3.490 \(\int \frac{x^8}{(a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.19494, antiderivative size = 149, 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 = {446, 89, 78, 63, 208} \[ -\frac{a^2 d^2+2 b^2 c^2}{3 b^2 d \sqrt{c+d x^3} (b c-a d)^2}-\frac{a^2}{3 b^2 \left (a+b x^3\right ) \sqrt{c+d x^3} (b c-a d)}+\frac{a (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

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 78

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

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 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]

Rubi steps

\begin{align*} \int \frac{x^8}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2}{(a+b x)^2 (c+d x)^{3/2}} \, dx,x,x^3\right )\\ &=-\frac{a^2}{3 b^2 (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{1}{2} a (2 b c+a d)+b (b c-a d) x}{(a+b x) (c+d x)^{3/2}} \, dx,x,x^3\right )}{3 b^2 (b c-a d)}\\ &=-\frac{2 b^2 c^2+a^2 d^2}{3 b^2 d (b c-a d)^2 \sqrt{c+d x^3}}-\frac{a^2}{3 b^2 (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}-\frac{(a (4 b c-a d)) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{6 b (b c-a d)^2}\\ &=-\frac{2 b^2 c^2+a^2 d^2}{3 b^2 d (b c-a d)^2 \sqrt{c+d x^3}}-\frac{a^2}{3 b^2 (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}-\frac{(a (4 b c-a d)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d x^3}\right )}{3 b d (b c-a d)^2}\\ &=-\frac{2 b^2 c^2+a^2 d^2}{3 b^2 d (b c-a d)^2 \sqrt{c+d x^3}}-\frac{a^2}{3 b^2 (b c-a d) \left (a+b x^3\right ) \sqrt{c+d x^3}}+\frac{a (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.278391, size = 134, normalized size = 0.9 \[ \frac{\frac{a (4 b c-a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{(b c-a d)^{5/2}}-\frac{\sqrt{b} \left (a^2 d \left (c+d x^3\right )+2 a b c^2+2 b^2 c^2 x^3\right )}{d \left (a+b x^3\right ) \sqrt{c+d x^3} (b c-a d)^2}}{3 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.053, size = 978, normalized size = 6.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(b*x^3+a)^2/(d*x^3+c)^(3/2),x)

[Out]

-2/3/b^2/d/(d*x^3+c)^(1/2)-2*a/b^2*(-2/3/(a*d-b*c)/((x^3+1/d*c)*d)^(1/2)-1/3*I/d^2*b*2^(1/2)*sum(1/(-a*d+b*c)/
(a*d-b*c)*(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(
d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d
^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2
)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d
^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2
)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*3^(1/2)
/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a)))+a^2
/b^2*(-2/3*d/(a*d-b*c)^2/((x^3+1/d*c)*d)^(1/2)-1/3*b/(a*d-b*c)^2*(d*x^3+c)^(1/2)/(b*x^3+a)+1/2*I/d*b*2^(1/2)*s
um(1/(a*d-b*c)^3*(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^
(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1
/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d^2*c)^(1/3)*_alpha*3^(1/2)*d-I
*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/
2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2),1/2*b/d*(2*I*(-d^2*c)^(1/3)
*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/3)*_alpha-3*c*d)/(a*d-b*c),(I*
3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b+a
)))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.89976, size = 1470, normalized size = 9.87 \begin{align*} \left [-\frac{{\left ({\left (4 \, a b^{2} c d^{2} - a^{2} b d^{3}\right )} x^{6} + 4 \, a^{2} b c^{2} d - a^{3} c d^{2} +{\left (4 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{3}\right )} \sqrt{b^{2} c - a b d} \log \left (\frac{b d x^{3} + 2 \, b c - a d - 2 \, \sqrt{d x^{3} + c} \sqrt{b^{2} c - a b d}}{b x^{3} + a}\right ) + 2 \,{\left (2 \, a b^{3} c^{3} - a^{2} b^{2} c^{2} d - a^{3} b c d^{2} +{\left (2 \, b^{4} c^{3} - 2 \, a b^{3} c^{2} d + a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{6 \,{\left (a b^{5} c^{4} d - 3 \, a^{2} b^{4} c^{3} d^{2} + 3 \, a^{3} b^{3} c^{2} d^{3} - a^{4} b^{2} c d^{4} +{\left (b^{6} c^{3} d^{2} - 3 \, a b^{5} c^{2} d^{3} + 3 \, a^{2} b^{4} c d^{4} - a^{3} b^{3} d^{5}\right )} x^{6} +{\left (b^{6} c^{4} d - 2 \, a b^{5} c^{3} d^{2} + 2 \, a^{3} b^{3} c d^{4} - a^{4} b^{2} d^{5}\right )} x^{3}\right )}}, -\frac{{\left ({\left (4 \, a b^{2} c d^{2} - a^{2} b d^{3}\right )} x^{6} + 4 \, a^{2} b c^{2} d - a^{3} c d^{2} +{\left (4 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{3}\right )} \sqrt{-b^{2} c + a b d} \arctan \left (\frac{\sqrt{d x^{3} + c} \sqrt{-b^{2} c + a b d}}{b d x^{3} + b c}\right ) +{\left (2 \, a b^{3} c^{3} - a^{2} b^{2} c^{2} d - a^{3} b c d^{2} +{\left (2 \, b^{4} c^{3} - 2 \, a b^{3} c^{2} d + a^{2} b^{2} c d^{2} - a^{3} b d^{3}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{3 \,{\left (a b^{5} c^{4} d - 3 \, a^{2} b^{4} c^{3} d^{2} + 3 \, a^{3} b^{3} c^{2} d^{3} - a^{4} b^{2} c d^{4} +{\left (b^{6} c^{3} d^{2} - 3 \, a b^{5} c^{2} d^{3} + 3 \, a^{2} b^{4} c d^{4} - a^{3} b^{3} d^{5}\right )} x^{6} +{\left (b^{6} c^{4} d - 2 \, a b^{5} c^{3} d^{2} + 2 \, a^{3} b^{3} c d^{4} - a^{4} b^{2} d^{5}\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

[-1/6*(((4*a*b^2*c*d^2 - a^2*b*d^3)*x^6 + 4*a^2*b*c^2*d - a^3*c*d^2 + (4*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3
)*x^3)*sqrt(b^2*c - a*b*d)*log((b*d*x^3 + 2*b*c - a*d - 2*sqrt(d*x^3 + c)*sqrt(b^2*c - a*b*d))/(b*x^3 + a)) +
2*(2*a*b^3*c^3 - a^2*b^2*c^2*d - a^3*b*c*d^2 + (2*b^4*c^3 - 2*a*b^3*c^2*d + a^2*b^2*c*d^2 - a^3*b*d^3)*x^3)*sq
rt(d*x^3 + c))/(a*b^5*c^4*d - 3*a^2*b^4*c^3*d^2 + 3*a^3*b^3*c^2*d^3 - a^4*b^2*c*d^4 + (b^6*c^3*d^2 - 3*a*b^5*c
^2*d^3 + 3*a^2*b^4*c*d^4 - a^3*b^3*d^5)*x^6 + (b^6*c^4*d - 2*a*b^5*c^3*d^2 + 2*a^3*b^3*c*d^4 - a^4*b^2*d^5)*x^
3), -1/3*(((4*a*b^2*c*d^2 - a^2*b*d^3)*x^6 + 4*a^2*b*c^2*d - a^3*c*d^2 + (4*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*
d^3)*x^3)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x^3 + b*c)) + (2*a*b^3*c^3 - a
^2*b^2*c^2*d - a^3*b*c*d^2 + (2*b^4*c^3 - 2*a*b^3*c^2*d + a^2*b^2*c*d^2 - a^3*b*d^3)*x^3)*sqrt(d*x^3 + c))/(a*
b^5*c^4*d - 3*a^2*b^4*c^3*d^2 + 3*a^3*b^3*c^2*d^3 - a^4*b^2*c*d^4 + (b^6*c^3*d^2 - 3*a*b^5*c^2*d^3 + 3*a^2*b^4
*c*d^4 - a^3*b^3*d^5)*x^6 + (b^6*c^4*d - 2*a*b^5*c^3*d^2 + 2*a^3*b^3*c*d^4 - a^4*b^2*d^5)*x^3)]

________________________________________________________________________________________

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(x**8/(b*x**3+a)**2/(d*x**3+c)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.13833, size = 263, normalized size = 1.77 \begin{align*} -\frac{{\left (4 \, a b c - a^{2} d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{3 \,{\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} \sqrt{-b^{2} c + a b d}} - \frac{2 \,{\left (d x^{3} + c\right )} b^{2} c^{2} - 2 \, b^{2} c^{3} + 2 \, a b c^{2} d +{\left (d x^{3} + c\right )} a^{2} d^{2}}{3 \,{\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )}{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} b - \sqrt{d x^{3} + c} b c + \sqrt{d x^{3} + c} a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

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