3.888 \(\int \frac{x^{15}}{(a+b x^8) \sqrt{c+d x^8}} \, dx\)

Optimal. Leaf size=74 \[ \frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{4 b^{3/2} \sqrt{b c-a d}}+\frac{\sqrt{c+d x^8}}{4 b d} \]

[Out]

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

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Rubi [A]  time = 0.0610699, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {446, 80, 63, 208} \[ \frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{4 b^{3/2} \sqrt{b c-a d}}+\frac{\sqrt{c+d x^8}}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[x^15/((a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

Sqrt[c + d*x^8]/(4*b*d) + (a*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^8])/Sqrt[b*c - a*d]])/(4*b^(3/2)*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 80

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

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{x^{15}}{\left (a+b x^8\right ) \sqrt{c+d x^8}} \, dx &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{x}{(a+b x) \sqrt{c+d x}} \, dx,x,x^8\right )\\ &=\frac{\sqrt{c+d x^8}}{4 b d}-\frac{a \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^8\right )}{8 b}\\ &=\frac{\sqrt{c+d x^8}}{4 b d}-\frac{a \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 b d}\\ &=\frac{\sqrt{c+d x^8}}{4 b d}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{4 b^{3/2} \sqrt{b c-a d}}\\ \end{align*}

Mathematica [A]  time = 0.0629486, size = 72, normalized size = 0.97 \[ \frac{1}{4} \left (\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{b^{3/2} \sqrt{b c-a d}}+\frac{\sqrt{c+d x^8}}{b d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^15/((a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

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

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{15}}{b{x}^{8}+a}{\frac{1}{\sqrt{d{x}^{8}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^15/(b*x^8+a)/(d*x^8+c)^(1/2),x)

[Out]

int(x^15/(b*x^8+a)/(d*x^8+c)^(1/2),x)

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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(x^15/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.30984, size = 437, normalized size = 5.91 \begin{align*} \left [\frac{\sqrt{b^{2} c - a b d} a d \log \left (\frac{b d x^{8} + 2 \, b c - a d + 2 \, \sqrt{d x^{8} + c} \sqrt{b^{2} c - a b d}}{b x^{8} + a}\right ) + 2 \, \sqrt{d x^{8} + c}{\left (b^{2} c - a b d\right )}}{8 \,{\left (b^{3} c d - a b^{2} d^{2}\right )}}, -\frac{\sqrt{-b^{2} c + a b d} a d \arctan \left (\frac{\sqrt{d x^{8} + c} \sqrt{-b^{2} c + a b d}}{b d x^{8} + b c}\right ) - \sqrt{d x^{8} + c}{\left (b^{2} c - a b d\right )}}{4 \,{\left (b^{3} c d - a b^{2} d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{15}}{\left (a + b x^{8}\right ) \sqrt{c + d x^{8}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**15/(b*x**8+a)/(d*x**8+c)**(1/2),x)

[Out]

Integral(x**15/((a + b*x**8)*sqrt(c + d*x**8)), x)

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Giac [A]  time = 1.12297, size = 86, normalized size = 1.16 \begin{align*} -\frac{\frac{a d \arctan \left (\frac{\sqrt{d x^{8} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{\sqrt{-b^{2} c + a b d} b} - \frac{\sqrt{d x^{8} + c}}{b}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

-1/4*(a*d*arctan(sqrt(d*x^8 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b) - sqrt(d*x^8 + c)/b)/d