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

Optimal. Leaf size=97 \[ \frac{32 c^2 \sqrt{c+d x^3}}{3 d^3}-\frac{32 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{\sqrt{3} d^3}-\frac{10 c \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3} \]

[Out]

(32*c^2*Sqrt[c + d*x^3])/(3*d^3) - (10*c*(c + d*x^3)^(3/2))/(9*d^3) + (2*(c + d*x^3)^(5/2))/(15*d^3) - (32*c^(
5/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(Sqrt[3]*d^3)

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Rubi [A]  time = 0.093991, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {446, 88, 50, 63, 203} \[ \frac{32 c^2 \sqrt{c+d x^3}}{3 d^3}-\frac{32 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{\sqrt{3} d^3}-\frac{10 c \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*c^2*Sqrt[c + d*x^3])/(3*d^3) - (10*c*(c + d*x^3)^(3/2))/(9*d^3) + (2*(c + d*x^3)^(5/2))/(15*d^3) - (32*c^(
5/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])])/(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 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 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}}{4 c+d x^3} \, 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 \sqrt{c+d x}}{d^2}+\frac{(c+d x)^{3/2}}{d^2}+\frac{16 c^2 \sqrt{c+d x}}{d^2 (4 c+d x)}\right ) \, dx,x,x^3\right )\\ &=-\frac{10 c \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3}+\frac{\left (16 c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{4 c+d x} \, dx,x,x^3\right )}{3 d^2}\\ &=\frac{32 c^2 \sqrt{c+d x^3}}{3 d^3}-\frac{10 c \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac{\left (16 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{c+d x} (4 c+d x)} \, dx,x,x^3\right )}{d^2}\\ &=\frac{32 c^2 \sqrt{c+d x^3}}{3 d^3}-\frac{10 c \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac{\left (32 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{3 c+x^2} \, dx,x,\sqrt{c+d x^3}\right )}{d^3}\\ &=\frac{32 c^2 \sqrt{c+d x^3}}{3 d^3}-\frac{10 c \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac{2 \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac{32 c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{\sqrt{3} d^3}\\ \end{align*}

Mathematica [A]  time = 0.0618643, size = 77, normalized size = 0.79 \[ \frac{2 \sqrt{c+d x^3} \left (218 c^2-19 c d x^3+3 d^2 x^6\right )-480 \sqrt{3} c^{5/2} \tan ^{-1}\left (\frac{\sqrt{c+d x^3}}{\sqrt{3} \sqrt{c}}\right )}{45 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c + d*x^3]*(218*c^2 - 19*c*d*x^3 + 3*d^2*x^6) - 480*Sqrt[3]*c^(5/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sq
rt[c])])/(45*d^3)

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Maple [C]  time = 0.187, size = 506, normalized size = 5.2 \begin{align*}{\frac{1}{{d}^{2}} \left ( d \left ({\frac{2\,{x}^{6}}{15}\sqrt{d{x}^{3}+c}}+{\frac{2\,c{x}^{3}}{45\,d}\sqrt{d{x}^{3}+c}}-{\frac{4\,{c}^{2}}{45\,{d}^{2}}\sqrt{d{x}^{3}+c}} \right ) -{\frac{8\,c}{9\,d} \left ( d{x}^{3}+c \right ) ^{{\frac{3}{2}}}} \right ) }+16\,{\frac{{c}^{2}}{{d}^{2}} \left ( 2/3\,{\frac{\sqrt{d{x}^{3}+c}}{d}}+{\frac{i/3\sqrt{2}}{{d}^{3}}\sum _{{\it \_alpha}={\it RootOf} \left ( d{{\it \_Z}}^{3}+4\,c \right ) }{\frac{\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 ) ^{2/3}+2\,{{\it \_alpha}}^{2}{d}^{2}-\sqrt [3]{-{d}^{2}c}{\it \_alpha}\,d- \left ( -{d}^{2}c \right ) ^{2/3} \right ) }{\sqrt{d{x}^{3}+c}}\sqrt{{\frac{i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{-i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{d}{-3\,\sqrt [3]{-{d}^{2}c}+i\sqrt{3}\sqrt [3]{-{d}^{2}c}} \left ( x-{\frac{\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}\sqrt{{\frac{-i/2d}{\sqrt [3]{-{d}^{2}c}} \left ( 2\,x+{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}+\sqrt [3]{-{d}^{2}c}}{d}} \right ) }}{\it EllipticPi} \left ( 1/3\,\sqrt{3}\sqrt{{\frac{id\sqrt{3}}{\sqrt [3]{-{d}^{2}c}} \left ( x+1/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}-{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) }},1/6\,{\frac{2\,i\sqrt [3]{-{d}^{2}c}\sqrt{3}{{\it \_alpha}}^{2}d-i \left ( -{d}^{2}c \right ) ^{2/3}\sqrt{3}{\it \_alpha}+i\sqrt{3}cd-3\, \left ( -{d}^{2}c \right ) ^{2/3}{\it \_alpha}-3\,cd}{cd}},\sqrt{{\frac{i\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d} \left ( -3/2\,{\frac{\sqrt [3]{-{d}^{2}c}}{d}}+{\frac{i/2\sqrt{3}\sqrt [3]{-{d}^{2}c}}{d}} \right ) ^{-1}}} \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d^2*(d*(2/15*x^6*(d*x^3+c)^(1/2)+2/45/d*c*x^3*(d*x^3+c)^(1/2)-4/45*c^2*(d*x^3+c)^(1/2)/d^2)-8/9*c/d*(d*x^3+c
)^(3/2))+16*c^2/d^2*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/d^3*2^(1/2)*sum((-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+c)^(1/2)/(d*x^3+4*c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10169, size = 397, normalized size = 4.09 \begin{align*} \left [\frac{2 \,{\left (120 \, \sqrt{3} \sqrt{-c} c^{2} \log \left (\frac{d x^{3} - 2 \, \sqrt{3} \sqrt{d x^{3} + c} \sqrt{-c} - 2 \, c}{d x^{3} + 4 \, c}\right ) +{\left (3 \, d^{2} x^{6} - 19 \, c d x^{3} + 218 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, d^{3}}, -\frac{2 \,{\left (240 \, \sqrt{3} c^{\frac{5}{2}} \arctan \left (\frac{\sqrt{3} \sqrt{d x^{3} + c}}{3 \, \sqrt{c}}\right ) -{\left (3 \, d^{2} x^{6} - 19 \, c d x^{3} + 218 \, c^{2}\right )} \sqrt{d x^{3} + c}\right )}}{45 \, d^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/45*(120*sqrt(3)*sqrt(-c)*c^2*log((d*x^3 - 2*sqrt(3)*sqrt(d*x^3 + c)*sqrt(-c) - 2*c)/(d*x^3 + 4*c)) + (3*d^2
*x^6 - 19*c*d*x^3 + 218*c^2)*sqrt(d*x^3 + c))/d^3, -2/45*(240*sqrt(3)*c^(5/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 +
c)/sqrt(c)) - (3*d^2*x^6 - 19*c*d*x^3 + 218*c^2)*sqrt(d*x^3 + c))/d^3]

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Sympy [A]  time = 55.3091, size = 85, normalized size = 0.88 \begin{align*} \frac{2 \left (- \frac{16 \sqrt{3} c^{\frac{5}{2}} \operatorname{atan}{\left (\frac{\sqrt{3} \sqrt{c + d x^{3}}}{3 \sqrt{c}} \right )}}{3} + \frac{16 c^{2} \sqrt{c + d x^{3}}}{3} - \frac{5 c \left (c + d x^{3}\right )^{\frac{3}{2}}}{9} + \frac{\left (c + d x^{3}\right )^{\frac{5}{2}}}{15}\right )}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-16*sqrt(3)*c**(5/2)*atan(sqrt(3)*sqrt(c + d*x**3)/(3*sqrt(c)))/3 + 16*c**2*sqrt(c + d*x**3)/3 - 5*c*(c + d
*x**3)**(3/2)/9 + (c + d*x**3)**(5/2)/15)/d**3

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Giac [A]  time = 1.11766, size = 111, normalized size = 1.14 \begin{align*} -\frac{32 \, \sqrt{3} c^{\frac{5}{2}} \arctan \left (\frac{\sqrt{3} \sqrt{d x^{3} + c}}{3 \, \sqrt{c}}\right )}{3 \, d^{3}} + \frac{2 \,{\left (3 \,{\left (d x^{3} + c\right )}^{\frac{5}{2}} d^{12} - 25 \,{\left (d x^{3} + c\right )}^{\frac{3}{2}} c d^{12} + 240 \, \sqrt{d x^{3} + c} c^{2} d^{12}\right )}}{45 \, d^{15}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32/3*sqrt(3)*c^(5/2)*arctan(1/3*sqrt(3)*sqrt(d*x^3 + c)/sqrt(c))/d^3 + 2/45*(3*(d*x^3 + c)^(5/2)*d^12 - 25*(d
*x^3 + c)^(3/2)*c*d^12 + 240*sqrt(d*x^3 + c)*c^2*d^12)/d^15