3.949 \(\int \frac{(a+b x^2)^{3/2}}{x^5 \sqrt{c+d x^2}} \, dx\)

Optimal. Leaf size=131 \[ -\frac{3 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)}{8 c^2 x^2}-\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 \sqrt{a} c^{5/2}}-\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 c x^4} \]

[Out]

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

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Rubi [A]  time = 0.106027, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {446, 94, 93, 208} \[ -\frac{3 \sqrt{a+b x^2} \sqrt{c+d x^2} (b c-a d)}{8 c^2 x^2}-\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 \sqrt{a} c^{5/2}}-\frac{\left (a+b x^2\right )^{3/2} \sqrt{c+d x^2}}{4 c x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/(x^5*Sqrt[c + d*x^2]),x]

[Out]

(-3*(b*c - a*d)*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(8*c^2*x^2) - ((a + b*x^2)^(3/2)*Sqrt[c + d*x^2])/(4*c*x^4) -
(3*(b*c - a*d)^2*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c + d*x^2])])/(8*Sqrt[a]*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 94

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

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

Mathematica [A]  time = 0.0707517, size = 110, normalized size = 0.84 \[ \frac{\sqrt{a+b x^2} \sqrt{c+d x^2} \left (-2 a c+3 a d x^2-5 b c x^2\right )}{8 c^2 x^4}-\frac{3 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{8 \sqrt{a} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/(x^5*Sqrt[c + d*x^2]),x]

[Out]

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

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Maple [B]  time = 0.014, size = 352, normalized size = 2.7 \begin{align*} -{\frac{1}{16\,{c}^{2}{x}^{4}}\sqrt{b{x}^{2}+a}\sqrt{d{x}^{2}+c} \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}-6\,\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}+10\,\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 ){\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((b*x^2+a)^(3/2)/x^5/(d*x^2+c)^(1/2),x)

[Out]

-1/16*(b*x^2+a)^(1/2)*(d*x^2+c)^(1/2)/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^2-6*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)+10*(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))/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/(a*c)^(1/2)
/x^4

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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((b*x^2+a)^(3/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.64661, size = 792, normalized size = 6.05 \begin{align*} \left [\frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + 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} +{\left (5 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{32 \, a c^{3} x^{4}}, \frac{3 \,{\left (b^{2} c^{2} - 2 \, a b c d + 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} +{\left (5 \, a b c^{2} - 3 \, a^{2} c d\right )} x^{2}\right )} \sqrt{b x^{2} + a} \sqrt{d x^{2} + c}}{16 \, a c^{3} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

[1/32*(3*(b^2*c^2 - 2*a*b*c*d + 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 + (5*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(b*x^2 + a)*sqrt(d*x^2 + c))/(a*c^3*x^4), 1/16*(3*(b^2*c^2 - 2*a*b
*c*d + 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 + (5*a*b*c^2 - 3*a^2*c*d)*x^2)*sqrt(b*x^2 +
a)*sqrt(d*x^2 + c))/(a*c^3*x^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x^{2}\right )^{\frac{3}{2}}}{x^{5} \sqrt{c + d x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/x**5/(d*x**2+c)**(1/2),x)

[Out]

Integral((a + b*x**2)**(3/2)/(x**5*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((b*x^2+a)^(3/2)/x^5/(d*x^2+c)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError