3.891 \(\int \frac{1}{x^9 (a+b x^8) \sqrt{c+d x^8}} \, dx\)

Optimal. Leaf size=117 \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{4 a^2 \sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^8}}{\sqrt{c}}\right )}{8 a^2 c^{3/2}}-\frac{\sqrt{c+d x^8}}{8 a c x^8} \]

[Out]

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

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Rubi [A]  time = 0.112717, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 103, 156, 63, 208} \[ -\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{4 a^2 \sqrt{b c-a d}}+\frac{(a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^8}}{\sqrt{c}}\right )}{8 a^2 c^{3/2}}-\frac{\sqrt{c+d x^8}}{8 a c x^8} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^9*(a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

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

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 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, 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^9 \left (a+b x^8\right ) \sqrt{c+d x^8}} \, dx &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x) \sqrt{c+d x}} \, dx,x,x^8\right )\\ &=-\frac{\sqrt{c+d x^8}}{8 a c x^8}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (2 b c+a d)+\frac{b d x}{2}}{x (a+b x) \sqrt{c+d x}} \, dx,x,x^8\right )}{8 a c}\\ &=-\frac{\sqrt{c+d x^8}}{8 a c x^8}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^8\right )}{8 a^2}-\frac{(2 b c+a d) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^8\right )}{16 a^2 c}\\ &=-\frac{\sqrt{c+d x^8}}{8 a c x^8}+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x^8}\right )}{4 a^2 d}-\frac{(2 b c+a d) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^8}\right )}{8 a^2 c d}\\ &=-\frac{\sqrt{c+d x^8}}{8 a c x^8}+\frac{(2 b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c+d x^8}}{\sqrt{c}}\right )}{8 a^2 c^{3/2}}-\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{4 a^2 \sqrt{b c-a d}}\\ \end{align*}

Mathematica [A]  time = 0.116601, size = 151, normalized size = 1.29 \[ \frac{b^{3/2} \sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{4 a^2 (a d-b c)}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{c+d x^8}}{\sqrt{c}}\right )}{4 a^2 \sqrt{c}}+\frac{d \tanh ^{-1}\left (\frac{\sqrt{c+d x^8}}{\sqrt{c}}\right )}{8 a c^{3/2}}-\frac{\sqrt{c+d x^8}}{8 a c x^8} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^9*(a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

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

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{9} \left ( b{x}^{8}+a \right ) }{\frac{1}{\sqrt{d{x}^{8}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^9/(b*x^8+a)/(d*x^8+c)^(1/2),x)

[Out]

int(1/x^9/(b*x^8+a)/(d*x^8+c)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{8} + a\right )} \sqrt{d x^{8} + c} x^{9}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^8 + a)*sqrt(d*x^8 + c)*x^9), x)

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Fricas [A]  time = 1.44823, size = 1264, normalized size = 10.8 \begin{align*} \left [\frac{2 \, b c^{2} x^{8} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b d x^{8} + 2 \, b c - a d - 2 \, \sqrt{d x^{8} + c}{\left (b c - a d\right )} \sqrt{\frac{b}{b c - a d}}}{b x^{8} + a}\right ) +{\left (2 \, b c + a d\right )} \sqrt{c} x^{8} \log \left (\frac{d x^{8} + 2 \, \sqrt{d x^{8} + c} \sqrt{c} + 2 \, c}{x^{8}}\right ) - 2 \, \sqrt{d x^{8} + c} a c}{16 \, a^{2} c^{2} x^{8}}, -\frac{4 \, b c^{2} x^{8} \sqrt{-\frac{b}{b c - a d}} \arctan \left (-\frac{\sqrt{d x^{8} + c}{\left (b c - a d\right )} \sqrt{-\frac{b}{b c - a d}}}{b d x^{8} + b c}\right ) -{\left (2 \, b c + a d\right )} \sqrt{c} x^{8} \log \left (\frac{d x^{8} + 2 \, \sqrt{d x^{8} + c} \sqrt{c} + 2 \, c}{x^{8}}\right ) + 2 \, \sqrt{d x^{8} + c} a c}{16 \, a^{2} c^{2} x^{8}}, \frac{b c^{2} x^{8} \sqrt{\frac{b}{b c - a d}} \log \left (\frac{b d x^{8} + 2 \, b c - a d - 2 \, \sqrt{d x^{8} + c}{\left (b c - a d\right )} \sqrt{\frac{b}{b c - a d}}}{b x^{8} + a}\right ) -{\left (2 \, b c + a d\right )} \sqrt{-c} x^{8} \arctan \left (\frac{\sqrt{d x^{8} + c} \sqrt{-c}}{c}\right ) - \sqrt{d x^{8} + c} a c}{8 \, a^{2} c^{2} x^{8}}, -\frac{2 \, b c^{2} x^{8} \sqrt{-\frac{b}{b c - a d}} \arctan \left (-\frac{\sqrt{d x^{8} + c}{\left (b c - a d\right )} \sqrt{-\frac{b}{b c - a d}}}{b d x^{8} + b c}\right ) +{\left (2 \, b c + a d\right )} \sqrt{-c} x^{8} \arctan \left (\frac{\sqrt{d x^{8} + c} \sqrt{-c}}{c}\right ) + \sqrt{d x^{8} + c} a c}{8 \, a^{2} c^{2} x^{8}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(2*b*c^2*x^8*sqrt(b/(b*c - a*d))*log((b*d*x^8 + 2*b*c - a*d - 2*sqrt(d*x^8 + c)*(b*c - a*d)*sqrt(b/(b*c
- a*d)))/(b*x^8 + a)) + (2*b*c + a*d)*sqrt(c)*x^8*log((d*x^8 + 2*sqrt(d*x^8 + c)*sqrt(c) + 2*c)/x^8) - 2*sqrt(
d*x^8 + c)*a*c)/(a^2*c^2*x^8), -1/16*(4*b*c^2*x^8*sqrt(-b/(b*c - a*d))*arctan(-sqrt(d*x^8 + c)*(b*c - a*d)*sqr
t(-b/(b*c - a*d))/(b*d*x^8 + b*c)) - (2*b*c + a*d)*sqrt(c)*x^8*log((d*x^8 + 2*sqrt(d*x^8 + c)*sqrt(c) + 2*c)/x
^8) + 2*sqrt(d*x^8 + c)*a*c)/(a^2*c^2*x^8), 1/8*(b*c^2*x^8*sqrt(b/(b*c - a*d))*log((b*d*x^8 + 2*b*c - a*d - 2*
sqrt(d*x^8 + c)*(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b*x^8 + a)) - (2*b*c + a*d)*sqrt(-c)*x^8*arctan(sqrt(d*x^8 +
c)*sqrt(-c)/c) - sqrt(d*x^8 + c)*a*c)/(a^2*c^2*x^8), -1/8*(2*b*c^2*x^8*sqrt(-b/(b*c - a*d))*arctan(-sqrt(d*x^
8 + c)*(b*c - a*d)*sqrt(-b/(b*c - a*d))/(b*d*x^8 + b*c)) + (2*b*c + a*d)*sqrt(-c)*x^8*arctan(sqrt(d*x^8 + c)*s
qrt(-c)/c) + sqrt(d*x^8 + c)*a*c)/(a^2*c^2*x^8)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**9/(b*x**8+a)/(d*x**8+c)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.16155, size = 159, normalized size = 1.36 \begin{align*} \frac{1}{8} \, d^{2}{\left (\frac{2 \, b^{2} \arctan \left (\frac{\sqrt{d x^{8} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} a^{2} d^{2}} - \frac{{\left (2 \, b c + a d\right )} \arctan \left (\frac{\sqrt{d x^{8} + c}}{\sqrt{-c}}\right )}{a^{2} \sqrt{-c} c d^{2}} - \frac{\sqrt{d x^{8} + c}}{a c d^{2} x^{8}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^9/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

1/8*d^2*(2*b^2*arctan(sqrt(d*x^8 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^2*d^2) - (2*b*c + a*d)*a
rctan(sqrt(d*x^8 + c)/sqrt(-c))/(a^2*sqrt(-c)*c*d^2) - sqrt(d*x^8 + c)/(a*c*d^2*x^8))