3.908 \(\int \frac {x^{23}}{(a+b x^8)^2 \sqrt {c+d x^8}} \, dx\)

Optimal. Leaf size=123 \[ -\frac {a^2 \sqrt {c+d x^8}}{8 b^2 \left (a+b x^8\right ) (b c-a d)}+\frac {a (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^8}}{\sqrt {b c-a d}}\right )}{8 b^{5/2} (b c-a d)^{3/2}}+\frac {\sqrt {c+d x^8}}{4 b^2 d} \]

[Out]

1/8*a*(-3*a*d+4*b*c)*arctanh(b^(1/2)*(d*x^8+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(5/2)/(-a*d+b*c)^(3/2)+1/4*(d*x^8+c)^
(1/2)/b^2/d-1/8*a^2*(d*x^8+c)^(1/2)/b^2/(-a*d+b*c)/(b*x^8+a)

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Rubi [A]  time = 0.14, antiderivative size = 123, 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 = {446, 89, 80, 63, 208} \[ -\frac {a^2 \sqrt {c+d x^8}}{8 b^2 \left (a+b x^8\right ) (b c-a d)}+\frac {a (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^8}}{\sqrt {b c-a d}}\right )}{8 b^{5/2} (b c-a d)^{3/2}}+\frac {\sqrt {c+d x^8}}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[x^23/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

Sqrt[c + d*x^8]/(4*b^2*d) - (a^2*Sqrt[c + d*x^8])/(8*b^2*(b*c - a*d)*(a + b*x^8)) + (a*(4*b*c - 3*a*d)*ArcTanh
[(Sqrt[b]*Sqrt[c + d*x^8])/Sqrt[b*c - a*d]])/(8*b^(5/2)*(b*c - a*d)^(3/2))

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 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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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]

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]]

Rubi steps

\begin {align*} \int \frac {x^{23}}{\left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx &=\frac {1}{8} \operatorname {Subst}\left (\int \frac {x^2}{(a+b x)^2 \sqrt {c+d x}} \, dx,x,x^8\right )\\ &=-\frac {a^2 \sqrt {c+d x^8}}{8 b^2 (b c-a d) \left (a+b x^8\right )}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{2} a (2 b c-a d)+b (b c-a d) x}{(a+b x) \sqrt {c+d x}} \, dx,x,x^8\right )}{8 b^2 (b c-a d)}\\ &=\frac {\sqrt {c+d x^8}}{4 b^2 d}-\frac {a^2 \sqrt {c+d x^8}}{8 b^2 (b c-a d) \left (a+b x^8\right )}-\frac {(a (4 b c-3 a d)) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^8\right )}{16 b^2 (b c-a d)}\\ &=\frac {\sqrt {c+d x^8}}{4 b^2 d}-\frac {a^2 \sqrt {c+d x^8}}{8 b^2 (b c-a d) \left (a+b x^8\right )}-\frac {(a (4 b c-3 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^2 d (b c-a d)}\\ &=\frac {\sqrt {c+d x^8}}{4 b^2 d}-\frac {a^2 \sqrt {c+d x^8}}{8 b^2 (b c-a d) \left (a+b x^8\right )}+\frac {a (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^8}}{\sqrt {b c-a d}}\right )}{8 b^{5/2} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 107, normalized size = 0.87 \[ \frac {1}{8} \left (\frac {\sqrt {c+d x^8} \left (\frac {a^2}{\left (a+b x^8\right ) (a d-b c)}+\frac {2}{d}\right )}{b^2}+\frac {a (4 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^8}}{\sqrt {b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^23/((a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

((Sqrt[c + d*x^8]*(2/d + a^2/((-(b*c) + a*d)*(a + b*x^8))))/b^2 + (a*(4*b*c - 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c +
d*x^8])/Sqrt[b*c - a*d]])/(b^(5/2)*(b*c - a*d)^(3/2)))/8

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fricas [B]  time = 0.77, size = 475, normalized size = 3.86 \[ \left [\frac {{\left ({\left (4 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{8} + 4 \, a^{2} b c d - 3 \, a^{3} d^{2}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{8} + 2 \, b c - a d + 2 \, \sqrt {d x^{8} + c} \sqrt {b^{2} c - a b d}}{b x^{8} + a}\right ) + 2 \, {\left (2 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{8} + 2 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2}\right )} \sqrt {d x^{8} + c}}{16 \, {\left (a b^{5} c^{2} d - 2 \, a^{2} b^{4} c d^{2} + a^{3} b^{3} d^{3} + {\left (b^{6} c^{2} d - 2 \, a b^{5} c d^{2} + a^{2} b^{4} d^{3}\right )} x^{8}\right )}}, -\frac {{\left ({\left (4 \, a b^{2} c d - 3 \, a^{2} b d^{2}\right )} x^{8} + 4 \, a^{2} b c d - 3 \, a^{3} d^{2}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{8} + c} \sqrt {-b^{2} c + a b d}}{b d x^{8} + b c}\right ) - {\left (2 \, {\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{8} + 2 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2}\right )} \sqrt {d x^{8} + c}}{8 \, {\left (a b^{5} c^{2} d - 2 \, a^{2} b^{4} c d^{2} + a^{3} b^{3} d^{3} + {\left (b^{6} c^{2} d - 2 \, a b^{5} c d^{2} + a^{2} b^{4} d^{3}\right )} x^{8}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^23/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(((4*a*b^2*c*d - 3*a^2*b*d^2)*x^8 + 4*a^2*b*c*d - 3*a^3*d^2)*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*(2*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^8 + 2
*a*b^3*c^2 - 5*a^2*b^2*c*d + 3*a^3*b*d^2)*sqrt(d*x^8 + c))/(a*b^5*c^2*d - 2*a^2*b^4*c*d^2 + a^3*b^3*d^3 + (b^6
*c^2*d - 2*a*b^5*c*d^2 + a^2*b^4*d^3)*x^8), -1/8*(((4*a*b^2*c*d - 3*a^2*b*d^2)*x^8 + 4*a^2*b*c*d - 3*a^3*d^2)*
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)) - (2*(b^4*c^2 - 2*a*b^3*c*d
+ a^2*b^2*d^2)*x^8 + 2*a*b^3*c^2 - 5*a^2*b^2*c*d + 3*a^3*b*d^2)*sqrt(d*x^8 + c))/(a*b^5*c^2*d - 2*a^2*b^4*c*d^
2 + a^3*b^3*d^3 + (b^6*c^2*d - 2*a*b^5*c*d^2 + a^2*b^4*d^3)*x^8)]

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giac [A]  time = 0.18, size = 134, normalized size = 1.09 \[ -\frac {\sqrt {d x^{8} + c} a^{2} d}{8 \, {\left (b^{3} c - a b^{2} d\right )} {\left ({\left (d x^{8} + c\right )} b - b c + a d\right )}} - \frac {{\left (4 \, a b c - 3 \, a^{2} d\right )} \arctan \left (\frac {\sqrt {d x^{8} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{8 \, {\left (b^{3} c - a b^{2} d\right )} \sqrt {-b^{2} c + a b d}} + \frac {\sqrt {d x^{8} + c}}{4 \, b^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^23/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.47, size = 0, normalized size = 0.00 \[ \int \frac {x^{23}}{\left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^23/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)

[Out]

int(x^23/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^23/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 5.02, size = 144, normalized size = 1.17 \[ \frac {\sqrt {d\,x^8+c}}{4\,b^2\,d}-\frac {a\,\mathrm {atan}\left (\frac {a\,\sqrt {b}\,\sqrt {d\,x^8+c}\,\left (3\,a\,d-4\,b\,c\right )}{\left (3\,a^2\,d-4\,a\,b\,c\right )\,\sqrt {a\,d-b\,c}}\right )\,\left (3\,a\,d-4\,b\,c\right )}{8\,b^{5/2}\,{\left (a\,d-b\,c\right )}^{3/2}}+\frac {a^2\,d\,\sqrt {d\,x^8+c}}{2\,\left (a\,d-b\,c\right )\,\left (4\,b^3\,\left (d\,x^8+c\right )-4\,b^3\,c+4\,a\,b^2\,d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^23/((a + b*x^8)^2*(c + d*x^8)^(1/2)),x)

[Out]

(c + d*x^8)^(1/2)/(4*b^2*d) - (a*atan((a*b^(1/2)*(c + d*x^8)^(1/2)*(3*a*d - 4*b*c))/((3*a^2*d - 4*a*b*c)*(a*d
- b*c)^(1/2)))*(3*a*d - 4*b*c))/(8*b^(5/2)*(a*d - b*c)^(3/2)) + (a^2*d*(c + d*x^8)^(1/2))/(2*(a*d - b*c)*(4*b^
3*(c + d*x^8) - 4*b^3*c + 4*a*b^2*d))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**23/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)

[Out]

Timed out

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