}+a \right )^{2}}-\frac {63 \ln \left (b x -\sqrt {-a b}\right ) f \,a^{3}}{16 b^{5} \sqrt {-a b}}+\frac {35 \ln \left (b x -\sqrt {-a b}\right ) a^{2} e}{16 b^{4} \sqrt {-a b}}-\frac {15 \ln \left (b x -\sqrt {-a b}\right ) a d}{16 b^{3} \sqrt {-a b}}+\frac {3 \ln \left (b x -\sqrt {-a b}\right ) c}{16 b^{2} \sqrt {-a b}}+\frac {63 \ln \left (-b x -\sqrt {-a b}\right ) f \,a^{3}}{16 b^{5} \sqrt {-a b}}-\frac {35 \ln \left (-b x -\sqrt {-a b}\right ) a^{2} e}{16 b^{4} \sqrt {-a b}}+\frac {15 \ln \left (-b x -\sqrt {-a b}\right ) a d}{16 b^{3} \sqrt {-a b}}-\frac {3 \ln \left (-b x -\sqrt {-a b}\right ) c}{16 b^{2} \sqrt {-a b}}\)	\(351\)

[In]

int(x^4*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/b^5*(1/5*f*x^5*b^2-a*b*f*x^3+1/3*b^2*e*x^3+6*a^2*f*x-3*a*b*e*x+b^2*d*x)-1/b^5*(((-17/8*a^3*b*f+13/8*a^2*e*b^
2-9/8*a*b^3*d+5/8*b^4*c)*x^3-1/8*a*(15*a^3*f-11*a^2*b*e+7*a*b^2*d-3*b^3*c)*x)/(b*x^2+a)^2+1/8*(63*a^3*f-35*a^2
*b*e+15*a*b^2*d-3*b^3*c)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 614, normalized size of antiderivative = 2.97 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\left [\frac {48 \, a b^{5} f x^{9} + 16 \, {\left (5 \, a b^{5} e - 9 \, a^{2} b^{4} f\right )} x^{7} + 16 \, {\left (15 \, a b^{5} d - 35 \, a^{2} b^{4} e + 63 \, a^{3} b^{3} f\right )} x^{5} - 50 \, {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 35 \, a^{3} b^{3} e - 63 \, a^{4} b^{2} f\right )} x^{3} + 15 \, {\left (3 \, a^{2} b^{3} c - 15 \, a^{3} b^{2} d + 35 \, a^{4} b e - 63 \, a^{5} f + {\left (3 \, b^{5} c - 15 \, a b^{4} d + 35 \, a^{2} b^{3} e - 63 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (3 \, a b^{4} c - 15 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 63 \, a^{4} b f\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) - 30 \, {\left (3 \, a^{2} b^{4} c - 15 \, a^{3} b^{3} d + 35 \, a^{4} b^{2} e - 63 \, a^{5} b f\right )} x}{240 \, {\left (a b^{8} x^{4} + 2 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}}, \frac {24 \, a b^{5} f x^{9} + 8 \, {\left (5 \, a b^{5} e - 9 \, a^{2} b^{4} f\right )} x^{7} + 8 \, {\left (15 \, a b^{5} d - 35 \, a^{2} b^{4} e + 63 \, a^{3} b^{3} f\right )} x^{5} - 25 \, {\left (3 \, a b^{5} c - 15 \, a^{2} b^{4} d + 35 \, a^{3} b^{3} e - 63 \, a^{4} b^{2} f\right )} x^{3} + 15 \, {\left (3 \, a^{2} b^{3} c - 15 \, a^{3} b^{2} d + 35 \, a^{4} b e - 63 \, a^{5} f + {\left (3 \, b^{5} c - 15 \, a b^{4} d + 35 \, a^{2} b^{3} e - 63 \, a^{3} b^{2} f\right )} x^{4} + 2 \, {\left (3 \, a b^{4} c - 15 \, a^{2} b^{3} d + 35 \, a^{3} b^{2} e - 63 \, a^{4} b f\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - 15 \, {\left (3 \, a^{2} b^{4} c - 15 \, a^{3} b^{3} d + 35 \, a^{4} b^{2} e - 63 \, a^{5} b f\right )} x}{120 \, {\left (a b^{8} x^{4} + 2 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}}\right ] \]

[In]

integrate(x^4*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/240*(48*a*b^5*f*x^9 + 16*(5*a*b^5*e - 9*a^2*b^4*f)*x^7 + 16*(15*a*b^5*d - 35*a^2*b^4*e + 63*a^3*b^3*f)*x^5
- 50*(3*a*b^5*c - 15*a^2*b^4*d + 35*a^3*b^3*e - 63*a^4*b^2*f)*x^3 + 15*(3*a^2*b^3*c - 15*a^3*b^2*d + 35*a^4*b*
e - 63*a^5*f + (3*b^5*c - 15*a*b^4*d + 35*a^2*b^3*e - 63*a^3*b^2*f)*x^4 + 2*(3*a*b^4*c - 15*a^2*b^3*d + 35*a^3
*b^2*e - 63*a^4*b*f)*x^2)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 30*(3*a^2*b^4*c - 15*a^3*
b^3*d + 35*a^4*b^2*e - 63*a^5*b*f)*x)/(a*b^8*x^4 + 2*a^2*b^7*x^2 + a^3*b^6), 1/120*(24*a*b^5*f*x^9 + 8*(5*a*b^
5*e - 9*a^2*b^4*f)*x^7 + 8*(15*a*b^5*d - 35*a^2*b^4*e + 63*a^3*b^3*f)*x^5 - 25*(3*a*b^5*c - 15*a^2*b^4*d + 35*
a^3*b^3*e - 63*a^4*b^2*f)*x^3 + 15*(3*a^2*b^3*c - 15*a^3*b^2*d + 35*a^4*b*e - 63*a^5*f + (3*b^5*c - 15*a*b^4*d
+ 35*a^2*b^3*e - 63*a^3*b^2*f)*x^4 + 2*(3*a*b^4*c - 15*a^2*b^3*d + 35*a^3*b^2*e - 63*a^4*b*f)*x^2)*sqrt(a*b)*
arctan(sqrt(a*b)*x/a) - 15*(3*a^2*b^4*c - 15*a^3*b^3*d + 35*a^4*b^2*e - 63*a^5*b*f)*x)/(a*b^8*x^4 + 2*a^2*b^7*
x^2 + a^3*b^6)]

Sympy [A] (verification not implemented)

Time = 15.94 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.35 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=x^{3} \left (- \frac {a f}{b^{4}} + \frac {e}{3 b^{3}}\right ) + x \left (\frac {6 a^{2} f}{b^{5}} - \frac {3 a e}{b^{4}} + \frac {d}{b^{3}}\right ) + \frac {\sqrt {- \frac {1}{a b^{11}}} \cdot \left (63 a^{3} f - 35 a^{2} b e + 15 a b^{2} d - 3 b^{3} c\right ) \log {\left (- a b^{5} \sqrt {- \frac {1}{a b^{11}}} + x \right )}}{16} - \frac {\sqrt {- \frac {1}{a b^{11}}} \cdot \left (63 a^{3} f - 35 a^{2} b e + 15 a b^{2} d - 3 b^{3} c\right ) \log {\left (a b^{5} \sqrt {- \frac {1}{a b^{11}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (17 a^{3} b f - 13 a^{2} b^{2} e + 9 a b^{3} d - 5 b^{4} c\right ) + x \left (15 a^{4} f - 11 a^{3} b e + 7 a^{2} b^{2} d - 3 a b^{3} c\right )}{8 a^{2} b^{5} + 16 a b^{6} x^{2} + 8 b^{7} x^{4}} + \frac {f x^{5}}{5 b^{3}} \]

[In]

integrate(x**4*(f*x**6+e*x**4+d*x**2+c)/(b*x**2+a)**3,x)

[Out]

x**3*(-a*f/b**4 + e/(3*b**3)) + x*(6*a**2*f/b**5 - 3*a*e/b**4 + d/b**3) + sqrt(-1/(a*b**11))*(63*a**3*f - 35*a
**2*b*e + 15*a*b**2*d - 3*b**3*c)*log(-a*b**5*sqrt(-1/(a*b**11)) + x)/16 - sqrt(-1/(a*b**11))*(63*a**3*f - 35*
a**2*b*e + 15*a*b**2*d - 3*b**3*c)*log(a*b**5*sqrt(-1/(a*b**11)) + x)/16 + (x**3*(17*a**3*b*f - 13*a**2*b**2*e
+ 9*a*b**3*d - 5*b**4*c) + x*(15*a**4*f - 11*a**3*b*e + 7*a**2*b**2*d - 3*a*b**3*c))/(8*a**2*b**5 + 16*a*b**6
*x**2 + 8*b**7*x**4) + f*x**5/(5*b**3)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.93 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=-\frac {{\left (5 \, b^{4} c - 9 \, a b^{3} d + 13 \, a^{2} b^{2} e - 17 \, a^{3} b f\right )} x^{3} + {\left (3 \, a b^{3} c - 7 \, a^{2} b^{2} d + 11 \, a^{3} b e - 15 \, a^{4} f\right )} x}{8 \, {\left (b^{7} x^{4} + 2 \, a b^{6} x^{2} + a^{2} b^{5}\right )}} + \frac {{\left (3 \, b^{3} c - 15 \, a b^{2} d + 35 \, a^{2} b e - 63 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} + \frac {3 \, b^{2} f x^{5} + 5 \, {\left (b^{2} e - 3 \, a b f\right )} x^{3} + 15 \, {\left (b^{2} d - 3 \, a b e + 6 \, a^{2} f\right )} x}{15 \, b^{5}} \]

[In]

integrate(x^4*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

-1/8*((5*b^4*c - 9*a*b^3*d + 13*a^2*b^2*e - 17*a^3*b*f)*x^3 + (3*a*b^3*c - 7*a^2*b^2*d + 11*a^3*b*e - 15*a^4*f
)*x)/(b^7*x^4 + 2*a*b^6*x^2 + a^2*b^5) + 1/8*(3*b^3*c - 15*a*b^2*d + 35*a^2*b*e - 63*a^3*f)*arctan(b*x/sqrt(a*
b))/(sqrt(a*b)*b^5) + 1/15*(3*b^2*f*x^5 + 5*(b^2*e - 3*a*b*f)*x^3 + 15*(b^2*d - 3*a*b*e + 6*a^2*f)*x)/b^5

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.94 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=\frac {{\left (3 \, b^{3} c - 15 \, a b^{2} d + 35 \, a^{2} b e - 63 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{5}} - \frac {5 \, b^{4} c x^{3} - 9 \, a b^{3} d x^{3} + 13 \, a^{2} b^{2} e x^{3} - 17 \, a^{3} b f x^{3} + 3 \, a b^{3} c x - 7 \, a^{2} b^{2} d x + 11 \, a^{3} b e x - 15 \, a^{4} f x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{5}} + \frac {3 \, b^{12} f x^{5} + 5 \, b^{12} e x^{3} - 15 \, a b^{11} f x^{3} + 15 \, b^{12} d x - 45 \, a b^{11} e x + 90 \, a^{2} b^{10} f x}{15 \, b^{15}} \]

[In]

integrate(x^4*(f*x^6+e*x^4+d*x^2+c)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

1/8*(3*b^3*c - 15*a*b^2*d + 35*a^2*b*e - 63*a^3*f)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/8*(5*b^4*c*x^3 -
9*a*b^3*d*x^3 + 13*a^2*b^2*e*x^3 - 17*a^3*b*f*x^3 + 3*a*b^3*c*x - 7*a^2*b^2*d*x + 11*a^3*b*e*x - 15*a^4*f*x)/(
(b*x^2 + a)^2*b^5) + 1/15*(3*b^12*f*x^5 + 5*b^12*e*x^3 - 15*a*b^11*f*x^3 + 15*b^12*d*x - 45*a*b^11*e*x + 90*a^
2*b^10*f*x)/b^15

Mupad [B] (verification not implemented)

Time = 5.62 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\left (a+b x^2\right )^3} \, dx=x^3\,\left (\frac {e}{3\,b^3}-\frac {a\,f}{b^4}\right )-x\,\left (\frac {3\,a^2\,f}{b^5}-\frac {d}{b^3}+\frac {3\,a\,\left (\frac {e}{b^3}-\frac {3\,a\,f}{b^4}\right )}{b}\right )-\frac {x^3\,\left (-\frac {17\,f\,a^3\,b}{8}+\frac {13\,e\,a^2\,b^2}{8}-\frac {9\,d\,a\,b^3}{8}+\frac {5\,c\,b^4}{8}\right )-x\,\left (\frac {15\,f\,a^4}{8}-\frac {11\,e\,a^3\,b}{8}+\frac {7\,d\,a^2\,b^2}{8}-\frac {3\,c\,a\,b^3}{8}\right )}{a^2\,b^5+2\,a\,b^6\,x^2+b^7\,x^4}+\frac {f\,x^5}{5\,b^3}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (-63\,f\,a^3+35\,e\,a^2\,b-15\,d\,a\,b^2+3\,c\,b^3\right )}{8\,\sqrt {a}\,b^{11/2}} \]

[In]

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

[Out]

x^3*(e/(3*b^3) - (a*f)/b^4) - x*((3*a^2*f)/b^5 - d/b^3 + (3*a*(e/b^3 - (3*a*f)/b^4))/b) - (x^3*((5*b^4*c)/8 +
(13*a^2*b^2*e)/8 - (9*a*b^3*d)/8 - (17*a^3*b*f)/8) - x*((15*a^4*f)/8 + (7*a^2*b^2*d)/8 - (3*a*b^3*c)/8 - (11*a
^3*b*e)/8))/(a^2*b^5 + b^7*x^4 + 2*a*b^6*x^2) + (f*x^5)/(5*b^3) + (atan((b^(1/2)*x)/a^(1/2))*(3*b^3*c - 63*a^3
*f - 15*a*b^2*d + 35*a^2*b*e))/(8*a^(1/2)*b^(11/2))