3.6.52 \(\int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx\) [552]

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

[Out]

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

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Rubi [A]
time = 0.21, antiderivative size = 140, normalized size of antiderivative = 1.00, 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 = {2186, 711} \begin {gather*} -\frac {2 d \left (a+b x^3\right )^{3/2}}{9 b^3 c^2}+\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 \sqrt {a+b x^3} \left (2 a c^2-d^2\right )}{3 b^3 c^4}+\frac {\left (a+b x^3\right )^2}{6 b^3 c}-\frac {x^3 \left (2 a c^2-d^2\right )}{3 b^2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((2*a*c^2 - d^2)*x^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 711

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]

Rule 2186

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]

Rubi steps

\begin {align*} \int \frac {x^8}{a c+b c x^3+d \sqrt {a+b x^3}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^2}{a c+b c x+d \sqrt {a+b x}} \, dx,x,x^3\right )\\ &=\frac {2 \text {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 \text {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*}

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Mathematica [A]
time = 0.16, size = 125, normalized size = 0.89 \begin {gather*} \frac {-4 c d \sqrt {a+b x^3} \left (-5 a c^2+3 d^2+b c^2 x^3\right )+3 c^2 \left (-3 a^2 c^2+2 a \left (d^2-b c^2 x^3\right )+b x^3 \left (2 d^2+b c^2 x^3\right )\right )+12 \left (-a c^2+d^2\right )^2 \log \left (d+c \sqrt {a+b x^3}\right )}{18 b^3 c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*c*d*Sqrt[a + b*x^3]*(-5*a*c^2 + 3*d^2 + b*c^2*x^3) + 3*c^2*(-3*a^2*c^2 + 2*a*(d^2 - b*c^2*x^3) + b*x^3*(2*
d^2 + b*c^2*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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.18, size = 647, normalized size = 4.62 Too large to display

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,method=_RETURNVERBOSE)

[Out]

d*(-2/9/b^3/c^2*(b*x^3+a)^(3/2)+2/3*a^2/b^3/d^2*(b*x^3+a)^(1/2)-(a^2*c^4-2*a*c^2*d^2+d^4)/d^2/c^2/b^2*(2/3/c^2
/b*(b*x^3+a)^(1/2)+1/3*I/b^3/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))*Elli
pticPi(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/b*c^2*(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*d^2/b
^3*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 = 0.29, size = 125, normalized size = 0.89 \begin {gather*} \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 {gather*}

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 = 0.39, size = 191, normalized size = 1.36 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{a c + b c x^{3} + d \sqrt {a + b x^{3}}}\, dx \end {gather*}

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 = 3.19, size = 156, normalized size = 1.11 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 3.74, size = 200, normalized size = 1.43 \begin {gather*} \frac {\left (\frac {2\,d\,\left (a\,c^2-d^2\right )}{b^2\,c^4}+\frac {4\,a\,d}{3\,b^2\,c^2}\right )\,\sqrt {b\,x^3+a}}{3\,b}+\frac {x^6}{6\,b\,c}-\frac {x^3\,\left (a\,c^2-d^2\right )}{3\,b^2\,c^3}+\frac {\ln \left (\frac {d+c\,\sqrt {b\,x^3+a}}{d-c\,\sqrt {b\,x^3+a}}\right )\,{\left (a\,c^2-d^2\right )}^2}{3\,b^3\,c^5}+\frac {\ln \left (b\,c^2\,x^3+a\,c^2-d^2\right )\,\left (a^2\,c^4-2\,a\,c^2\,d^2+d^4\right )}{3\,b^3\,c^5}-\frac {2\,d\,x^3\,\sqrt {b\,x^3+a}}{9\,b^2\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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