3.5.71 \(\int \frac {1}{x^2 (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\)

Optimal. Leaf size=169 \[ \frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )}-\frac {15 \sqrt {b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \left (a+b x^2\right )}{8 a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.07, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1112, 290, 325, 205} \begin {gather*} -\frac {15 \left (a+b x^2\right )}{8 a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \sqrt {a^2+2 a b x^2+b^2 x^4} \left (a+b x^2\right )}-\frac {15 \sqrt {b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

5/(8*a^2*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*x*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (15*(a +
b*x^2))/(8*a^3*x*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (15*Sqrt[b]*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a
^(7/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])

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 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )^3} \, dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (5 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )^2} \, dx}{4 a \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (15 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^2 \left (a b+b^2 x^2\right )} \, dx}{8 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \left (a+b x^2\right )}{8 a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (15 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {5}{8 a^2 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \left (a+b x^2\right )}{8 a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \sqrt {b} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 93, normalized size = 0.55 \begin {gather*} \frac {-\sqrt {a} \left (8 a^2+25 a b x^2+15 b^2 x^4\right )-15 \sqrt {b} x \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} x \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

(-(Sqrt[a]*(8*a^2 + 25*a*b*x^2 + 15*b^2*x^4)) - 15*Sqrt[b]*x*(a + b*x^2)^2*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(
7/2)*x*(a + b*x^2)*Sqrt[(a + b*x^2)^2])

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IntegrateAlgebraic [A]  time = 11.67, size = 89, normalized size = 0.53 \begin {gather*} \frac {\left (a+b x^2\right ) \left (\frac {-8 a^2-25 a b x^2-15 b^2 x^4}{8 a^3 x \left (a+b x^2\right )^2}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2}}\right )}{\sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]

[Out]

((a + b*x^2)*((-8*a^2 - 25*a*b*x^2 - 15*b^2*x^4)/(8*a^3*x*(a + b*x^2)^2) - (15*Sqrt[b]*ArcTan[(Sqrt[b]*x)/Sqrt
[a]])/(8*a^(7/2))))/Sqrt[(a + b*x^2)^2]

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fricas [A]  time = 1.30, size = 202, normalized size = 1.20 \begin {gather*} \left [-\frac {30 \, b^{2} x^{4} + 50 \, a b x^{2} - 15 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 16 \, a^{2}}{16 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}, -\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.02, size = 119, normalized size = 0.70 \begin {gather*} -\frac {\left (15 b^{3} x^{5} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+30 a \,b^{2} x^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+15 \sqrt {a b}\, b^{2} x^{4}+15 a^{2} b x \arctan \left (\frac {b x}{\sqrt {a b}}\right )+25 \sqrt {a b}\, a b \,x^{2}+8 \sqrt {a b}\, a^{2}\right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} a^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)

[Out]

-1/8*(15*arctan(1/(a*b)^(1/2)*b*x)*x^5*b^3+15*(a*b)^(1/2)*x^4*b^2+30*arctan(1/(a*b)^(1/2)*b*x)*x^3*a*b^2+25*(a
*b)^(1/2)*x^2*a*b+15*arctan(1/(a*b)^(1/2)*b*x)*x*a^2*b+8*(a*b)^(1/2)*a^2)*(b*x^2+a)/(a*b)^(1/2)/x/a^3/((b*x^2+
a)^2)^(3/2)

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maxima [A]  time = 2.98, size = 71, normalized size = 0.42 \begin {gather*} -\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} - \frac {15 \, b \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(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")

[Out]

-1/8*(15*b^2*x^4 + 25*a*b*x^2 + 8*a^2)/(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x) - 15/8*b*arctan(b*x/sqrt(a*b))/(sqr
t(a*b)*a^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)),x)

[Out]

int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)

[Out]

Integral(1/(x**2*((a + b*x**2)**2)**(3/2)), x)

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