3.2.40 \(\int \frac {c+d x^2+e x^4+f x^6}{x^6 (a+b x^2)^3} \, dx\) [140]

Optimal. Leaf size=196 \[ -\frac {c}{5 a^3 x^5}+\frac {3 b c-a d}{3 a^4 x^3}-\frac {6 b^2 c-3 a b d+a^2 e}{a^5 x}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}-\frac {\left (63 b^3 c-35 a b^2 d+15 a^2 b e-3 a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2} \sqrt {b}} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1819, 1816, 211} \begin {gather*} \frac {3 b c-a d}{3 a^4 x^3}-\frac {c}{5 a^3 x^5}-\frac {a^2 e-3 a b d+6 b^2 c}{a^5 x}-\frac {\text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-3 a^3 f+15 a^2 b e-35 a b^2 d+63 b^3 c\right )}{8 a^{11/2} \sqrt {b}}-\frac {x \left (-3 a^3 f+7 a^2 b e-11 a b^2 d+15 b^3 c\right )}{8 a^5 \left (a+b x^2\right )}-\frac {x \left (a^3 (-f)+a^2 b e-a b^2 d+b^3 c\right )}{4 a^4 \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*c/(a^3*x^5) + (3*b*c - a*d)/(3*a^4*x^3) - (6*b^2*c - 3*a*b*d + a^2*e)/(a^5*x) - ((b^3*c - a*b^2*d + a^2*b
*e - a^3*f)*x)/(4*a^4*(a + b*x^2)^2) - ((15*b^3*c - 11*a*b^2*d + 7*a^2*b*e - 3*a^3*f)*x)/(8*a^5*(a + b*x^2)) -
((63*b^3*c - 35*a*b^2*d + 15*a^2*b*e - 3*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(11/2)*Sqrt[b])

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 1816

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

Rule 1819

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

Rubi steps

\begin {align*} \int \frac {c+d x^2+e x^4+f x^6}{x^6 \left (a+b x^2\right )^3} \, dx &=-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {\int \frac {-4 c+4 \left (\frac {b c}{a}-d\right ) x^2-\frac {4 \left (b^2 c-a b d+a^2 e\right ) x^4}{a^2}+\frac {3 \left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x^6}{a^3}}{x^6 \left (a+b x^2\right )^2} \, dx}{4 a}\\ &=-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}+\frac {\int \frac {8 c-8 \left (\frac {2 b c}{a}-d\right ) x^2+8 \left (\frac {3 b^2 c}{a^2}-\frac {2 b d}{a}+e\right ) x^4-\frac {\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x^6}{a^3}}{x^6 \left (a+b x^2\right )} \, dx}{8 a^2}\\ &=-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}+\frac {\int \left (\frac {8 c}{a x^6}+\frac {8 (-3 b c+a d)}{a^2 x^4}+\frac {8 \left (6 b^2 c-3 a b d+a^2 e\right )}{a^3 x^2}+\frac {-63 b^3 c+35 a b^2 d-15 a^2 b e+3 a^3 f}{a^3 \left (a+b x^2\right )}\right ) \, dx}{8 a^2}\\ &=-\frac {c}{5 a^3 x^5}+\frac {3 b c-a d}{3 a^4 x^3}-\frac {6 b^2 c-3 a b d+a^2 e}{a^5 x}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}-\frac {\left (63 b^3 c-35 a b^2 d+15 a^2 b e-3 a^3 f\right ) \int \frac {1}{a+b x^2} \, dx}{8 a^5}\\ &=-\frac {c}{5 a^3 x^5}+\frac {3 b c-a d}{3 a^4 x^3}-\frac {6 b^2 c-3 a b d+a^2 e}{a^5 x}-\frac {\left (b^3 c-a b^2 d+a^2 b e-a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}-\frac {\left (15 b^3 c-11 a b^2 d+7 a^2 b e-3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}-\frac {\left (63 b^3 c-35 a b^2 d+15 a^2 b e-3 a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2} \sqrt {b}}\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 196, normalized size = 1.00 \begin {gather*} -\frac {c}{5 a^3 x^5}+\frac {3 b c-a d}{3 a^4 x^3}+\frac {-6 b^2 c+3 a b d-a^2 e}{a^5 x}+\frac {\left (-b^3 c+a b^2 d-a^2 b e+a^3 f\right ) x}{4 a^4 \left (a+b x^2\right )^2}+\frac {\left (-15 b^3 c+11 a b^2 d-7 a^2 b e+3 a^3 f\right ) x}{8 a^5 \left (a+b x^2\right )}+\frac {\left (-63 b^3 c+35 a b^2 d-15 a^2 b e+3 a^3 f\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{11/2} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*c/(a^3*x^5) + (3*b*c - a*d)/(3*a^4*x^3) + (-6*b^2*c + 3*a*b*d - a^2*e)/(a^5*x) + ((-(b^3*c) + a*b^2*d - a
^2*b*e + a^3*f)*x)/(4*a^4*(a + b*x^2)^2) + ((-15*b^3*c + 11*a*b^2*d - 7*a^2*b*e + 3*a^3*f)*x)/(8*a^5*(a + b*x^
2)) + ((-63*b^3*c + 35*a*b^2*d - 15*a^2*b*e + 3*a^3*f)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(11/2)*Sqrt[b])

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Maple [A]
time = 0.16, size = 176, normalized size = 0.90

method	result	size
default	\(\frac {\frac {\left (\frac {3}{8} a^{3} b f -\frac {7}{8} a^{2} e \,b^{2}+\frac {11}{8} a d \,b^{3}-\frac {15}{8} c \,b^{4}\right ) x^{3}+\frac {a \left (5 a^{3} f -9 a^{2} b e +13 a \,b^{2} d -17 b^{3} c \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (3 a^{3} f -15 a^{2} b e +35 a \,b^{2} d -63 b^{3} c \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{a^{5}}-\frac {c}{5 a^{3} x^{5}}-\frac {a d -3 b c}{3 a^{4} x^{3}}-\frac {a^{2} e -3 a b d +6 b^{2} c}{a^{5} x}\)	\(176\)
risch	\(\frac {\frac {b \left (3 a^{3} f -15 a^{2} b e +35 a \,b^{2} d -63 b^{3} c \right ) x^{8}}{8 a^{5}}+\frac {5 \left (3 a^{3} f -15 a^{2} b e +35 a \,b^{2} d -63 b^{3} c \right ) x^{6}}{24 a^{4}}-\frac {\left (15 a^{2} e -35 a b d +63 b^{2} c \right ) x^{4}}{15 a^{3}}-\frac {\left (5 a d -9 b c \right ) x^{2}}{15 a^{2}}-\frac {c}{5 a}}{x^{5} \left (b \,x^{2}+a \right )^{2}}-\frac {3 \ln \left (-\sqrt {-a b}\, x +a \right ) f}{16 \sqrt {-a b}\, a^{2}}+\frac {15 \ln \left (-\sqrt {-a b}\, x +a \right ) b e}{16 \sqrt {-a b}\, a^{3}}-\frac {35 \ln \left (-\sqrt {-a b}\, x +a \right ) b^{2} d}{16 \sqrt {-a b}\, a^{4}}+\frac {63 \ln \left (-\sqrt {-a b}\, x +a \right ) b^{3} c}{16 \sqrt {-a b}\, a^{5}}+\frac {3 \ln \left (-\sqrt {-a b}\, x -a \right ) f}{16 \sqrt {-a b}\, a^{2}}-\frac {15 \ln \left (-\sqrt {-a b}\, x -a \right ) b e}{16 \sqrt {-a b}\, a^{3}}+\frac {35 \ln \left (-\sqrt {-a b}\, x -a \right ) b^{2} d}{16 \sqrt {-a b}\, a^{4}}-\frac {63 \ln \left (-\sqrt {-a b}\, x -a \right ) b^{3} c}{16 \sqrt {-a b}\, a^{5}}\)	\(350\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 206, normalized size = 1.05 \begin {gather*} -\frac {15 \, {\left (63 \, b^{4} c - 35 \, a b^{3} d - 3 \, a^{3} b f + 15 \, a^{2} b^{2} e\right )} x^{8} + 25 \, {\left (63 \, a b^{3} c - 35 \, a^{2} b^{2} d - 3 \, a^{4} f + 15 \, a^{3} b e\right )} x^{6} + 24 \, a^{4} c + 8 \, {\left (63 \, a^{2} b^{2} c - 35 \, a^{3} b d + 15 \, a^{4} e\right )} x^{4} - 8 \, {\left (9 \, a^{3} b c - 5 \, a^{4} d\right )} x^{2}}{120 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}} - \frac {{\left (63 \, b^{3} c - 35 \, a b^{2} d - 3 \, a^{3} f + 15 \, a^{2} b e\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*(63*b^4*c - 35*a*b^3*d - 3*a^3*b*f + 15*a^2*b^2*e)*x^8 + 25*(63*a*b^3*c - 35*a^2*b^2*d - 3*a^4*f +
15*a^3*b*e)*x^6 + 24*a^4*c + 8*(63*a^2*b^2*c - 35*a^3*b*d + 15*a^4*e)*x^4 - 8*(9*a^3*b*c - 5*a^4*d)*x^2)/(a^5*
b^2*x^9 + 2*a^6*b*x^7 + a^7*x^5) - 1/8*(63*b^3*c - 35*a*b^2*d - 3*a^3*f + 15*a^2*b*e)*arctan(b*x/sqrt(a*b))/(s
qrt(a*b)*a^5)

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Fricas [A]
time = 5.35, size = 668, normalized size = 3.41 \begin {gather*} \left [-\frac {30 \, {\left (63 \, a b^{5} c - 35 \, a^{2} b^{4} d - 3 \, a^{4} b^{2} f\right )} x^{8} + 48 \, a^{5} b c + 50 \, {\left (63 \, a^{2} b^{4} c - 35 \, a^{3} b^{3} d - 3 \, a^{5} b f\right )} x^{6} + 112 \, {\left (9 \, a^{3} b^{3} c - 5 \, a^{4} b^{2} d\right )} x^{4} - 16 \, {\left (9 \, a^{4} b^{2} c - 5 \, a^{5} b d\right )} x^{2} + 15 \, {\left ({\left (63 \, b^{5} c - 35 \, a b^{4} d - 3 \, a^{3} b^{2} f\right )} x^{9} + 2 \, {\left (63 \, a b^{4} c - 35 \, a^{2} b^{3} d - 3 \, a^{4} b f\right )} x^{7} + {\left (63 \, a^{2} b^{3} c - 35 \, a^{3} b^{2} d - 3 \, a^{5} f\right )} x^{5} + 15 \, {\left (a^{2} b^{3} x^{9} + 2 \, a^{3} b^{2} x^{7} + a^{4} b x^{5}\right )} e\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 30 \, {\left (15 \, a^{3} b^{3} x^{8} + 25 \, a^{4} b^{2} x^{6} + 8 \, a^{5} b x^{4}\right )} e}{240 \, {\left (a^{6} b^{3} x^{9} + 2 \, a^{7} b^{2} x^{7} + a^{8} b x^{5}\right )}}, -\frac {15 \, {\left (63 \, a b^{5} c - 35 \, a^{2} b^{4} d - 3 \, a^{4} b^{2} f\right )} x^{8} + 24 \, a^{5} b c + 25 \, {\left (63 \, a^{2} b^{4} c - 35 \, a^{3} b^{3} d - 3 \, a^{5} b f\right )} x^{6} + 56 \, {\left (9 \, a^{3} b^{3} c - 5 \, a^{4} b^{2} d\right )} x^{4} - 8 \, {\left (9 \, a^{4} b^{2} c - 5 \, a^{5} b d\right )} x^{2} + 15 \, {\left ({\left (63 \, b^{5} c - 35 \, a b^{4} d - 3 \, a^{3} b^{2} f\right )} x^{9} + 2 \, {\left (63 \, a b^{4} c - 35 \, a^{2} b^{3} d - 3 \, a^{4} b f\right )} x^{7} + {\left (63 \, a^{2} b^{3} c - 35 \, a^{3} b^{2} d - 3 \, a^{5} f\right )} x^{5} + 15 \, {\left (a^{2} b^{3} x^{9} + 2 \, a^{3} b^{2} x^{7} + a^{4} b x^{5}\right )} e\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 15 \, {\left (15 \, a^{3} b^{3} x^{8} + 25 \, a^{4} b^{2} x^{6} + 8 \, a^{5} b x^{4}\right )} e}{120 \, {\left (a^{6} b^{3} x^{9} + 2 \, a^{7} b^{2} x^{7} + a^{8} b x^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/240*(30*(63*a*b^5*c - 35*a^2*b^4*d - 3*a^4*b^2*f)*x^8 + 48*a^5*b*c + 50*(63*a^2*b^4*c - 35*a^3*b^3*d - 3*a
^5*b*f)*x^6 + 112*(9*a^3*b^3*c - 5*a^4*b^2*d)*x^4 - 16*(9*a^4*b^2*c - 5*a^5*b*d)*x^2 + 15*((63*b^5*c - 35*a*b^
4*d - 3*a^3*b^2*f)*x^9 + 2*(63*a*b^4*c - 35*a^2*b^3*d - 3*a^4*b*f)*x^7 + (63*a^2*b^3*c - 35*a^3*b^2*d - 3*a^5*
f)*x^5 + 15*(a^2*b^3*x^9 + 2*a^3*b^2*x^7 + a^4*b*x^5)*e)*sqrt(-a*b)*log((b*x^2 + 2*sqrt(-a*b)*x - a)/(b*x^2 +
a)) + 30*(15*a^3*b^3*x^8 + 25*a^4*b^2*x^6 + 8*a^5*b*x^4)*e)/(a^6*b^3*x^9 + 2*a^7*b^2*x^7 + a^8*b*x^5), -1/120*
(15*(63*a*b^5*c - 35*a^2*b^4*d - 3*a^4*b^2*f)*x^8 + 24*a^5*b*c + 25*(63*a^2*b^4*c - 35*a^3*b^3*d - 3*a^5*b*f)*
x^6 + 56*(9*a^3*b^3*c - 5*a^4*b^2*d)*x^4 - 8*(9*a^4*b^2*c - 5*a^5*b*d)*x^2 + 15*((63*b^5*c - 35*a*b^4*d - 3*a^
3*b^2*f)*x^9 + 2*(63*a*b^4*c - 35*a^2*b^3*d - 3*a^4*b*f)*x^7 + (63*a^2*b^3*c - 35*a^3*b^2*d - 3*a^5*f)*x^5 + 1
5*(a^2*b^3*x^9 + 2*a^3*b^2*x^7 + a^4*b*x^5)*e)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + 15*(15*a^3*b^3*x^8 + 25*a^4*b
^2*x^6 + 8*a^5*b*x^4)*e)/(a^6*b^3*x^9 + 2*a^7*b^2*x^7 + a^8*b*x^5)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 1.41, size = 198, normalized size = 1.01 \begin {gather*} -\frac {{\left (63 \, b^{3} c - 35 \, a b^{2} d - 3 \, a^{3} f + 15 \, a^{2} b e\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{5}} - \frac {15 \, b^{4} c x^{3} - 11 \, a b^{3} d x^{3} - 3 \, a^{3} b f x^{3} + 7 \, a^{2} b^{2} x^{3} e + 17 \, a b^{3} c x - 13 \, a^{2} b^{2} d x - 5 \, a^{4} f x + 9 \, a^{3} b x e}{8 \, {\left (b x^{2} + a\right )}^{2} a^{5}} - \frac {90 \, b^{2} c x^{4} - 45 \, a b d x^{4} + 15 \, a^{2} x^{4} e - 15 \, a b c x^{2} + 5 \, a^{2} d x^{2} + 3 \, a^{2} c}{15 \, a^{5} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.04, size = 192, normalized size = 0.98 \begin {gather*} -\frac {\frac {c}{5\,a}+\frac {5\,x^6\,\left (-3\,f\,a^3+15\,e\,a^2\,b-35\,d\,a\,b^2+63\,c\,b^3\right )}{24\,a^4}+\frac {x^2\,\left (5\,a\,d-9\,b\,c\right )}{15\,a^2}+\frac {x^4\,\left (15\,e\,a^2-35\,d\,a\,b+63\,c\,b^2\right )}{15\,a^3}+\frac {b\,x^8\,\left (-3\,f\,a^3+15\,e\,a^2\,b-35\,d\,a\,b^2+63\,c\,b^3\right )}{8\,a^5}}{a^2\,x^5+2\,a\,b\,x^7+b^2\,x^9}-\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (-3\,f\,a^3+15\,e\,a^2\,b-35\,d\,a\,b^2+63\,c\,b^3\right )}{8\,a^{11/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (c/(5*a) + (5*x^6*(63*b^3*c - 3*a^3*f - 35*a*b^2*d + 15*a^2*b*e))/(24*a^4) + (x^2*(5*a*d - 9*b*c))/(15*a^2)
+ (x^4*(63*b^2*c + 15*a^2*e - 35*a*b*d))/(15*a^3) + (b*x^8*(63*b^3*c - 3*a^3*f - 35*a*b^2*d + 15*a^2*b*e))/(8*
a^5))/(a^2*x^5 + b^2*x^9 + 2*a*b*x^7) - (atan((b^(1/2)*x)/a^(1/2))*(63*b^3*c - 3*a^3*f - 35*a*b^2*d + 15*a^2*b
*e))/(8*a^(11/2)*b^(1/2))

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