3.2.3 \(\int \frac {A+B x^2}{(a+b x^2)^3} \, dx\) [103]

Optimal. Leaf size=92 \[ \frac {(A b-a B) x}{4 a b \left (a+b x^2\right )^2}+\frac {(3 A b+a B) x}{8 a^2 b \left (a+b x^2\right )}+\frac {(3 A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {393, 205, 211} \begin {gather*} \frac {(a B+3 A b) \text {ArcTan}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}}+\frac {x (a B+3 A b)}{8 a^2 b \left (a+b x^2\right )}+\frac {x (A b-a B)}{4 a b \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(a + b*x^2)^3,x]

[Out]

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

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 393

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 84, normalized size = 0.91 \begin {gather*} \frac {x \left (-a^2 B+3 A b^2 x^2+a b \left (5 A+B x^2\right )\right )}{8 a^2 b \left (a+b x^2\right )^2}+\frac {(3 A b+a B) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 77, normalized size = 0.84

method	result	size
default	\(\frac {\frac {\left (3 A b +B a \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A b -B a \right ) x}{8 a b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {\left (3 A b +B a \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a^{2} b \sqrt {a b}}\)	\(77\)
risch	\(\frac {\frac {\left (3 A b +B a \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A b -B a \right ) x}{8 a b}}{\left (b \,x^{2}+a \right )^{2}}-\frac {3 \ln \left (b x +\sqrt {-a b}\right ) A}{16 \sqrt {-a b}\, a^{2}}-\frac {\ln \left (b x +\sqrt {-a b}\right ) B}{16 \sqrt {-a b}\, b a}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right ) A}{16 \sqrt {-a b}\, a^{2}}+\frac {\ln \left (-b x +\sqrt {-a b}\right ) B}{16 \sqrt {-a b}\, b a}\)	\(147\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.54, size = 92, normalized size = 1.00 \begin {gather*} \frac {{\left (B a b + 3 \, A b^{2}\right )} x^{3} - {\left (B a^{2} - 5 \, A a b\right )} x}{8 \, {\left (a^{2} b^{3} x^{4} + 2 \, a^{3} b^{2} x^{2} + a^{4} b\right )}} + \frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.58, size = 300, normalized size = 3.26 \begin {gather*} \left [\frac {2 \, {\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} - {\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{4} + B a^{3} + 3 \, A a^{2} b + 2 \, {\left (B a^{2} b + 3 \, 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 ) - 2 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{16 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}, \frac {{\left (B a^{2} b^{2} + 3 \, A a b^{3}\right )} x^{3} + {\left ({\left (B a b^{2} + 3 \, A b^{3}\right )} x^{4} + B a^{3} + 3 \, A a^{2} b + 2 \, {\left (B a^{2} b + 3 \, A a b^{2}\right )} x^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) - {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x}{8 \, {\left (a^{3} b^{4} x^{4} + 2 \, a^{4} b^{3} x^{2} + a^{5} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.39, size = 150, normalized size = 1.63 \begin {gather*} - \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \cdot \left (3 A b + B a\right ) \log {\left (- a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \cdot \left (3 A b + B a\right ) \log {\left (a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 A b^{2} + B a b\right ) + x \left (5 A a b - B a^{2}\right )}{8 a^{4} b + 16 a^{3} b^{2} x^{2} + 8 a^{2} b^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.47, size = 78, normalized size = 0.85 \begin {gather*} \frac {{\left (B a + 3 \, A b\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} b} + \frac {B a b x^{3} + 3 \, A b^{2} x^{3} - B a^{2} x + 5 \, A a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.08, size = 82, normalized size = 0.89 \begin {gather*} \frac {\frac {x^3\,\left (3\,A\,b+B\,a\right )}{8\,a^2}+\frac {x\,\left (5\,A\,b-B\,a\right )}{8\,a\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )\,\left (3\,A\,b+B\,a\right )}{8\,a^{5/2}\,b^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(a + b*x^2)^3,x)

[Out]

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

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