3.5.71 \(\int \frac {x^5 (c+d x^3)^{3/2}}{(a+b x^3)^2} \, dx\) [471]

Optimal. Leaf size=163 \[ \frac {(2 b c-5 a d) \sqrt {c+d x^3}}{3 b^3}+\frac {(2 b c-5 a d) \left (c+d x^3\right )^{3/2}}{9 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{5/2}}{3 b (b c-a d) \left (a+b x^3\right )}-\frac {(2 b c-5 a d) \sqrt {b c-a d} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{7/2}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {457, 79, 52, 65, 214} \begin {gather*} -\frac {(2 b c-5 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^{7/2}}+\frac {\sqrt {c+d x^3} (2 b c-5 a d)}{3 b^3}+\frac {\left (c+d x^3\right )^{3/2} (2 b c-5 a d)}{9 b^2 (b c-a d)}+\frac {a \left (c+d x^3\right )^{5/2}}{3 b \left (a+b x^3\right ) (b c-a d)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^5*(c + d*x^3)^(3/2))/(a + b*x^3)^2,x]

[Out]

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

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 79

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

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

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Mathematica [A]
time = 0.25, size = 125, normalized size = 0.77 \begin {gather*} \frac {\sqrt {c+d x^3} \left (-15 a^2 d+a b \left (11 c-10 d x^3\right )+2 b^2 x^3 \left (4 c+d x^3\right )\right )}{9 b^3 \left (a+b x^3\right )}-\frac {(2 b c-5 a d) \sqrt {-b c+a d} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{3 b^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^5*(c + d*x^3)^(3/2))/(a + b*x^3)^2,x]

[Out]

(Sqrt[c + d*x^3]*(-15*a^2*d + a*b*(11*c - 10*d*x^3) + 2*b^2*x^3*(4*c + d*x^3)))/(9*b^3*(a + b*x^3)) - ((2*b*c
- 5*a*d)*Sqrt[-(b*c) + a*d]*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/(3*b^(7/2))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.43, size = 983, normalized size = 6.03

method	result	size
elliptic	\(-\frac {a \left (a d -b c \right ) \sqrt {d \,x^{3}+c}}{3 b^{3} \left (b \,x^{3}+a \right )}+\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9 b^{2}}+\frac {2 \left (-\frac {2 d \left (a d -b c \right )}{b^{3}}-\frac {2 c d}{3 b^{2}}\right ) \sqrt {d \,x^{3}+c}}{3 d}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-5 a^{2} d^{2}+7 a b c d -2 b^{2} c^{2}\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}}\)	\(539\)
risch	\(\text {Expression too large to display}\)	\(964\)
default	\(\text {Expression too large to display}\)	\(983\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(2/9*d/b*x^3*(d*x^3+c)^(1/2)+2/3*(-d*(a*d-2*b*c)/b^2-2/3*c*d/b)/d*(d*x^3+c)^(1/2)+1/3*I/b^2/d^2*2^(1/2)*su
m((-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*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))*Ellipti
cPi(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)*_al
pha-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/b*(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*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
)))

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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^5*(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 detail

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Fricas [A]
time = 2.11, size = 314, normalized size = 1.93 \begin {gather*} \left [-\frac {3 \, {\left ({\left (2 \, b^{2} c - 5 \, a b d\right )} x^{3} + 2 \, a b c - 5 \, a^{2} d\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 (2 \, b^{2} d x^{6} + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{3} + 11 \, a b c - 15 \, a^{2} d\right )} \sqrt {d x^{3} + c}}{18 \, {\left (b^{4} x^{3} + a b^{3}\right )}}, -\frac {3 \, {\left ({\left (2 \, b^{2} c - 5 \, a b d\right )} x^{3} + 2 \, a b c - 5 \, a^{2} d\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 (2 \, b^{2} d x^{6} + 2 \, {\left (4 \, b^{2} c - 5 \, a b d\right )} x^{3} + 11 \, a b c - 15 \, a^{2} d\right )} \sqrt {d x^{3} + c}}{9 \, {\left (b^{4} x^{3} + a b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

[-1/18*(3*((2*b^2*c - 5*a*b*d)*x^3 + 2*a*b*c - 5*a^2*d)*sqrt((b*c - a*d)/b)*log((b*d*x^3 + 2*b*c - a*d + 2*sqr
t(d*x^3 + c)*b*sqrt((b*c - a*d)/b))/(b*x^3 + a)) - 2*(2*b^2*d*x^6 + 2*(4*b^2*c - 5*a*b*d)*x^3 + 11*a*b*c - 15*
a^2*d)*sqrt(d*x^3 + c))/(b^4*x^3 + a*b^3), -1/9*(3*((2*b^2*c - 5*a*b*d)*x^3 + 2*a*b*c - 5*a^2*d)*sqrt(-(b*c -
a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)) - (2*b^2*d*x^6 + 2*(4*b^2*c - 5*a*b*d)*x^3
+ 11*a*b*c - 15*a^2*d)*sqrt(d*x^3 + c))/(b^4*x^3 + a*b^3)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{5} \left (c + d x^{3}\right )^{\frac {3}{2}}}{\left (a + b x^{3}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(d*x**3+c)**(3/2)/(b*x**3+a)**2,x)

[Out]

Integral(x**5*(c + d*x**3)**(3/2)/(a + b*x**3)**2, x)

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Giac [A]
time = 1.20, size = 173, normalized size = 1.06 \begin {gather*} \frac {{\left (2 \, b^{2} c^{2} - 7 \, a b c d + 5 \, a^{2} 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^{3}} + \frac {\sqrt {d x^{3} + c} a b c d - \sqrt {d x^{3} + c} a^{2} d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )} b^{3}} + \frac {2 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} b^{4} + 3 \, \sqrt {d x^{3} + c} b^{4} c - 6 \, \sqrt {d x^{3} + c} a b^{3} d\right )}}{9 \, b^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 7.38, size = 229, normalized size = 1.40 \begin {gather*} \frac {2\,d\,x^3\,\sqrt {d\,x^3+c}}{9\,b^2}-\frac {\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 {4\,c\,d}{3\,b^2}\right )}{3\,d}+\frac {a\,\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\,\left (b\,x^3+a\right )}+\frac {\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 )\,\sqrt {a\,d-b\,c}\,\left (5\,a\,d-2\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^5*(c + d*x^3)^(3/2))/(a + b*x^3)^2,x)

[Out]

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

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