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

Optimal. Leaf size=168 \[ \frac{9 x \left (a-b x^2\right )}{32 a^3 \sqrt{a+b x^2} \sqrt{a^2-b^2 x^4}}+\frac{x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt{a^2-b^2 x^4}}+\frac{19 \sqrt{a+b x^2} \sqrt{a-b x^2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a-b x^2}}\right )}{32 \sqrt{2} a^3 \sqrt{b} \sqrt{a^2-b^2 x^4}} \]

[Out]

(x*(a - b*x^2))/(8*a^2*(a + b*x^2)^(3/2)*Sqrt[a^2 - b^2*x^4]) + (9*x*(a - b*x^2))/(32*a^3*Sqrt[a + b*x^2]*Sqrt
[a^2 - b^2*x^4]) + (19*Sqrt[a - b*x^2]*Sqrt[a + b*x^2]*ArcTan[(Sqrt[2]*Sqrt[b]*x)/Sqrt[a - b*x^2]])/(32*Sqrt[2
]*a^3*Sqrt[b]*Sqrt[a^2 - b^2*x^4])

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Rubi [A]  time = 0.0862913, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {1152, 414, 527, 12, 377, 205} \[ \frac{9 x \left (a-b x^2\right )}{32 a^3 \sqrt{a+b x^2} \sqrt{a^2-b^2 x^4}}+\frac{x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt{a^2-b^2 x^4}}+\frac{19 \sqrt{a+b x^2} \sqrt{a-b x^2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a-b x^2}}\right )}{32 \sqrt{2} a^3 \sqrt{b} \sqrt{a^2-b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(a - b*x^2))/(8*a^2*(a + b*x^2)^(3/2)*Sqrt[a^2 - b^2*x^4]) + (9*x*(a - b*x^2))/(32*a^3*Sqrt[a + b*x^2]*Sqrt
[a^2 - b^2*x^4]) + (19*Sqrt[a - b*x^2]*Sqrt[a + b*x^2]*ArcTan[(Sqrt[2]*Sqrt[b]*x)/Sqrt[a - b*x^2]])/(32*Sqrt[2
]*a^3*Sqrt[b]*Sqrt[a^2 - b^2*x^4])

Rule 1152

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + c*x^4)^FracPart[p]/((d + e*x
^2)^FracPart[p]*(a/d + (c*x^2)/e)^FracPart[p]), Int[(d + e*x^2)^(p + q)*(a/d + (c*x^2)/e)^p, x], x] /; FreeQ[{
a, c, d, e, p, q}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

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]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^2\right )^{5/2} \sqrt{a^2-b^2 x^4}} \, dx &=\frac{\left (\sqrt{a-b x^2} \sqrt{a+b x^2}\right ) \int \frac{1}{\sqrt{a-b x^2} \left (a+b x^2\right )^3} \, dx}{\sqrt{a^2-b^2 x^4}}\\ &=\frac{x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt{a^2-b^2 x^4}}-\frac{\left (\sqrt{a-b x^2} \sqrt{a+b x^2}\right ) \int \frac{-7 a b+2 b^2 x^2}{\sqrt{a-b x^2} \left (a+b x^2\right )^2} \, dx}{8 a^2 b \sqrt{a^2-b^2 x^4}}\\ &=\frac{x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt{a^2-b^2 x^4}}+\frac{9 x \left (a-b x^2\right )}{32 a^3 \sqrt{a+b x^2} \sqrt{a^2-b^2 x^4}}+\frac{\left (\sqrt{a-b x^2} \sqrt{a+b x^2}\right ) \int \frac{19 a^2 b^2}{\sqrt{a-b x^2} \left (a+b x^2\right )} \, dx}{32 a^4 b^2 \sqrt{a^2-b^2 x^4}}\\ &=\frac{x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt{a^2-b^2 x^4}}+\frac{9 x \left (a-b x^2\right )}{32 a^3 \sqrt{a+b x^2} \sqrt{a^2-b^2 x^4}}+\frac{\left (19 \sqrt{a-b x^2} \sqrt{a+b x^2}\right ) \int \frac{1}{\sqrt{a-b x^2} \left (a+b x^2\right )} \, dx}{32 a^2 \sqrt{a^2-b^2 x^4}}\\ &=\frac{x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt{a^2-b^2 x^4}}+\frac{9 x \left (a-b x^2\right )}{32 a^3 \sqrt{a+b x^2} \sqrt{a^2-b^2 x^4}}+\frac{\left (19 \sqrt{a-b x^2} \sqrt{a+b x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 a b x^2} \, dx,x,\frac{x}{\sqrt{a-b x^2}}\right )}{32 a^2 \sqrt{a^2-b^2 x^4}}\\ &=\frac{x \left (a-b x^2\right )}{8 a^2 \left (a+b x^2\right )^{3/2} \sqrt{a^2-b^2 x^4}}+\frac{9 x \left (a-b x^2\right )}{32 a^3 \sqrt{a+b x^2} \sqrt{a^2-b^2 x^4}}+\frac{19 \sqrt{a-b x^2} \sqrt{a+b x^2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a-b x^2}}\right )}{32 \sqrt{2} a^3 \sqrt{b} \sqrt{a^2-b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.108498, size = 123, normalized size = 0.73 \[ \frac{\sqrt{a^2-b^2 x^4} \left (2 \sqrt{b} x \sqrt{a-b x^2} \left (13 a+9 b x^2\right )+19 \sqrt{2} \left (a+b x^2\right )^2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} x}{\sqrt{a-b x^2}}\right )\right )}{64 a^3 \sqrt{b} \sqrt{a-b x^2} \left (a+b x^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 - b^2*x^4]*(2*Sqrt[b]*x*Sqrt[a - b*x^2]*(13*a + 9*b*x^2) + 19*Sqrt[2]*(a + b*x^2)^2*ArcTan[(Sqrt[2]*
Sqrt[b]*x)/Sqrt[a - b*x^2]]))/(64*a^3*Sqrt[b]*Sqrt[a - b*x^2]*(a + b*x^2)^(5/2))

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Maple [B]  time = 0.054, size = 711, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(-b^2*x^4+a^2)^(1/2)*b^(9/2)*(19*ln(2*b*(2^(1/2)*a^(1/2)*(-b*x^2+a)^(1/2)-(-a*b)^(1/2)*x+a)/(b*x-(-a*b)^
(1/2)))*2^(1/2)*x^4*b^(5/2)*a^(1/2)-19*ln(2*b*(2^(1/2)*a^(1/2)*(-b*x^2+a)^(1/2)+(-a*b)^(1/2)*x+a)/(b*x+(-a*b)^
(1/2)))*2^(1/2)*x^4*b^(5/2)*a^(1/2)+38*ln(2*b*(2^(1/2)*a^(1/2)*(-b*x^2+a)^(1/2)-(-a*b)^(1/2)*x+a)/(b*x-(-a*b)^
(1/2)))*2^(1/2)*x^2*a^(3/2)*b^(3/2)-38*ln(2*b*(2^(1/2)*a^(1/2)*(-b*x^2+a)^(1/2)+(-a*b)^(1/2)*x+a)/(b*x+(-a*b)^
(1/2)))*2^(1/2)*x^2*a^(3/2)*b^(3/2)-16*arctan(b^(1/2)*x/((-b*x+(a*b)^(1/2))/b*(b*x+(a*b)^(1/2)))^(1/2))*x^4*b^
2*(-a*b)^(1/2)+16*arctan(x*b^(1/2)/(-b*x^2+a)^(1/2))*x^4*b^2*(-a*b)^(1/2)-36*b^(3/2)*(-a*b)^(1/2)*(-b*x^2+a)^(
1/2)*x^3+19*ln(2*b*(2^(1/2)*a^(1/2)*(-b*x^2+a)^(1/2)-(-a*b)^(1/2)*x+a)/(b*x-(-a*b)^(1/2)))*2^(1/2)*a^(5/2)*b^(
1/2)-19*ln(2*b*(2^(1/2)*a^(1/2)*(-b*x^2+a)^(1/2)+(-a*b)^(1/2)*x+a)/(b*x+(-a*b)^(1/2)))*2^(1/2)*a^(5/2)*b^(1/2)
-32*arctan(b^(1/2)*x/((-b*x+(a*b)^(1/2))/b*(b*x+(a*b)^(1/2)))^(1/2))*x^2*a*b*(-a*b)^(1/2)+32*arctan(x*b^(1/2)/
(-b*x^2+a)^(1/2))*x^2*a*b*(-a*b)^(1/2)-52*b^(1/2)*a*(-a*b)^(1/2)*(-b*x^2+a)^(1/2)*x-16*arctan(b^(1/2)*x/((-b*x
+(a*b)^(1/2))/b*(b*x+(a*b)^(1/2)))^(1/2))*a^2*(-a*b)^(1/2)+16*arctan(x*b^(1/2)/(-b*x^2+a)^(1/2))*a^2*(-a*b)^(1
/2))/(b*x^2+a)^(1/2)/(-b*x^2+a)^(1/2)/(-a*b)^(1/2)/((-a*b)^(1/2)+(a*b)^(1/2))^3/(-(-a*b)^(1/2)+(a*b)^(1/2))^3/
(b*x+(-a*b)^(1/2))^2/(b*x-(-a*b)^(1/2))^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-b^{2} x^{4} + a^{2}}{\left (b x^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.87349, size = 801, normalized size = 4.77 \begin{align*} \left [-\frac{19 \, \sqrt{2}{\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt{-b} \log \left (-\frac{3 \, b^{2} x^{4} + 2 \, a b x^{2} - 2 \, \sqrt{2} \sqrt{-b^{2} x^{4} + a^{2}} \sqrt{b x^{2} + a} \sqrt{-b} x - a^{2}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}\right ) - 4 \, \sqrt{-b^{2} x^{4} + a^{2}}{\left (9 \, b^{2} x^{3} + 13 \, a b x\right )} \sqrt{b x^{2} + a}}{128 \,{\left (a^{3} b^{4} x^{6} + 3 \, a^{4} b^{3} x^{4} + 3 \, a^{5} b^{2} x^{2} + a^{6} b\right )}}, -\frac{19 \, \sqrt{2}{\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt{b} \arctan \left (\frac{\sqrt{2} \sqrt{-b^{2} x^{4} + a^{2}} \sqrt{b x^{2} + a} \sqrt{b}}{2 \,{\left (b^{2} x^{3} + a b x\right )}}\right ) - 2 \, \sqrt{-b^{2} x^{4} + a^{2}}{\left (9 \, b^{2} x^{3} + 13 \, a b x\right )} \sqrt{b x^{2} + a}}{64 \,{\left (a^{3} b^{4} x^{6} + 3 \, a^{4} b^{3} x^{4} + 3 \, a^{5} b^{2} x^{2} + a^{6} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/128*(19*sqrt(2)*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(-b)*log(-(3*b^2*x^4 + 2*a*b*x^2 - 2*sqrt(
2)*sqrt(-b^2*x^4 + a^2)*sqrt(b*x^2 + a)*sqrt(-b)*x - a^2)/(b^2*x^4 + 2*a*b*x^2 + a^2)) - 4*sqrt(-b^2*x^4 + a^2
)*(9*b^2*x^3 + 13*a*b*x)*sqrt(b*x^2 + a))/(a^3*b^4*x^6 + 3*a^4*b^3*x^4 + 3*a^5*b^2*x^2 + a^6*b), -1/64*(19*sqr
t(2)*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(b)*arctan(1/2*sqrt(2)*sqrt(-b^2*x^4 + a^2)*sqrt(b*x^2 +
a)*sqrt(b)/(b^2*x^3 + a*b*x)) - 2*sqrt(-b^2*x^4 + a^2)*(9*b^2*x^3 + 13*a*b*x)*sqrt(b*x^2 + a))/(a^3*b^4*x^6 +
3*a^4*b^3*x^4 + 3*a^5*b^2*x^2 + a^6*b)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- \left (- a + b x^{2}\right ) \left (a + b x^{2}\right )} \left (a + b x^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(-(-a + b*x**2)*(a + b*x**2))*(a + b*x**2)**(5/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{-b^{2} x^{4} + a^{2}}{\left (b x^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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