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

Optimal. Leaf size=87 \[ \frac{a x (b c-a d)}{2 b^3 \left (a+b x^2\right )}+\frac{x (b c-2 a d)}{b^3}-\frac{\sqrt{a} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}+\frac{d x^3}{3 b^2} \]

[Out]

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

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Rubi [A]  time = 0.071397, antiderivative size = 87, 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 = {455, 1153, 205} \[ \frac{a x (b c-a d)}{2 b^3 \left (a+b x^2\right )}+\frac{x (b c-2 a d)}{b^3}-\frac{\sqrt{a} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}+\frac{d x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 455

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[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 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^4 \left (c+d x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac{a (b c-a d) x}{2 b^3 \left (a+b x^2\right )}-\frac{\int \frac{a (b c-a d)-2 b (b c-a d) x^2-2 b^2 d x^4}{a+b x^2} \, dx}{2 b^3}\\ &=\frac{a (b c-a d) x}{2 b^3 \left (a+b x^2\right )}-\frac{\int \left (-2 (b c-2 a d)-2 b d x^2+\frac{3 a b c-5 a^2 d}{a+b x^2}\right ) \, dx}{2 b^3}\\ &=\frac{(b c-2 a d) x}{b^3}+\frac{d x^3}{3 b^2}+\frac{a (b c-a d) x}{2 b^3 \left (a+b x^2\right )}-\frac{(a (3 b c-5 a d)) \int \frac{1}{a+b x^2} \, dx}{2 b^3}\\ &=\frac{(b c-2 a d) x}{b^3}+\frac{d x^3}{3 b^2}+\frac{a (b c-a d) x}{2 b^3 \left (a+b x^2\right )}-\frac{\sqrt{a} (3 b c-5 a d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0753117, size = 89, normalized size = 1.02 \[ \frac{x \left (a b c-a^2 d\right )}{2 b^3 \left (a+b x^2\right )}+\frac{x (b c-2 a d)}{b^3}+\frac{\sqrt{a} (5 a d-3 b c) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{7/2}}+\frac{d x^3}{3 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 105, normalized size = 1.2 \begin{align*}{\frac{d{x}^{3}}{3\,{b}^{2}}}-2\,{\frac{adx}{{b}^{3}}}+{\frac{cx}{{b}^{2}}}-{\frac{{a}^{2}dx}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{axc}{2\,{b}^{2} \left ( b{x}^{2}+a \right ) }}+{\frac{5\,{a}^{2}d}{2\,{b}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{3\,ac}{2\,{b}^{2}}\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^4*(d*x^2+c)/(b*x^2+a)^2,x)

[Out]

1/3*d*x^3/b^2-2/b^3*a*d*x+1/b^2*x*c-1/2*a^2/b^3*x/(b*x^2+a)*d+1/2*a/b^2*x/(b*x^2+a)*c+5/2*a^2/b^3/(a*b)^(1/2)*
arctan(b*x/(a*b)^(1/2))*d-3/2*a/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(x^4*(d*x^2+c)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55801, size = 513, normalized size = 5.9 \begin{align*} \left [\frac{4 \, b^{2} d x^{5} + 4 \,{\left (3 \, b^{2} c - 5 \, a b d\right )} x^{3} - 3 \,{\left (3 \, a b c - 5 \, a^{2} d +{\left (3 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{2} + 2 \, b x \sqrt{-\frac{a}{b}} - a}{b x^{2} + a}\right ) + 6 \,{\left (3 \, a b c - 5 \, a^{2} d\right )} x}{12 \,{\left (b^{4} x^{2} + a b^{3}\right )}}, \frac{2 \, b^{2} d x^{5} + 2 \,{\left (3 \, b^{2} c - 5 \, a b d\right )} x^{3} - 3 \,{\left (3 \, a b c - 5 \, a^{2} d +{\left (3 \, b^{2} c - 5 \, a b d\right )} x^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right ) + 3 \,{\left (3 \, a b c - 5 \, a^{2} d\right )} x}{6 \,{\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.766974, size = 128, normalized size = 1.47 \begin{align*} - \frac{x \left (a^{2} d - a b c\right )}{2 a b^{3} + 2 b^{4} x^{2}} - \frac{\sqrt{- \frac{a}{b^{7}}} \left (5 a d - 3 b c\right ) \log{\left (- b^{3} \sqrt{- \frac{a}{b^{7}}} + x \right )}}{4} + \frac{\sqrt{- \frac{a}{b^{7}}} \left (5 a d - 3 b c\right ) \log{\left (b^{3} \sqrt{- \frac{a}{b^{7}}} + x \right )}}{4} + \frac{d x^{3}}{3 b^{2}} - \frac{x \left (2 a d - b c\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.11273, size = 119, normalized size = 1.37 \begin{align*} -\frac{{\left (3 \, a b c - 5 \, a^{2} d\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{3}} + \frac{a b c x - a^{2} d x}{2 \,{\left (b x^{2} + a\right )} b^{3}} + \frac{b^{4} d x^{3} + 3 \, b^{4} c x - 6 \, a b^{3} d x}{3 \, b^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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