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

Optimal. Leaf size=140 \[ -\frac{x^3 \left (2 a c^2-d^2\right )}{3 b^2 c^3}+\frac{2 d \sqrt{a+b x^3} \left (2 a c^2-d^2\right )}{3 b^3 c^4}+\frac{2 \left (a c^2-d^2\right )^2 \log \left (c \sqrt{a+b x^3}+d\right )}{3 b^3 c^5}-\frac{2 d \left (a+b x^3\right )^{3/2}}{9 b^3 c^2}+\frac{\left (a+b x^3\right )^2}{6 b^3 c} \]

[Out]

-((2*a*c^2 - d^2)*x^3)/(3*b^2*c^3) + (2*d*(2*a*c^2 - d^2)*Sqrt[a + b*x^3])/(3*b^3*c^4) - (2*d*(a + b*x^3)^(3/2
))/(9*b^3*c^2) + (a + b*x^3)^2/(6*b^3*c) + (2*(a*c^2 - d^2)^2*Log[d + c*Sqrt[a + b*x^3]])/(3*b^3*c^5)

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Rubi [A]  time = 0.299107, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2155, 697} \[ -\frac{x^3 \left (2 a c^2-d^2\right )}{3 b^2 c^3}+\frac{2 d \sqrt{a+b x^3} \left (2 a c^2-d^2\right )}{3 b^3 c^4}+\frac{2 \left (a c^2-d^2\right )^2 \log \left (c \sqrt{a+b x^3}+d\right )}{3 b^3 c^5}-\frac{2 d \left (a+b x^3\right )^{3/2}}{9 b^3 c^2}+\frac{\left (a+b x^3\right )^2}{6 b^3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a*c^2 - d^2)*x^3)/(3*b^2*c^3) + (2*d*(2*a*c^2 - d^2)*Sqrt[a + b*x^3])/(3*b^3*c^4) - (2*d*(a + b*x^3)^(3/2
))/(9*b^3*c^2) + (a + b*x^3)^2/(6*b^3*c) + (2*(a*c^2 - d^2)^2*Log[d + c*Sqrt[a + b*x^3]])/(3*b^3*c^5)

Rule 2155

Int[(x_)^(m_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[1/n, Subst[Int
[x^((m + 1)/n - 1)/(c + d*x + e*Sqrt[a + b*x]), x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[b*c
- a*d, 0] && IntegerQ[(m + 1)/n]

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{x^8}{a c+b c x^3+d \sqrt{a+b x^3}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2}{a c+b c x+d \sqrt{a+b x}} \, dx,x,x^3\right )\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{\left (a-x^2\right )^2}{d+c x} \, dx,x,\sqrt{a+b x^3}\right )}{3 b^3}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (\frac{2 a c^2 d-d^3}{c^4}-\frac{\left (2 a c^2-d^2\right ) x}{c^3}-\frac{d x^2}{c^2}+\frac{x^3}{c}+\frac{\left (a c^2-d^2\right )^2}{c^4 (d+c x)}\right ) \, dx,x,\sqrt{a+b x^3}\right )}{3 b^3}\\ &=-\frac{\left (2 a c^2-d^2\right ) x^3}{3 b^2 c^3}+\frac{2 d \left (2 a c^2-d^2\right ) \sqrt{a+b x^3}}{3 b^3 c^4}-\frac{2 d \left (a+b x^3\right )^{3/2}}{9 b^3 c^2}+\frac{\left (a+b x^3\right )^2}{6 b^3 c}+\frac{2 \left (a c^2-d^2\right )^2 \log \left (d+c \sqrt{a+b x^3}\right )}{3 b^3 c^5}\\ \end{align*}

Mathematica [A]  time = 0.181143, size = 126, normalized size = 0.9 \[ \frac{c \left (a \left (20 c^2 d \sqrt{a+b x^3}-6 b c^3 x^3\right )+2 b c d x^3 \left (3 d-2 c \sqrt{a+b x^3}\right )-12 d^3 \sqrt{a+b x^3}+3 b^2 c^3 x^6\right )+12 \left (d^2-a c^2\right )^2 \log \left (c \sqrt{a+b x^3}+d\right )}{18 b^3 c^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(3*b^2*c^3*x^6 - 12*d^3*Sqrt[a + b*x^3] + 2*b*c*d*x^3*(3*d - 2*c*Sqrt[a + b*x^3]) + a*(-6*b*c^3*x^3 + 20*c^
2*d*Sqrt[a + b*x^3])) + 12*(-(a*c^2) + d^2)^2*Log[d + c*Sqrt[a + b*x^3]])/(18*b^3*c^5)

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Maple [C]  time = 0.099, size = 1473, normalized size = 10.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/9*d*(b*x^3+a)^(3/2)/b^3/c^2+4/3*d/b^3/c^2*(b*x^3+a)^(1/2)*a-2/3/b^3/c^4*d^3*(b*x^3+a)^(1/2)-1/3*I/b^5/d*2^(
1/2)*sum((-a*b^2)^(1/3)*(1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)-I*3^(1/2)*(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)*(b*
(x-1/b*(-a*b^2)^(1/3))/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)))^(1/2)*(-1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)+
I*3^(1/2)*(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*(I*(-a*b^2)^(1/3)*3^(1/2)*_alpha*b-I*(-a*b^2)
^(2/3)*3^(1/2)+2*_alpha^2*b^2-(-a*b^2)^(1/3)*_alpha*b-(-a*b^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b
^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),-1/2*c^2/b*(2*I*(-a*b^2)^(1/3)*3^(1/
2)*_alpha^2*b-I*(-a*b^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*a*b-3*(-a*b^2)^(2/3)*_alpha-3*a*b)/d^2,(I*3^(1/2)/b*(-
a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b*c^2+a*c^2-d^2
))*a^2+2/3*I*d/b^5/c^2*2^(1/2)*sum((-a*b^2)^(1/3)*(1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)-I*3^(1/2)*(-a*b^2)^(1/3)))
/(-a*b^2)^(1/3))^(1/2)*(b*(x-1/b*(-a*b^2)^(1/3))/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)))^(1/2)*(-1/2*I*b
*(2*x+1/b*((-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*(I*(-a*b^2)^(1/3)*3
^(1/2)*_alpha*b-I*(-a*b^2)^(2/3)*3^(1/2)+2*_alpha^2*b^2-(-a*b^2)^(1/3)*_alpha*b-(-a*b^2)^(2/3))*EllipticPi(1/3
*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),-1/2*c^2/b
*(2*I*(-a*b^2)^(1/3)*3^(1/2)*_alpha^2*b-I*(-a*b^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*a*b-3*(-a*b^2)^(2/3)*_alpha-
3*a*b)/d^2,(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)),_alpha=R
ootOf(_Z^3*b*c^2+a*c^2-d^2))*a-1/3*I/b^5/c^4*d^3*2^(1/2)*sum((-a*b^2)^(1/3)*(1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)-
I*3^(1/2)*(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)*(b*(x-1/b*(-a*b^2)^(1/3))/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^
2)^(1/3)))^(1/2)*(-1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)
^(1/2)*(I*(-a*b^2)^(1/3)*3^(1/2)*_alpha*b-I*(-a*b^2)^(2/3)*3^(1/2)+2*_alpha^2*b^2-(-a*b^2)^(1/3)*_alpha*b-(-a*
b^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^
2)^(1/3))^(1/2),-1/2*c^2/b*(2*I*(-a*b^2)^(1/3)*3^(1/2)*_alpha^2*b-I*(-a*b^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*a*
b-3*(-a*b^2)^(2/3)*_alpha-3*a*b)/d^2,(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^
2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b*c^2+a*c^2-d^2))-1/3*a/c/b^2*x^3+1/3*a^2/c/b^3*ln(b*c^2*x^3+a*c^2-d^2)-2
/3*a/c^3/b^3*d^2*ln(b*c^2*x^3+a*c^2-d^2)+1/6/b/c*x^6+1/3/b^2/c^3*x^3*d^2+1/3/b^3/c^5*d^4*ln(b*c^2*x^3+a*c^2-d^
2)

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Maxima [A]  time = 1.31592, size = 169, normalized size = 1.21 \begin{align*} \frac{\frac{3 \,{\left (b x^{3} + a\right )}^{2} c^{3} - 4 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} c^{2} d - 6 \,{\left (2 \, a c^{3} - c d^{2}\right )}{\left (b x^{3} + a\right )} + 12 \,{\left (2 \, a c^{2} d - d^{3}\right )} \sqrt{b x^{3} + a}}{c^{4}} + \frac{12 \,{\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt{b x^{3} + a} c + d\right )}{c^{5}}}{18 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*((3*(b*x^3 + a)^2*c^3 - 4*(b*x^3 + a)^(3/2)*c^2*d - 6*(2*a*c^3 - c*d^2)*(b*x^3 + a) + 12*(2*a*c^2*d - d^3
)*sqrt(b*x^3 + a))/c^4 + 12*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(sqrt(b*x^3 + a)*c + d)/c^5)/b^3

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Fricas [A]  time = 1.28078, size = 409, normalized size = 2.92 \begin{align*} \frac{3 \, b^{2} c^{4} x^{6} - 6 \,{\left (a b c^{4} - b c^{2} d^{2}\right )} x^{3} + 6 \,{\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (b c^{2} x^{3} + a c^{2} - d^{2}\right ) + 6 \,{\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt{b x^{3} + a} c + d\right ) - 6 \,{\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left (\sqrt{b x^{3} + a} c - d\right ) - 4 \,{\left (b c^{3} d x^{3} - 5 \, a c^{3} d + 3 \, c d^{3}\right )} \sqrt{b x^{3} + a}}{18 \, b^{3} c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(3*b^2*c^4*x^6 - 6*(a*b*c^4 - b*c^2*d^2)*x^3 + 6*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(b*c^2*x^3 + a*c^2 - d^
2) + 6*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(sqrt(b*x^3 + a)*c + d) - 6*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(sqrt(b*x
^3 + a)*c - d) - 4*(b*c^3*d*x^3 - 5*a*c^3*d + 3*c*d^3)*sqrt(b*x^3 + a))/(b^3*c^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{a c + b c x^{3} + d \sqrt{a + b x^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13094, size = 211, normalized size = 1.51 \begin{align*} \frac{2 \,{\left (a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4}\right )} \log \left ({\left | \sqrt{b x^{3} + a} c + d \right |}\right )}{3 \, b^{3} c^{5}} + \frac{3 \,{\left (b x^{3} + a\right )}^{2} b^{9} c^{3} - 12 \,{\left (b x^{3} + a\right )} a b^{9} c^{3} - 4 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} b^{9} c^{2} d + 24 \, \sqrt{b x^{3} + a} a b^{9} c^{2} d + 6 \,{\left (b x^{3} + a\right )} b^{9} c d^{2} - 12 \, \sqrt{b x^{3} + a} b^{9} d^{3}}{18 \, b^{12} c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a^2*c^4 - 2*a*c^2*d^2 + d^4)*log(abs(sqrt(b*x^3 + a)*c + d))/(b^3*c^5) + 1/18*(3*(b*x^3 + a)^2*b^9*c^3 -
12*(b*x^3 + a)*a*b^9*c^3 - 4*(b*x^3 + a)^(3/2)*b^9*c^2*d + 24*sqrt(b*x^3 + a)*a*b^9*c^2*d + 6*(b*x^3 + a)*b^9*
c*d^2 - 12*sqrt(b*x^3 + a)*b^9*d^3)/(b^12*c^4)