3.74 \(\int \frac{x^6 (A+B x^2)}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=110 \[ -\frac{a^2 x (A b-a B)}{2 b^4 \left (a+b x^2\right )}+\frac{a^{3/2} (5 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{9/2}}+\frac{x^3 (A b-2 a B)}{3 b^3}-\frac{a x (2 A b-3 a B)}{b^4}+\frac{B x^5}{5 b^2} \]

[Out]

-((a*(2*A*b - 3*a*B)*x)/b^4) + ((A*b - 2*a*B)*x^3)/(3*b^3) + (B*x^5)/(5*b^2) - (a^2*(A*b - a*B)*x)/(2*b^4*(a +
b*x^2)) + (a^(3/2)*(5*A*b - 7*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(9/2))

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Rubi [A]  time = 0.113795, antiderivative size = 110, 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, 1810, 205} \[ -\frac{a^2 x (A b-a B)}{2 b^4 \left (a+b x^2\right )}+\frac{a^{3/2} (5 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{9/2}}+\frac{x^3 (A b-2 a B)}{3 b^3}-\frac{a x (2 A b-3 a B)}{b^4}+\frac{B x^5}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

-((a*(2*A*b - 3*a*B)*x)/b^4) + ((A*b - 2*a*B)*x^3)/(3*b^3) + (B*x^5)/(5*b^2) - (a^2*(A*b - a*B)*x)/(2*b^4*(a +
b*x^2)) + (a^(3/2)*(5*A*b - 7*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(9/2))

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 1810

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]

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^6 \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=-\frac{a^2 (A b-a B) x}{2 b^4 \left (a+b x^2\right )}-\frac{\int \frac{-a^2 (A b-a B)+2 a b (A b-a B) x^2-2 b^2 (A b-a B) x^4-2 b^3 B x^6}{a+b x^2} \, dx}{2 b^4}\\ &=-\frac{a^2 (A b-a B) x}{2 b^4 \left (a+b x^2\right )}-\frac{\int \left (2 a (2 A b-3 a B)-2 b (A b-2 a B) x^2-2 b^2 B x^4+\frac{-5 a^2 A b+7 a^3 B}{a+b x^2}\right ) \, dx}{2 b^4}\\ &=-\frac{a (2 A b-3 a B) x}{b^4}+\frac{(A b-2 a B) x^3}{3 b^3}+\frac{B x^5}{5 b^2}-\frac{a^2 (A b-a B) x}{2 b^4 \left (a+b x^2\right )}+\frac{\left (a^2 (5 A b-7 a B)\right ) \int \frac{1}{a+b x^2} \, dx}{2 b^4}\\ &=-\frac{a (2 A b-3 a B) x}{b^4}+\frac{(A b-2 a B) x^3}{3 b^3}+\frac{B x^5}{5 b^2}-\frac{a^2 (A b-a B) x}{2 b^4 \left (a+b x^2\right )}+\frac{a^{3/2} (5 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0853155, size = 111, normalized size = 1.01 \[ -\frac{x \left (a^2 A b-a^3 B\right )}{2 b^4 \left (a+b x^2\right )}-\frac{a^{3/2} (7 a B-5 A b) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{9/2}}+\frac{x^3 (A b-2 a B)}{3 b^3}+\frac{a x (3 a B-2 A b)}{b^4}+\frac{B x^5}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

(a*(-2*A*b + 3*a*B)*x)/b^4 + ((A*b - 2*a*B)*x^3)/(3*b^3) + (B*x^5)/(5*b^2) - ((a^2*A*b - a^3*B)*x)/(2*b^4*(a +
b*x^2)) - (a^(3/2)*(-5*A*b + 7*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*b^(9/2))

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Maple [A]  time = 0.008, size = 132, normalized size = 1.2 \begin{align*}{\frac{B{x}^{5}}{5\,{b}^{2}}}+{\frac{A{x}^{3}}{3\,{b}^{2}}}-{\frac{2\,B{x}^{3}a}{3\,{b}^{3}}}-2\,{\frac{aAx}{{b}^{3}}}+3\,{\frac{{a}^{2}Bx}{{b}^{4}}}-{\frac{{a}^{2}Ax}{2\,{b}^{3} \left ( b{x}^{2}+a \right ) }}+{\frac{{a}^{3}xB}{2\,{b}^{4} \left ( b{x}^{2}+a \right ) }}+{\frac{5\,A{a}^{2}}{2\,{b}^{3}}\arctan \left ({bx{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{7\,B{a}^{3}}{2\,{b}^{4}}\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^6*(B*x^2+A)/(b*x^2+a)^2,x)

[Out]

1/5*B*x^5/b^2+1/3/b^2*A*x^3-2/3/b^3*B*x^3*a-2/b^3*a*A*x+3/b^4*a^2*B*x-1/2*a^2/b^3*x/(b*x^2+a)*A+1/2*a^3/b^4*x/
(b*x^2+a)*B+5/2*a^2/b^3/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*A-7/2*a^3/b^4/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))*
B

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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^6*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.28166, size = 637, normalized size = 5.79 \begin{align*} \left [\frac{12 \, B b^{3} x^{7} - 4 \,{\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 20 \,{\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{3} - 15 \,{\left (7 \, B a^{3} - 5 \, A a^{2} b +{\left (7 \, B a^{2} b - 5 \, A a b^{2}\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 ) + 30 \,{\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} x}{60 \,{\left (b^{5} x^{2} + a b^{4}\right )}}, \frac{6 \, B b^{3} x^{7} - 2 \,{\left (7 \, B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 10 \,{\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{3} - 15 \,{\left (7 \, B a^{3} - 5 \, A a^{2} b +{\left (7 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{2}\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right ) + 15 \,{\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} x}{30 \,{\left (b^{5} x^{2} + a b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/60*(12*B*b^3*x^7 - 4*(7*B*a*b^2 - 5*A*b^3)*x^5 + 20*(7*B*a^2*b - 5*A*a*b^2)*x^3 - 15*(7*B*a^3 - 5*A*a^2*b +
(7*B*a^2*b - 5*A*a*b^2)*x^2)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)) + 30*(7*B*a^3 - 5*A*a
^2*b)*x)/(b^5*x^2 + a*b^4), 1/30*(6*B*b^3*x^7 - 2*(7*B*a*b^2 - 5*A*b^3)*x^5 + 10*(7*B*a^2*b - 5*A*a*b^2)*x^3 -
15*(7*B*a^3 - 5*A*a^2*b + (7*B*a^2*b - 5*A*a*b^2)*x^2)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) + 15*(7*B*a^3 - 5*A*
a^2*b)*x)/(b^5*x^2 + a*b^4)]

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Sympy [A]  time = 0.853715, size = 206, normalized size = 1.87 \begin{align*} \frac{B x^{5}}{5 b^{2}} + \frac{x \left (- A a^{2} b + B a^{3}\right )}{2 a b^{4} + 2 b^{5} x^{2}} + \frac{\sqrt{- \frac{a^{3}}{b^{9}}} \left (- 5 A b + 7 B a\right ) \log{\left (- \frac{b^{4} \sqrt{- \frac{a^{3}}{b^{9}}} \left (- 5 A b + 7 B a\right )}{- 5 A a b + 7 B a^{2}} + x \right )}}{4} - \frac{\sqrt{- \frac{a^{3}}{b^{9}}} \left (- 5 A b + 7 B a\right ) \log{\left (\frac{b^{4} \sqrt{- \frac{a^{3}}{b^{9}}} \left (- 5 A b + 7 B a\right )}{- 5 A a b + 7 B a^{2}} + x \right )}}{4} - \frac{x^{3} \left (- A b + 2 B a\right )}{3 b^{3}} + \frac{x \left (- 2 A a b + 3 B a^{2}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(B*x**2+A)/(b*x**2+a)**2,x)

[Out]

B*x**5/(5*b**2) + x*(-A*a**2*b + B*a**3)/(2*a*b**4 + 2*b**5*x**2) + sqrt(-a**3/b**9)*(-5*A*b + 7*B*a)*log(-b**
4*sqrt(-a**3/b**9)*(-5*A*b + 7*B*a)/(-5*A*a*b + 7*B*a**2) + x)/4 - sqrt(-a**3/b**9)*(-5*A*b + 7*B*a)*log(b**4*
sqrt(-a**3/b**9)*(-5*A*b + 7*B*a)/(-5*A*a*b + 7*B*a**2) + x)/4 - x**3*(-A*b + 2*B*a)/(3*b**3) + x*(-2*A*a*b +
3*B*a**2)/b**4

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Giac [A]  time = 1.15127, size = 155, normalized size = 1.41 \begin{align*} -\frac{{\left (7 \, B a^{3} - 5 \, A a^{2} b\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{4}} + \frac{B a^{3} x - A a^{2} b x}{2 \,{\left (b x^{2} + a\right )} b^{4}} + \frac{3 \, B b^{8} x^{5} - 10 \, B a b^{7} x^{3} + 5 \, A b^{8} x^{3} + 45 \, B a^{2} b^{6} x - 30 \, A a b^{7} x}{15 \, b^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

-1/2*(7*B*a^3 - 5*A*a^2*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^4) + 1/2*(B*a^3*x - A*a^2*b*x)/((b*x^2 + a)*b^4)
+ 1/15*(3*B*b^8*x^5 - 10*B*a*b^7*x^3 + 5*A*b^8*x^3 + 45*B*a^2*b^6*x - 30*A*a*b^7*x)/b^10