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

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

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Rubi [A]  time = 0.09, 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 = {465, 471, 12, 377, 205} \begin {gather*} \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)} \end {gather*}

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 12

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

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]

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 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]

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*}

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Mathematica [A]  time = 0.55, size = 124, normalized size = 1.33 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 1.82, size = 145, normalized size = 1.56 \begin {gather*} \frac {c \tan ^{-1}\left (\frac {b \sqrt {d} x^8}{\sqrt {a} \sqrt {b c-a d}}+\frac {b x^4 \sqrt {c+d x^8}}{\sqrt {a} \sqrt {b c-a d}}+\frac {\sqrt {a} \sqrt {d}}{\sqrt {b c-a d}}\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(x^4*Sqrt[c + d*x^8])/((b*c - a*d)*(a + b*x^8)) + (c*ArcTan[(Sqrt[a]*Sqrt[d])/Sqrt[b*c - a*d] + (b*Sqrt[d
]*x^8)/(Sqrt[a]*Sqrt[b*c - a*d]) + (b*x^4*Sqrt[c + d*x^8])/(Sqrt[a]*Sqrt[b*c - a*d])])/(8*Sqrt[a]*(b*c - a*d)^
(3/2))

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fricas [B]  time = 0.54, size = 426, normalized size = 4.58 \begin {gather*} \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 {gather*}

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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giac [B]  time = 1.59, size = 244, normalized size = 2.62 \begin {gather*} \frac {c \sqrt {d} \arctan \left (-\frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b - b c + 2 \, a d}{2 \, \sqrt {a b c d - a^{2} d^{2}}}\right )}{8 \, \sqrt {a b c d - a^{2} d^{2}} {\left (b c - a d\right )}} + \frac {{\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c \sqrt {d} - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d^{\frac {3}{2}} - b c^{2} \sqrt {d}}{4 \, {\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b - 2 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c + 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d + b c^{2}\right )} {\left (b^{2} c - a b d\right )}} \end {gather*}

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

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maple [F]  time = 0.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11}}{\left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{11}}{{\left (b x^{8} + a\right )}^{2} \sqrt {d x^{8} + c}}\,{d x} \end {gather*}

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{11}}{{\left (b\,x^8+a\right )}^2\,\sqrt {d\,x^8+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  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**11/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)

[Out]

Timed out

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