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

Optimal. Leaf size=154 \[ \frac{2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}+\frac{2 a^2 \sqrt{c+d x^3} (b c-a d)}{3 b^4}-\frac{2 a^2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{9/2}}-\frac{2 \left (c+d x^3\right )^{5/2} (a d+b c)}{15 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{7/2}}{21 b d^2} \]

[Out]

(2*a^2*(b*c - a*d)*Sqrt[c + d*x^3])/(3*b^4) + (2*a^2*(c + d*x^3)^(3/2))/(9*b^3) - (2*(b*c + a*d)*(c + d*x^3)^(
5/2))/(15*b^2*d^2) + (2*(c + d*x^3)^(7/2))/(21*b*d^2) - (2*a^2*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x
^3])/Sqrt[b*c - a*d]])/(3*b^(9/2))

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Rubi [A]  time = 0.15477, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 88, 50, 63, 208} \[ \frac{2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}+\frac{2 a^2 \sqrt{c+d x^3} (b c-a d)}{3 b^4}-\frac{2 a^2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{9/2}}-\frac{2 \left (c+d x^3\right )^{5/2} (a d+b c)}{15 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{7/2}}{21 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(b*c - a*d)*Sqrt[c + d*x^3])/(3*b^4) + (2*a^2*(c + d*x^3)^(3/2))/(9*b^3) - (2*(b*c + a*d)*(c + d*x^3)^(
5/2))/(15*b^2*d^2) + (2*(c + d*x^3)^(7/2))/(21*b*d^2) - (2*a^2*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x
^3])/Sqrt[b*c - a*d]])/(3*b^(9/2))

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 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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 \left (c+d x^3\right )^{3/2}}{a+b x^3} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2 (c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \left (\frac{(-b c-a d) (c+d x)^{3/2}}{b^2 d}+\frac{a^2 (c+d x)^{3/2}}{b^2 (a+b x)}+\frac{(c+d x)^{5/2}}{b d}\right ) \, dx,x,x^3\right )\\ &=-\frac{2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{7/2}}{21 b d^2}+\frac{a^2 \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )}{3 b^2}\\ &=\frac{2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac{2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{7/2}}{21 b d^2}+\frac{\left (a^2 (b c-a d)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{a+b x} \, dx,x,x^3\right )}{3 b^3}\\ &=\frac{2 a^2 (b c-a d) \sqrt{c+d x^3}}{3 b^4}+\frac{2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac{2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{7/2}}{21 b d^2}+\frac{\left (a^2 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,x^3\right )}{3 b^4}\\ &=\frac{2 a^2 (b c-a d) \sqrt{c+d x^3}}{3 b^4}+\frac{2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac{2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{7/2}}{21 b d^2}+\frac{\left (2 a^2 (b c-a d)^2\right ) \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^4 d}\\ &=\frac{2 a^2 (b c-a d) \sqrt{c+d x^3}}{3 b^4}+\frac{2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac{2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac{2 \left (c+d x^3\right )^{7/2}}{21 b d^2}-\frac{2 a^2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{3 b^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.151854, size = 145, normalized size = 0.94 \[ \frac{2 \left (105 a^2 (b c-a d) \left (\frac{\sqrt{c+d x^3}}{b}-\frac{\sqrt{b c-a d} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d x^3}}{\sqrt{b c-a d}}\right )}{b^{3/2}}\right )+35 a^2 \left (c+d x^3\right )^{3/2}-\frac{21 b \left (c+d x^3\right )^{5/2} (a d+b c)}{d^2}+\frac{15 b^2 \left (c+d x^3\right )^{7/2}}{d^2}\right )}{315 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(35*a^2*(c + d*x^3)^(3/2) - (21*b*(b*c + a*d)*(c + d*x^3)^(5/2))/d^2 + (15*b^2*(c + d*x^3)^(7/2))/d^2 + 105
*a^2*(b*c - a*d)*(Sqrt[c + d*x^3]/b - (Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(
3/2))))/(315*b^3)

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Maple [C]  time = 0.033, size = 605, normalized size = 3.9 \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)^(3/2)/(b*x^3+a),x)

[Out]

1/b^2*(b*(2/21*d*x^9*(d*x^3+c)^(1/2)+16/105*c*x^6*(d*x^3+c)^(1/2)+2/105/d*c^2*x^3*(d*x^3+c)^(1/2)-4/105/d^2*c^
3*(d*x^3+c)^(1/2))-2/15*a/d*(d*x^3+c)^(5/2))+a^2/b^2*(2/9*d/b*x^3*(d*x^3+c)^(1/2)+2/3*(-d*(a*d-2*b*c)/b^2-2/3*
d/b*c)/d*(d*x^3+c)^(1/2)+1/3*I/b^2/d^2*2^(1/2)*sum((-a^2*d^2+2*a*b*c*d-b^2*c^2)/(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)))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.71122, size = 883, normalized size = 5.73 \begin{align*} \left [-\frac{105 \,{\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{\frac{b c - a d}{b}} \log \left (\frac{b d x^{3} + 2 \, b c - a d + 2 \, \sqrt{d x^{3} + c} b \sqrt{\frac{b c - a d}{b}}}{b x^{3} + a}\right ) - 2 \,{\left (15 \, b^{3} d^{3} x^{9} + 3 \,{\left (8 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{6} - 6 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 140 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} +{\left (3 \, b^{3} c^{2} d - 42 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{3}\right )} \sqrt{d x^{3} + c}}{315 \, b^{4} d^{2}}, -\frac{2 \,{\left (105 \,{\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt{-\frac{b c - a d}{b}} \arctan \left (-\frac{\sqrt{d x^{3} + c} b \sqrt{-\frac{b c - a d}{b}}}{b c - a d}\right ) -{\left (15 \, b^{3} d^{3} x^{9} + 3 \,{\left (8 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{6} - 6 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 140 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} +{\left (3 \, b^{3} c^{2} d - 42 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{3}\right )} \sqrt{d x^{3} + c}\right )}}{315 \, b^{4} d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/315*(105*(a^2*b*c*d^2 - a^3*d^3)*sqrt((b*c - a*d)/b)*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 + c)*b*sqrt
((b*c - a*d)/b))/(b*x^3 + a)) - 2*(15*b^3*d^3*x^9 + 3*(8*b^3*c*d^2 - 7*a*b^2*d^3)*x^6 - 6*b^3*c^3 - 21*a*b^2*c
^2*d + 140*a^2*b*c*d^2 - 105*a^3*d^3 + (3*b^3*c^2*d - 42*a*b^2*c*d^2 + 35*a^2*b*d^3)*x^3)*sqrt(d*x^3 + c))/(b^
4*d^2), -2/315*(105*(a^2*b*c*d^2 - a^3*d^3)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b
)/(b*c - a*d)) - (15*b^3*d^3*x^9 + 3*(8*b^3*c*d^2 - 7*a*b^2*d^3)*x^6 - 6*b^3*c^3 - 21*a*b^2*c^2*d + 140*a^2*b*
c*d^2 - 105*a^3*d^3 + (3*b^3*c^2*d - 42*a*b^2*c*d^2 + 35*a^2*b*d^3)*x^3)*sqrt(d*x^3 + c))/(b^4*d^2)]

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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)**(3/2)/(b*x**3+a),x)

[Out]

Timed out

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Giac [A]  time = 1.13043, size = 261, normalized size = 1.69 \begin{align*} \frac{2 \,{\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\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^{4}} + \frac{2 \,{\left (15 \,{\left (d x^{3} + c\right )}^{\frac{7}{2}} b^{6} d^{12} - 21 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} b^{6} c d^{12} - 21 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} a b^{5} d^{13} + 35 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} a^{2} b^{4} d^{14} + 105 \, \sqrt{d x^{3} + c} a^{2} b^{4} c d^{14} - 105 \, \sqrt{d x^{3} + c} a^{3} b^{3} d^{15}\right )}}{315 \, b^{7} d^{14}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)
*b^4) + 2/315*(15*(d*x^3 + c)^(7/2)*b^6*d^12 - 21*(d*x^3 + c)^(5/2)*b^6*c*d^12 - 21*(d*x^3 + c)^(5/2)*a*b^5*d^
13 + 35*(d*x^3 + c)^(3/2)*a^2*b^4*d^14 + 105*sqrt(d*x^3 + c)*a^2*b^4*c*d^14 - 105*sqrt(d*x^3 + c)*a^3*b^3*d^15
)/(b^7*d^14)