3.895 \(\int \frac{1}{x^5 (a+b x^8) \sqrt{c+d x^8}} \, dx\)

Optimal. Leaf size=80 \[ -\frac{b \tan ^{-1}\left (\frac{x^4 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^8}}\right )}{4 a^{3/2} \sqrt{b c-a d}}-\frac{\sqrt{c+d x^8}}{4 a c x^4} \]

[Out]

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

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Rubi [A]  time = 0.0873839, antiderivative size = 80, normalized size of antiderivative = 1., 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, 480, 12, 377, 205} \[ -\frac{b \tan ^{-1}\left (\frac{x^4 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^8}}\right )}{4 a^{3/2} \sqrt{b c-a d}}-\frac{\sqrt{c+d x^8}}{4 a c x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 480

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 12

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

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 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]

Rubi steps

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

Mathematica [C]  time = 0.699402, size = 179, normalized size = 2.24 \[ -\frac{\left (\frac{d x^8}{c}+1\right ) \left (\frac{4 x^8 \left (c+d x^8\right ) (b c-a d) \, _2F_1\left (2,2;\frac{5}{2};\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )}{3 c^2 \left (a+b x^8\right )}+\frac{\left (c+2 d x^8\right ) \sin ^{-1}\left (\sqrt{\frac{x^8 (b c-a d)}{c \left (a+b x^8\right )}}\right )}{c \sqrt{\frac{a x^8 \left (c+d x^8\right ) (b c-a d)}{c^2 \left (a+b x^8\right )^2}}}\right )}{4 x^4 \left (a+b x^8\right ) \sqrt{c+d x^8}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((1 + (d*x^8)/c)*(((c + 2*d*x^8)*ArcSin[Sqrt[((b*c - a*d)*x^8)/(c*(a + b*x^8))]])/(c*Sqrt[(a*(b*c - a*d)*x^8*
(c + d*x^8))/(c^2*(a + b*x^8)^2)]) + (4*(b*c - a*d)*x^8*(c + d*x^8)*Hypergeometric2F1[2, 2, 5/2, ((b*c - a*d)*
x^8)/(c*(a + b*x^8))])/(3*c^2*(a + b*x^8))))/(4*x^4*(a + b*x^8)*Sqrt[c + d*x^8])

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5} \left ( b{x}^{8}+a \right ) }{\frac{1}{\sqrt{d{x}^{8}+c}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(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^5), x)

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Fricas [B]  time = 1.42864, size = 705, normalized size = 8.81 \begin{align*} \left [-\frac{\sqrt{-a b c + a^{2} d} b c x^{4} \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 \, \sqrt{d x^{8} + c}{\left (a b c - a^{2} d\right )}}{16 \,{\left (a^{2} b c^{2} - a^{3} c d\right )} x^{4}}, -\frac{\sqrt{a b c - a^{2} d} b c x^{4} \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 \, \sqrt{d x^{8} + c}{\left (a b c - a^{2} d\right )}}{8 \,{\left (a^{2} b c^{2} - a^{3} c d\right )} x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**5*(a + b*x**8)*sqrt(c + d*x**8)), x)

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Giac [A]  time = 1.21025, size = 86, normalized size = 1.08 \begin{align*} \frac{\frac{b c \arctan \left (\frac{a \sqrt{d + \frac{c}{x^{8}}}}{\sqrt{a b c - a^{2} d}}\right )}{\sqrt{a b c - a^{2} d} a} - \frac{\sqrt{d + \frac{c}{x^{8}}}}{a}}{4 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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