3.9.96 \(\int \frac {1}{x^{13} (a+b x^8) \sqrt {c+d x^8}} \, dx\) [896]

Optimal. Leaf size=115 \[ -\frac {\sqrt {c+d x^8}}{12 a c x^{12}}+\frac {(3 b c+2 a d) \sqrt {c+d x^8}}{12 a^2 c^2 x^4}+\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{4 a^{5/2} \sqrt {b c-a d}} \]

[Out]

1/4*b^2*arctan(x^4*(-a*d+b*c)^(1/2)/a^(1/2)/(d*x^8+c)^(1/2))/a^(5/2)/(-a*d+b*c)^(1/2)-1/12*(d*x^8+c)^(1/2)/a/c
/x^12+1/12*(2*a*d+3*b*c)*(d*x^8+c)^(1/2)/a^2/c^2/x^4

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Rubi [A]
time = 0.10, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {476, 491, 597, 12, 385, 211} \begin {gather*} \frac {b^2 \text {ArcTan}\left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{4 a^{5/2} \sqrt {b c-a d}}+\frac {\sqrt {c+d x^8} (2 a d+3 b c)}{12 a^2 c^2 x^4}-\frac {\sqrt {c+d x^8}}{12 a c x^{12}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^13*(a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

-1/12*Sqrt[c + d*x^8]/(a*c*x^12) + ((3*b*c + 2*a*d)*Sqrt[c + d*x^8])/(12*a^2*c^2*x^4) + (b^2*ArcTan[(Sqrt[b*c
- a*d]*x^4)/(Sqrt[a]*Sqrt[c + d*x^8])])/(4*a^(5/2)*Sqrt[b*c - a*d])

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 491

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*e*(m + 1))), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 597

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

Rubi steps

\begin {align*} \int \frac {1}{x^{13} \left (a+b x^8\right ) \sqrt {c+d x^8}} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {1}{x^4 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )\\ &=-\frac {\sqrt {c+d x^8}}{12 a c x^{12}}+\frac {\text {Subst}\left (\int \frac {-3 b c-2 a d-2 b d x^2}{x^2 \left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )}{12 a c}\\ &=-\frac {\sqrt {c+d x^8}}{12 a c x^{12}}+\frac {(3 b c+2 a d) \sqrt {c+d x^8}}{12 a^2 c^2 x^4}-\frac {\text {Subst}\left (\int -\frac {3 b^2 c^2}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )}{12 a^2 c^2}\\ &=-\frac {\sqrt {c+d x^8}}{12 a c x^{12}}+\frac {(3 b c+2 a d) \sqrt {c+d x^8}}{12 a^2 c^2 x^4}+\frac {b^2 \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )}{4 a^2}\\ &=-\frac {\sqrt {c+d x^8}}{12 a c x^{12}}+\frac {(3 b c+2 a d) \sqrt {c+d x^8}}{12 a^2 c^2 x^4}+\frac {b^2 \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x^4}{\sqrt {c+d x^8}}\right )}{4 a^2}\\ &=-\frac {\sqrt {c+d x^8}}{12 a c x^{12}}+\frac {(3 b c+2 a d) \sqrt {c+d x^8}}{12 a^2 c^2 x^4}+\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{4 a^{5/2} \sqrt {b c-a d}}\\ \end {align*}

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Mathematica [A]
time = 1.50, size = 121, normalized size = 1.05 \begin {gather*} \frac {\sqrt {c+d x^8} \left (-a c+3 b c x^8+2 a d x^8\right )}{12 a^2 c^2 x^{12}}+\frac {b^2 \tan ^{-1}\left (\frac {a \sqrt {d}+b x^4 \left (\sqrt {d} x^4+\sqrt {c+d x^8}\right )}{\sqrt {a} \sqrt {b c-a d}}\right )}{4 a^{5/2} \sqrt {b c-a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^13*(a + b*x^8)*Sqrt[c + d*x^8]),x]

[Out]

(Sqrt[c + d*x^8]*(-(a*c) + 3*b*c*x^8 + 2*a*d*x^8))/(12*a^2*c^2*x^12) + (b^2*ArcTan[(a*Sqrt[d] + b*x^4*(Sqrt[d]
*x^4 + Sqrt[c + d*x^8]))/(Sqrt[a]*Sqrt[b*c - a*d])])/(4*a^(5/2)*Sqrt[b*c - a*d])

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{13} \left (b \,x^{8}+a \right ) \sqrt {d \,x^{8}+c}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x)

[Out]

int(1/x^13/(b*x^8+a)/(d*x^8+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(1/x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^8 + a)*sqrt(d*x^8 + c)*x^13), x)

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Fricas [A]
time = 3.01, size = 416, normalized size = 3.62 \begin {gather*} \left [-\frac {3 \, \sqrt {-a b c + a^{2} d} b^{2} c^{2} x^{12} \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 ) - 4 \, {\left ({\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x^{8} - a^{2} b c^{2} + a^{3} c d\right )} \sqrt {d x^{8} + c}}{48 \, {\left (a^{3} b c^{3} - a^{4} c^{2} d\right )} x^{12}}, \frac {3 \, \sqrt {a b c - a^{2} d} b^{2} c^{2} x^{12} \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 ) + 2 \, {\left ({\left (3 \, a b^{2} c^{2} - a^{2} b c d - 2 \, a^{3} d^{2}\right )} x^{8} - a^{2} b c^{2} + a^{3} c d\right )} \sqrt {d x^{8} + c}}{24 \, {\left (a^{3} b c^{3} - a^{4} c^{2} d\right )} x^{12}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*sqrt(-a*b*c + a^2*d)*b^2*c^2*x^12*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)) - 4*((3*a*b^2*c^2 - a^2*b*c*d - 2*a^3*d^2)*x^8 - a^2*b*c^2 + a^3*c*d)*sqrt(d*x^8 + c))/((a^3*b*c^
3 - a^4*c^2*d)*x^12), 1/24*(3*sqrt(a*b*c - a^2*d)*b^2*c^2*x^12*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)) + 2*((3*a*b^2*c^2 - a^2*b*c*d
- 2*a^3*d^2)*x^8 - a^2*b*c^2 + a^3*c*d)*sqrt(d*x^8 + c))/((a^3*b*c^3 - a^4*c^2*d)*x^12)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{13} \left (a + b x^{8}\right ) \sqrt {c + d x^{8}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**13/(b*x**8+a)/(d*x**8+c)**(1/2),x)

[Out]

Integral(1/(x**13*(a + b*x**8)*sqrt(c + d*x**8)), x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 205 vs. \(2 (95) = 190\).
time = 4.42, size = 205, normalized size = 1.78 \begin {gather*} -\frac {1}{12} \, d^{\frac {5}{2}} {\left (\frac {3 \, b^{2} \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 )}{\sqrt {a b c d - a^{2} d^{2}} a^{2} d^{2}} + \frac {2 \, {\left (3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b - 6 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c - 6 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a d + 3 \, b c^{2} + 2 \, a c d\right )}}{{\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} - c\right )}^{3} a^{2} d^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^13/(b*x^8+a)/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

-1/12*d^(5/2)*(3*b^2*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)*a^2*d^2) + 2*(3*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*b - 6*(sqrt(d)*x^4 - sqrt(d*x^8 + c)
)^2*b*c - 6*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*d + 3*b*c^2 + 2*a*c*d)/(((sqrt(d)*x^4 - sqrt(d*x^8 + c))^2 - c
)^3*a^2*d^2))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^{13}\,\left (b\,x^8+a\right )\,\sqrt {d\,x^8+c}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^13*(a + b*x^8)*(c + d*x^8)^(1/2)),x)

[Out]

int(1/(x^13*(a + b*x^8)*(c + d*x^8)^(1/2)), x)

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