3.3.59 \(\int \frac {x^8 \sqrt {c+d x^3}}{4 c+d x^3} \, dx\) [259]

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

[Out]

-10/9*c*(d*x^3+c)^(3/2)/d^3+2/15*(d*x^3+c)^(5/2)/d^3-32/3*c^(5/2)*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^(1/2))/
d^3*3^(1/2)+32/3*c^2*(d*x^3+c)^(1/2)/d^3

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Rubi [A]
time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 = {457, 90, 52, 65, 209} \begin {gather*} -\frac {32 c^{5/2} \text {ArcTan}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{\sqrt {3} 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} \end {gather*}

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 52

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 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}}{4 c+d x^3} \, 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 \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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.07, size = 78, normalized size = 0.80 \begin {gather*} \frac {2 \sqrt {c+d x^3} \left (218 c^2-19 c d x^3+3 d^2 x^6\right )}{45 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 {gather*}

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))/(45*d^3) - (32*c^(5/2)*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*
Sqrt[c])])/(Sqrt[3]*d^3)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 1.22, size = 503, normalized size = 5.19 Too large to display

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,method=_RETURNVERBOSE)

[Out]

1/d*(2/15*x^6*(d*x^3+c)^(1/2)+2/45*c/d*x^3*(d*x^3+c)^(1/2)-4/45*c^2*(d*x^3+c)^(1/2)/d^2)-8/9*c*(d*x^3+c)^(3/2)
/d^3+16*c^2/d^2*(2/3*(d*x^3+c)^(1/2)/d+1/3*I/d^3*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 = 0.49, size = 69, normalized size = 0.71 \begin {gather*} -\frac {2 \, {\left (240 \, \sqrt {3} c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {3} \sqrt {d x^{3} + c}}{3 \, \sqrt {c}}\right ) - 3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} + 25 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c - 240 \, \sqrt {d x^{3} + c} c^{2}\right )}}{45 \, d^{3}} \end {gather*}

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]

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

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Fricas [A]
time = 2.24, size = 156, normalized size = 1.61 \begin {gather*} \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 {gather*}

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 = 11.85, size = 85, normalized size = 0.88 \begin {gather*} \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 {gather*}

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 = 1.07, size = 82, normalized size = 0.85 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 4.54, size = 109, normalized size = 1.12 \begin {gather*} \frac {436\,c^2\,\sqrt {d\,x^3+c}}{45\,d^3}+\frac {2\,x^6\,\sqrt {d\,x^3+c}}{15\,d}-\frac {38\,c\,x^3\,\sqrt {d\,x^3+c}}{45\,d^2}+\frac {\sqrt {3}\,c^{5/2}\,\ln \left (\frac {2\,\sqrt {3}\,c-\sqrt {3}\,d\,x^3+\sqrt {c}\,\sqrt {d\,x^3+c}\,6{}\mathrm {i}}{d\,x^3+4\,c}\right )\,16{}\mathrm {i}}{3\,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]

(436*c^2*(c + d*x^3)^(1/2))/(45*d^3) + (2*x^6*(c + d*x^3)^(1/2))/(15*d) - (38*c*x^3*(c + d*x^3)^(1/2))/(45*d^2
) + (3^(1/2)*c^(5/2)*log((2*3^(1/2)*c + c^(1/2)*(c + d*x^3)^(1/2)*6i - 3^(1/2)*d*x^3)/(4*c + d*x^3))*16i)/(3*d
^3)

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