3.914 \(\int \frac{x^{11}}{(a+b x^8)^2 \sqrt{c+d x^8}} \, dx\)

Optimal. Leaf size=93 \[ \frac{c \tan ^{-1}\left (\frac{x^4 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^8}}\right )}{8 \sqrt{a} (b c-a d)^{3/2}}-\frac{x^4 \sqrt{c+d x^8}}{8 \left (a+b x^8\right ) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0856842, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {465, 471, 12, 377, 205} \[ \frac{c \tan ^{-1}\left (\frac{x^4 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^8}}\right )}{8 \sqrt{a} (b c-a d)^{3/2}}-\frac{x^4 \sqrt{c+d x^8}}{8 \left (a+b x^8\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x^11/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

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

Rule 465

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1,
n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /;
FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^{11}}{\left (a+b x^8\right )^2 \sqrt{c+d x^8}} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^2\right )^2 \sqrt{c+d x^2}} \, dx,x,x^4\right )\\ &=-\frac{x^4 \sqrt{c+d x^8}}{8 (b c-a d) \left (a+b x^8\right )}+\frac{\operatorname{Subst}\left (\int \frac{c}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx,x,x^4\right )}{8 (b c-a d)}\\ &=-\frac{x^4 \sqrt{c+d x^8}}{8 (b c-a d) \left (a+b x^8\right )}+\frac{c \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right ) \sqrt{c+d x^2}} \, dx,x,x^4\right )}{8 (b c-a d)}\\ &=-\frac{x^4 \sqrt{c+d x^8}}{8 (b c-a d) \left (a+b x^8\right )}+\frac{c \operatorname{Subst}\left (\int \frac{1}{a-(-b c+a d) x^2} \, dx,x,\frac{x^4}{\sqrt{c+d x^8}}\right )}{8 (b c-a d)}\\ &=-\frac{x^4 \sqrt{c+d x^8}}{8 (b c-a d) \left (a+b x^8\right )}+\frac{c \tan ^{-1}\left (\frac{\sqrt{b c-a d} x^4}{\sqrt{a} \sqrt{c+d x^8}}\right )}{8 \sqrt{a} (b c-a d)^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.472277, size = 124, normalized size = 1.33 \[ \frac{\sqrt{c+d x^8} \left (-\frac{x^8 (b c-a d)}{a+b x^8}-\frac{c \sqrt{x^8 \left (\frac{d}{c}-\frac{b}{a}\right )} \tanh ^{-1}\left (\frac{\sqrt{x^8 \left (\frac{d}{c}-\frac{b}{a}\right )}}{\sqrt{\frac{d x^8}{c}+1}}\right )}{\sqrt{\frac{d x^8}{c}+1}}\right )}{8 x^4 (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

(Sqrt[c + d*x^8]*(-(((b*c - a*d)*x^8)/(a + b*x^8)) - (c*Sqrt[(-(b/a) + d/c)*x^8]*ArcTanh[Sqrt[(-(b/a) + d/c)*x
^8]/Sqrt[1 + (d*x^8)/c]])/Sqrt[1 + (d*x^8)/c]))/(8*(b*c - a*d)^2*x^4)

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Maple [F]  time = 0.042, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{11}}{ \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^11/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)

[Out]

int(x^11/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{11}}{{\left (b x^{8} + a\right )}^{2} \sqrt{d x^{8} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^11/((b*x^8 + a)^2*sqrt(d*x^8 + c)), x)

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Fricas [B]  time = 2.0048, size = 890, normalized size = 9.57 \begin{align*} \left [-\frac{4 \, \sqrt{d x^{8} + c}{\left (a b c - a^{2} d\right )} x^{4} -{\left (b c x^{8} + a c\right )} \sqrt{-a b c + a^{2} d} \log \left (\frac{{\left (b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} x^{16} - 2 \,{\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{8} + a^{2} c^{2} + 4 \,{\left ({\left (b c - 2 \, a d\right )} x^{12} - a c x^{4}\right )} \sqrt{d x^{8} + c} \sqrt{-a b c + a^{2} d}}{b^{2} x^{16} + 2 \, a b x^{8} + a^{2}}\right )}{32 \,{\left ({\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{8} + a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}}, -\frac{2 \, \sqrt{d x^{8} + c}{\left (a b c - a^{2} d\right )} x^{4} -{\left (b c x^{8} + a c\right )} \sqrt{a b c - a^{2} d} \arctan \left (\frac{{\left ({\left (b c - 2 \, a d\right )} x^{8} - a c\right )} \sqrt{d x^{8} + c} \sqrt{a b c - a^{2} d}}{2 \,{\left ({\left (a b c d - a^{2} d^{2}\right )} x^{12} +{\left (a b c^{2} - a^{2} c d\right )} x^{4}\right )}}\right )}{16 \,{\left ({\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{8} + a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/32*(4*sqrt(d*x^8 + c)*(a*b*c - a^2*d)*x^4 - (b*c*x^8 + a*c)*sqrt(-a*b*c + a^2*d)*log(((b^2*c^2 - 8*a*b*c*d
+ 8*a^2*d^2)*x^16 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^8 + a^2*c^2 + 4*((b*c - 2*a*d)*x^12 - a*c*x^4)*sqrt(d*x^8 + c
)*sqrt(-a*b*c + a^2*d))/(b^2*x^16 + 2*a*b*x^8 + a^2)))/((a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^8 + a^2*b^2*
c^2 - 2*a^3*b*c*d + a^4*d^2), -1/16*(2*sqrt(d*x^8 + c)*(a*b*c - a^2*d)*x^4 - (b*c*x^8 + a*c)*sqrt(a*b*c - a^2*
d)*arctan(1/2*((b*c - 2*a*d)*x^8 - a*c)*sqrt(d*x^8 + c)*sqrt(a*b*c - a^2*d)/((a*b*c*d - a^2*d^2)*x^12 + (a*b*c
^2 - a^2*c*d)*x^4)))/((a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^8 + a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*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**11/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.32557, size = 124, normalized size = 1.33 \begin{align*} -\frac{1}{8} \, c{\left (\frac{\arctan \left (\frac{a \sqrt{d + \frac{c}{x^{8}}}}{\sqrt{a b c - a^{2} d}}\right )}{\sqrt{a b c - a^{2} d}{\left (b c - a d\right )}} + \frac{\sqrt{d + \frac{c}{x^{8}}}}{{\left (b c + a{\left (d + \frac{c}{x^{8}}\right )} - a d\right )}{\left (b c - a d\right )}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

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