3.3.24 \(\int \frac {x^8 (c+d x^3)^{3/2}}{8 c-d x^3} \, dx\)

Optimal. Leaf size=109 \[ \frac {1152 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3}-\frac {384 c^3 \sqrt {c+d x^3}}{d^3}-\frac {128 c^2 \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac {14 c \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^3} \]

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Rubi [A]  time = 0.10, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {446, 88, 50, 63, 206} \begin {gather*} -\frac {384 c^3 \sqrt {c+d x^3}}{d^3}-\frac {128 c^2 \left (c+d x^3\right )^{3/2}}{9 d^3}+\frac {1152 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3}-\frac {14 c \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^8*(c + d*x^3)^(3/2))/(8*c - d*x^3),x]

[Out]

(-384*c^3*Sqrt[c + d*x^3])/d^3 - (128*c^2*(c + d*x^3)^(3/2))/(9*d^3) - (14*c*(c + d*x^3)^(5/2))/(15*d^3) - (2*
(c + d*x^3)^(7/2))/(21*d^3) + (1152*c^(7/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^3

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 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 206

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

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]]

Rubi steps

\begin {align*} \int \frac {x^8 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2 (c+d x)^{3/2}}{8 c-d x} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (-\frac {7 c (c+d x)^{3/2}}{d^2}+\frac {64 c^2 (c+d x)^{3/2}}{d^2 (8 c-d x)}-\frac {(c+d x)^{5/2}}{d^2}\right ) \, dx,x,x^3\right )\\ &=-\frac {14 c \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^3}+\frac {\left (64 c^2\right ) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{8 c-d x} \, dx,x,x^3\right )}{3 d^2}\\ &=-\frac {128 c^2 \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac {14 c \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^3}+\frac {\left (192 c^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{8 c-d x} \, dx,x,x^3\right )}{d^2}\\ &=-\frac {384 c^3 \sqrt {c+d x^3}}{d^3}-\frac {128 c^2 \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac {14 c \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^3}+\frac {\left (1728 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{(8 c-d x) \sqrt {c+d x}} \, dx,x,x^3\right )}{d^2}\\ &=-\frac {384 c^3 \sqrt {c+d x^3}}{d^3}-\frac {128 c^2 \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac {14 c \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^3}+\frac {\left (3456 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{9 c-x^2} \, dx,x,\sqrt {c+d x^3}\right )}{d^3}\\ &=-\frac {384 c^3 \sqrt {c+d x^3}}{d^3}-\frac {128 c^2 \left (c+d x^3\right )^{3/2}}{9 d^3}-\frac {14 c \left (c+d x^3\right )^{5/2}}{15 d^3}-\frac {2 \left (c+d x^3\right )^{7/2}}{21 d^3}+\frac {1152 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 81, normalized size = 0.74 \begin {gather*} \frac {362880 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-2 \sqrt {c+d x^3} \left (62882 c^3+2579 c^2 d x^3+192 c d^2 x^6+15 d^3 x^9\right )}{315 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(c + d*x^3)^(3/2))/(8*c - d*x^3),x]

[Out]

(-2*Sqrt[c + d*x^3]*(62882*c^3 + 2579*c^2*d*x^3 + 192*c*d^2*x^6 + 15*d^3*x^9) + 362880*c^(7/2)*ArcTanh[Sqrt[c
+ d*x^3]/(3*Sqrt[c])])/(315*d^3)

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IntegrateAlgebraic [A]  time = 0.08, size = 82, normalized size = 0.75 \begin {gather*} \frac {1152 c^{7/2} \tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^3}-\frac {2 \sqrt {c+d x^3} \left (62882 c^3+2579 c^2 d x^3+192 c d^2 x^6+15 d^3 x^9\right )}{315 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^8*(c + d*x^3)^(3/2))/(8*c - d*x^3),x]

[Out]

(-2*Sqrt[c + d*x^3]*(62882*c^3 + 2579*c^2*d*x^3 + 192*c*d^2*x^6 + 15*d^3*x^9))/(315*d^3) + (1152*c^(7/2)*ArcTa
nh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^3

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fricas [A]  time = 0.96, size = 169, normalized size = 1.55 \begin {gather*} \left [\frac {2 \, {\left (90720 \, c^{\frac {7}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - {\left (15 \, d^{3} x^{9} + 192 \, c d^{2} x^{6} + 2579 \, c^{2} d x^{3} + 62882 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{315 \, d^{3}}, -\frac {2 \, {\left (181440 \, \sqrt {-c} c^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) + {\left (15 \, d^{3} x^{9} + 192 \, c d^{2} x^{6} + 2579 \, c^{2} d x^{3} + 62882 \, c^{3}\right )} \sqrt {d x^{3} + c}\right )}}{315 \, d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[2/315*(90720*c^(7/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 - 8*c)) - (15*d^3*x^9 + 192*c*d^2*
x^6 + 2579*c^2*d*x^3 + 62882*c^3)*sqrt(d*x^3 + c))/d^3, -2/315*(181440*sqrt(-c)*c^3*arctan(1/3*sqrt(d*x^3 + c)
*sqrt(-c)/c) + (15*d^3*x^9 + 192*c*d^2*x^6 + 2579*c^2*d*x^3 + 62882*c^3)*sqrt(d*x^3 + c))/d^3]

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giac [A]  time = 0.16, size = 100, normalized size = 0.92 \begin {gather*} -\frac {1152 \, c^{4} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{3}} - \frac {2 \, {\left (15 \, {\left (d x^{3} + c\right )}^{\frac {7}{2}} d^{18} + 147 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c d^{18} + 2240 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{2} d^{18} + 60480 \, \sqrt {d x^{3} + c} c^{3} d^{18}\right )}}{315 \, d^{21}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1152*c^4*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^3) - 2/315*(15*(d*x^3 + c)^(7/2)*d^18 + 147*(d*x^3
+ c)^(5/2)*c*d^18 + 2240*(d*x^3 + c)^(3/2)*c^2*d^18 + 60480*sqrt(d*x^3 + c)*c^3*d^18)/d^21

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maple [C]  time = 0.16, size = 541, normalized size = 4.96 \begin {gather*} -\frac {64 \left (\frac {2 \sqrt {d \,x^{3}+c}\, x^{3}}{9}+\frac {56 \sqrt {d \,x^{3}+c}\, c}{9 d}+\frac {3 i c \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, -\frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}-8 c \right )}{18 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{d^{3} \sqrt {d \,x^{3}+c}}\right ) c^{2}}{d^{2}}-\frac {\left (\frac {2 \sqrt {d \,x^{3}+c}\, d \,x^{9}}{21}+\frac {16 \sqrt {d \,x^{3}+c}\, c \,x^{6}}{105}+\frac {2 \sqrt {d \,x^{3}+c}\, c^{2} x^{3}}{105 d}-\frac {4 \sqrt {d \,x^{3}+c}\, c^{3}}{105 d^{2}}\right ) d +\frac {16 \left (d \,x^{3}+c \right )^{\frac {5}{2}} c}{15 d}}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d^2*(d*(2/21*(d*x^3+c)^(1/2)*d*x^9+16/105*(d*x^3+c)^(1/2)*c*x^6+2/105*(d*x^3+c)^(1/2)*c^2/d*x^3-4/105*(d*x^
3+c)^(1/2)*c^3/d^2)+16/15*c/d*(d*x^3+c)^(5/2))-64*c^2/d^2*(2/9*(d*x^3+c)^(1/2)*x^3+56/9*(d*x^3+c)^(1/2)*c/d+3*
I*c/d^3*2^(1/2)*sum((-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)
^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+(I*3^(1/2)*(-c
*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2)^(1/3)*3^(1/2
)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(
-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),-1/18*(2*I*(-c*d^2)^(1/3)*3^(1
/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/c/d,(I*3^(1/2)*(-c
*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*d-8*c)))

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maxima [A]  time = 1.34, size = 96, normalized size = 0.88 \begin {gather*} -\frac {2 \, {\left (90720 \, c^{\frac {7}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 15 \, {\left (d x^{3} + c\right )}^{\frac {7}{2}} + 147 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} c + 2240 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c^{2} + 60480 \, \sqrt {d x^{3} + c} c^{3}\right )}}{315 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*(90720*c^(7/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3*sqrt(c))) + 15*(d*x^3 + c)^(7/2)
+ 147*(d*x^3 + c)^(5/2)*c + 2240*(d*x^3 + c)^(3/2)*c^2 + 60480*sqrt(d*x^3 + c)*c^3)/d^3

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mupad [B]  time = 3.50, size = 115, normalized size = 1.06 \begin {gather*} \frac {576\,c^{7/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^3}-\frac {2\,x^9\,\sqrt {d\,x^3+c}}{21}-\frac {125764\,c^3\,\sqrt {d\,x^3+c}}{315\,d^3}-\frac {128\,c\,x^6\,\sqrt {d\,x^3+c}}{105\,d}-\frac {5158\,c^2\,x^3\,\sqrt {d\,x^3+c}}{315\,d^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(576*c^(7/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3)))/d^3 - (2*x^9*(c + d*x^3)^(1/2))/
21 - (125764*c^3*(c + d*x^3)^(1/2))/(315*d^3) - (128*c*x^6*(c + d*x^3)^(1/2))/(105*d) - (5158*c^2*x^3*(c + d*x
^3)^(1/2))/(315*d^2)

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sympy [A]  time = 108.79, size = 110, normalized size = 1.01 \begin {gather*} - \frac {1152 c^{4} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{d^{3} \sqrt {- c}} - \frac {384 c^{3} \sqrt {c + d x^{3}}}{d^{3}} - \frac {128 c^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 d^{3}} - \frac {14 c \left (c + d x^{3}\right )^{\frac {5}{2}}}{15 d^{3}} - \frac {2 \left (c + d x^{3}\right )^{\frac {7}{2}}}{21 d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1152*c**4*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(d**3*sqrt(-c)) - 384*c**3*sqrt(c + d*x**3)/d**3 - 128*c**2*(c
+ d*x**3)**(3/2)/(9*d**3) - 14*c*(c + d*x**3)**(5/2)/(15*d**3) - 2*(c + d*x**3)**(7/2)/(21*d**3)

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