3.270 \(\int \frac{x^8}{\sqrt{c+d x^3} (4 c+d x^3)} \, dx\)

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

[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)

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Rubi [A]  time = 0.0741554, antiderivative size = 78, normalized size of antiderivative = 1., 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 = {446, 88, 63, 203} \[ \frac{32 c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{3 \sqrt{3} d^3}-\frac{10 c \sqrt{c+d x^3}}{3 d^3}+\frac{2 \left (c+d x^3\right )^{3/2}}{9 d^3} \]

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 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 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 203

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

Rubi steps

\begin{align*} \int \frac{x^8}{\sqrt{c+d x^3} \left (4 c+d x^3\right )} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{c+d x} (4 c+d x)} \, dx,x,x^3\right )\\ &=\frac{1}{3} \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [A]  time = 0.0538965, size = 65, normalized size = 0.83 \[ \frac{32 \sqrt{3} c^{3/2} \tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )+2 \left (d x^3-14 c\right ) \sqrt{c+d x^3}}{9 d^3} \]

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]  time = 0.035, size = 467, normalized size = 6. \begin{align*}{\frac{1}{{d}^{2}} \left ( d \left ({\frac{2\,{x}^{3}}{9\,d}\sqrt{d{x}^{3}+c}}-{\frac{4\,c}{9\,{d}^{2}}\sqrt{d{x}^{3}+c}} \right ) -{\frac{8\,c}{3\,d}\sqrt{d{x}^{3}+c}} \right ) }-{\frac{{\frac{16\,i}{9}}c\sqrt{2}}{{d}^{5}}\sum _{{\it \_alpha}={\it RootOf} \left ( d{{\it \_Z}}^{3}+4\,c \right ) }{\sqrt [3]{-{d}^{2}c}\sqrt{{{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( -i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}\sqrt{{d \left ( x-{\frac{1}{d}\sqrt [3]{-{d}^{2}c}} \right ) \left ( -3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c} \right ) ^{-1}}}\sqrt{{-{\frac{i}{2}}d \left ( 2\,x+{\frac{1}{d} \left ( i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c} \right ) } \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}} \left ( i\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,\sqrt{3}d-i\sqrt{3} \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}} \right ){\it EllipticPi} \left ({\frac{\sqrt{3}}{3}\sqrt{{id\sqrt{3} \left ( x+{\frac{1}{2\,d}\sqrt [3]{-{d}^{2}c}}-{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ){\frac{1}{\sqrt [3]{-{d}^{2}c}}}}}},{\frac{1}{6\,cd} \left ( 2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{{\frac{2}{3}}}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd \right ) },\sqrt{{\frac{i\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c} \left ( -{\frac{3}{2\,d}\sqrt [3]{-{d}^{2}c}}+{\frac{{\frac{i}{2}}\sqrt{3}}{d}\sqrt [3]{-{d}^{2}c}} \right ) ^{-1}}} \right ){\frac{1}{\sqrt{d{x}^{3}+c}}}}} \end{align*}

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)

[Out]

1/d^2*(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)-16/9*I*c/d^5*2^(1/2)*s
um((-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),1/6/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)/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*d+4*c))

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.53041, size = 332, normalized size = 4.26 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{8}}{\sqrt{c + d x^{3}} \left (4 c + d x^{3}\right )}\, dx \end{align*}

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 = 1.10584, size = 86, normalized size = 1.1 \begin{align*} \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{align*}

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