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

Optimal. Leaf size=153 \[ -\frac {c}{a^3 x}-\frac {x \left (-\frac {a^2 f}{b^2}+\frac {b c}{a}+\frac {a e}{b}-d\right )}{4 a \left (a+b x^2\right )^2}-\frac {x \left (5 a^3 f-a^2 b e-3 a b^2 d+7 b^3 c\right )}{8 a^3 b^2 \left (a+b x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-3 a^3 f-a^2 b e-3 a b^2 d+15 b^3 c\right )}{8 a^{7/2} b^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1805, 1259, 453, 205} \[ -\frac {x \left (-a^2 b e+5 a^3 f-3 a b^2 d+7 b^3 c\right )}{8 a^3 b^2 \left (a+b x^2\right )}-\frac {x \left (-\frac {a^2 f}{b^2}+\frac {b c}{a}+\frac {a e}{b}-d\right )}{4 a \left (a+b x^2\right )^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-a^2 b e-3 a^3 f-3 a b^2 d+15 b^3 c\right )}{8 a^{7/2} b^{5/2}}-\frac {c}{a^3 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 1259

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[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(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] && ILtQ[m/2, 0]

Rule 1805

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

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Mathematica [A]  time = 0.13, size = 155, normalized size = 1.01 \[ -\frac {c}{a^3 x}-\frac {x \left (5 a^3 f-a^2 b e-3 a b^2 d+7 b^3 c\right )}{8 a^3 b^2 \left (a+b x^2\right )}+\frac {x \left (a^3 f-a^2 b e+a b^2 d-b^3 c\right )}{4 a^2 b^2 \left (a+b x^2\right )^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (3 a^3 f+a^2 b e+3 a b^2 d-15 b^3 c\right )}{8 a^{7/2} b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.63, size = 517, normalized size = 3.38 \[ \left [-\frac {16 \, a^{3} b^{3} c + 2 \, {\left (15 \, a b^{5} c - 3 \, a^{2} b^{4} d - a^{3} b^{3} e + 5 \, a^{4} b^{2} f\right )} x^{4} + 2 \, {\left (25 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + a^{4} b^{2} e + 3 \, a^{5} b f\right )} x^{2} - {\left ({\left (15 \, b^{5} c - 3 \, a b^{4} d - a^{2} b^{3} e - 3 \, a^{3} b^{2} f\right )} x^{5} + 2 \, {\left (15 \, a b^{4} c - 3 \, a^{2} b^{3} d - a^{3} b^{2} e - 3 \, a^{4} b f\right )} x^{3} + {\left (15 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d - a^{4} b e - 3 \, a^{5} f\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{4} b^{5} x^{5} + 2 \, a^{5} b^{4} x^{3} + a^{6} b^{3} x\right )}}, -\frac {8 \, a^{3} b^{3} c + {\left (15 \, a b^{5} c - 3 \, a^{2} b^{4} d - a^{3} b^{3} e + 5 \, a^{4} b^{2} f\right )} x^{4} + {\left (25 \, a^{2} b^{4} c - 5 \, a^{3} b^{3} d + a^{4} b^{2} e + 3 \, a^{5} b f\right )} x^{2} + {\left ({\left (15 \, b^{5} c - 3 \, a b^{4} d - a^{2} b^{3} e - 3 \, a^{3} b^{2} f\right )} x^{5} + 2 \, {\left (15 \, a b^{4} c - 3 \, a^{2} b^{3} d - a^{3} b^{2} e - 3 \, a^{4} b f\right )} x^{3} + {\left (15 \, a^{2} b^{3} c - 3 \, a^{3} b^{2} d - a^{4} b e - 3 \, a^{5} f\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{4} b^{5} x^{5} + 2 \, a^{5} b^{4} x^{3} + a^{6} b^{3} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(16*a^3*b^3*c + 2*(15*a*b^5*c - 3*a^2*b^4*d - a^3*b^3*e + 5*a^4*b^2*f)*x^4 + 2*(25*a^2*b^4*c - 5*a^3*b^
3*d + a^4*b^2*e + 3*a^5*b*f)*x^2 - ((15*b^5*c - 3*a*b^4*d - a^2*b^3*e - 3*a^3*b^2*f)*x^5 + 2*(15*a*b^4*c - 3*a
^2*b^3*d - a^3*b^2*e - 3*a^4*b*f)*x^3 + (15*a^2*b^3*c - 3*a^3*b^2*d - a^4*b*e - 3*a^5*f)*x)*sqrt(-a*b)*log((b*
x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^4*b^5*x^5 + 2*a^5*b^4*x^3 + a^6*b^3*x), -1/8*(8*a^3*b^3*c + (15*a*b
^5*c - 3*a^2*b^4*d - a^3*b^3*e + 5*a^4*b^2*f)*x^4 + (25*a^2*b^4*c - 5*a^3*b^3*d + a^4*b^2*e + 3*a^5*b*f)*x^2 +
((15*b^5*c - 3*a*b^4*d - a^2*b^3*e - 3*a^3*b^2*f)*x^5 + 2*(15*a*b^4*c - 3*a^2*b^3*d - a^3*b^2*e - 3*a^4*b*f)*
x^3 + (15*a^2*b^3*c - 3*a^3*b^2*d - a^4*b*e - 3*a^5*f)*x)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^4*b^5*x^5 + 2*a^
5*b^4*x^3 + a^6*b^3*x)]

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giac [A]  time = 0.45, size = 153, normalized size = 1.00 \[ -\frac {c}{a^{3} x} - \frac {{\left (15 \, b^{3} c - 3 \, a b^{2} d - 3 \, a^{3} f - a^{2} b e\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} b^{2}} - \frac {7 \, b^{4} c x^{3} - 3 \, a b^{3} d x^{3} + 5 \, a^{3} b f x^{3} - a^{2} b^{2} x^{3} e + 9 \, a b^{3} c x - 5 \, a^{2} b^{2} d x + 3 \, a^{4} f x + a^{3} b x e}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 237, normalized size = 1.55 \[ \frac {e \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a}+\frac {3 b d \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{2}}-\frac {7 b^{2} c \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} a^{3}}-\frac {5 f \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b}-\frac {3 a f x}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}+\frac {5 d x}{8 \left (b \,x^{2}+a \right )^{2} a}-\frac {9 b c x}{8 \left (b \,x^{2}+a \right )^{2} a^{2}}-\frac {e x}{8 \left (b \,x^{2}+a \right )^{2} b}+\frac {e \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a b}+\frac {3 d \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{2}}-\frac {15 b c \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{3}}+\frac {3 f \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{2}}-\frac {c}{a^{3} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.99, size = 161, normalized size = 1.05 \[ -\frac {8 \, a^{2} b^{2} c + {\left (15 \, b^{4} c - 3 \, a b^{3} d - a^{2} b^{2} e + 5 \, a^{3} b f\right )} x^{4} + {\left (25 \, a b^{3} c - 5 \, a^{2} b^{2} d + a^{3} b e + 3 \, a^{4} f\right )} x^{2}}{8 \, {\left (a^{3} b^{4} x^{5} + 2 \, a^{4} b^{3} x^{3} + a^{5} b^{2} x\right )}} - \frac {{\left (15 \, b^{3} c - 3 \, a b^{2} d - a^{2} b e - 3 \, a^{3} f\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*a^2*b^2*c + (15*b^4*c - 3*a*b^3*d - a^2*b^2*e + 5*a^3*b*f)*x^4 + (25*a*b^3*c - 5*a^2*b^2*d + a^3*b*e +
3*a^4*f)*x^2)/(a^3*b^4*x^5 + 2*a^4*b^3*x^3 + a^5*b^2*x) - 1/8*(15*b^3*c - 3*a*b^2*d - a^2*b*e - 3*a^3*f)*arct
an(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b^2)

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mupad [B]  time = 1.09, size = 149, normalized size = 0.97 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (3\,f\,a^3+e\,a^2\,b+3\,d\,a\,b^2-15\,c\,b^3\right )}{8\,a^{7/2}\,b^{5/2}}-\frac {\frac {c}{a}+\frac {x^4\,\left (5\,f\,a^3-e\,a^2\,b-3\,d\,a\,b^2+15\,c\,b^3\right )}{8\,a^3\,b}+\frac {x^2\,\left (3\,f\,a^3+e\,a^2\,b-5\,d\,a\,b^2+25\,c\,b^3\right )}{8\,a^2\,b^2}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((b^(1/2)*x)/a^(1/2))*(3*a^3*f - 15*b^3*c + 3*a*b^2*d + a^2*b*e))/(8*a^(7/2)*b^(5/2)) - (c/a + (x^4*(15*b
^3*c + 5*a^3*f - 3*a*b^2*d - a^2*b*e))/(8*a^3*b) + (x^2*(25*b^3*c + 3*a^3*f - 5*a*b^2*d + a^2*b*e))/(8*a^2*b^2
))/(a^2*x + b^2*x^5 + 2*a*b*x^3)

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sympy [A]  time = 26.56, size = 250, normalized size = 1.63 \[ - \frac {\sqrt {- \frac {1}{a^{7} b^{5}}} \left (3 a^{3} f + a^{2} b e + 3 a b^{2} d - 15 b^{3} c\right ) \log {\left (- a^{4} b^{2} \sqrt {- \frac {1}{a^{7} b^{5}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{7} b^{5}}} \left (3 a^{3} f + a^{2} b e + 3 a b^{2} d - 15 b^{3} c\right ) \log {\left (a^{4} b^{2} \sqrt {- \frac {1}{a^{7} b^{5}}} + x \right )}}{16} + \frac {- 8 a^{2} b^{2} c + x^{4} \left (- 5 a^{3} b f + a^{2} b^{2} e + 3 a b^{3} d - 15 b^{4} c\right ) + x^{2} \left (- 3 a^{4} f - a^{3} b e + 5 a^{2} b^{2} d - 25 a b^{3} c\right )}{8 a^{5} b^{2} x + 16 a^{4} b^{3} x^{3} + 8 a^{3} b^{4} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-1/(a**7*b**5))*(3*a**3*f + a**2*b*e + 3*a*b**2*d - 15*b**3*c)*log(-a**4*b**2*sqrt(-1/(a**7*b**5)) + x)/
16 + sqrt(-1/(a**7*b**5))*(3*a**3*f + a**2*b*e + 3*a*b**2*d - 15*b**3*c)*log(a**4*b**2*sqrt(-1/(a**7*b**5)) +
x)/16 + (-8*a**2*b**2*c + x**4*(-5*a**3*b*f + a**2*b**2*e + 3*a*b**3*d - 15*b**4*c) + x**2*(-3*a**4*f - a**3*b
*e + 5*a**2*b**2*d - 25*a*b**3*c))/(8*a**5*b**2*x + 16*a**4*b**3*x**3 + 8*a**3*b**4*x**5)

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