\(\int \frac {x^5}{(a+b x^3)^2 \sqrt {c+d x^3}} \, dx\) [654]

Optimal result

Integrand size = 24, antiderivative size = 99 \[ \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {a \sqrt {c+d x^3}}{3 b (b c-a d) \left (a+b x^3\right )}-\frac {(2 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 b^{3/2} (b c-a d)^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01 \[ \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\frac {a \sqrt {b} \sqrt {c+d x^3}}{(b c-a d) \left (a+b x^3\right )}-\frac {(2 b c-a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{(-b c+a d)^{3/2}}}{3 b^{3/2}} \] Input:

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

Output:

((a*Sqrt[b]*Sqrt[c + d*x^3])/((b*c - a*d)*(a + b*x^3)) - ((2*b*c - a*d)*Ar 
cTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b*c) + a*d]])/(-(b*c) + a*d)^(3/2))/ 
(3*b^(3/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {948, 87, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^3}{\left (b x^3+a\right )^2 \sqrt {d x^3+c}}dx^3\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {1}{3} \left (\frac {(2 b c-a d) \int \frac {1}{\left (b x^3+a\right ) \sqrt {d x^3+c}}dx^3}{2 b (b c-a d)}+\frac {a \sqrt {c+d x^3}}{b \left (a+b x^3\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {(2 b c-a d) \int \frac {1}{\frac {b x^6}{d}+a-\frac {b c}{d}}d\sqrt {d x^3+c}}{b d (b c-a d)}+\frac {a \sqrt {c+d x^3}}{b \left (a+b x^3\right ) (b c-a d)}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {a \sqrt {c+d x^3}}{b \left (a+b x^3\right ) (b c-a d)}-\frac {(2 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{b^{3/2} (b c-a d)^{3/2}}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(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] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 221

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 948

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. 
), x_Symbol] :> Simp[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]]
 

Maple [A] (verified)

Time = 1.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {-\frac {a \sqrt {d \,x^{3}+c}}{b \,x^{3}+a}+\frac {\left (a d -2 b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}}{3 \left (a d -b c \right ) b}\)	\(83\)
default	\(\frac {2 \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b \sqrt {\left (a d -b c \right ) b}}-\frac {a \left (\frac {\sqrt {d \,x^{3}+c}}{b \,x^{3}+a}+\frac {d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 b \left (a d -b c \right )}\)	\(117\)
elliptic	\(-\frac {a \sqrt {d \,x^{3}+c}}{3 \left (a d -b c \right ) b \left (b \,x^{3}+a \right )}+\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\left (-a d +2 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 ) \operatorname {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}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+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 )^{2} \sqrt {d \,x^{3}+c}}\right )}{6 b \,d^{2}}\)	\(473\)

Input:

int(x^5/(b*x^3+a)^2/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/3/(a*d-b*c)/b*(-a*(d*x^3+c)^(1/2)/(b*x^3+a)+(a*d-2*b*c)/((a*d-b*c)*b)^(1 
/2)*arctan(b*(d*x^3+c)^(1/2)/((a*d-b*c)*b)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 348, normalized size of antiderivative = 3.52 \[ \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\left [\frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{3} + 2 \, a b c - a^{2} d\right )} \sqrt {b^{2} c - a b d} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) + 2 \, {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{3} + c}}{6 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{3}\right )}}, \frac {{\left ({\left (2 \, b^{2} c - a b d\right )} x^{3} + 2 \, a b c - a^{2} d\right )} \sqrt {-b^{2} c + a b d} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) + {\left (a b^{2} c - a^{2} b d\right )} \sqrt {d x^{3} + c}}{3 \, {\left (a b^{4} c^{2} - 2 \, a^{2} b^{3} c d + a^{3} b^{2} d^{2} + {\left (b^{5} c^{2} - 2 \, a b^{4} c d + a^{2} b^{3} d^{2}\right )} x^{3}\right )}}\right ] \] Input:

integrate(x^5/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/6*(((2*b^2*c - a*b*d)*x^3 + 2*a*b*c - a^2*d)*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*(a*b^2*c - a^2*b*d)*sqrt(d*x^3 + c))/(a*b^4*c^2 - 2*a^2*b^3*c*d + a^3 
*b^2*d^2 + (b^5*c^2 - 2*a*b^4*c*d + a^2*b^3*d^2)*x^3), 1/3*(((2*b^2*c - a* 
b*d)*x^3 + 2*a*b*c - a^2*d)*sqrt(-b^2*c + a*b*d)*arctan(sqrt(d*x^3 + c)*sq 
rt(-b^2*c + a*b*d)/(b*d*x^3 + b*c)) + (a*b^2*c - a^2*b*d)*sqrt(d*x^3 + c)) 
/(a*b^4*c^2 - 2*a^2*b^3*c*d + a^3*b^2*d^2 + (b^5*c^2 - 2*a*b^4*c*d + a^2*b 
^3*d^2)*x^3)]
 

Sympy [F]

\[ \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {x^{5}}{\left (a + b x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \] Input:

integrate(x**5/(b*x**3+a)**2/(d*x**3+c)**(1/2),x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^5/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.17 \[ \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\frac {\sqrt {d x^{3} + c} a d^{2}}{{\left (b^{2} c - a b d\right )} {\left ({\left (d x^{3} + c\right )} b - b c + a d\right )}} + \frac {{\left (2 \, b c d - a d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{{\left (b^{2} c - a b d\right )} \sqrt {-b^{2} c + a b d}}}{3 \, d} \] Input:

integrate(x^5/(b*x^3+a)^2/(d*x^3+c)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 5.73 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.12 \[ \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {2\,a\,\sqrt {d\,x^3+c}}{3\,\left (b\,x^3+a\right )\,\left (2\,b^2\,c-2\,a\,b\,d\right )}+\frac {\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (a\,d-2\,b\,c\right )\,1{}\mathrm {i}}{6\,b^{3/2}\,{\left (a\,d-b\,c\right )}^{3/2}} \] Input:

int(x^5/((a + b*x^3)^2*(c + d*x^3)^(1/2)),x)
 

Output:

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

Reduce [F]

\[ \int \frac {x^5}{\left (a+b x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b^{2} d \,x^{9}+2 a b d \,x^{6}+b^{2} c \,x^{6}+a^{2} d \,x^{3}+2 a b c \,x^{3}+a^{2} c}d x \] Input:

int(x^5/(b*x^3+a)^2/(d*x^3+c)^(1/2),x)
 

Output:

int((sqrt(c + d*x**3)*x**5)/(a**2*c + a**2*d*x**3 + 2*a*b*c*x**3 + 2*a*b*d 
*x**6 + b**2*c*x**6 + b**2*d*x**9),x)