3.5.60 \(\int \frac {x^8 \sqrt {c+d x^3}}{(a+b x^3)^2} \, dx\) [460]

Optimal. Leaf size=161 \[ -\frac {a (4 b c-5 a d) \sqrt {c+d x^3}}{3 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}+\frac {a (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2} \sqrt {b c-a d}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {457, 91, 81, 52, 65, 214} \begin {gather*} -\frac {a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 \left (a+b x^3\right ) (b c-a d)}+\frac {a (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2} \sqrt {b c-a d}}-\frac {a \sqrt {c+d x^3} (4 b c-5 a d)}{3 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(a*(4*b*c - 5*a*d)*Sqrt[c + d*x^3])/(b^3*(b*c - a*d)) + (2*(c + d*x^3)^(3/2))/(9*b^2*d) - (a^2*(c + d*x^3
)^(3/2))/(3*b^2*(b*c - a*d)*(a + b*x^3)) + (a*(4*b*c - 5*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d
]])/(3*b^(7/2)*Sqrt[b*c - a*d])

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 81

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 91

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 214

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

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}}{\left (a+b x^3\right )^2} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {x^2 \sqrt {c+d x}}{(a+b x)^2} \, dx,x,x^3\right )\\ &=-\frac {a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}+\frac {\text {Subst}\left (\int \frac {\sqrt {c+d x} \left (-\frac {1}{2} a (2 b c-3 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 )^{3/2}}{9 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac {(a (4 b c-5 a d)) \text {Subst}\left (\int \frac {\sqrt {c+d x}}{a+b x} \, dx,x,x^3\right )}{6 b^2 (b c-a d)}\\ &=-\frac {a (4 b c-5 a d) \sqrt {c+d x^3}}{3 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac {(a (4 b c-5 a d)) \text {Subst}\left (\int \frac {1}{(a+b x) \sqrt {c+d x}} \, dx,x,x^3\right )}{6 b^3}\\ &=-\frac {a (4 b c-5 a d) \sqrt {c+d x^3}}{3 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}-\frac {(a (4 b c-5 a d)) \text {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^3 d}\\ &=-\frac {a (4 b c-5 a d) \sqrt {c+d x^3}}{3 b^3 (b c-a d)}+\frac {2 \left (c+d x^3\right )^{3/2}}{9 b^2 d}-\frac {a^2 \left (c+d x^3\right )^{3/2}}{3 b^2 (b c-a d) \left (a+b x^3\right )}+\frac {a (4 b c-5 a d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2} \sqrt {b c-a d}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 126, normalized size = 0.78 \begin {gather*} \frac {\sqrt {c+d x^3} \left (-15 a^2 d+2 a b \left (c-5 d x^3\right )+2 b^2 x^3 \left (c+d x^3\right )\right )}{9 b^3 d \left (a+b x^3\right )}+\frac {a (-4 b c+5 a d) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{7/2} \sqrt {-b c+a d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.44, size = 917, normalized size = 5.70

method	result	size
elliptic	\(-\frac {a^{2} \sqrt {d \,x^{3}+c}}{3 b^{3} \left (b \,x^{3}+a \right )}+\frac {2 x^{3} \sqrt {d \,x^{3}+c}}{9 b^{2}}+\frac {2 \left (-\frac {2 a d -b c}{b^{3}}-\frac {2 c}{3 b^{2}}\right ) \sqrt {d \,x^{3}+c}}{3 d}-\frac {i a \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (5 a d -4 b c \right ) \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \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 )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \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 )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\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 {b \left (2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha ^{2} d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \underline {\hspace {1.25 ex}}\alpha +i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d \right )}{2 d \left (a d -b c \right )}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \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 )}}\right )}{2 \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}\right )}{6 b^{3} d^{2}}\)	\(518\)
default	\(\text {Expression too large to display}\)	\(917\)
risch	\(\text {Expression too large to display}\)	\(942\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(d*x^3+c)^(1/2)/(b*x^3+a)^2,x,method=_RETURNVERBOSE)

[Out]

2/9*(d*x^3+c)^(3/2)/b^2/d-2*a/b^2*(2/3*(d*x^3+c)^(1/2)/b+1/3*I/b/d^2*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),1/2*b/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)/(a*d-b*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*b+a)))+a^2/b^2*(-1/3*(d*x^3+c)^(1/2)/b/(b*x^3+a)-
1/6*I/b/d*2^(1/2)*sum(1/(a*d-b*c)*(-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),1/2*b/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)/(a*d-b*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)),_alph
a=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)^(1/2)/(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 detail

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Fricas [A]
time = 2.32, size = 469, normalized size = 2.91 \begin {gather*} \left [-\frac {3 \, {\left (4 \, a^{2} b c d - 5 \, a^{3} d^{2} + {\left (4 \, a b^{2} c d - 5 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) - 2 \, {\left (2 \, {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + 2 \, a b^{3} c^{2} - 17 \, a^{2} b^{2} c d + 15 \, a^{3} b d^{2} + 2 \, {\left (b^{4} c^{2} - 6 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{18 \, {\left (a b^{5} c d - a^{2} b^{4} d^{2} + {\left (b^{6} c d - a b^{5} d^{2}\right )} x^{3}\right )}}, -\frac {3 \, {\left (4 \, a^{2} b c d - 5 \, a^{3} d^{2} + {\left (4 \, a b^{2} c d - 5 \, a^{2} b d^{2}\right )} x^{3}\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - {\left (2 \, {\left (b^{4} c d - a b^{3} d^{2}\right )} x^{6} + 2 \, a b^{3} c^{2} - 17 \, a^{2} b^{2} c d + 15 \, a^{3} b d^{2} + 2 \, {\left (b^{4} c^{2} - 6 \, a b^{3} c d + 5 \, a^{2} b^{2} d^{2}\right )} x^{3}\right )} \sqrt {d x^{3} + c}}{9 \, {\left (a b^{5} c d - a^{2} b^{4} d^{2} + {\left (b^{6} c d - a b^{5} d^{2}\right )} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/18*(3*(4*a^2*b*c*d - 5*a^3*d^2 + (4*a*b^2*c*d - 5*a^2*b*d^2)*x^3)*sqrt(b^2*c - a*b*d)*log((b*d*x^3 + 2*b*c
- a*d - 2*sqrt(d*x^3 + c)*sqrt(b^2*c - a*b*d))/(b*x^3 + a)) - 2*(2*(b^4*c*d - a*b^3*d^2)*x^6 + 2*a*b^3*c^2 -
17*a^2*b^2*c*d + 15*a^3*b*d^2 + 2*(b^4*c^2 - 6*a*b^3*c*d + 5*a^2*b^2*d^2)*x^3)*sqrt(d*x^3 + c))/(a*b^5*c*d - a
^2*b^4*d^2 + (b^6*c*d - a*b^5*d^2)*x^3), -1/9*(3*(4*a^2*b*c*d - 5*a^3*d^2 + (4*a*b^2*c*d - 5*a^2*b*d^2)*x^3)*s
qrt(-b^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sqrt(-b^2*c + a*b*d)/(b*d*x^3 + b*c)) - (2*(b^4*c*d - a*b^3*d^2)*x^
6 + 2*a*b^3*c^2 - 17*a^2*b^2*c*d + 15*a^3*b*d^2 + 2*(b^4*c^2 - 6*a*b^3*c*d + 5*a^2*b^2*d^2)*x^3)*sqrt(d*x^3 +
c))/(a*b^5*c*d - a^2*b^4*d^2 + (b^6*c*d - a*b^5*d^2)*x^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{8} \sqrt {c + d x^{3}}}{\left (a + b x^{3}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.21, size = 136, normalized size = 0.84 \begin {gather*} -\frac {\sqrt {d x^{3} + c} a^{2} d}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{3}} - \frac {{\left (4 \, a b c - 5 \, a^{2} d\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^{3}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{4} d^{2} - 6 \, \sqrt {d x^{3} + c} a b^{3} d^{3}\right )}}{9 \, b^{6} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(d*x^3 + c)*a^2*d/(((d*x^3 + c)*b - b*c + a*d)*b^3) - 1/3*(4*a*b*c - 5*a^2*d)*arctan(sqrt(d*x^3 + c)*
b/sqrt(-b^2*c + a*b*d))/(sqrt(-b^2*c + a*b*d)*b^3) + 2/9*((d*x^3 + c)^(3/2)*b^4*d^2 - 6*sqrt(d*x^3 + c)*a*b^3*
d^3)/(b^6*d^3)

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Mupad [B]
time = 6.84, size = 202, normalized size = 1.25 \begin {gather*} \frac {2\,x^3\,\sqrt {d\,x^3+c}}{9\,b^2}-\frac {\sqrt {d\,x^3+c}\,\left (\frac {4\,c}{3\,b^2}-\frac {2\,b^2\,c-2\,a\,b\,d}{b^4}+\frac {2\,a\,d}{b^3}\right )}{3\,d}+\frac {a^2\,\left (\frac {2\,a\,d}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}-\frac {2\,b\,c}{3\,\left (2\,b^2\,c-2\,a\,b\,d\right )}\right )\,\sqrt {d\,x^3+c}}{b^2\,\left (b\,x^3+a\right )}+\frac {a\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (5\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{7/2}\,\sqrt {a\,d-b\,c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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