3.909 \(\int \frac{x^{15}}{(a+b x^8)^2 \sqrt{c+d x^8}} \, dx\)

Optimal. Leaf size=99 \[ \frac{a \sqrt{c+d x^8}}{8 b \left (a+b x^8\right ) (b c-a d)}-\frac{(2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{8 b^{3/2} (b c-a d)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0829682, antiderivative size = 99, 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, 78, 63, 208} \[ \frac{a \sqrt{c+d x^8}}{8 b \left (a+b x^8\right ) (b c-a d)}-\frac{(2 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{8 b^{3/2} (b c-a d)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^15/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

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

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 78

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

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

Mathematica [A]  time = 0.0878775, size = 98, normalized size = 0.99 \[ \frac{\frac{a \sqrt{b} \sqrt{c+d x^8}}{\left (a+b x^8\right ) (b c-a d)}+\frac{(a d-2 b c) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^8}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{3/2}}}{8 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^15/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

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

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{15}}{ \left ( b{x}^{8}+a \right ) ^{2}}{\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)^2/(d*x^8+c)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64851, size = 724, normalized size = 7.31 \begin{align*} \left [\frac{{\left ({\left (2 \, b^{2} c - a b d\right )} x^{8} + 2 \, a b c - a^{2} d\right )} \sqrt{b^{2} c - a b 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 (a b^{2} c - a^{2} b d\right )}}{16 \,{\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{8} + a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2}\right )}}, \frac{{\left ({\left (2 \, b^{2} c - a b d\right )} x^{8} + 2 \, a b c - a^{2} d\right )} \sqrt{-b^{2} c + a b 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 (a b^{2} c - a^{2} b d\right )}}{8 \,{\left ({\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{8} + a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} 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)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(((2*b^2*c - a*b*d)*x^8 + 2*a*b*c - a^2*d)*sqrt(b^2*c - a*b*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)*(a*b^2*c - a^2*b*d))/((b^5*c^2 - 2*a*b^4*c*d + a^2
*b^3*d^2)*x^8 + a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2), 1/8*(((2*b^2*c - a*b*d)*x^8 + 2*a*b*c - a^2*d)*sqrt(
-b^2*c + a*b*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)*(a*b^2*c - a^2*
b*d))/((b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*x^8 + a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2)]

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

[Out]

Timed out

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Giac [A]  time = 1.4907, size = 157, normalized size = 1.59 \begin{align*} \frac{\frac{\sqrt{d x^{8} + c} a d^{2}}{{\left (b^{2} c - a b d\right )}{\left ({\left (d x^{8} + c\right )} b - b c + a d\right )}} + \frac{{\left (2 \, b c d - a d^{2}\right )} \arctan \left (\frac{\sqrt{d x^{8} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{{\left (b^{2} c - a b d\right )} \sqrt{-b^{2} c + a b d}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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