3.270 \(\int \frac{c+d x^2}{x^4 (a+b x^2)^2} \, dx\)

Optimal. Leaf size=90 \[ \frac{b x (b c-a d)}{2 a^3 \left (a+b x^2\right )}+\frac{2 b c-a d}{a^3 x}+\frac{\sqrt{b} (5 b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{c}{3 a^2 x^3} \]

[Out]

-c/(3*a^2*x^3) + (2*b*c - a*d)/(a^3*x) + (b*(b*c - a*d)*x)/(2*a^3*(a + b*x^2)) + (Sqrt[b]*(5*b*c - 3*a*d)*ArcT
an[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2))

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Rubi [A]  time = 0.106381, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {456, 1261, 205} \[ \frac{b x (b c-a d)}{2 a^3 \left (a+b x^2\right )}+\frac{2 b c-a d}{a^3 x}+\frac{\sqrt{b} (5 b c-3 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{c}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-c/(3*a^2*x^3) + (2*b*c - a*d)/(a^3*x) + (b*(b*c - a*d)*x)/(2*a^3*(a + b*x^2)) + (Sqrt[b]*(5*b*c - 3*a*d)*ArcT
an[(Sqrt[b]*x)/Sqrt[a]])/(2*a^(7/2))

Rule 456

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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

Mathematica [A]  time = 0.070273, size = 90, normalized size = 1. \[ -\frac{b x (a d-b c)}{2 a^3 \left (a+b x^2\right )}+\frac{2 b c-a d}{a^3 x}-\frac{\sqrt{b} (3 a d-5 b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 a^{7/2}}-\frac{c}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 110, normalized size = 1.2 \begin{align*} -{\frac{c}{3\,{a}^{2}{x}^{3}}}-{\frac{d}{{a}^{2}x}}+2\,{\frac{bc}{{a}^{3}x}}-{\frac{bdx}{2\,{a}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{{b}^{2}cx}{2\,{a}^{3} \left ( b{x}^{2}+a \right ) }}-{\frac{3\,bd}{2\,{a}^{2}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,{b}^{2}c}{2\,{a}^{3}}\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((d*x^2+c)/x^4/(b*x^2+a)^2,x)

[Out]

-1/3*c/a^2/x^3-1/a^2/x*d+2/a^3/x*b*c-1/2/a^2*b*x/(b*x^2+a)*d+1/2/a^3*b^2*x/(b*x^2+a)*c-3/2/a^2*b/(a*b)^(1/2)*a
rctan(b*x/(a*b)^(1/2))*d+5/2/a^3*b^2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*c

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52035, size = 529, normalized size = 5.88 \begin{align*} \left [\frac{6 \,{\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} - 4 \, a^{2} c + 4 \,{\left (5 \, a b c - 3 \, a^{2} d\right )} x^{2} - 3 \,{\left ({\left (5 \, b^{2} c - 3 \, a b d\right )} x^{5} +{\left (5 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x^{2} - 2 \, a x \sqrt{-\frac{b}{a}} - a}{b x^{2} + a}\right )}{12 \,{\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, \frac{3 \,{\left (5 \, b^{2} c - 3 \, a b d\right )} x^{4} - 2 \, a^{2} c + 2 \,{\left (5 \, a b c - 3 \, a^{2} d\right )} x^{2} + 3 \,{\left ({\left (5 \, b^{2} c - 3 \, a b d\right )} x^{5} +{\left (5 \, a b c - 3 \, a^{2} d\right )} x^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (x \sqrt{\frac{b}{a}}\right )}{6 \,{\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(6*(5*b^2*c - 3*a*b*d)*x^4 - 4*a^2*c + 4*(5*a*b*c - 3*a^2*d)*x^2 - 3*((5*b^2*c - 3*a*b*d)*x^5 + (5*a*b*c
- 3*a^2*d)*x^3)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^3*b*x^5 + a^4*x^3), 1/6*(3*(5*
b^2*c - 3*a*b*d)*x^4 - 2*a^2*c + 2*(5*a*b*c - 3*a^2*d)*x^2 + 3*((5*b^2*c - 3*a*b*d)*x^5 + (5*a*b*c - 3*a^2*d)*
x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^3*b*x^5 + a^4*x^3)]

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Sympy [B]  time = 0.795143, size = 184, normalized size = 2.04 \begin{align*} \frac{\sqrt{- \frac{b}{a^{7}}} \left (3 a d - 5 b c\right ) \log{\left (- \frac{a^{4} \sqrt{- \frac{b}{a^{7}}} \left (3 a d - 5 b c\right )}{3 a b d - 5 b^{2} c} + x \right )}}{4} - \frac{\sqrt{- \frac{b}{a^{7}}} \left (3 a d - 5 b c\right ) \log{\left (\frac{a^{4} \sqrt{- \frac{b}{a^{7}}} \left (3 a d - 5 b c\right )}{3 a b d - 5 b^{2} c} + x \right )}}{4} - \frac{2 a^{2} c + x^{4} \left (9 a b d - 15 b^{2} c\right ) + x^{2} \left (6 a^{2} d - 10 a b c\right )}{6 a^{4} x^{3} + 6 a^{3} b x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(-b/a**7)*(3*a*d - 5*b*c)*log(-a**4*sqrt(-b/a**7)*(3*a*d - 5*b*c)/(3*a*b*d - 5*b**2*c) + x)/4 - sqrt(-b/a*
*7)*(3*a*d - 5*b*c)*log(a**4*sqrt(-b/a**7)*(3*a*d - 5*b*c)/(3*a*b*d - 5*b**2*c) + x)/4 - (2*a**2*c + x**4*(9*a
*b*d - 15*b**2*c) + x**2*(6*a**2*d - 10*a*b*c))/(6*a**4*x**3 + 6*a**3*b*x**5)

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Giac [A]  time = 1.16169, size = 116, normalized size = 1.29 \begin{align*} \frac{{\left (5 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{3}} + \frac{b^{2} c x - a b d x}{2 \,{\left (b x^{2} + a\right )} a^{3}} + \frac{6 \, b c x^{2} - 3 \, a d x^{2} - a c}{3 \, a^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(5*b^2*c - 3*a*b*d)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/2*(b^2*c*x - a*b*d*x)/((b*x^2 + a)*a^3) + 1/
3*(6*b*c*x^2 - 3*a*d*x^2 - a*c)/(a^3*x^3)