3.656 \(\int \frac {1}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\)

Optimal. Leaf size=213 \[ \frac {7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 \left (a+b x^2\right )^5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{9/2} \sqrt {b} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 x \left (a+b x^2\right )^4}{128 a^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \]

[Out]

1/8*x*(b*x^2+a)/a/(b^2*x^4+2*a*b*x^2+a^2)^(5/2)+7/48*x*(b*x^2+a)^2/a^2/(b^2*x^4+2*a*b*x^2+a^2)^(5/2)+35/192*x*
(b*x^2+a)^3/a^3/(b^2*x^4+2*a*b*x^2+a^2)^(5/2)+35/128*x*(b*x^2+a)^4/a^4/(b^2*x^4+2*a*b*x^2+a^2)^(5/2)+35/128*(b
*x^2+a)^5*arctan(x*b^(1/2)/a^(1/2))/a^(9/2)/(b^2*x^4+2*a*b*x^2+a^2)^(5/2)/b^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 213, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {1088, 199, 205} \[ \frac {35 x \left (a+b x^2\right )^4}{128 a^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 \left (a+b x^2\right )^5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{9/2} \sqrt {b} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-5/2),x]

[Out]

(x*(a + b*x^2))/(8*a*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)) + (7*x*(a + b*x^2)^2)/(48*a^2*(a^2 + 2*a*b*x^2 + b^2*x
^4)^(5/2)) + (35*x*(a + b*x^2)^3)/(192*a^3*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)) + (35*x*(a + b*x^2)^4)/(128*a^4*
(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)) + (35*(a + b*x^2)^5*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(128*a^(9/2)*Sqrt[b]*(a^2
+ 2*a*b*x^2 + b^2*x^4)^(5/2))

Rule 199

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] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 1088

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^p/(b + 2*c*x^2)^(2*p), In
t[(b + 2*c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx &=\frac {\left (2 a b+2 b^2 x^2\right )^5 \int \frac {1}{\left (2 a b+2 b^2 x^2\right )^5} \, dx}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac {x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {\left (7 \left (2 a b+2 b^2 x^2\right )^5\right ) \int \frac {1}{\left (2 a b+2 b^2 x^2\right )^4} \, dx}{16 a b \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac {x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {\left (35 \left (2 a b+2 b^2 x^2\right )^5\right ) \int \frac {1}{\left (2 a b+2 b^2 x^2\right )^3} \, dx}{192 a^2 b^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac {x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {\left (35 \left (2 a b+2 b^2 x^2\right )^5\right ) \int \frac {1}{\left (2 a b+2 b^2 x^2\right )^2} \, dx}{512 a^3 b^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac {x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 x \left (a+b x^2\right )^4}{128 a^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {\left (35 \left (2 a b+2 b^2 x^2\right )^5\right ) \int \frac {1}{2 a b+2 b^2 x^2} \, dx}{2048 a^4 b^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ &=\frac {x \left (a+b x^2\right )}{8 a \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {7 x \left (a+b x^2\right )^2}{48 a^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 x \left (a+b x^2\right )^3}{192 a^3 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 x \left (a+b x^2\right )^4}{128 a^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}+\frac {35 \left (a+b x^2\right )^5 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{9/2} \sqrt {b} \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 105, normalized size = 0.49 \[ \frac {\sqrt {a} \sqrt {b} x \left (279 a^3+511 a^2 b x^2+385 a b^2 x^4+105 b^3 x^6\right )+105 \left (a+b x^2\right )^4 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{384 a^{9/2} \sqrt {b} \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-5/2),x]

[Out]

(Sqrt[a]*Sqrt[b]*x*(279*a^3 + 511*a^2*b*x^2 + 385*a*b^2*x^4 + 105*b^3*x^6) + 105*(a + b*x^2)^4*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(384*a^(9/2)*Sqrt[b]*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])

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fricas [A]  time = 1.12, size = 320, normalized size = 1.50 \[ \left [\frac {210 \, a b^{4} x^{7} + 770 \, a^{2} b^{3} x^{5} + 1022 \, a^{3} b^{2} x^{3} + 558 \, a^{4} b x - 105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{768 \, {\left (a^{5} b^{5} x^{8} + 4 \, a^{6} b^{4} x^{6} + 6 \, a^{7} b^{3} x^{4} + 4 \, a^{8} b^{2} x^{2} + a^{9} b\right )}}, \frac {105 \, a b^{4} x^{7} + 385 \, a^{2} b^{3} x^{5} + 511 \, a^{3} b^{2} x^{3} + 279 \, a^{4} b x + 105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{384 \, {\left (a^{5} b^{5} x^{8} + 4 \, a^{6} b^{4} x^{6} + 6 \, a^{7} b^{3} x^{4} + 4 \, a^{8} b^{2} x^{2} + a^{9} b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")

[Out]

[1/768*(210*a*b^4*x^7 + 770*a^2*b^3*x^5 + 1022*a^3*b^2*x^3 + 558*a^4*b*x - 105*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*
b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^5*b^5*x^8 + 4*a^6*b^
4*x^6 + 6*a^7*b^3*x^4 + 4*a^8*b^2*x^2 + a^9*b), 1/384*(105*a*b^4*x^7 + 385*a^2*b^3*x^5 + 511*a^3*b^2*x^3 + 279
*a^4*b*x + 105*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a
^5*b^5*x^8 + 4*a^6*b^4*x^6 + 6*a^7*b^3*x^4 + 4*a^8*b^2*x^2 + a^9*b)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.01, size = 169, normalized size = 0.79 \[ \frac {\left (105 b^{4} x^{8} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+420 a \,b^{3} x^{6} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+105 \sqrt {a b}\, b^{3} x^{7}+630 a^{2} b^{2} x^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+385 \sqrt {a b}\, a \,b^{2} x^{5}+420 a^{3} b \,x^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+511 \sqrt {a b}\, a^{2} b \,x^{3}+105 a^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+279 \sqrt {a b}\, a^{3} x \right ) \left (b \,x^{2}+a \right )}{384 \sqrt {a b}\, \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {5}{2}} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)

[Out]

1/384*(105*b^4*x^8*arctan(1/(a*b)^(1/2)*b*x)+105*(a*b)^(1/2)*b^3*x^7+420*a*b^3*x^6*arctan(1/(a*b)^(1/2)*b*x)+3
85*(a*b)^(1/2)*a*b^2*x^5+630*a^2*b^2*x^4*arctan(1/(a*b)^(1/2)*b*x)+511*(a*b)^(1/2)*a^2*b*x^3+420*a^3*b*x^2*arc
tan(1/(a*b)^(1/2)*b*x)+279*(a*b)^(1/2)*a^3*x+105*a^4*arctan(1/(a*b)^(1/2)*b*x))*(b*x^2+a)/(a*b)^(1/2)/a^4/((b*
x^2+a)^2)^(5/2)

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maxima [A]  time = 3.05, size = 102, normalized size = 0.48 \[ \frac {105 \, b^{3} x^{7} + 385 \, a b^{2} x^{5} + 511 \, a^{2} b x^{3} + 279 \, a^{3} x}{384 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} + \frac {35 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")

[Out]

1/384*(105*b^3*x^7 + 385*a*b^2*x^5 + 511*a^2*b*x^3 + 279*a^3*x)/(a^4*b^4*x^8 + 4*a^5*b^3*x^6 + 6*a^6*b^2*x^4 +
4*a^7*b*x^2 + a^8) + 35/128*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2),x)

[Out]

int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)

[Out]

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-5/2), x)

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