\(\int \frac {A+B x^2}{x^4 (a+b x^2)^3} \, dx\) [105]

Optimal result

Integrand size = 20, antiderivative size = 117 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^3} \, dx=-\frac {A}{3 a^3 x^3}+\frac {3 A b-a B}{a^4 x}+\frac {b (A b-a B) x}{4 a^3 \left (a+b x^2\right )^2}+\frac {b (11 A b-7 a B) x}{8 a^4 \left (a+b x^2\right )}+\frac {5 \sqrt {b} (7 A b-3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}} \]

[Out]

-1/3*A/a^3/x^3+(3*A*b-B*a)/a^4/x+1/4*b*(A*b-B*a)*x/a^3/(b*x^2+a)^2+1/8*b*(11*A*b-7*B*a)*x/a^4/(b*x^2+a)+5/8*(7
*A*b-3*B*a)*arctan(x*b^(1/2)/a^(1/2))*b^(1/2)/a^(9/2)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {467, 1273, 1275, 211} \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^3} \, dx=\frac {5 \sqrt {b} (7 A b-3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}}+\frac {b x (11 A b-7 a B)}{8 a^4 \left (a+b x^2\right )}+\frac {3 A b-a B}{a^4 x}+\frac {b x (A b-a B)}{4 a^3 \left (a+b x^2\right )^2}-\frac {A}{3 a^3 x^3} \]

[In]

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

[Out]

-1/3*A/(a^3*x^3) + (3*A*b - a*B)/(a^4*x) + (b*(A*b - a*B)*x)/(4*a^3*(a + b*x^2)^2) + (b*(11*A*b - 7*a*B)*x)/(8
*a^4*(a + b*x^2)) + (5*Sqrt[b]*(7*A*b - 3*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(9/2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 467

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 1273

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/(d + e*x^2))*(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))], 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 1275

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]

Rubi steps \begin{align*} \text {integral}& = \frac {b (A b-a B) x}{4 a^3 \left (a+b x^2\right )^2}-\frac {1}{4} b \int \frac {-\frac {4 A}{a b}+\frac {4 (A b-a B) x^2}{a^2 b}-\frac {3 (A b-a B) x^4}{a^3}}{x^4 \left (a+b x^2\right )^2} \, dx \\ & = \frac {b (A b-a B) x}{4 a^3 \left (a+b x^2\right )^2}+\frac {b (11 A b-7 a B) x}{8 a^4 \left (a+b x^2\right )}-\frac {\int \frac {-8 a A b+8 b (2 A b-a B) x^2-\frac {b^2 (11 A b-7 a B) x^4}{a}}{x^4 \left (a+b x^2\right )} \, dx}{8 a^3 b} \\ & = \frac {b (A b-a B) x}{4 a^3 \left (a+b x^2\right )^2}+\frac {b (11 A b-7 a B) x}{8 a^4 \left (a+b x^2\right )}-\frac {\int \left (-\frac {8 A b}{x^4}-\frac {8 b (-3 A b+a B)}{a x^2}+\frac {5 b^2 (-7 A b+3 a B)}{a \left (a+b x^2\right )}\right ) \, dx}{8 a^3 b} \\ & = -\frac {A}{3 a^3 x^3}+\frac {3 A b-a B}{a^4 x}+\frac {b (A b-a B) x}{4 a^3 \left (a+b x^2\right )^2}+\frac {b (11 A b-7 a B) x}{8 a^4 \left (a+b x^2\right )}+\frac {(5 b (7 A b-3 a B)) \int \frac {1}{a+b x^2} \, dx}{8 a^4} \\ & = -\frac {A}{3 a^3 x^3}+\frac {3 A b-a B}{a^4 x}+\frac {b (A b-a B) x}{4 a^3 \left (a+b x^2\right )^2}+\frac {b (11 A b-7 a B) x}{8 a^4 \left (a+b x^2\right )}+\frac {5 \sqrt {b} (7 A b-3 a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^3} \, dx=\frac {105 A b^3 x^6+a^2 b x^2 \left (56 A-75 B x^2\right )+5 a b^2 x^4 \left (35 A-9 B x^2\right )-8 a^3 \left (A+3 B x^2\right )}{24 a^4 x^3 \left (a+b x^2\right )^2}+\frac {5 \sqrt {b} (7 A b-3 a B) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}} \]

[In]

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

[Out]

(105*A*b^3*x^6 + a^2*b*x^2*(56*A - 75*B*x^2) + 5*a*b^2*x^4*(35*A - 9*B*x^2) - 8*a^3*(A + 3*B*x^2))/(24*a^4*x^3
*(a + b*x^2)^2) + (5*Sqrt[b]*(7*A*b - 3*a*B)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(9/2))

Maple [A] (verified)

Time = 2.54 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.84

method	result	size
default	\(-\frac {A}{3 a^{3} x^{3}}-\frac {-3 A b +B a}{a^{4} x}+\frac {b \left (\frac {\left (\frac {11}{8} b^{2} A -\frac {7}{8} a b B \right ) x^{3}+\frac {a \left (13 A b -9 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {5 \left (7 A b -3 B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{4}}\)	\(98\)
risch	\(\frac {\frac {5 b^{2} \left (7 A b -3 B a \right ) x^{6}}{8 a^{4}}+\frac {25 b \left (7 A b -3 B a \right ) x^{4}}{24 a^{3}}+\frac {\left (7 A b -3 B a \right ) x^{2}}{3 a^{2}}-\frac {A}{3 a}}{x^{3} \left (b \,x^{2}+a \right )^{2}}+\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{9} \textit {\_Z}^{2}+49 A^{2} b^{3}-42 A B a \,b^{2}+9 B^{2} a^{2} b \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{9}+98 A^{2} b^{3}-84 A B a \,b^{2}+18 B^{2} a^{2} b \right ) x +\left (-7 A \,a^{5} b +3 B \,a^{6}\right ) \textit {\_R} \right )\right )}{16}\)	\(172\)

[In]

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

[Out]

-1/3*A/a^3/x^3-(-3*A*b+B*a)/a^4/x+1/a^4*b*(((11/8*b^2*A-7/8*a*b*B)*x^3+1/8*a*(13*A*b-9*B*a)*x)/(b*x^2+a)^2+5/8
*(7*A*b-3*B*a)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 368, normalized size of antiderivative = 3.15 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^3} \, dx=\left [-\frac {30 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{6} + 50 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{4} + 16 \, A a^{3} + 16 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{2} + 15 \, {\left ({\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{7} + 2 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{5} + {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{48 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}, -\frac {15 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{6} + 25 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{4} + 8 \, A a^{3} + 8 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{2} + 15 \, {\left ({\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{7} + 2 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{5} + {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}\right ] \]

[In]

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

[Out]

[-1/48*(30*(3*B*a*b^2 - 7*A*b^3)*x^6 + 50*(3*B*a^2*b - 7*A*a*b^2)*x^4 + 16*A*a^3 + 16*(3*B*a^3 - 7*A*a^2*b)*x^
2 + 15*((3*B*a*b^2 - 7*A*b^3)*x^7 + 2*(3*B*a^2*b - 7*A*a*b^2)*x^5 + (3*B*a^3 - 7*A*a^2*b)*x^3)*sqrt(-b/a)*log(
(b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3), -1/24*(15*(3*B*a*b^2 - 7*A
*b^3)*x^6 + 25*(3*B*a^2*b - 7*A*a*b^2)*x^4 + 8*A*a^3 + 8*(3*B*a^3 - 7*A*a^2*b)*x^2 + 15*((3*B*a*b^2 - 7*A*b^3)
*x^7 + 2*(3*B*a^2*b - 7*A*a*b^2)*x^5 + (3*B*a^3 - 7*A*a^2*b)*x^3)*sqrt(b/a)*arctan(x*sqrt(b/a)))/(a^4*b^2*x^7
+ 2*a^5*b*x^5 + a^6*x^3)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (109) = 218\).

Time = 0.42 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.93 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^3} \, dx=\frac {5 \sqrt {- \frac {b}{a^{9}}} \left (- 7 A b + 3 B a\right ) \log {\left (- \frac {5 a^{5} \sqrt {- \frac {b}{a^{9}}} \left (- 7 A b + 3 B a\right )}{- 35 A b^{2} + 15 B a b} + x \right )}}{16} - \frac {5 \sqrt {- \frac {b}{a^{9}}} \left (- 7 A b + 3 B a\right ) \log {\left (\frac {5 a^{5} \sqrt {- \frac {b}{a^{9}}} \left (- 7 A b + 3 B a\right )}{- 35 A b^{2} + 15 B a b} + x \right )}}{16} + \frac {- 8 A a^{3} + x^{6} \cdot \left (105 A b^{3} - 45 B a b^{2}\right ) + x^{4} \cdot \left (175 A a b^{2} - 75 B a^{2} b\right ) + x^{2} \cdot \left (56 A a^{2} b - 24 B a^{3}\right )}{24 a^{6} x^{3} + 48 a^{5} b x^{5} + 24 a^{4} b^{2} x^{7}} \]

[In]

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

[Out]

5*sqrt(-b/a**9)*(-7*A*b + 3*B*a)*log(-5*a**5*sqrt(-b/a**9)*(-7*A*b + 3*B*a)/(-35*A*b**2 + 15*B*a*b) + x)/16 -
5*sqrt(-b/a**9)*(-7*A*b + 3*B*a)*log(5*a**5*sqrt(-b/a**9)*(-7*A*b + 3*B*a)/(-35*A*b**2 + 15*B*a*b) + x)/16 + (
-8*A*a**3 + x**6*(105*A*b**3 - 45*B*a*b**2) + x**4*(175*A*a*b**2 - 75*B*a**2*b) + x**2*(56*A*a**2*b - 24*B*a**
3))/(24*a**6*x**3 + 48*a**5*b*x**5 + 24*a**4*b**2*x**7)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.09 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^3} \, dx=-\frac {15 \, {\left (3 \, B a b^{2} - 7 \, A b^{3}\right )} x^{6} + 25 \, {\left (3 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{4} + 8 \, A a^{3} + 8 \, {\left (3 \, B a^{3} - 7 \, A a^{2} b\right )} x^{2}}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}} - \frac {5 \, {\left (3 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \]

[In]

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

[Out]

-1/24*(15*(3*B*a*b^2 - 7*A*b^3)*x^6 + 25*(3*B*a^2*b - 7*A*a*b^2)*x^4 + 8*A*a^3 + 8*(3*B*a^3 - 7*A*a^2*b)*x^2)/
(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3) - 5/8*(3*B*a*b - 7*A*b^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^3} \, dx=-\frac {5 \, {\left (3 \, B a b - 7 \, A b^{2}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} - \frac {7 \, B a b^{2} x^{3} - 11 \, A b^{3} x^{3} + 9 \, B a^{2} b x - 13 \, A a b^{2} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{4}} - \frac {3 \, B a x^{2} - 9 \, A b x^{2} + A a}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

-5/8*(3*B*a*b - 7*A*b^2)*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4) - 1/8*(7*B*a*b^2*x^3 - 11*A*b^3*x^3 + 9*B*a^2*b
*x - 13*A*a*b^2*x)/((b*x^2 + a)^2*a^4) - 1/3*(3*B*a*x^2 - 9*A*b*x^2 + A*a)/(a^4*x^3)

Mupad [B] (verification not implemented)

Time = 5.20 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x^2}{x^4 \left (a+b x^2\right )^3} \, dx=\frac {\frac {x^2\,\left (7\,A\,b-3\,B\,a\right )}{3\,a^2}-\frac {A}{3\,a}+\frac {5\,b^2\,x^6\,\left (7\,A\,b-3\,B\,a\right )}{8\,a^4}+\frac {25\,b\,x^4\,\left (7\,A\,b-3\,B\,a\right )}{24\,a^3}}{a^2\,x^3+2\,a\,b\,x^5+b^2\,x^7}+\frac {5\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (7\,A\,b-3\,B\,a\right )}{8\,a^{9/2}} \]

[In]

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

[Out]

((x^2*(7*A*b - 3*B*a))/(3*a^2) - A/(3*a) + (5*b^2*x^6*(7*A*b - 3*B*a))/(8*a^4) + (25*b*x^4*(7*A*b - 3*B*a))/(2
4*a^3))/(a^2*x^3 + b^2*x^7 + 2*a*b*x^5) + (5*b^(1/2)*atan((b^(1/2)*x)/a^(1/2))*(7*A*b - 3*B*a))/(8*a^(9/2))