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

Optimal. Leaf size=68 \[ -\frac{1}{b \sqrt{a+b x^2}}-\frac{1}{3 b \left (a+b x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a-b}}\right )}{\sqrt{a-b}} \]

[Out]

-1/(3*b*(a + b*x^2)^(3/2)) - 1/(b*Sqrt[a + b*x^2]) - ArcTanh[Sqrt[a + b*x^2]/Sqrt[a - b]]/Sqrt[a - b]

________________________________________________________________________________________

Rubi [A]  time = 0.509217, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 58, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {6, 6715, 897, 1261, 207} \[ -\frac{1}{b \sqrt{a+b x^2}}-\frac{1}{3 b \left (a+b x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a-b}}\right )}{\sqrt{a-b}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(1 + a + a^2 + x^2 + a*x^2 + b*x^2 + 2*a*b*x^2 + b*x^4 + b^2*x^4))/((1 + x^2)*(a + b*x^2)^(5/2)),x]

[Out]

-1/(3*b*(a + b*x^2)^(3/2)) - 1/(b*Sqrt[a + b*x^2]) - ArcTanh[Sqrt[a + b*x^2]/Sqrt[a - b]]/Sqrt[a - b]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0734214, size = 63, normalized size = 0.93 \[ \frac{-3 a-3 b x^2-1}{3 b \left (a+b x^2\right )^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a-b}}\right )}{\sqrt{a-b}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(1 + a + a^2 + x^2 + a*x^2 + b*x^2 + 2*a*b*x^2 + b*x^4 + b^2*x^4))/((1 + x^2)*(a + b*x^2)^(5/2)),
x]

[Out]

(-1 - 3*a - 3*b*x^2)/(3*b*(a + b*x^2)^(3/2)) - ArcTanh[Sqrt[a + b*x^2]/Sqrt[a - b]]/Sqrt[a - b]

________________________________________________________________________________________

Maple [B]  time = 0.02, size = 314, normalized size = 4.6 \begin{align*} -{b{x}^{2} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{{x}^{2} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{4\,a}{3} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{a}{b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{b}{3} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{1}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{{a}^{2}}{ \left ( -b+a \right ) ^{2}}\arctan \left ({\sqrt{b{x}^{2}+a}{\frac{1}{\sqrt{b-a}}}} \right ){\frac{1}{\sqrt{b-a}}}}-2\,{\frac{ab}{ \left ( -b+a \right ) ^{2}\sqrt{b-a}}\arctan \left ({\frac{\sqrt{b{x}^{2}+a}}{\sqrt{b-a}}} \right ) }+{\frac{{b}^{2}}{ \left ( -b+a \right ) ^{2}}\arctan \left ({\sqrt{b{x}^{2}+a}{\frac{1}{\sqrt{b-a}}}} \right ){\frac{1}{\sqrt{b-a}}}}+{\frac{{a}^{2}}{ \left ( -b+a \right ) ^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-2\,{\frac{ab}{ \left ( -b+a \right ) ^{2}\sqrt{b{x}^{2}+a}}}+{\frac{{b}^{2}}{ \left ( -b+a \right ) ^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{{a}^{2}}{-3\,b+3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,ab}{-3\,b+3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{{b}^{2}}{-3\,b+3\,a} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2*b/(b*x^2+a)^(3/2)-x^2/(b*x^2+a)^(3/2)-4/3*a/(b*x^2+a)^(3/2)-a/b/(b*x^2+a)^(3/2)+1/3*b/(b*x^2+a)^(3/2)-1/3
/b/(b*x^2+a)^(3/2)+1/(-b+a)^2/(b-a)^(1/2)*arctan((b*x^2+a)^(1/2)/(b-a)^(1/2))*a^2-2/(-b+a)^2/(b-a)^(1/2)*arcta
n((b*x^2+a)^(1/2)/(b-a)^(1/2))*a*b+1/(-b+a)^2/(b-a)^(1/2)*arctan((b*x^2+a)^(1/2)/(b-a)^(1/2))*b^2+1/(-b+a)^2/(
b*x^2+a)^(1/2)*a^2-2/(-b+a)^2/(b*x^2+a)^(1/2)*a*b+1/(-b+a)^2/(b*x^2+a)^(1/2)*b^2+1/3/(-b+a)/(b*x^2+a)^(3/2)*a^
2-2/3/(-b+a)/(b*x^2+a)^(3/2)*a*b+1/3/(-b+a)/(b*x^2+a)^(3/2)*b^2

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.60535, size = 799, normalized size = 11.75 \begin{align*} \left [\frac{3 \,{\left (b^{3} x^{4} + 2 \, a b^{2} x^{2} + a^{2} b\right )} \sqrt{a - b} \log \left (\frac{b^{2} x^{4} + 2 \,{\left (4 \, a b - 3 \, b^{2}\right )} x^{2} - 4 \,{\left (b x^{2} + 2 \, a - b\right )} \sqrt{b x^{2} + a} \sqrt{a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{x^{4} + 2 \, x^{2} + 1}\right ) - 4 \,{\left (3 \,{\left (a b - b^{2}\right )} x^{2} + 3 \, a^{2} -{\left (3 \, a + 1\right )} b + a\right )} \sqrt{b x^{2} + a}}{12 \,{\left ({\left (a b^{3} - b^{4}\right )} x^{4} + a^{3} b - a^{2} b^{2} + 2 \,{\left (a^{2} b^{2} - a b^{3}\right )} x^{2}\right )}}, -\frac{3 \,{\left (b^{3} x^{4} + 2 \, a b^{2} x^{2} + a^{2} b\right )} \sqrt{-a + b} \arctan \left (-\frac{{\left (b x^{2} + 2 \, a - b\right )} \sqrt{b x^{2} + a} \sqrt{-a + b}}{2 \,{\left ({\left (a b - b^{2}\right )} x^{2} + a^{2} - a b\right )}}\right ) + 2 \,{\left (3 \,{\left (a b - b^{2}\right )} x^{2} + 3 \, a^{2} -{\left (3 \, a + 1\right )} b + a\right )} \sqrt{b x^{2} + a}}{6 \,{\left ({\left (a b^{3} - b^{4}\right )} x^{4} + a^{3} b - a^{2} b^{2} + 2 \,{\left (a^{2} b^{2} - a b^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*(b^3*x^4 + 2*a*b^2*x^2 + a^2*b)*sqrt(a - b)*log((b^2*x^4 + 2*(4*a*b - 3*b^2)*x^2 - 4*(b*x^2 + 2*a - b
)*sqrt(b*x^2 + a)*sqrt(a - b) + 8*a^2 - 8*a*b + b^2)/(x^4 + 2*x^2 + 1)) - 4*(3*(a*b - b^2)*x^2 + 3*a^2 - (3*a
+ 1)*b + a)*sqrt(b*x^2 + a))/((a*b^3 - b^4)*x^4 + a^3*b - a^2*b^2 + 2*(a^2*b^2 - a*b^3)*x^2), -1/6*(3*(b^3*x^4
+ 2*a*b^2*x^2 + a^2*b)*sqrt(-a + b)*arctan(-1/2*(b*x^2 + 2*a - b)*sqrt(b*x^2 + a)*sqrt(-a + b)/((a*b - b^2)*x
^2 + a^2 - a*b)) + 2*(3*(a*b - b^2)*x^2 + 3*a^2 - (3*a + 1)*b + a)*sqrt(b*x^2 + a))/((a*b^3 - b^4)*x^4 + a^3*b
- a^2*b^2 + 2*(a^2*b^2 - a*b^3)*x^2)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.18646, size = 70, normalized size = 1.03 \begin{align*} \frac{\arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a + b}}\right )}{\sqrt{-a + b}} - \frac{3 \, b x^{2} + 3 \, a + 1}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sqrt(b*x^2 + a)/sqrt(-a + b))/sqrt(-a + b) - 1/3*(3*b*x^2 + 3*a + 1)/((b*x^2 + a)^(3/2)*b)