3.359 \(\int \frac{1}{x^3 (a+\frac{b}{c+d x^2})^{3/2}} \, dx\)

Optimal. Leaf size=146 \[ \frac{3 b d}{2 (a c+b)^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}-\frac{c+d x^2}{2 x^2 (a c+b) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}-\frac{3 b \sqrt{c} d \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a c+b}}\right )}{2 (a c+b)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.449395, antiderivative size = 174, normalized size of antiderivative = 1.19, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6722, 1975, 446, 94, 93, 208} \[ \frac{3 b d}{2 (a c+b)^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{c+d x^2}{2 x^2 (a c+b) \sqrt{a+\frac{b}{c+d x^2}}}-\frac{3 b \sqrt{c} d \sqrt{a \left (c+d x^2\right )+b} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a \left (c+d x^2\right )+b}}\right )}{2 (a c+b)^{5/2} \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b/(c + d*x^2))^(3/2)),x]

[Out]

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

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

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

Mathematica [A]  time = 0.354509, size = 186, normalized size = 1.27 \[ -\frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (\sqrt{a c+b} \left (c+d x^2\right ) \sqrt{a \left (c+d x^2\right )+b} \left (a c \left (c+d x^2\right )+b \left (c-2 d x^2\right )\right )+3 b \sqrt{c} d x^2 \sqrt{c+d x^2} \left (a \left (c+d x^2\right )+b\right ) \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a c+a d x^2+b}}\right )\right )}{2 x^2 (a c+b)^{5/2} \left (a \left (c+d x^2\right )+b\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b/(c + d*x^2))^(3/2)),x]

[Out]

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

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Maple [B]  time = 0.018, size = 1088, normalized size = 7.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b/(d*x^2+c))^(3/2),x)

[Out]

-1/4*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*
c)^(1/2)*x^6*a^2*d^3+3*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^
2+b*c)^(1/2)+2*b*c)/x^2)*x^4*a^2*b*c^2*d^2-6*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)
*x^4*a^2*c*d^2+3*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)
^(1/2)+2*b*c)/x^2)*x^4*a*b^2*c*d^2-4*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)*x^4*a*b
*d^2+3*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)+2*b
*c)/x^2)*x^2*a^2*b*c^3*d-4*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)*x^2*a^2*c^2*d+6*l
n((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)+2*b*c)/x^2)
*x^2*a*b^2*c^2*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(1/2)*x^2*a*d-6*(a*d^2*x^4+2*a*
c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)*x^2*a*b*c*d-4*(a*c^2+b*c)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))
^(1/2)*x^2*a*b*c*d+3*ln((2*a*c*d*x^2+b*d*x^2+2*c^2*a+2*(a*c^2+b*c)^(1/2)*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+
b*c)^(1/2)+2*b*c)/x^2)*x^2*b^3*c*d-2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*(a*c^2+b*c)^(1/2)*x^2*b^2
*d-4*(a*c^2+b*c)^(1/2)*((d*x^2+c)*(a*d*x^2+a*c+b))^(1/2)*x^2*b^2*d+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)
^(3/2)*(a*c^2+b*c)^(1/2)*a*c+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*(a*c^2+b*c)^(1/2)*b)/((d*x^2+c)
*(a*d*x^2+a*c+b))^(1/2)/(a*c+b)^3/(a*d*x^2+a*c+b)/x^2/(a*c^2+b*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^3/(a+b/(d*x^2+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.64759, size = 1277, normalized size = 8.75 \begin{align*} \left [\frac{3 \,{\left (a b d^{2} x^{4} +{\left (a b c + b^{2}\right )} d x^{2}\right )} \sqrt{\frac{c}{a c + b}} \log \left (\frac{{\left (8 \, a^{2} c^{2} + 8 \, a b c + b^{2}\right )} d^{2} x^{4} + 8 \, a^{2} c^{4} + 16 \, a b c^{3} + 8 \, b^{2} c^{2} + 8 \,{\left (2 \, a^{2} c^{3} + 3 \, a b c^{2} + b^{2} c\right )} d x^{2} - 4 \,{\left ({\left (2 \, a^{2} c^{2} + 3 \, a b c + b^{2}\right )} d^{2} x^{4} + 2 \, a^{2} c^{4} + 4 \, a b c^{3} + 2 \, b^{2} c^{2} +{\left (4 \, a^{2} c^{3} + 7 \, a b c^{2} + 3 \, b^{2} c\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}} \sqrt{\frac{c}{a c + b}}}{x^{4}}\right ) - 4 \,{\left ({\left (a c - 2 \, b\right )} d^{2} x^{4} + a c^{3} +{\left (2 \, a c^{2} - b c\right )} d x^{2} + b c^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{8 \,{\left ({\left (a^{3} c^{2} + 2 \, a^{2} b c + a b^{2}\right )} d x^{4} +{\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} + 3 \, a b^{2} c + b^{3}\right )} x^{2}\right )}}, \frac{3 \,{\left (a b d^{2} x^{4} +{\left (a b c + b^{2}\right )} d x^{2}\right )} \sqrt{-\frac{c}{a c + b}} \arctan \left (\frac{{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}} \sqrt{-\frac{c}{a c + b}}}{2 \,{\left (a c d x^{2} + a c^{2} + b c\right )}}\right ) - 2 \,{\left ({\left (a c - 2 \, b\right )} d^{2} x^{4} + a c^{3} +{\left (2 \, a c^{2} - b c\right )} d x^{2} + b c^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{4 \,{\left ({\left (a^{3} c^{2} + 2 \, a^{2} b c + a b^{2}\right )} d x^{4} +{\left (a^{3} c^{3} + 3 \, a^{2} b c^{2} + 3 \, a b^{2} c + b^{3}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(a*b*d^2*x^4 + (a*b*c + b^2)*d*x^2)*sqrt(c/(a*c + b))*log(((8*a^2*c^2 + 8*a*b*c + b^2)*d^2*x^4 + 8*a^2
*c^4 + 16*a*b*c^3 + 8*b^2*c^2 + 8*(2*a^2*c^3 + 3*a*b*c^2 + b^2*c)*d*x^2 - 4*((2*a^2*c^2 + 3*a*b*c + b^2)*d^2*x
^4 + 2*a^2*c^4 + 4*a*b*c^3 + 2*b^2*c^2 + (4*a^2*c^3 + 7*a*b*c^2 + 3*b^2*c)*d*x^2)*sqrt((a*d*x^2 + a*c + b)/(d*
x^2 + c))*sqrt(c/(a*c + b)))/x^4) - 4*((a*c - 2*b)*d^2*x^4 + a*c^3 + (2*a*c^2 - b*c)*d*x^2 + b*c^2)*sqrt((a*d*
x^2 + a*c + b)/(d*x^2 + c)))/((a^3*c^2 + 2*a^2*b*c + a*b^2)*d*x^4 + (a^3*c^3 + 3*a^2*b*c^2 + 3*a*b^2*c + b^3)*
x^2), 1/4*(3*(a*b*d^2*x^4 + (a*b*c + b^2)*d*x^2)*sqrt(-c/(a*c + b))*arctan(1/2*((2*a*c + b)*d*x^2 + 2*a*c^2 +
2*b*c)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c))*sqrt(-c/(a*c + b))/(a*c*d*x^2 + a*c^2 + b*c)) - 2*((a*c - 2*b)*d^
2*x^4 + a*c^3 + (2*a*c^2 - b*c)*d*x^2 + b*c^2)*sqrt((a*d*x^2 + a*c + b)/(d*x^2 + c)))/((a^3*c^2 + 2*a^2*b*c +
a*b^2)*d*x^4 + (a^3*c^3 + 3*a^2*b*c^2 + 3*a*b^2*c + b^3)*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (\frac{a c + a d x^{2} + b}{c + d x^{2}}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{d x^{2} + c}\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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