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

Optimal. Leaf size=177 \[ \frac{b d^2 (4 a c+b) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{\sqrt{a c+b}}\right )}{8 c^{3/2} (a c+b)^{5/2}}+\frac{d (4 a c+b) \left (c+d x^2\right ) \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{8 c x^2 (a c+b)^2}-\frac{\left (c+d x^2\right )^2 \sqrt{\frac{a c+a d x^2+b}{c+d x^2}}}{4 c x^4 (a c+b)} \]

[Out]

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

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Rubi [A]  time = 0.472157, antiderivative size = 218, normalized size of antiderivative = 1.23, number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6722, 1975, 446, 96, 94, 93, 208} \[ \frac{b d^2 (4 a c+b) \sqrt{a \left (c+d x^2\right )+b} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a \left (c+d x^2\right )+b}}\right )}{8 c^{3/2} (a c+b)^{5/2} \sqrt{c+d x^2} \sqrt{a+\frac{b}{c+d x^2}}}+\frac{d (4 a c+b) \left (a \left (c+d x^2\right )+b\right )}{8 c x^2 (a c+b)^2 \sqrt{a+\frac{b}{c+d x^2}}}-\frac{\left (c+d x^2\right ) \left (a \left (c+d x^2\right )+b\right )}{4 c x^4 (a c+b) \sqrt{a+\frac{b}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + 4*a*c)*d*(b + a*(c + d*x^2)))/(8*c*(b + a*c)^2*x^2*Sqrt[a + b/(c + d*x^2)]) - ((c + d*x^2)*(b + a*(c + d
*x^2)))/(4*c*(b + a*c)*x^4*Sqrt[a + b/(c + d*x^2)]) + (b*(b + 4*a*c)*d^2*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)])])/(8*c^(3/2)*(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 96

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[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 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, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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

Mathematica [A]  time = 0.324367, size = 191, normalized size = 1.08 \[ \frac{\sqrt{\frac{a c+a d x^2+b}{c+d x^2}} \left (b d^2 x^4 (4 a c+b) \sqrt{c+d x^2} \tanh ^{-1}\left (\frac{\sqrt{a c+b} \sqrt{c+d x^2}}{\sqrt{c} \sqrt{a c+a d x^2+b}}\right )-\sqrt{c} \sqrt{a c+b} \left (c+d x^2\right ) \sqrt{a \left (c+d x^2\right )+b} \left (2 a c \left (c-d x^2\right )+b \left (2 c+d x^2\right )\right )\right )}{8 c^{3/2} x^4 (a c+b)^{5/2} \sqrt{a \left (c+d x^2\right )+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.016, size = 922, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((a*d*x^2+a*c+b)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-12*a^2*d^3*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*
x^6*c*(a*c^2+b*c)^(3/2)+4*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^3*b*c^5*d^2-2*a*d^3*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*x^6*b*(a
*c^2+b*c)^(3/2)+9*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^2*c^4*d^2-20*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*a^2*c^2*d^2*x^4*(a*
c^2+b*c)^(3/2)+6*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^3*c^3*d^2-12*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*a*c*d^2*b*x^4*(a*c^2+b
*c)^(3/2)+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*b^4*c^2*d^2-2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(1/2)*b^2*d^2*x^4*(a*c^2+b*c)^(3/2)+12
*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+a*c^2+b*c)^(3/2)*a*c*d*x^2*(a*c^2+b*c)^(3/2)+2*(a*d^2*x^4+2*a*c*d*x^2+b*d*x^2+
a*c^2+b*c)^(3/2)*b*d*x^2*(a*c^2+b*c)^(3/2)-4*(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)^(3/2)
*a*c^2-4*(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)^(3/2)*b*c)/((d*x^2+c)*(a*d*x^2+a*c+b))^(1
/2)/(a*c+b)^3/c^2/x^4/(a*c^2+b*c)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.95101, size = 1251, normalized size = 7.07 \begin{align*} \left [\frac{{\left (4 \, a b c + b^{2}\right )} \sqrt{a c^{2} + b c} d^{2} x^{4} \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 c + b\right )} d^{2} x^{4} + 2 \, a c^{3} +{\left (4 \, a c^{2} + 3 \, b c\right )} d x^{2} + 2 \, b c^{2}\right )} \sqrt{a c^{2} + b c} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{x^{4}}\right ) - 4 \,{\left (2 \, a^{2} c^{5} -{\left (2 \, a^{2} c^{3} + a b c^{2} - b^{2} c\right )} d^{2} x^{4} + 4 \, a b c^{4} + 2 \, b^{2} c^{3} + 3 \,{\left (a b c^{3} + b^{2} c^{2}\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{32 \,{\left (a^{3} c^{5} + 3 \, a^{2} b c^{4} + 3 \, a b^{2} c^{3} + b^{3} c^{2}\right )} x^{4}}, -\frac{{\left (4 \, a b c + b^{2}\right )} \sqrt{-a c^{2} - b c} d^{2} x^{4} \arctan \left (\frac{{\left ({\left (2 \, a c + b\right )} d x^{2} + 2 \, a c^{2} + 2 \, b c\right )} \sqrt{-a c^{2} - b c} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{2 \,{\left (a^{2} c^{3} + 2 \, a b c^{2} +{\left (a^{2} c^{2} + a b c\right )} d x^{2} + b^{2} c\right )}}\right ) + 2 \,{\left (2 \, a^{2} c^{5} -{\left (2 \, a^{2} c^{3} + a b c^{2} - b^{2} c\right )} d^{2} x^{4} + 4 \, a b c^{4} + 2 \, b^{2} c^{3} + 3 \,{\left (a b c^{3} + b^{2} c^{2}\right )} d x^{2}\right )} \sqrt{\frac{a d x^{2} + a c + b}{d x^{2} + c}}}{16 \,{\left (a^{3} c^{5} + 3 \, a^{2} b c^{4} + 3 \, a b^{2} c^{3} + b^{3} c^{2}\right )} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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