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

Optimal. Leaf size=209 \[ \frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

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Rubi [A]  time = 0.08, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

7/(8*a^2*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + 1/(4*a*x^3*(a + b*x^2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) - (35*
(a + b*x^2))/(24*a^3*x^3*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]) + (35*b*(a + b*x^2))/(8*a^4*x*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4]) + (35*b^(3/2)*(a + b*x^2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(9/2)*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^4 \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^4 \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^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (7 b \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^4 \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 {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (35 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{x^4 \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 {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {\left (35 b \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^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {\left (35 b^2 \left (a b+b^2 x^2\right )\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{8 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ &=\frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b^{3/2} \left (a+b x^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 105, normalized size = 0.50 \begin {gather*} \frac {\sqrt {a} \left (-8 a^3+56 a^2 b x^2+175 a b^2 x^4+105 b^3 x^6\right )+105 b^{3/2} x^3 \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{24 a^{9/2} x^3 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*(-8*a^3 + 56*a^2*b*x^2 + 175*a*b^2*x^4 + 105*b^3*x^6) + 105*b^(3/2)*x^3*(a + b*x^2)^2*ArcTan[(Sqrt[b]
*x)/Sqrt[a]])/(24*a^(9/2)*x^3*(a + b*x^2)*Sqrt[(a + b*x^2)^2])

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IntegrateAlgebraic [A]  time = 19.09, size = 100, normalized size = 0.48 \begin {gather*} \frac {\left (a+b x^2\right ) \left (\frac {35 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2}}+\frac {-8 a^3+56 a^2 b x^2+175 a b^2 x^4+105 b^3 x^6}{24 a^4 x^3 \left (a+b x^2\right )^2}\right )}{\sqrt {\left (a+b x^2\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x^2)*((-8*a^3 + 56*a^2*b*x^2 + 175*a*b^2*x^4 + 105*b^3*x^6)/(24*a^4*x^3*(a + b*x^2)^2) + (35*b^(3/2)*A
rcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(9/2))))/Sqrt[(a + b*x^2)^2]

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fricas [A]  time = 2.22, size = 238, normalized size = 1.14 \begin {gather*} \left [\frac {210 \, b^{3} x^{6} + 350 \, a b^{2} x^{4} + 112 \, a^{2} b x^{2} - 16 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b 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 {105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b 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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(210*b^3*x^6 + 350*a*b^2*x^4 + 112*a^2*b*x^2 - 16*a^3 + 105*(b^3*x^7 + 2*a*b^2*x^5 + 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*(105*b^3*x^6 +
175*a*b^2*x^4 + 56*a^2*b*x^2 - 8*a^3 + 105*(b^3*x^7 + 2*a*b^2*x^5 + 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)]

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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^4/(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 = 139, normalized size = 0.67 \begin {gather*} \frac {\left (105 b^{4} x^{7} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+210 a \,b^{3} x^{5} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+105 \sqrt {a b}\, b^{3} x^{6}+105 a^{2} b^{2} x^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )+175 \sqrt {a b}\, a \,b^{2} x^{4}+56 \sqrt {a b}\, a^{2} b \,x^{2}-8 \sqrt {a b}\, a^{3}\right ) \left (b \,x^{2}+a \right )}{24 \sqrt {a b}\, \left (\left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}} a^{4} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(105*arctan(1/(a*b)^(1/2)*b*x)*x^7*b^4+105*(a*b)^(1/2)*x^6*b^3+210*arctan(1/(a*b)^(1/2)*b*x)*x^5*a*b^3+17
5*(a*b)^(1/2)*x^4*a*b^2+105*arctan(1/(a*b)^(1/2)*b*x)*x^3*a^2*b^2+56*(a*b)^(1/2)*x^2*a^2*b-8*(a*b)^(1/2)*a^3)*
(b*x^2+a)/(a*b)^(1/2)/x^3/a^4/((b*x^2+a)^2)^(3/2)

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maxima [A]  time = 3.03, size = 86, normalized size = 0.41 \begin {gather*} \frac {105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3}}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(105*b^3*x^6 + 175*a*b^2*x^4 + 56*a^2*b*x^2 - 8*a^3)/(a^4*b^2*x^7 + 2*a^5*b*x^5 + a^6*x^3) + 35/8*b^2*arc
tan(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 \begin {gather*} \int \frac {1}{x^4\,{\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^4*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)),x)

[Out]

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

[Out]

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

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