\(\int \frac {1}{x^4 (a^2+2 a b x^2+b^2 x^4)^{3/2}} \, dx\) [644]

Optimal result

Integrand size = 26, antiderivative size = 209 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b^{3/2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[Out]

7/8/a^2/x^3/((b*x^2+a)^2)^(1/2)+1/4/a/x^3/(b*x^2+a)/((b*x^2+a)^2)^(1/2)-35/24*(b*x^2+a)/a^3/x^3/((b*x^2+a)^2)^
(1/2)+35/8*b*(b*x^2+a)/a^4/x/((b*x^2+a)^2)^(1/2)+35/8*b^(3/2)*(b*x^2+a)*arctan(x*b^(1/2)/a^(1/2))/a^(9/2)/((b*
x^2+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1126, 296, 331, 211} \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {1}{4 a x^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {7}{8 a^2 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b^{3/2} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{9/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 b \left (a+b x^2\right )}{8 a^4 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {35 \left (a+b x^2\right )}{24 a^3 x^3 \sqrt {a^2+2 a b x^2+b^2 x^4}} \]

[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 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 296

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] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

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 1126

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {a} \left (-8 a^3+56 a^2 b x^2+175 a b^2 x^4+105 b^3 x^6\right )+105 b^{3/2} x^3 \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{24 a^{9/2} x^3 \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \]

[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])

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.67

method	result	size
default	\(-\frac {\left (-105 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{4} x^{7}-105 \sqrt {a b}\, b^{3} x^{6}-210 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{3} x^{5}-175 \sqrt {a b}\, a \,b^{2} x^{4}-105 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b^{2} x^{3}-56 \sqrt {a b}\, a^{2} b \,x^{2}+8 \sqrt {a b}\, a^{3}\right ) \left (b \,x^{2}+a \right )}{24 \sqrt {a b}\, x^{3} a^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\)	\(139\)
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {35 b^{3} x^{6}}{8 a^{4}}+\frac {175 b^{2} x^{4}}{24 a^{3}}+\frac {7 b \,x^{2}}{3 a^{2}}-\frac {1}{3 a}\right )}{\left (b \,x^{2}+a \right )^{3} x^{3}}+\frac {35 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, b \ln \left (-b x -\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) a^{5}}-\frac {35 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, b \ln \left (-b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) a^{5}}\)	\(153\)

[In]

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

[Out]

-1/24*(-105*arctan(b*x/(a*b)^(1/2))*b^4*x^7-105*(a*b)^(1/2)*b^3*x^6-210*arctan(b*x/(a*b)^(1/2))*a*b^3*x^5-175*
(a*b)^(1/2)*a*b^2*x^4-105*arctan(b*x/(a*b)^(1/2))*a^2*b^2*x^3-56*(a*b)^(1/2)*a^2*b*x^2+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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.14 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [\frac {210 \, b^{3} x^{6} + 350 \, a b^{2} x^{4} + 112 \, a^{2} b x^{2} - 16 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b x^{3}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{48 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}, \frac {105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3} + 105 \, {\left (b^{3} x^{7} + 2 \, a b^{2} x^{5} + a^{2} b x^{3}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}}\right ] \]

[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)]

Sympy [F]

\[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{4} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.41 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {105 \, b^{3} x^{6} + 175 \, a b^{2} x^{4} + 56 \, a^{2} b x^{2} - 8 \, a^{3}}{24 \, {\left (a^{4} b^{2} x^{7} + 2 \, a^{5} b x^{5} + a^{6} x^{3}\right )}} + \frac {35 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {35 \, b^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {11 \, b^{3} x^{3} + 13 \, a b^{2} x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {9 \, b x^{2} - a}{3 \, a^{4} x^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \]

[In]

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

[Out]

35/8*b^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4*sgn(b*x^2 + a)) + 1/8*(11*b^3*x^3 + 13*a*b^2*x)/((b*x^2 + a)^2*a
^4*sgn(b*x^2 + a)) + 1/3*(9*b*x^2 - a)/(a^4*x^3*sgn(b*x^2 + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x^4\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \]

[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)