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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.184842, antiderivative size = 149, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {465, 472, 583, 12, 377, 205} \[ -\frac{\sqrt{c+d x^8} (3 b c-2 a d)}{8 a^2 c x^4 (b c-a d)}-\frac{b (3 b c-4 a d) \tan ^{-1}\left (\frac{x^4 \sqrt{b c-a d}}{\sqrt{a} \sqrt{c+d x^8}}\right )}{8 a^{5/2} (b c-a d)^{3/2}}+\frac{b \sqrt{c+d x^8}}{8 a x^4 \left (a+b x^8\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b*c - 2*a*d)*Sqrt[c + d*x^8])/(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 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 472

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 583

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]

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 )^2 \sqrt{c+d x^8}} \, dx &=\frac{1}{4} \operatorname{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{\operatorname{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{\operatorname{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)) \operatorname{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)) \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 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*}

Mathematica [C]  time = 1.57073, size = 869, normalized size = 5.83 \[ -\frac{\sqrt{d x^8+c} \left (120 d^2 \sin ^{-1}\left (\sqrt{\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}}\right ) x^{16}+96 d^2 \left (\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )^{5/2} \sqrt{\frac{a \left (d x^8+c\right )}{c \left (b x^8+a\right )}} \, _2F_1\left (2,3;\frac{7}{2};\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right ) x^{16}+32 d^2 \left (\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )^{5/2} \sqrt{\frac{a \left (d x^8+c\right )}{c \left (b x^8+a\right )}} \text{HypergeometricPFQ}\left (\{2,2,3\},\left \{1,\frac{7}{2}\right \},\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right ) x^{16}-120 d^2 \sqrt{\frac{a (b c-a d) x^8 \left (d x^8+c\right )}{c^2 \left (b x^8+a\right )^2}} x^{16}+180 c d \sin ^{-1}\left (\sqrt{\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}}\right ) x^8+160 c d \left (\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )^{5/2} \sqrt{\frac{a \left (d x^8+c\right )}{c \left (b x^8+a\right )}} \, _2F_1\left (2,3;\frac{7}{2};\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right ) x^8+64 c d \left (\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )^{5/2} \sqrt{\frac{a \left (d x^8+c\right )}{c \left (b x^8+a\right )}} \text{HypergeometricPFQ}\left (\{2,2,3\},\left \{1,\frac{7}{2}\right \},\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right ) x^8-180 c d \sqrt{\frac{a (b c-a d) x^8 \left (d x^8+c\right )}{c^2 \left (b x^8+a\right )^2}} x^8+45 c^2 \sin ^{-1}\left (\sqrt{\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}}\right )+64 c^2 \left (\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )^{5/2} \sqrt{\frac{a \left (d x^8+c\right )}{c \left (b x^8+a\right )}} \, _2F_1\left (2,3;\frac{7}{2};\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )+32 c^2 \left (\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )^{5/2} \sqrt{\frac{a \left (d x^8+c\right )}{c \left (b x^8+a\right )}} \text{HypergeometricPFQ}\left (\{2,2,3\},\left \{1,\frac{7}{2}\right \},\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )-45 c^2 \sqrt{\frac{a (b c-a d) x^8 \left (d x^8+c\right )}{c^2 \left (b x^8+a\right )^2}}\right )}{120 c^3 x^4 \left (\frac{(b c-a d) x^8}{c \left (b x^8+a\right )}\right )^{3/2} \left (b x^8+a\right )^2 \sqrt{\frac{a \left (d x^8+c\right )}{c \left (b x^8+a\right )}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(Sqrt[c + d*x^8]*(-45*c^2*Sqrt[(a*(b*c - a*d)*x^8*(c + d*x^8))/(c^2*(a + b*x^8)^2)] - 180*c*d*x^8*Sqrt[(a*(b*
c - a*d)*x^8*(c + d*x^8))/(c^2*(a + b*x^8)^2)] - 120*d^2*x^16*Sqrt[(a*(b*c - a*d)*x^8*(c + d*x^8))/(c^2*(a + b
*x^8)^2)] + 45*c^2*ArcSin[Sqrt[((b*c - a*d)*x^8)/(c*(a + b*x^8))]] + 180*c*d*x^8*ArcSin[Sqrt[((b*c - a*d)*x^8)
/(c*(a + b*x^8))]] + 120*d^2*x^16*ArcSin[Sqrt[((b*c - a*d)*x^8)/(c*(a + b*x^8))]] + 64*c^2*(((b*c - a*d)*x^8)/
(c*(a + b*x^8)))^(5/2)*Sqrt[(a*(c + d*x^8))/(c*(a + b*x^8))]*Hypergeometric2F1[2, 3, 7/2, ((b*c - a*d)*x^8)/(c
*(a + b*x^8))] + 160*c*d*x^8*(((b*c - a*d)*x^8)/(c*(a + b*x^8)))^(5/2)*Sqrt[(a*(c + d*x^8))/(c*(a + b*x^8))]*H
ypergeometric2F1[2, 3, 7/2, ((b*c - a*d)*x^8)/(c*(a + b*x^8))] + 96*d^2*x^16*(((b*c - a*d)*x^8)/(c*(a + b*x^8)
))^(5/2)*Sqrt[(a*(c + d*x^8))/(c*(a + b*x^8))]*Hypergeometric2F1[2, 3, 7/2, ((b*c - a*d)*x^8)/(c*(a + b*x^8))]
+ 32*c^2*(((b*c - a*d)*x^8)/(c*(a + b*x^8)))^(5/2)*Sqrt[(a*(c + d*x^8))/(c*(a + b*x^8))]*HypergeometricPFQ[{2
, 2, 3}, {1, 7/2}, ((b*c - a*d)*x^8)/(c*(a + b*x^8))] + 64*c*d*x^8*(((b*c - a*d)*x^8)/(c*(a + b*x^8)))^(5/2)*S
qrt[(a*(c + d*x^8))/(c*(a + b*x^8))]*HypergeometricPFQ[{2, 2, 3}, {1, 7/2}, ((b*c - a*d)*x^8)/(c*(a + b*x^8))]
+ 32*d^2*x^16*(((b*c - a*d)*x^8)/(c*(a + b*x^8)))^(5/2)*Sqrt[(a*(c + d*x^8))/(c*(a + b*x^8))]*HypergeometricP
FQ[{2, 2, 3}, {1, 7/2}, ((b*c - a*d)*x^8)/(c*(a + b*x^8))]))/(120*c^3*x^4*(((b*c - a*d)*x^8)/(c*(a + b*x^8)))^
(3/2)*(a + b*x^8)^2*Sqrt[(a*(c + d*x^8))/(c*(a + b*x^8))])

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{5} \left ( b{x}^{8}+a \right ) ^{2}}{\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)^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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{8} + a\right )}^{2} \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)^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]  time = 2.49406, size = 1268, normalized size = 8.51 \begin{align*} \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{align*}

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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.21622, size = 181, normalized size = 1.21 \begin{align*} -\frac{b^{2} c \sqrt{d + \frac{c}{x^{8}}}}{8 \,{\left (a^{2} b c - a^{3} d\right )}{\left (b c + a{\left (d + \frac{c}{x^{8}}\right )} - a d\right )}} + \frac{{\left (3 \, b^{2} c - 4 \, a b d\right )} \arctan \left (\frac{a \sqrt{d + \frac{c}{x^{8}}}}{\sqrt{a b c - a^{2} d}}\right )}{8 \,{\left (a^{2} b c - a^{3} d\right )} \sqrt{a b c - a^{2} d}} - \frac{\sqrt{d + \frac{c}{x^{8}}}}{4 \, a^{2} c} \end{align*}

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