3.10.16 \(\int \frac {1}{x^5 (a+b x^8)^2 \sqrt {c+d x^8}} \, dx\) [916]

Optimal. Leaf size=149 \[ -\frac {(3 b c-2 a d) \sqrt {c+d x^8}}{8 a^2 c (b c-a d) x^4}+\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^4 \left (a+b x^8\right )}-\frac {b (3 b c-4 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{8 a^{5/2} (b c-a d)^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 149, 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, 483, 597, 12, 385, 211} \begin {gather*} -\frac {b (3 b c-4 a d) \text {ArcTan}\left (\frac {x^4 \sqrt {b c-a d}}{\sqrt {a} \sqrt {c+d x^8}}\right )}{8 a^{5/2} (b c-a d)^{3/2}}-\frac {\sqrt {c+d x^8} (3 b c-2 a d)}{8 a^2 c x^4 (b c-a d)}+\frac {b \sqrt {c+d x^8}}{8 a x^4 \left (a+b x^8\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

-1/8*((3*b*c - 2*a*d)*Sqrt[c + d*x^8])/(a^2*c*(b*c - a*d)*x^4) + (b*Sqrt[c + d*x^8])/(8*a*(b*c - a*d)*x^4*(a +
b*x^8)) - (b*(3*b*c - 4*a*d)*ArcTan[(Sqrt[b*c - a*d]*x^4)/(Sqrt[a]*Sqrt[c + d*x^8])])/(8*a^(5/2)*(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 483

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*(e*
x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a*d)
*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*(b*c - a*d)*(p + 1) + d*b*(m + n
*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ
[p, -1] && IntBinomialQ[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^5 \left (a+b x^8\right )^2 \sqrt {c+d x^8}} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^2 \sqrt {c+d x^2}} \, dx,x,x^4\right )\\ &=\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^4 \left (a+b x^8\right )}-\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 )}{8 a (b c-a d)}\\ &=-\frac {(3 b c-2 a d) \sqrt {c+d x^8}}{8 a^2 c (b c-a d) x^4}+\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^4 \left (a+b x^8\right )}-\frac {\text {Subst}\left (\int \frac {b c (3 b c-4 a d)}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )}{8 a^2 c (b c-a d)}\\ &=-\frac {(3 b c-2 a d) \sqrt {c+d x^8}}{8 a^2 c (b c-a d) x^4}+\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^4 \left (a+b x^8\right )}-\frac {(b (3 b c-4 a d)) \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right ) \sqrt {c+d x^2}} \, dx,x,x^4\right )}{8 a^2 (b c-a d)}\\ &=-\frac {(3 b c-2 a d) \sqrt {c+d x^8}}{8 a^2 c (b c-a d) x^4}+\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^4 \left (a+b x^8\right )}-\frac {(b (3 b c-4 a d)) \text {Subst}\left (\int \frac {1}{a-(-b c+a d) x^2} \, dx,x,\frac {x^4}{\sqrt {c+d x^8}}\right )}{8 a^2 (b c-a d)}\\ &=-\frac {(3 b c-2 a d) \sqrt {c+d x^8}}{8 a^2 c (b c-a d) x^4}+\frac {b \sqrt {c+d x^8}}{8 a (b c-a d) x^4 \left (a+b x^8\right )}-\frac {b (3 b c-4 a d) \tan ^{-1}\left (\frac {\sqrt {b c-a d} x^4}{\sqrt {a} \sqrt {c+d x^8}}\right )}{8 a^{5/2} (b c-a d)^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 1.43, size = 157, normalized size = 1.05 \begin {gather*} \frac {\sqrt {c+d x^8} \left (2 a b c-2 a^2 d+3 b^2 c x^8-2 a b d x^8\right )}{8 a^2 c (-b c+a d) x^4 \left (a+b x^8\right )}-\frac {b (3 b c-4 a d) \tan ^{-1}\left (\frac {a \sqrt {d}+b \sqrt {d} x^8+b x^4 \sqrt {c+d x^8}}{\sqrt {a} \sqrt {b c-a d}}\right )}{8 a^{5/2} (b c-a d)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^8)^2*Sqrt[c + d*x^8]),x]

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{5} \left (b \,x^{8}+a \right )^{2} \sqrt {d \,x^{8}+c}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^8+a)^2/(d*x^8+c)^(1/2),x)

[Out]

int(1/x^5/(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*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^8 + a)^2*sqrt(d*x^8 + c)*x^5), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (129) = 258\).
time = 3.01, size = 612, normalized size = 4.11 \begin {gather*} \left [-\frac {{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{12} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{4}\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 ) + 4 \, {\left ({\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{8} + 2 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d + 2 \, a^{4} d^{2}\right )} \sqrt {d x^{8} + c}}{32 \, {\left ({\left (a^{3} b^{3} c^{3} - 2 \, a^{4} b^{2} c^{2} d + a^{5} b c d^{2}\right )} x^{12} + {\left (a^{4} b^{2} c^{3} - 2 \, a^{5} b c^{2} d + a^{6} c d^{2}\right )} x^{4}\right )}}, -\frac {{\left ({\left (3 \, b^{3} c^{2} - 4 \, a b^{2} c d\right )} x^{12} + {\left (3 \, a b^{2} c^{2} - 4 \, a^{2} b c d\right )} x^{4}\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 ) + 2 \, {\left ({\left (3 \, a b^{3} c^{2} - 5 \, a^{2} b^{2} c d + 2 \, a^{3} b d^{2}\right )} x^{8} + 2 \, a^{2} b^{2} c^{2} - 4 \, a^{3} b c d + 2 \, a^{4} d^{2}\right )} \sqrt {d x^{8} + c}}{16 \, {\left ({\left (a^{3} b^{3} c^{3} - 2 \, a^{4} b^{2} c^{2} d + a^{5} b c d^{2}\right )} x^{12} + {\left (a^{4} b^{2} c^{3} - 2 \, a^{5} b c^{2} d + a^{6} c d^{2}\right )} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="fricas")

[Out]

[-1/32*(((3*b^3*c^2 - 4*a*b^2*c*d)*x^12 + (3*a*b^2*c^2 - 4*a^2*b*c*d)*x^4)*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)*sqr
t(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^3*c^2 - 5*a^2*b^2*c*d + 2*a^3*b*d
^2)*x^8 + 2*a^2*b^2*c^2 - 4*a^3*b*c*d + 2*a^4*d^2)*sqrt(d*x^8 + c))/((a^3*b^3*c^3 - 2*a^4*b^2*c^2*d + a^5*b*c*
d^2)*x^12 + (a^4*b^2*c^3 - 2*a^5*b*c^2*d + a^6*c*d^2)*x^4), -1/16*(((3*b^3*c^2 - 4*a*b^2*c*d)*x^12 + (3*a*b^2*
c^2 - 4*a^2*b*c*d)*x^4)*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)) + 2*((3*a*b^3*c^2 - 5*a^2*b^2*c*d + 2*a^3*b*d^2)*
x^8 + 2*a^2*b^2*c^2 - 4*a^3*b*c*d + 2*a^4*d^2)*sqrt(d*x^8 + c))/((a^3*b^3*c^3 - 2*a^4*b^2*c^2*d + a^5*b*c*d^2)
*x^12 + (a^4*b^2*c^3 - 2*a^5*b*c^2*d + a^6*c*d^2)*x^4)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{5} \left (a + b x^{8}\right )^{2} \sqrt {c + d x^{8}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**8+a)**2/(d*x**8+c)**(1/2),x)

[Out]

Integral(1/(x**5*(a + b*x**8)**2*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. 418 vs. \(2 (129) = 258\).
time = 2.62, size = 418, normalized size = 2.81 \begin {gather*} \frac {1}{8} \, d^{\frac {5}{2}} {\left (\frac {{\left (3 \, b^{2} c - 4 \, a b d\right )} \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 )}{{\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {a b c d - a^{2} d^{2}}} + \frac {2 \, {\left (3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b^{2} c - 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} a b d - 6 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b^{2} c^{2} + 14 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a b c d - 8 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a^{2} d^{2} + 3 \, b^{2} c^{3} - 2 \, a b c^{2} d\right )}}{{\left ({\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{6} b - 3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} b c + 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{4} a d + 3 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} b c^{2} - 4 \, {\left (\sqrt {d} x^{4} - \sqrt {d x^{8} + c}\right )}^{2} a c d - b c^{3}\right )} {\left (a^{2} b c d^{2} - a^{3} d^{3}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^8+a)^2/(d*x^8+c)^(1/2),x, algorithm="giac")

[Out]

1/8*d^(5/2)*((3*b^2*c - 4*a*b*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))/((a^2*b*c*d^2 - a^3*d^3)*sqrt(a*b*c*d - a^2*d^2)) + 2*(3*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*b^2*c -
4*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*a*b*d - 6*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*b^2*c^2 + 14*(sqrt(d)*x^4 - sq
rt(d*x^8 + c))^2*a*b*c*d - 8*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a^2*d^2 + 3*b^2*c^3 - 2*a*b*c^2*d)/(((sqrt(d)*x
^4 - sqrt(d*x^8 + c))^6*b - 3*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*b*c + 4*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^4*a*d
+ 3*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*b*c^2 - 4*(sqrt(d)*x^4 - sqrt(d*x^8 + c))^2*a*c*d - b*c^3)*(a^2*b*c*d^2
- a^3*d^3)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^5\,{\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(1/(x^5*(a + b*x^8)^2*(c + d*x^8)^(1/2)),x)

[Out]

int(1/(x^5*(a + b*x^8)^2*(c + d*x^8)^(1/2)), x)

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