3.2.4 \(\int \frac {A+B x^2}{x^2 (a+b x^2)^3} \, dx\) [104]

Optimal. Leaf size=97 \[ -\frac {A}{a^3 x}-\frac {(A b-a B) x}{4 a^2 \left (a+b x^2\right )^2}-\frac {(7 A b-3 a B) x}{8 a^3 \left (a+b x^2\right )}-\frac {3 (5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {467, 464, 211} \begin {gather*} -\frac {3 (5 A b-a B) \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}}-\frac {x (7 A b-3 a B)}{8 a^3 \left (a+b x^2\right )}-\frac {A}{a^3 x}-\frac {x (A b-a B)}{4 a^2 \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

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])

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 96, normalized size = 0.99 \begin {gather*} -\frac {A}{a^3 x}+\frac {(-A b+a B) x}{4 a^2 \left (a+b x^2\right )^2}+\frac {(-7 A b+3 a B) x}{8 a^3 \left (a+b x^2\right )}+\frac {3 (-5 A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 82, normalized size = 0.85

method	result	size
default	\(-\frac {\frac {\left (\frac {7}{8} b^{2} A -\frac {3}{8} a b B \right ) x^{3}+\frac {a \left (9 A b -5 B a \right ) x}{8}}{\left (b \,x^{2}+a \right )^{2}}+\frac {3 \left (5 A b -B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}}{a^{3}}-\frac {A}{a^{3} x}\)	\(82\)
risch	\(\frac {-\frac {3 b \left (5 A b -B a \right ) x^{4}}{8 a^{3}}-\frac {5 \left (5 A b -B a \right ) x^{2}}{8 a^{2}}-\frac {A}{a}}{x \left (b \,x^{2}+a \right )^{2}}+\frac {3 \left (\munderset {\textit {\_R} =\RootOf \left (a^{7} b \,\textit {\_Z}^{2}+25 A^{2} b^{2}-10 A B a b +B^{2} a^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{7} b +50 A^{2} b^{2}-20 A B a b +2 B^{2} a^{2}\right ) x +\left (5 a^{4} b A -a^{5} B \right ) \textit {\_R} \right )\right )}{16}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 96, normalized size = 0.99 \begin {gather*} \frac {3 \, {\left (B a b - 5 \, A b^{2}\right )} x^{4} - 8 \, A a^{2} + 5 \, {\left (B a^{2} - 5 \, A a b\right )} x^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} + \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.69, size = 324, normalized size = 3.34 \begin {gather*} \left [-\frac {16 \, A a^{3} b - 6 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 10 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}, -\frac {8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{4} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{2} - 3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{5} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{3} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{4} b^{3} x^{5} + 2 \, a^{5} b^{2} x^{3} + a^{6} b x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 194 vs. \(2 (94) = 188\).
time = 0.37, size = 194, normalized size = 2.00 \begin {gather*} - \frac {3 \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right ) \log {\left (- \frac {3 a^{4} \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right ) \log {\left (\frac {3 a^{4} \sqrt {- \frac {1}{a^{7} b}} \left (- 5 A b + B a\right )}{- 15 A b + 3 B a} + x \right )}}{16} + \frac {- 8 A a^{2} + x^{4} \left (- 15 A b^{2} + 3 B a b\right ) + x^{2} \left (- 25 A a b + 5 B a^{2}\right )}{8 a^{5} x + 16 a^{4} b x^{3} + 8 a^{3} b^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*sqrt(-1/(a**7*b))*(-5*A*b + B*a)*log(-3*a**4*sqrt(-1/(a**7*b))*(-5*A*b + B*a)/(-15*A*b + 3*B*a) + x)/16 + 3
*sqrt(-1/(a**7*b))*(-5*A*b + B*a)*log(3*a**4*sqrt(-1/(a**7*b))*(-5*A*b + B*a)/(-15*A*b + 3*B*a) + x)/16 + (-8*
A*a**2 + x**4*(-15*A*b**2 + 3*B*a*b) + x**2*(-25*A*a*b + 5*B*a**2))/(8*a**5*x + 16*a**4*b*x**3 + 8*a**3*b**2*x
**5)

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Giac [A]
time = 1.23, size = 82, normalized size = 0.85 \begin {gather*} \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} - \frac {A}{a^{3} x} + \frac {3 \, B a b x^{3} - 7 \, A b^{2} x^{3} + 5 \, B a^{2} x - 9 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.10, size = 113, normalized size = 1.16 \begin {gather*} -\frac {\frac {A}{a}+\frac {5\,x^2\,\left (5\,A\,b-B\,a\right )}{8\,a^2}+\frac {3\,b\,x^4\,\left (5\,A\,b-B\,a\right )}{8\,a^3}}{a^2\,x+2\,a\,b\,x^3+b^2\,x^5}-\frac {3\,\mathrm {atan}\left (\frac {3\,\sqrt {b}\,x\,\left (5\,A\,b-B\,a\right )}{\sqrt {a}\,\left (15\,A\,b-3\,B\,a\right )}\right )\,\left (5\,A\,b-B\,a\right )}{8\,a^{7/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (A/a + (5*x^2*(5*A*b - B*a))/(8*a^2) + (3*b*x^4*(5*A*b - B*a))/(8*a^3))/(a^2*x + b^2*x^5 + 2*a*b*x^3) - (3*a
tan((3*b^(1/2)*x*(5*A*b - B*a))/(a^(1/2)*(15*A*b - 3*B*a)))*(5*A*b - B*a))/(8*a^(7/2)*b^(1/2))

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