3.974 \(\int \frac{1}{x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=149 \[ -\frac{\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 a^{5/2} c^{5/2}}+\frac{3 \sqrt{a+b x^2} \sqrt{c+d x^2} (a d+b c)}{8 a^2 c^2 x^2}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{4 a c x^4} \]

[Out]

-(Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(4*a*c*x^4) + (3*(b*c + a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(8*a^2*c^2*x^
2) - ((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*a^(
5/2)*c^(5/2))

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Rubi [A]  time = 0.13681, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {446, 103, 151, 12, 93, 208} \[ -\frac{\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 a^{5/2} c^{5/2}}+\frac{3 \sqrt{a+b x^2} \sqrt{c+d x^2} (a d+b c)}{8 a^2 c^2 x^2}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{4 a c x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(4*a*c*x^4) + (3*(b*c + a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(8*a^2*c^2*x^
2) - ((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*a^(
5/2)*c^(5/2))

Rule 446

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 12

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

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{x^5 \sqrt{a+b x^2} \sqrt{c+d x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{4 a c x^4}-\frac{\operatorname{Subst}\left (\int \frac{\frac{3}{2} (b c+a d)+b d x}{x^2 \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )}{4 a c}\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{4 a c x^4}+\frac{3 (b c+a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{8 a^2 c^2 x^2}+\frac{\operatorname{Subst}\left (\int \frac{3 b^2 c^2+2 a b c d+3 a^2 d^2}{4 x \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )}{4 a^2 c^2}\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{4 a c x^4}+\frac{3 (b c+a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{8 a^2 c^2 x^2}+\frac{\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x} \sqrt{c+d x}} \, dx,x,x^2\right )}{16 a^2 c^2}\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{4 a c x^4}+\frac{3 (b c+a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{8 a^2 c^2 x^2}+\frac{\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a+c x^2} \, dx,x,\frac{\sqrt{a+b x^2}}{\sqrt{c+d x^2}}\right )}{8 a^2 c^2}\\ &=-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{4 a c x^4}+\frac{3 (b c+a d) \sqrt{a+b x^2} \sqrt{c+d x^2}}{8 a^2 c^2 x^2}-\frac{\left (3 b^2 c^2+2 a b c d+3 a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 a^{5/2} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0800869, size = 126, normalized size = 0.85 \[ \frac{\sqrt{a+b x^2} \sqrt{c+d x^2} \left (-2 a c+3 a d x^2+3 b c x^2\right )}{8 a^2 c^2 x^4}-\frac{\left (3 a^2 d^2+2 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 a^{5/2} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-2*a*c + 3*b*c*x^2 + 3*a*d*x^2))/(8*a^2*c^2*x^4) - ((3*b^2*c^2 + 2*a*b*c*d +
3*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*a^(5/2)*c^(5/2))

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Maple [B]  time = 0.022, size = 355, normalized size = 2.4 \begin{align*} -{\frac{1}{16\,{a}^{2}{c}^{2}{x}^{4}} \left ( 3\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{4}{a}^{2}{d}^{2}+2\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{4}abcd+3\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){x}^{4}{b}^{2}{c}^{2}-6\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}da{x}^{2}\sqrt{ac}-6\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}bc{x}^{2}\sqrt{ac}+4\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}ac\sqrt{ac} \right ) \sqrt{d{x}^{2}+c}\sqrt{b{x}^{2}+a}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/a^2/c^2*(3*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^4*a^2*d^2
+2*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^4*a*b*c*d+3*ln((a*d*x^2
+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+2*a*c)/x^2)*x^4*b^2*c^2-6*(b*d*x^4+a*d*x^2+b*c*x^2+
a*c)^(1/2)*d*a*x^2*(a*c)^(1/2)-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*b*c*x^2*(a*c)^(1/2)+4*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*a*c*(a*c)^(1/2))*(d*x^2+c)^(1/2)*(b*x^2+a)^(1/2)/x^4/(a*c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c
)^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.58152, size = 798, normalized size = 5.36 \begin{align*} \left [\frac{{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt{a c} x^{4} \log \left (\frac{{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \,{\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \,{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{a c}}{x^{4}}\right ) - 4 \,{\left (2 \, a^{2} c^{2} - 3 \,{\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{32 \, a^{3} c^{3} x^{4}}, \frac{{\left (3 \, b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \sqrt{-a c} x^{4} \arctan \left (\frac{{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c} \sqrt{-a c}}{2 \,{\left (a b c d x^{4} + a^{2} c^{2} +{\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )}}\right ) - 2 \,{\left (2 \, a^{2} c^{2} - 3 \,{\left (a b c^{2} + a^{2} c d\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{16 \, a^{3} c^{3} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*((3*b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*sqrt(a*c)*x^4*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2
+ 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(a*c))/x^4) - 4*
(2*a^2*c^2 - 3*(a*b*c^2 + a^2*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a^3*c^3*x^4), 1/16*((3*b^2*c^2 + 2*a
*b*c*d + 3*a^2*d^2)*sqrt(-a*c)*x^4*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c)*sqrt(-
a*c)/(a*b*c*d*x^4 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x^2)) - 2*(2*a^2*c^2 - 3*(a*b*c^2 + a^2*c*d)*x^2)*sqrt(b*x^2
+ a)*sqrt(d*x^2 + c))/(a^3*c^3*x^4)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError