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

Optimal. Leaf size=123 \[ -\frac{a^2 \sqrt{c+d x^3}}{3 b^2 \left (a+b x^3\right ) (b c-a d)}+\frac{a (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{5/2} (b c-a d)^{3/2}}+\frac{2 \sqrt{c+d x^3}}{3 b^2 d} \]

[Out]

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

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Rubi [A]  time = 0.143646, antiderivative size = 123, 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, 80, 63, 208} \[ -\frac{a^2 \sqrt{c+d x^3}}{3 b^2 \left (a+b x^3\right ) (b c-a d)}+\frac{a (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{5/2} (b c-a d)^{3/2}}+\frac{2 \sqrt{c+d x^3}}{3 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 80

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

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

Mathematica [A]  time = 0.251641, size = 107, normalized size = 0.87 \[ \frac{1}{3} \left (\frac{\sqrt{c+d x^3} \left (\frac{a^2}{\left (a+b x^3\right ) (a d-b c)}+\frac{2}{d}\right )}{b^2}+\frac{a (4 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{b^{5/2} (b c-a d)^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.037, size = 911, normalized size = 7.4 \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)^(1/2),x)

[Out]

2/3*(d*x^3+c)^(1/2)/b^2/d+2/3*I*a/b^2/d^2*2^(1/2)*sum(1/(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*(1/3/(a*d-b*c)*(d*x^3+c)^(1/2)/(b*x^3+a)-1/6*I/d*2
^(1/2)*sum(1/(a*d-b*c)^2*(-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)))

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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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.73021, size = 983, normalized size = 7.99 \begin{align*} \left [\frac{{\left (4 \, a^{2} b c d - 3 \, a^{3} d^{2} +{\left (4 \, a b^{2} c d - 3 \, a^{2} b d^{2}\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^{2} - 5 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2} + 2 \,{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{6 \,{\left (a b^{5} c^{2} d - 2 \, a^{2} b^{4} c d^{2} + a^{3} b^{3} d^{3} +{\left (b^{6} c^{2} d - 2 \, a b^{5} c d^{2} + a^{2} b^{4} d^{3}\right )} x^{3}\right )}}, -\frac{{\left (4 \, a^{2} b c d - 3 \, a^{3} d^{2} +{\left (4 \, a b^{2} c d - 3 \, a^{2} b d^{2}\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^{2} - 5 \, a^{2} b^{2} c d + 3 \, a^{3} b d^{2} + 2 \,{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{3 \,{\left (a b^{5} c^{2} d - 2 \, a^{2} b^{4} c d^{2} + a^{3} b^{3} d^{3} +{\left (b^{6} c^{2} d - 2 \, a b^{5} c d^{2} + a^{2} b^{4} d^{3}\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)^(1/2),x, algorithm="fricas")

[Out]

[1/6*((4*a^2*b*c*d - 3*a^3*d^2 + (4*a*b^2*c*d - 3*a^2*b*d^2)*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^2 - 5*a^2*b^2*c*d + 3*a^3*b*d^2 + 2*(b
^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^3)*sqrt(d*x^3 + c))/(a*b^5*c^2*d - 2*a^2*b^4*c*d^2 + a^3*b^3*d^3 + (b^6*
c^2*d - 2*a*b^5*c*d^2 + a^2*b^4*d^3)*x^3), -1/3*((4*a^2*b*c*d - 3*a^3*d^2 + (4*a*b^2*c*d - 3*a^2*b*d^2)*x^3)*s
qrt(-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^2 - 5*a^2*b^2*c*
d + 3*a^3*b*d^2 + 2*(b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2)*x^3)*sqrt(d*x^3 + c))/(a*b^5*c^2*d - 2*a^2*b^4*c*d^2
+ a^3*b^3*d^3 + (b^6*c^2*d - 2*a*b^5*c*d^2 + a^2*b^4*d^3)*x^3)]

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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(x**8/(b*x**3+a)**2/(d*x**3+c)**(1/2),x)

[Out]

Timed out

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Giac [A]  time = 1.11259, size = 181, normalized size = 1.47 \begin{align*} -\frac{\sqrt{d x^{3} + c} a^{2} d}{3 \,{\left (b^{3} c - a b^{2} d\right )}{\left ({\left (d x^{3} + c\right )} b - b c + a d\right )}} - \frac{{\left (4 \, a b c - 3 \, 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 - a b^{2} d\right )} \sqrt{-b^{2} c + a b d}} + \frac{2 \, \sqrt{d x^{3} + c}}{3 \, b^{2} d} \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)^(1/2),x, algorithm="giac")

[Out]

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