3.3.70 \(\int \frac {x^8}{\sqrt {c+d x^3} (4 c+d x^3)} \, dx\) [270]

Optimal. Leaf size=78 \[ -\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {32 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^3} \]

[Out]

2/9*(d*x^3+c)^(3/2)/d^3+32/9*c^(3/2)*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1/2))/d^3*3^(1/2)-10/3*c*(d*x^3+c)^
(1/2)/d^3

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Rubi [A]
time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {457, 90, 65, 209} \begin {gather*} \frac {32 c^{3/2} \text {ArcTan}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^3}-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^8/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

(-10*c*Sqrt[c + d*x^3])/(3*d^3) + (2*(c + d*x^3)^(3/2))/(9*d^3) + (32*c^(3/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*
Sqrt[c])])/(3*Sqrt[3]*d^3)

Rule 65

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 90

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 457

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]]

Rubi steps

\begin {align*} \int \frac {x^8}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^2}{\sqrt {c+d x} (4 c+d x)} \, dx,x,x^3\right )\\ &=\frac {1}{3} \text {Subst}\left (\int \left (-\frac {5 c}{d^2 \sqrt {c+d x}}+\frac {\sqrt {c+d x}}{d^2}+\frac {16 c^2}{d^2 \sqrt {c+d x} (4 c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {\left (16 c^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {c+d x} (4 c+d x)} \, dx,x,x^3\right )}{3 d^2}\\ &=-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {\left (32 c^2\right ) \text {Subst}\left (\int \frac {1}{3 c+x^2} \, dx,x,\sqrt {c+d x^3}\right )}{3 d^3}\\ &=-\frac {10 c \sqrt {c+d x^3}}{3 d^3}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {32 c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3 \sqrt {3} d^3}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 65, normalized size = 0.83 \begin {gather*} \frac {2 \left (-14 c+d x^3\right ) \sqrt {c+d x^3}+32 \sqrt {3} c^{3/2} \tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{9 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^8/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

(2*(-14*c + d*x^3)*Sqrt[c + d*x^3] + 32*Sqrt[3]*c^(3/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(9*d^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.39, size = 464, normalized size = 5.95 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/(d*x^3+4*c)/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/d*(2/9/d*x^3*(d*x^3+c)^(1/2)-4/9*c*(d*x^3+c)^(1/2)/d^2)-8/3*c*(d*x^3+c)^(1/2)/d^3-16/9*I*c/d^5*2^(1/2)*sum((
-c*d^2)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)*(d*(x-1/d*(-
c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)
+(-c*d^2)^(1/3)))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-c*d^2)^
(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)
-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),1/6/d*(2*I*(-c*d^2)^(1/3)*3^(1/2)*_alpha^2*d-
I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3
/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*d+4*c))

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Maxima [A]
time = 0.50, size = 53, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (16 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + {\left (d x^{3} + c\right )}^{\frac {3}{2}} - 15 \, \sqrt {d x^{3} + c} c\right )}}{9 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

2/9*(16*sqrt(3)*c^(3/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) + (d*x^3 + c)^(3/2) - 15*sqrt(d*x^3 + c)*c
)/d^3

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Fricas [A]
time = 2.81, size = 129, normalized size = 1.65 \begin {gather*} \left [\frac {2 \, {\left (8 \, \sqrt {3} \sqrt {-c} c \log \left (\frac {d x^{3} + 2 \, \sqrt {3} \sqrt {d x^{3} + c} \sqrt {-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) + \sqrt {d x^{3} + c} {\left (d x^{3} - 14 \, c\right )}\right )}}{9 \, d^{3}}, \frac {2 \, {\left (16 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) + \sqrt {d x^{3} + c} {\left (d x^{3} - 14 \, c\right )}\right )}}{9 \, d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

[2/9*(8*sqrt(3)*sqrt(-c)*c*log((d*x^3 + 2*sqrt(3)*sqrt(d*x^3 + c)*sqrt(-c) - 2*c)/(d*x^3 + 4*c)) + sqrt(d*x^3
+ c)*(d*x^3 - 14*c))/d^3, 2/9*(16*sqrt(3)*c^(3/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c)) + sqrt(d*x^3 + c
)*(d*x^3 - 14*c))/d^3]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8}}{\sqrt {c + d x^{3}} \cdot \left (4 c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8/(d*x**3+4*c)/(d*x**3+c)**(1/2),x)

[Out]

Integral(x**8/(sqrt(c + d*x**3)*(4*c + d*x**3)), x)

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Giac [A]
time = 1.12, size = 64, normalized size = 0.82 \begin {gather*} \frac {32 \, \sqrt {3} c^{\frac {3}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right )}{9 \, d^{3}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{6} - 15 \, \sqrt {d x^{3} + c} c d^{6}\right )}}{9 \, d^{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

32/9*sqrt(3)*c^(3/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/d^3 + 2/9*((d*x^3 + c)^(3/2)*d^6 - 15*sqrt(d*
x^3 + c)*c*d^6)/d^9

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Mupad [B]
time = 5.38, size = 88, normalized size = 1.13 \begin {gather*} \frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,d^2}-\frac {28\,c\,\sqrt {d\,x^3+c}}{9\,d^3}+\frac {\sqrt {3}\,c^{3/2}\,\ln \left (\frac {\sqrt {3}\,d\,x^3-2\,\sqrt {3}\,c+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,16{}\mathrm {i}}{9\,d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8/((c + d*x^3)^(1/2)*(4*c + d*x^3)),x)

[Out]

(2*x^3*(c + d*x^3)^(1/2))/(9*d^2) - (28*c*(c + d*x^3)^(1/2))/(9*d^3) + (3^(1/2)*c^(3/2)*log((c^(1/2)*(c + d*x^
3)^(1/2)*6i - 2*3^(1/2)*c + 3^(1/2)*d*x^3)/(4*c + d*x^3))*16i)/(9*d^3)

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