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

Optimal. Leaf size=189 \[ -\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 \left (a+b x^3\right ) (b c-a d)}+\frac {a (4 b c-7 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{9/2}}-\frac {a \sqrt {c+d x^3} (4 b c-7 a d)}{3 b^4}-\frac {a \left (c+d x^3\right )^{3/2} (4 b c-7 a d)}{9 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 b^2 d} \]

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Rubi [A]  time = 0.24, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 89, 80, 50, 63, 208} \begin {gather*} -\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 \left (a+b x^3\right ) (b c-a d)}-\frac {a \left (c+d x^3\right )^{3/2} (4 b c-7 a d)}{9 b^3 (b c-a d)}-\frac {a \sqrt {c+d x^3} (4 b c-7 a d)}{3 b^4}+\frac {a (4 b c-7 a d) \sqrt {b c-a d} \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}}{15 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(4*b*c - 7*a*d)*Sqrt[c + d*x^3])/(3*b^4) - (a*(4*b*c - 7*a*d)*(c + d*x^3)^(3/2))/(9*b^3*(b*c - a*d)) + (2*
(c + d*x^3)^(5/2))/(15*b^2*d) - (a^2*(c + d*x^3)^(5/2))/(3*b^2*(b*c - a*d)*(a + b*x^3)) + (a*(4*b*c - 7*a*d)*S
qrt[b*c - a*d]*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 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 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}}{\left (a+b x^3\right )^2} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {x^2 (c+d x)^{3/2}}{(a+b x)^2} \, dx,x,x^3\right )\\ &=-\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}+\frac {\operatorname {Subst}\left (\int \frac {(c+d x)^{3/2} \left (-\frac {1}{2} a (2 b c-5 a d)+b (b c-a d) x\right )}{a+b x} \, dx,x,x^3\right )}{3 b^2 (b c-a d)}\\ &=\frac {2 \left (c+d x^3\right )^{5/2}}{15 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac {(a (4 b c-7 a d)) \operatorname {Subst}\left (\int \frac {(c+d x)^{3/2}}{a+b x} \, dx,x,x^3\right )}{6 b^2 (b c-a d)}\\ &=-\frac {a (4 b c-7 a d) \left (c+d x^3\right )^{3/2}}{9 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac {(a (4 b c-7 a d)) \operatorname {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^3\right )}{6 b^3}\\ &=-\frac {a (4 b c-7 a d) \sqrt {c+d x^3}}{3 b^4}-\frac {a (4 b c-7 a d) \left (c+d x^3\right )^{3/2}}{9 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac {(a (4 b c-7 a d) (b c-a d)) \operatorname {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{6 b^4}\\ &=-\frac {a (4 b c-7 a d) \sqrt {c+d x^3}}{3 b^4}-\frac {a (4 b c-7 a d) \left (c+d x^3\right )^{3/2}}{9 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac {(a (4 b c-7 a d) (b c-a d)) \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 {a (4 b c-7 a d) \sqrt {c+d x^3}}{3 b^4}-\frac {a (4 b c-7 a d) \left (c+d x^3\right )^{3/2}}{9 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{5/2}}{15 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{5/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}+\frac {a (4 b c-7 a d) \sqrt {b c-a d} \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 = 0.19, size = 162, normalized size = 0.86 \begin {gather*} \frac {\sqrt {c+d x^3} \left (105 a^3 d^2+5 a^2 b d \left (14 d x^3-19 c\right )+2 a b^2 \left (3 c^2-34 c d x^3-7 d^2 x^6\right )+6 b^3 x^3 \left (c+d x^3\right )^2\right )}{45 b^4 d \left (a+b x^3\right )}+\frac {a (4 b c-7 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^3]*(105*a^3*d^2 + 6*b^3*x^3*(c + d*x^3)^2 + 5*a^2*b*d*(-19*c + 14*d*x^3) + 2*a*b^2*(3*c^2 - 34*c
*d*x^3 - 7*d^2*x^6)))/(45*b^4*d*(a + b*x^3)) + (a*(4*b*c - 7*a*d)*Sqrt[b*c - a*d]*ArcTanh[(Sqrt[b]*Sqrt[c + d*
x^3])/Sqrt[b*c - a*d]])/(3*b^(9/2))

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IntegrateAlgebraic [A]  time = 0.29, size = 211, normalized size = 1.12 \begin {gather*} \frac {\left (7 a^3 d^2-11 a^2 b c d+4 a b^2 c^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3} \sqrt {a d-b c}}{b c-a d}\right )}{3 b^{9/2} \sqrt {a d-b c}}+\frac {\sqrt {c+d x^3} \left (105 a^3 d^2-95 a^2 b c d+70 a^2 b d^2 x^3+6 a b^2 c^2-68 a b^2 c d x^3-14 a b^2 d^2 x^6+6 b^3 c^2 x^3+12 b^3 c d x^6+6 b^3 d^2 x^9\right )}{45 b^4 d \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d*x^3]*(6*a*b^2*c^2 - 95*a^2*b*c*d + 105*a^3*d^2 + 6*b^3*c^2*x^3 - 68*a*b^2*c*d*x^3 + 70*a^2*b*d^2*x
^3 + 12*b^3*c*d*x^6 - 14*a*b^2*d^2*x^6 + 6*b^3*d^2*x^9))/(45*b^4*d*(a + b*x^3)) + ((4*a*b^2*c^2 - 11*a^2*b*c*d
+ 7*a^3*d^2)*ArcTan[(Sqrt[b]*Sqrt[-(b*c) + a*d]*Sqrt[c + d*x^3])/(b*c - a*d)])/(3*b^(9/2)*Sqrt[-(b*c) + a*d])

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fricas [A]  time = 0.69, size = 443, normalized size = 2.34 \begin {gather*} \left [-\frac {15 \, {\left (4 \, a^{2} b c d - 7 \, a^{3} d^{2} + {\left (4 \, a b^{2} c d - 7 \, a^{2} b d^{2}\right )} x^{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 (6 \, b^{3} d^{2} x^{9} + 2 \, {\left (6 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{6} + 6 \, a b^{2} c^{2} - 95 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (3 \, b^{3} c^{2} - 34 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{90 \, {\left (b^{5} d x^{3} + a b^{4} d\right )}}, \frac {15 \, {\left (4 \, a^{2} b c d - 7 \, a^{3} d^{2} + {\left (4 \, a b^{2} c d - 7 \, a^{2} b d^{2}\right )} x^{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 (6 \, b^{3} d^{2} x^{9} + 2 \, {\left (6 \, b^{3} c d - 7 \, a b^{2} d^{2}\right )} x^{6} + 6 \, a b^{2} c^{2} - 95 \, a^{2} b c d + 105 \, a^{3} d^{2} + 2 \, {\left (3 \, b^{3} c^{2} - 34 \, a b^{2} c d + 35 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{45 \, {\left (b^{5} d x^{3} + a b^{4} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/90*(15*(4*a^2*b*c*d - 7*a^3*d^2 + (4*a*b^2*c*d - 7*a^2*b*d^2)*x^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*(6*b^3*d^2*x^9 + 2*(6*b^3*c*d - 7*a*b^2*d^
2)*x^6 + 6*a*b^2*c^2 - 95*a^2*b*c*d + 105*a^3*d^2 + 2*(3*b^3*c^2 - 34*a*b^2*c*d + 35*a^2*b*d^2)*x^3)*sqrt(d*x^
3 + c))/(b^5*d*x^3 + a*b^4*d), 1/45*(15*(4*a^2*b*c*d - 7*a^3*d^2 + (4*a*b^2*c*d - 7*a^2*b*d^2)*x^3)*sqrt(-(b*c
- a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) + (6*b^3*d^2*x^9 + 2*(6*b^3*c*d - 7*a*b
^2*d^2)*x^6 + 6*a*b^2*c^2 - 95*a^2*b*c*d + 105*a^3*d^2 + 2*(3*b^3*c^2 - 34*a*b^2*c*d + 35*a^2*b*d^2)*x^3)*sqrt
(d*x^3 + c))/(b^5*d*x^3 + a*b^4*d)]

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giac [A]  time = 0.18, size = 211, normalized size = 1.12 \begin {gather*} -\frac {{\left (4 \, a b^{2} c^{2} - 11 \, a^{2} b c d + 7 \, a^{3} 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 {\sqrt {d x^{3} + c} a^{2} b c d - \sqrt {d x^{3} + c} a^{3} d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{4}} + \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} b^{8} d^{4} - 10 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} a b^{7} d^{5} - 30 \, \sqrt {d x^{3} + c} a b^{7} c d^{5} + 45 \, \sqrt {d x^{3} + c} a^{2} b^{6} d^{6}\right )}}{45 \, b^{10} d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(4*a*b^2*c^2 - 11*a^2*b*c*d + 7*a^3*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) - 1/3*(sqrt(d*x^3 + c)*a^2*b*c*d - sqrt(d*x^3 + c)*a^3*d^2)/(((d*x^3 + c)*b - b*c + a*d)*b^4) + 2/45
*(3*(d*x^3 + c)^(5/2)*b^8*d^4 - 10*(d*x^3 + c)^(3/2)*a*b^7*d^5 - 30*sqrt(d*x^3 + c)*a*b^7*c*d^5 + 45*sqrt(d*x^
3 + c)*a^2*b^6*d^6)/(b^10*d^5)

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maple [C]  time = 0.37, size = 1003, normalized size = 5.31

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(d*x^3+c)^(5/2)/b^2/d-2*a/b^2*(2/9*(d*x^3+c)^(1/2)/b*d*x^3+2/3*(-2/3/b*c*d-(a*d-2*b*c)/b^2*d)*(d*x^3+c)^(
1/2)/d+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)))+a^2/b^2*(1/3*(a*d-b*c)/b^2*(d*x^3+c)^(1/2)/(b*x^3+a)+2/3*
d*(d*x^3+c)^(1/2)/b^2+1/2*I/d/b^2*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))*Elliptic
Pi(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)*_a
lpha)/(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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(d*x^3+c)^(3/2)/(b*x^3+a)^2,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 = 7.75, size = 331, normalized size = 1.75 \begin {gather*} \frac {\sqrt {d\,x^3+c}\,\left (\frac {2\,{\left (a\,d-b\,c\right )}^2}{b^4}+\frac {2\,c\,\left (\frac {2\,d\,\left (a\,d-2\,b\,c\right )}{b^3}+\frac {2\,a\,d^2}{b^3}+\frac {8\,c\,d}{5\,b^2}\right )}{3\,d}+\frac {2\,a\,\left (\frac {d\,\left (a\,d-2\,b\,c\right )}{b^3}+\frac {a\,d^2}{b^3}\right )}{b}\right )}{3\,d}+\frac {2\,d\,x^6\,\sqrt {d\,x^3+c}}{15\,b^2}-\frac {x^3\,\sqrt {d\,x^3+c}\,\left (\frac {2\,d\,\left (a\,d-2\,b\,c\right )}{b^3}+\frac {2\,a\,d^2}{b^3}+\frac {8\,c\,d}{5\,b^2}\right )}{9\,d}-\frac {a^2\,\left (\frac {2\,b\,c^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}+\frac {a\,\left (\frac {2\,a\,d^2}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}-\frac {4\,b\,c\,d}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}\right )}{b}\right )\,\sqrt {d\,x^3+c}}{b^2\,\left (b\,x^3+a\right )}+\frac {a\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\sqrt {a\,d-b\,c}\,\left (7\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((c + d*x^3)^(1/2)*((2*(a*d - b*c)^2)/b^4 + (2*c*((2*d*(a*d - 2*b*c))/b^3 + (2*a*d^2)/b^3 + (8*c*d)/(5*b^2)))/
(3*d) + (2*a*((d*(a*d - 2*b*c))/b^3 + (a*d^2)/b^3))/b))/(3*d) + (2*d*x^6*(c + d*x^3)^(1/2))/(15*b^2) - (x^3*(c
+ d*x^3)^(1/2)*((2*d*(a*d - 2*b*c))/b^3 + (2*a*d^2)/b^3 + (8*c*d)/(5*b^2)))/(9*d) + (a*log((a*d - 2*b*c + b^(
1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^(1/2)*2i - b*d*x^3)/(a + b*x^3))*(a*d - b*c)^(1/2)*(7*a*d - 4*b*c)*1i)/(6*b
^(9/2)) - (a^2*((2*b*c^2)/(3*(2*b^2*c - 2*a*b*d)) + (a*((2*a*d^2)/(3*(2*b^2*c - 2*a*b*d)) - (4*b*c*d)/(3*(2*b^
2*c - 2*a*b*d))))/b)*(c + d*x^3)^(1/2))/(b^2*(a + b*x^3))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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