3.101 \(\int \frac {x^4 (A+B x^2)}{(a+b x^2)^3} \, dx\)

Optimal. Leaf size=94 \[ \frac {3 (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}-\frac {x (5 A b-9 a B)}{8 b^3 \left (a+b x^2\right )}+\frac {a x (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}+\frac {B x}{b^3} \]

[Out]

B*x/b^3+1/4*a*(A*b-B*a)*x/b^3/(b*x^2+a)^2-1/8*(5*A*b-9*B*a)*x/b^3/(b*x^2+a)+3/8*(A*b-5*B*a)*arctan(x*b^(1/2)/a
^(1/2))/b^(7/2)/a^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {455, 1157, 388, 205} \[ -\frac {x (5 A b-9 a B)}{8 b^3 \left (a+b x^2\right )}+\frac {a x (A b-a B)}{4 b^3 \left (a+b x^2\right )^2}+\frac {3 (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}+\frac {B x}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(B*x)/b^3 + (a*(A*b - a*B)*x)/(4*b^3*(a + b*x^2)^2) - ((5*A*b - 9*a*B)*x)/(8*b^3*(a + b*x^2)) + (3*(A*b - 5*a*
B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2))

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \int \frac {x^4 \left (A+B x^2\right )}{\left (a+b x^2\right )^3} \, dx &=\frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {\int \frac {a (A b-a B)-4 b (A b-a B) x^2-4 b^2 B x^4}{\left (a+b x^2\right )^2} \, dx}{4 b^3}\\ &=\frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {(5 A b-9 a B) x}{8 b^3 \left (a+b x^2\right )}+\frac {\int \frac {a (3 A b-7 a B)+8 a b B x^2}{a+b x^2} \, dx}{8 a b^3}\\ &=\frac {B x}{b^3}+\frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {(5 A b-9 a B) x}{8 b^3 \left (a+b x^2\right )}+\frac {(3 (A b-5 a B)) \int \frac {1}{a+b x^2} \, dx}{8 b^3}\\ &=\frac {B x}{b^3}+\frac {a (A b-a B) x}{4 b^3 \left (a+b x^2\right )^2}-\frac {(5 A b-9 a B) x}{8 b^3 \left (a+b x^2\right )}+\frac {3 (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 91, normalized size = 0.97 \[ \frac {x \left (15 a^2 B+a \left (25 b B x^2-3 A b\right )+b^2 x^2 \left (8 B x^2-5 A\right )\right )}{8 b^3 \left (a+b x^2\right )^2}+\frac {3 (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x^2))/(a + b*x^2)^3,x]

[Out]

(x*(15*a^2*B + b^2*x^2*(-5*A + 8*B*x^2) + a*(-3*A*b + 25*b*B*x^2)))/(8*b^3*(a + b*x^2)^2) + (3*(A*b - 5*a*B)*A
rcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*Sqrt[a]*b^(7/2))

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fricas [A]  time = 0.46, size = 328, normalized size = 3.49 \[ \left [\frac {16 \, B a b^{3} x^{5} + 10 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} + 3 \, {\left ({\left (5 \, B a b^{2} - A b^{3}\right )} x^{4} + 5 \, B a^{3} - A a^{2} b + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + 6 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} x}{16 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}, \frac {8 \, B a b^{3} x^{5} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3} - 3 \, {\left ({\left (5 \, B a b^{2} - A b^{3}\right )} x^{4} + 5 \, B a^{3} - A a^{2} b + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + 3 \, {\left (5 \, B a^{3} b - A a^{2} b^{2}\right )} x}{8 \, {\left (a b^{6} x^{4} + 2 \, a^{2} b^{5} x^{2} + a^{3} b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(16*B*a*b^3*x^5 + 10*(5*B*a^2*b^2 - A*a*b^3)*x^3 + 3*((5*B*a*b^2 - A*b^3)*x^4 + 5*B*a^3 - A*a^2*b + 2*(5
*B*a^2*b - A*a*b^2)*x^2)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) + 6*(5*B*a^3*b - A*a^2*b^2)*
x)/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4), 1/8*(8*B*a*b^3*x^5 + 5*(5*B*a^2*b^2 - A*a*b^3)*x^3 - 3*((5*B*a*b^2 -
A*b^3)*x^4 + 5*B*a^3 - A*a^2*b + 2*(5*B*a^2*b - A*a*b^2)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) + 3*(5*B*a^3*b
- A*a^2*b^2)*x)/(a*b^6*x^4 + 2*a^2*b^5*x^2 + a^3*b^4)]

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giac [A]  time = 0.41, size = 80, normalized size = 0.85 \[ \frac {B x}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {9 \, B a b x^{3} - 5 \, A b^{2} x^{3} + 7 \, B a^{2} x - 3 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="giac")

[Out]

B*x/b^3 - 3/8*(5*B*a - A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/8*(9*B*a*b*x^3 - 5*A*b^2*x^3 + 7*B*a^2*x
- 3*A*a*b*x)/((b*x^2 + a)^2*b^3)

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maple [A]  time = 0.01, size = 122, normalized size = 1.30 \[ -\frac {5 A \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b}+\frac {9 B a \,x^{3}}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}-\frac {3 A a x}{8 \left (b \,x^{2}+a \right )^{2} b^{2}}+\frac {7 B \,a^{2} x}{8 \left (b \,x^{2}+a \right )^{2} b^{3}}+\frac {3 A \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{2}}-\frac {15 B a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{3}}+\frac {B x}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x^2+A)/(b*x^2+a)^3,x)

[Out]

B*x/b^3-5/8/b/(b*x^2+a)^2*A*x^3+9/8/b^2/(b*x^2+a)^2*B*x^3*a-3/8/b^2/(b*x^2+a)^2*a*A*x+7/8/b^3/(b*x^2+a)^2*a^2*
B*x+3/8/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*A-15/8/b^3/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)*B*a

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maxima [A]  time = 2.29, size = 94, normalized size = 1.00 \[ \frac {{\left (9 \, B a b - 5 \, A b^{2}\right )} x^{3} + {\left (7 \, B a^{2} - 3 \, A a b\right )} x}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} + \frac {B x}{b^{3}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(b*x^2+a)^3,x, algorithm="maxima")

[Out]

1/8*((9*B*a*b - 5*A*b^2)*x^3 + (7*B*a^2 - 3*A*a*b)*x)/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3) + B*x/b^3 - 3/8*(5*B*a
- A*b)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3)

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mupad [B]  time = 0.25, size = 92, normalized size = 0.98 \[ \frac {B\,x}{b^3}-\frac {x^3\,\left (\frac {5\,A\,b^2}{8}-\frac {9\,B\,a\,b}{8}\right )-x\,\left (\frac {7\,B\,a^2}{8}-\frac {3\,A\,a\,b}{8}\right )}{a^2\,b^3+2\,a\,b^4\,x^2+b^5\,x^4}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (A\,b-5\,B\,a\right )}{8\,\sqrt {a}\,b^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(A + B*x^2))/(a + b*x^2)^3,x)

[Out]

(B*x)/b^3 - (x^3*((5*A*b^2)/8 - (9*B*a*b)/8) - x*((7*B*a^2)/8 - (3*A*a*b)/8))/(a^2*b^3 + b^5*x^4 + 2*a*b^4*x^2
) + (3*atan((b^(1/2)*x)/a^(1/2))*(A*b - 5*B*a))/(8*a^(1/2)*b^(7/2))

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sympy [B]  time = 1.06, size = 194, normalized size = 2.06 \[ \frac {B x}{b^{3}} + \frac {3 \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right ) \log {\left (- \frac {3 a b^{3} \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right )}{- 3 A b + 15 B a} + x \right )}}{16} - \frac {3 \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right ) \log {\left (\frac {3 a b^{3} \sqrt {- \frac {1}{a b^{7}}} \left (- A b + 5 B a\right )}{- 3 A b + 15 B a} + x \right )}}{16} + \frac {x^{3} \left (- 5 A b^{2} + 9 B a b\right ) + x \left (- 3 A a b + 7 B a^{2}\right )}{8 a^{2} b^{3} + 16 a b^{4} x^{2} + 8 b^{5} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x**2+A)/(b*x**2+a)**3,x)

[Out]

B*x/b**3 + 3*sqrt(-1/(a*b**7))*(-A*b + 5*B*a)*log(-3*a*b**3*sqrt(-1/(a*b**7))*(-A*b + 5*B*a)/(-3*A*b + 15*B*a)
+ x)/16 - 3*sqrt(-1/(a*b**7))*(-A*b + 5*B*a)*log(3*a*b**3*sqrt(-1/(a*b**7))*(-A*b + 5*B*a)/(-3*A*b + 15*B*a)
+ x)/16 + (x**3*(-5*A*b**2 + 9*B*a*b) + x*(-3*A*a*b + 7*B*a**2))/(8*a**2*b**3 + 16*a*b**4*x**2 + 8*b**5*x**4)

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