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

Optimal. Leaf size=161 \[ -\frac{a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 \left (a+b x^3\right ) (b c-a d)}-\frac{a \sqrt{c+d x^3} (4 b c-5 a d)}{3 b^3 (b c-a d)}+\frac{a (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2} \sqrt{b c-a d}}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 b^2 d} \]

[Out]

-(a*(4*b*c - 5*a*d)*Sqrt[c + d*x^3])/(3*b^3*(b*c - a*d)) + (2*(c + d*x^3)^(3/2))/(9*b^2*d) - (a^2*(c + d*x^3)^
(3/2))/(3*b^2*(b*c - a*d)*(a + b*x^3)) + (a*(4*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]]
)/(3*b^(7/2)*Sqrt[b*c - a*d])

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Rubi [A]  time = 0.192206, antiderivative size = 161, 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 = {446, 89, 80, 50, 63, 208} \[ -\frac{a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 \left (a+b x^3\right ) (b c-a d)}-\frac{a \sqrt{c+d x^3} (4 b c-5 a d)}{3 b^3 (b c-a d)}+\frac{a (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2} \sqrt{b c-a d}}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 b^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(4*b*c - 5*a*d)*Sqrt[c + d*x^3])/(3*b^3*(b*c - a*d)) + (2*(c + d*x^3)^(3/2))/(9*b^2*d) - (a^2*(c + d*x^3)^
(3/2))/(3*b^2*(b*c - a*d)*(a + b*x^3)) + (a*(4*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]]
)/(3*b^(7/2)*Sqrt[b*c - a*d])

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 \sqrt{c+d x^3}}{\left (a+b x^3\right )^2} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2 \sqrt{c+d x}}{(a+b x)^2} \, dx,x,x^3\right )\\ &=-\frac{a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{c+d x} \left (-\frac{1}{2} a (2 b c-3 a d)+b (b c-a d) x\right )}{a+b x} \, dx,x,x^3\right )}{3 b^2 (b c-a d)}\\ &=\frac{2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}-\frac{a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac{(a (4 b c-5 a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{a+b x} \, dx,x,x^3\right )}{6 b^2 (b c-a d)}\\ &=-\frac{a (4 b c-5 a d) \sqrt{c+d x^3}}{3 b^3 (b c-a d)}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}-\frac{a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac{(a (4 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{6 b^3}\\ &=-\frac{a (4 b c-5 a d) \sqrt{c+d x^3}}{3 b^3 (b c-a d)}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}-\frac{a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac{(a (4 b c-5 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^3 d}\\ &=-\frac{a (4 b c-5 a d) \sqrt{c+d x^3}}{3 b^3 (b c-a d)}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}-\frac{a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}+\frac{a (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{7/2} \sqrt{b c-a d}}\\ \end{align*}

Mathematica [A]  time = 0.24721, size = 147, normalized size = 0.91 \[ \frac{-\frac{a^2 \left (c+d x^3\right )^{3/2}}{a+b x^3}+\frac{a (5 a d-4 b c) \left (\sqrt{b} \sqrt{c+d x^3}-\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )\right )}{b^{3/2}}+\frac{2 \left (c+d x^3\right )^{3/2} (b c-a d)}{3 d}}{3 b^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.037, size = 917, normalized size = 5.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/9*(d*x^3+c)^(3/2)/b^2/d-2*a/b^2*(2/3*(d*x^3+c)^(1/2)/b+1/3*I/b/d^2*2^(1/2)*sum((-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*(d*x^3+c)^(1/2)/b/(b*x^3+a)-
1/6*I/d/b*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)),_alph
a=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*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.9075, size = 975, normalized size = 6.06 \begin{align*} \left [-\frac{3 \,{\left (4 \, a^{2} b c d - 5 \, a^{3} d^{2} +{\left (4 \, a b^{2} c d - 5 \, 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 \,{\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + 2 \, a b^{3} c^{2} - 17 \, a^{2} b^{2} c d + 15 \, a^{3} b d^{2} + 2 \,{\left (b^{4} c^{2} - 6 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{18 \,{\left (a b^{5} c d - a^{2} b^{4} d^{2} +{\left (b^{6} c d - a b^{5} d^{2}\right )} x^{3}\right )}}, -\frac{3 \,{\left (4 \, a^{2} b c d - 5 \, a^{3} d^{2} +{\left (4 \, a b^{2} c d - 5 \, 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 \,{\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + 2 \, a b^{3} c^{2} - 17 \, a^{2} b^{2} c d + 15 \, a^{3} b d^{2} + 2 \,{\left (b^{4} c^{2} - 6 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{9 \,{\left (a b^{5} c d - a^{2} b^{4} d^{2} +{\left (b^{6} c d - a b^{5} d^{2}\right )} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

[-1/18*(3*(4*a^2*b*c*d - 5*a^3*d^2 + (4*a*b^2*c*d - 5*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*(b^4*c*d - a*b^3*d^2)*x^6 + 2*a*b^3*c^2 -
17*a^2*b^2*c*d + 15*a^3*b*d^2 + 2*(b^4*c^2 - 6*a*b^3*c*d + 5*a^2*b^2*d^2)*x^3)*sqrt(d*x^3 + c))/(a*b^5*c*d - a
^2*b^4*d^2 + (b^6*c*d - a*b^5*d^2)*x^3), -1/9*(3*(4*a^2*b*c*d - 5*a^3*d^2 + (4*a*b^2*c*d - 5*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*(b^4*c*d - a*b^3*d^2)*x^
6 + 2*a*b^3*c^2 - 17*a^2*b^2*c*d + 15*a^3*b*d^2 + 2*(b^4*c^2 - 6*a*b^3*c*d + 5*a^2*b^2*d^2)*x^3)*sqrt(d*x^3 +
c))/(a*b^5*c*d - a^2*b^4*d^2 + (b^6*c*d - a*b^5*d^2)*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*(d*x**3+c)**(1/2)/(b*x**3+a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.11787, size = 184, normalized size = 1.14 \begin{align*} -\frac{\sqrt{d x^{3} + c} a^{2} d}{3 \,{\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{3}} - \frac{{\left (4 \, a b c - 5 \, a^{2} d\right )} \arctan \left (\frac{\sqrt{d x^{3} + c} b}{\sqrt{-b^{2} c + a b d}}\right )}{3 \, \sqrt{-b^{2} c + a b d} b^{3}} + \frac{2 \,{\left ({\left (d x^{3} + c\right )}^{\frac{3}{2}} b^{4} d^{2} - 6 \, \sqrt{d x^{3} + c} a b^{3} d^{3}\right )}}{9 \, b^{6} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

-1/3*sqrt(d*x^3 + c)*a^2*d/(((d*x^3 + c)*b - b*c + a*d)*b^3) - 1/3*(4*a*b*c - 5*a^2*d)*arctan(sqrt(d*x^3 + c)*
b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3) + 2/9*((d*x^3 + c)^(3/2)*b^4*d^2 - 6*sqrt(d*x^3 + c)*a*b^3*
d^3)/(b^6*d^3)