3.368 \(\int \frac {x^8 (c+d x^3)^{3/2}}{a+b x^3} \, dx\)

Optimal. Leaf size=154 \[ -\frac {2 a^2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{9/2}}+\frac {2 a^2 \sqrt {c+d x^3} (b c-a d)}{3 b^4}+\frac {2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac {2 \left (c+d x^3\right )^{5/2} (a d+b c)}{15 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 b d^2} \]

[Out]

2/9*a^2*(d*x^3+c)^(3/2)/b^3-2/15*(a*d+b*c)*(d*x^3+c)^(5/2)/b^2/d^2+2/21*(d*x^3+c)^(7/2)/b/d^2-2/3*a^2*(-a*d+b*
c)^(3/2)*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b*c)^(1/2))/b^(9/2)+2/3*a^2*(-a*d+b*c)*(d*x^3+c)^(1/2)/b^4

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Rubi [A]  time = 0.15, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {446, 88, 50, 63, 208} \[ \frac {2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}+\frac {2 a^2 \sqrt {c+d x^3} (b c-a d)}{3 b^4}-\frac {2 a^2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{9/2}}-\frac {2 \left (c+d x^3\right )^{5/2} (a d+b c)}{15 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(b*c - a*d)*Sqrt[c + d*x^3])/(3*b^4) + (2*a^2*(c + d*x^3)^(3/2))/(9*b^3) - (2*(b*c + a*d)*(c + d*x^3)^(
5/2))/(15*b^2*d^2) + (2*(c + d*x^3)^(7/2))/(21*b*d^2) - (2*a^2*(b*c - a*d)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x
^3])/Sqrt[b*c - a*d]])/(3*b^(9/2))

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 208

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

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}}{a+b x^3} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2 (c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {(-b c-a d) (c+d x)^{3/2}}{b^2 d}+\frac {a^2 (c+d x)^{3/2}}{b^2 (a+b x)}+\frac {(c+d x)^{5/2}}{b d}\right ) \, dx,x,x^3\right )\\ &=-\frac {2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 b d^2}+\frac {a^2 \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )}{3 b^2}\\ &=\frac {2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac {2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 b d^2}+\frac {\left (a^2 (b c-a d)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^3\right )}{3 b^3}\\ &=\frac {2 a^2 (b c-a d) \sqrt {c+d x^3}}{3 b^4}+\frac {2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac {2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 b d^2}+\frac {\left (a^2 (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{3 b^4}\\ &=\frac {2 a^2 (b c-a d) \sqrt {c+d x^3}}{3 b^4}+\frac {2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac {2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 b d^2}+\frac {\left (2 a^2 (b c-a d)^2\right ) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b c}{d}+\frac {b x^2}{d}} \, dx,x,\sqrt {c+d x^3}\right )}{3 b^4 d}\\ &=\frac {2 a^2 (b c-a d) \sqrt {c+d x^3}}{3 b^4}+\frac {2 a^2 \left (c+d x^3\right )^{3/2}}{9 b^3}-\frac {2 (b c+a d) \left (c+d x^3\right )^{5/2}}{15 b^2 d^2}+\frac {2 \left (c+d x^3\right )^{7/2}}{21 b d^2}-\frac {2 a^2 (b c-a d)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 145, normalized size = 0.94 \[ \frac {2 \left (105 a^2 (b c-a d) \left (\frac {\sqrt {c+d x^3}}{b}-\frac {\sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2}}\right )+35 a^2 \left (c+d x^3\right )^{3/2}-\frac {21 b \left (c+d x^3\right )^{5/2} (a d+b c)}{d^2}+\frac {15 b^2 \left (c+d x^3\right )^{7/2}}{d^2}\right )}{315 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(35*a^2*(c + d*x^3)^(3/2) - (21*b*(b*c + a*d)*(c + d*x^3)^(5/2))/d^2 + (15*b^2*(c + d*x^3)^(7/2))/d^2 + 105
*a^2*(b*c - a*d)*(Sqrt[c + d*x^3]/b - (Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/b^(
3/2))))/(315*b^3)

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fricas [A]  time = 0.90, size = 410, normalized size = 2.66 \[ \left [-\frac {105 \, {\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d + 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right ) - 2 \, {\left (15 \, b^{3} d^{3} x^{9} + 3 \, {\left (8 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{6} - 6 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 140 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + {\left (3 \, b^{3} c^{2} d - 42 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{315 \, b^{4} d^{2}}, -\frac {2 \, {\left (105 \, {\left (a^{2} b c d^{2} - a^{3} d^{3}\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right ) - {\left (15 \, b^{3} d^{3} x^{9} + 3 \, {\left (8 \, b^{3} c d^{2} - 7 \, a b^{2} d^{3}\right )} x^{6} - 6 \, b^{3} c^{3} - 21 \, a b^{2} c^{2} d + 140 \, a^{2} b c d^{2} - 105 \, a^{3} d^{3} + {\left (3 \, b^{3} c^{2} d - 42 \, a b^{2} c d^{2} + 35 \, a^{2} b d^{3}\right )} x^{3}\right )} \sqrt {d x^{3} + c}\right )}}{315 \, b^{4} d^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/315*(105*(a^2*b*c*d^2 - a^3*d^3)*sqrt((b*c - a*d)/b)*log((b*d*x^3 + 2*b*c - a*d + 2*sqrt(d*x^3 + c)*b*sqrt
((b*c - a*d)/b))/(b*x^3 + a)) - 2*(15*b^3*d^3*x^9 + 3*(8*b^3*c*d^2 - 7*a*b^2*d^3)*x^6 - 6*b^3*c^3 - 21*a*b^2*c
^2*d + 140*a^2*b*c*d^2 - 105*a^3*d^3 + (3*b^3*c^2*d - 42*a*b^2*c*d^2 + 35*a^2*b*d^3)*x^3)*sqrt(d*x^3 + c))/(b^
4*d^2), -2/315*(105*(a^2*b*c*d^2 - a^3*d^3)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b
)/(b*c - a*d)) - (15*b^3*d^3*x^9 + 3*(8*b^3*c*d^2 - 7*a*b^2*d^3)*x^6 - 6*b^3*c^3 - 21*a*b^2*c^2*d + 140*a^2*b*
c*d^2 - 105*a^3*d^3 + (3*b^3*c^2*d - 42*a*b^2*c*d^2 + 35*a^2*b*d^3)*x^3)*sqrt(d*x^3 + c))/(b^4*d^2)]

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giac [A]  time = 0.17, size = 193, normalized size = 1.25 \[ \frac {2 \, {\left (a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} b^{4}} + \frac {2 \, {\left (15 \, {\left (d x^{3} + c\right )}^{\frac {7}{2}} b^{6} d^{12} - 21 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} b^{6} c d^{12} - 21 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} a b^{5} d^{13} + 35 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} a^{2} b^{4} d^{14} + 105 \, \sqrt {d x^{3} + c} a^{2} b^{4} c d^{14} - 105 \, \sqrt {d x^{3} + c} a^{3} b^{3} d^{15}\right )}}{315 \, b^{7} d^{14}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)
*b^4) + 2/315*(15*(d*x^3 + c)^(7/2)*b^6*d^12 - 21*(d*x^3 + c)^(5/2)*b^6*c*d^12 - 21*(d*x^3 + c)^(5/2)*a*b^5*d^
13 + 35*(d*x^3 + c)^(3/2)*a^2*b^4*d^14 + 105*sqrt(d*x^3 + c)*a^2*b^4*c*d^14 - 105*sqrt(d*x^3 + c)*a^3*b^3*d^15
)/(b^7*d^14)

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maple [C]  time = 0.36, size = 605, normalized size = 3.93 \[ \frac {\left (\frac {2 \sqrt {d \,x^{3}+c}\, d \,x^{3}}{9 b}+\frac {2 \left (-\frac {2 c d}{3 b}-\frac {\left (a d -2 b c \right ) d}{b^{2}}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \left (-a^{2} d^{2}+2 a b c d -b^{2} c^{2}\right ) \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 (\textit {\_Z}^{3} b +a \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \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 {\left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (\textit {\_Z}^{3} b +a \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (\textit {\_Z}^{3} b +a \right )\right ) b}{2 \left (a d -b c \right ) 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 )}{3 b^{2} d^{2} \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right ) a^{2}}{b^{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 ) b -\frac {2 \left (d \,x^{3}+c \right )^{\frac {5}{2}} a}{15 d}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(b*(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)-2/15*a/d*(d*x^3+c)^(5/2))+a^2/b^2*(2/9*d/b*x^3*(d*x^3+c)^(1/2)+2/3*(-d*(a*d-2*b*c)/b^2-2/3*
d/b*c)/d*(d*x^3+c)^(1/2)+1/3*I/b^2/d^2*2^(1/2)*sum((-a^2*d^2+2*a*b*c*d-b^2*c^2)/(a*d-b*c)*(-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),1/2*(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)/(a*d-b*c)*b/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*b+a)))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c positive or negative?

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mupad [B]  time = 6.12, size = 330, normalized size = 2.14 \[ \frac {2\,d\,x^9\,\sqrt {d\,x^3+c}}{21\,b}-\frac {\left (\frac {2\,a\,\left (\frac {c^2}{b}+\frac {a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}\right )}{b}+\frac {2\,c\,\left (\frac {2\,c^2}{b}+\frac {2\,a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}+\frac {4\,c\,\left (\frac {2\,a\,d^2}{b^2}-\frac {16\,c\,d}{7\,b}\right )}{5\,d}\right )}{3\,d}\right )\,\sqrt {d\,x^3+c}}{3\,d}+\frac {x^3\,\sqrt {d\,x^3+c}\,\left (\frac {2\,c^2}{b}+\frac {2\,a\,\left (\frac {a\,d^2}{b^2}-\frac {2\,c\,d}{b}\right )}{b}+\frac {4\,c\,\left (\frac {2\,a\,d^2}{b^2}-\frac {16\,c\,d}{7\,b}\right )}{5\,d}\right )}{9\,d}-\frac {x^6\,\sqrt {d\,x^3+c}\,\left (\frac {2\,a\,d^2}{b^2}-\frac {16\,c\,d}{7\,b}\right )}{15\,d}+\frac {a^2\,\ln \left (\frac {a^2\,d^2+2\,b^2\,c^2-a\,b\,d^2\,x^3+b^2\,c\,d\,x^3-3\,a\,b\,c\,d-\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,1{}\mathrm {i}}{3\,b^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*d*x^9*(c + d*x^3)^(1/2))/(21*b) - (((2*a*(c^2/b + (a*((a*d^2)/b^2 - (2*c*d)/b))/b))/b + (2*c*((2*c^2)/b + (
2*a*((a*d^2)/b^2 - (2*c*d)/b))/b + (4*c*((2*a*d^2)/b^2 - (16*c*d)/(7*b)))/(5*d)))/(3*d))*(c + d*x^3)^(1/2))/(3
*d) + (x^3*(c + d*x^3)^(1/2)*((2*c^2)/b + (2*a*((a*d^2)/b^2 - (2*c*d)/b))/b + (4*c*((2*a*d^2)/b^2 - (16*c*d)/(
7*b)))/(5*d)))/(9*d) - (x^6*(c + d*x^3)^(1/2)*((2*a*d^2)/b^2 - (16*c*d)/(7*b)))/(15*d) + (a^2*log((a^2*d^2 + 2
*b^2*c^2 - b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(3/2)*2i - a*b*d^2*x^3 + b^2*c*d*x^3 - 3*a*b*c*d)/(a + b*x^3)
)*(a*d - b*c)^(3/2)*1i)/(3*b^(9/2))

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sympy [A]  time = 129.92, size = 153, normalized size = 0.99 \[ \frac {2 a^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 b^{3}} + \frac {2 a^{2} \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 b^{5} \sqrt {\frac {a d - b c}{b}}} + \frac {2 \left (c + d x^{3}\right )^{\frac {7}{2}}}{21 b d^{2}} + \frac {\left (c + d x^{3}\right )^{\frac {5}{2}} \left (- 2 a d - 2 b c\right )}{15 b^{2} d^{2}} + \frac {\sqrt {c + d x^{3}} \left (- 2 a^{3} d + 2 a^{2} b c\right )}{3 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**2*(c + d*x**3)**(3/2)/(9*b**3) + 2*a**2*(a*d - b*c)**2*atan(sqrt(c + d*x**3)/sqrt((a*d - b*c)/b))/(3*b**5
*sqrt((a*d - b*c)/b)) + 2*(c + d*x**3)**(7/2)/(21*b*d**2) + (c + d*x**3)**(5/2)*(-2*a*d - 2*b*c)/(15*b**2*d**2
) + sqrt(c + d*x**3)*(-2*a**3*d + 2*a**2*b*c)/(3*b**4)

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