3.9.76 \(\int \frac {x^8}{(a+b x^6)^2 \sqrt {c+d x^6}} \, dx\) [876]

Optimal. Leaf size=93 \[ -\frac {x^3 \sqrt {c+d x^6}}{6 (b c-a d) \left (a+b x^6\right )}+\frac {c \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^3}{\sqrt {a} \sqrt {c+d x^6}}\right )}{6 \sqrt {a} (b c-a d)^{3/2}} \]

[Out]

1/6*c*arctan(x^3*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^6+c)^(1/2))/(-a*d+b*c)^(3/2)/a^(1/2)-1/6*x^3*(d*x^6+c)^(1/2)/(-
a*d+b*c)/(b*x^6+a)

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Rubi [A]
time = 0.06, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = {476, 482, 12, 385, 211} \begin {gather*} \frac {c \text {ArcTan}\left (\frac {x^3 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^6}}\right )}{6 \sqrt {a} (b c-a d)^{3/2}}-\frac {x^3 \sqrt {c+d x^6}}{6 \left (a+b x^6\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/((a + b*x^6)^2*Sqrt[c + d*x^6]),x]

[Out]

-1/6*(x^3*Sqrt[c + d*x^6])/((b*c - a*d)*(a + b*x^6)) + (c*ArcTan[(Sqrt[b*c - a*d]*x^3)/(Sqrt[a]*Sqrt[c + d*x^6
])])/(6*Sqrt[a]*(b*c - a*d)^(3/2))

Rule 12

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

Rule 211

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

Rule 385

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 476

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 482

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]

Rubi steps

\begin {align*} \int \frac {x^8}{\left (a+b x^6\right )^2 \sqrt {c+d x^6}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^2}{\left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx,x,x^3\right )\\ &=-\frac {x^3 \sqrt {c+d x^6}}{6 (b c-a d) \left (a+b x^6\right )}+\frac {\text {Subst}\left (\int \frac {c}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^3\right )}{6 (b c-a d)}\\ &=-\frac {x^3 \sqrt {c+d x^6}}{6 (b c-a d) \left (a+b x^6\right )}+\frac {c \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^3\right )}{6 (b c-a d)}\\ &=-\frac {x^3 \sqrt {c+d x^6}}{6 (b c-a d) \left (a+b x^6\right )}+\frac {c \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x^3}{\sqrt {c+d x^6}}\right )}{6 (b c-a d)}\\ &=-\frac {x^3 \sqrt {c+d x^6}}{6 (b c-a d) \left (a+b x^6\right )}+\frac {c \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^3}{\sqrt {a} \sqrt {c+d x^6}}\right )}{6 \sqrt {a} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.95, size = 112, normalized size = 1.20 \begin {gather*} \frac {1}{6} \left (-\frac {x^3 \sqrt {c+d x^6}}{(b c-a d) \left (a+b x^6\right )}+\frac {c \tan ^{-1}\left (\frac {a \sqrt {d}+b x^3 \left (\sqrt {d} x^3+\sqrt {c+d x^6}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{\sqrt {a} (b c-a d)^{3/2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/((a + b*x^6)^2*Sqrt[c + d*x^6]),x]

[Out]

(-((x^3*Sqrt[c + d*x^6])/((b*c - a*d)*(a + b*x^6))) + (c*ArcTan[(a*Sqrt[d] + b*x^3*(Sqrt[d]*x^3 + Sqrt[c + d*x
^6]))/(Sqrt[a]*Sqrt[b*c - a*d])])/(Sqrt[a]*(b*c - a*d)^(3/2)))/6

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Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {x^{8}}{\left (b \,x^{6}+a \right )^{2} \sqrt {d \,x^{6}+c}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^6+a)^2/(d*x^6+c)^(1/2),x)

[Out]

int(x^8/(b*x^6+a)^2/(d*x^6+c)^(1/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^8/((b*x^6 + a)^2*sqrt(d*x^6 + c)), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (77) = 154\).
time = 3.83, size = 426, normalized size = 4.58 \begin {gather*} \left [-\frac {4 \, \sqrt {d x^{6} + c} {\left (a b c - a^{2} d\right )} x^{3} - {\left (b c x^{6} + 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^{12} - 2 \, {\left (3 \, a b c^{2} - 4 \, a^{2} c d\right )} x^{6} + a^{2} c^{2} + 4 \, {\left ({\left (b c - 2 \, a d\right )} x^{9} - a c x^{3}\right )} \sqrt {d x^{6} + c} \sqrt {-a b c + a^{2} d}}{b^{2} x^{12} + 2 \, a b x^{6} + a^{2}}\right )}{24 \, {\left ({\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{6} + a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}}, -\frac {2 \, \sqrt {d x^{6} + c} {\left (a b c - a^{2} d\right )} x^{3} - {\left (b c x^{6} + a c\right )} \sqrt {a b c - a^{2} d} \arctan \left (\frac {{\left ({\left (b c - 2 \, a d\right )} x^{6} - a c\right )} \sqrt {d x^{6} + c} \sqrt {a b c - a^{2} d}}{2 \, {\left ({\left (a b c d - a^{2} d^{2}\right )} x^{9} + {\left (a b c^{2} - a^{2} c d\right )} x^{3}\right )}}\right )}{12 \, {\left ({\left (a b^{3} c^{2} - 2 \, a^{2} b^{2} c d + a^{3} b d^{2}\right )} x^{6} + a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/24*(4*sqrt(d*x^6 + c)*(a*b*c - a^2*d)*x^3 - (b*c*x^6 + 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^12 - 2*(3*a*b*c^2 - 4*a^2*c*d)*x^6 + a^2*c^2 + 4*((b*c - 2*a*d)*x^9 - a*c*x^3)*sqrt(d*x^6 + c)
*sqrt(-a*b*c + a^2*d))/(b^2*x^12 + 2*a*b*x^6 + a^2)))/((a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^6 + a^2*b^2*c
^2 - 2*a^3*b*c*d + a^4*d^2), -1/12*(2*sqrt(d*x^6 + c)*(a*b*c - a^2*d)*x^3 - (b*c*x^6 + a*c)*sqrt(a*b*c - a^2*d
)*arctan(1/2*((b*c - 2*a*d)*x^6 - a*c)*sqrt(d*x^6 + c)*sqrt(a*b*c - a^2*d)/((a*b*c*d - a^2*d^2)*x^9 + (a*b*c^2
- a^2*c*d)*x^3)))/((a*b^3*c^2 - 2*a^2*b^2*c*d + a^3*b*d^2)*x^6 + a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\left (a + b x^{6}\right )^{2} \sqrt {c + d x^{6}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(b*x**6+a)**2/(d*x**6+c)**(1/2),x)

[Out]

Integral(x**8/((a + b*x**6)**2*sqrt(c + d*x**6)), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^6+a)^2/(d*x^6+c)^(1/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^8}{{\left (b\,x^6+a\right )}^2\,\sqrt {d\,x^6+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((a + b*x^6)^2*(c + d*x^6)^(1/2)),x)

[Out]

int(x^8/((a + b*x^6)^2*(c + d*x^6)^(1/2)), x)

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