\(\int \frac {x^5 (c+d x^3)^{3/2}}{8 c-d x^3} \, dx\) [472]

Optimal result

Integrand size = 27, antiderivative size = 88 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {48 c^2 \sqrt {c+d x^3}}{d^2}-\frac {16 c \left (c+d x^3\right )^{3/2}}{9 d^2}-\frac {2 \left (c+d x^3\right )^{5/2}}{15 d^2}+\frac {144 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^2} \] Output:

-48*c^2*(d*x^3+c)^(1/2)/d^2-16/9*c*(d*x^3+c)^(3/2)/d^2-2/15*(d*x^3+c)^(5/2 
)/d^2+144*c^(5/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.81 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {2 \sqrt {c+d x^3} \left (1123 c^2+46 c d x^3+3 d^2 x^6\right )}{45 d^2}+\frac {144 c^{5/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d^2} \] Input:

Integrate[(x^5*(c + d*x^3)^(3/2))/(8*c - d*x^3),x]
 

Output:

(-2*Sqrt[c + d*x^3]*(1123*c^2 + 46*c*d*x^3 + 3*d^2*x^6))/(45*d^2) + (144*c 
^(5/2)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/d^2
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {948, 90, 60, 60, 73, 219}

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.

\(\displaystyle \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx\)

\(\Big \downarrow \) 948

\(\displaystyle \frac {1}{3} \int \frac {x^3 \left (d x^3+c\right )^{3/2}}{8 c-d x^3}dx^3\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {1}{3} \left (\frac {8 c \int \frac {\left (d x^3+c\right )^{3/2}}{8 c-d x^3}dx^3}{d}-\frac {2 \left (c+d x^3\right )^{5/2}}{5 d^2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{3} \left (\frac {8 c \left (9 c \int \frac {\sqrt {d x^3+c}}{8 c-d x^3}dx^3-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )}{d}-\frac {2 \left (c+d x^3\right )^{5/2}}{5 d^2}\right )\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{3} \left (\frac {8 c \left (9 c \left (9 c \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )}{d}-\frac {2 \left (c+d x^3\right )^{5/2}}{5 d^2}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {8 c \left (9 c \left (\frac {18 c \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )}{d}-\frac {2 \left (c+d x^3\right )^{5/2}}{5 d^2}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {8 c \left (9 c \left (\frac {6 \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{d}-\frac {2 \sqrt {c+d x^3}}{d}\right )-\frac {2 \left (c+d x^3\right )^{3/2}}{3 d}\right )}{d}-\frac {2 \left (c+d x^3\right )^{5/2}}{5 d^2}\right )\)

Input:

Int[(x^5*(c + d*x^3)^(3/2))/(8*c - d*x^3),x]
 

Output:

((-2*(c + d*x^3)^(5/2))/(5*d^2) + (8*c*((-2*(c + d*x^3)^(3/2))/(3*d) + 9*c 
*((-2*Sqrt[c + d*x^3])/d + (6*Sqrt[c]*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])] 
)/d)))/d)/3
 

Defintions of rubi rules used

rule 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

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 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 = 1.16 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.66

method	result	size
pseudoelliptic	\(\frac {6480 c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )+\left (-6 d^{2} x^{6}-92 c d \,x^{3}-2246 c^{2}\right ) \sqrt {d \,x^{3}+c}}{45 d^{2}}\)	\(58\)
risch	\(-\frac {2 \left (3 d^{2} x^{6}+46 c d \,x^{3}+1123 c^{2}\right ) \sqrt {d \,x^{3}+c}}{45 d^{2}}+\frac {144 c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{d^{2}}\)	\(60\)
default	\(-\frac {2 \left (d \,x^{3}+c \right )^{\frac {5}{2}}}{15 d^{2}}+\frac {16 c \left (81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}\right )}{9 d^{2}}\)	\(63\)
elliptic	\(-\frac {2 x^{6} \sqrt {d \,x^{3}+c}}{15}-\frac {92 c \,x^{3} \sqrt {d \,x^{3}+c}}{45 d}-\frac {2246 c^{2} \sqrt {d \,x^{3}+c}}{45 d^{2}}-\frac {24 i c^{2} \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}-8 c \right )}{\sum }\frac {\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 {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}{18 d c}, \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 \sqrt {d \,x^{3}+c}}\right )}{d^{4}}\)	\(463\)

Input:

int(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c),x,method=_RETURNVERBOSE)
 

Output:

1/45*(6480*c^(5/2)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))+(-6*d^2*x^6-92*c*d 
*x^3-2246*c^2)*(d*x^3+c)^(1/2))/d^2
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.64 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\left [\frac {2 \, {\left (1620 \, c^{\frac {5}{2}} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) - {\left (3 \, d^{2} x^{6} + 46 \, c d x^{3} + 1123 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, d^{2}}, -\frac {2 \, {\left (3240 \, \sqrt {-c} c^{2} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + {\left (3 \, d^{2} x^{6} + 46 \, c d x^{3} + 1123 \, c^{2}\right )} \sqrt {d x^{3} + c}\right )}}{45 \, d^{2}}\right ] \] Input:

integrate(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c),x, algorithm="fricas")
 

Output:

[2/45*(1620*c^(5/2)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c) + 10*c)/(d*x^3 
- 8*c)) - (3*d^2*x^6 + 46*c*d*x^3 + 1123*c^2)*sqrt(d*x^3 + c))/d^2, -2/45* 
(3240*sqrt(-c)*c^2*arctan(3*sqrt(-c)/sqrt(d*x^3 + c)) + (3*d^2*x^6 + 46*c* 
d*x^3 + 1123*c^2)*sqrt(d*x^3 + c))/d^2]
 

Sympy [A] (verification not implemented)

Time = 15.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\begin {cases} \frac {2 \left (- \frac {72 c^{3} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{3 \sqrt {- c}} \right )}}{d \sqrt {- c}} - \frac {24 c^{2} \sqrt {c + d x^{3}}}{d} - \frac {8 c \left (c + d x^{3}\right )^{\frac {3}{2}}}{9 d} - \frac {\left (c + d x^{3}\right )^{\frac {5}{2}}}{15 d}\right )}{d} & \text {for}\: d \neq 0 \\\frac {\sqrt {c} x^{6}}{48} & \text {otherwise} \end {cases} \] Input:

integrate(x**5*(d*x**3+c)**(3/2)/(-d*x**3+8*c),x)
 

Output:

Piecewise((2*(-72*c**3*atan(sqrt(c + d*x**3)/(3*sqrt(-c)))/(d*sqrt(-c)) - 
24*c**2*sqrt(c + d*x**3)/d - 8*c*(c + d*x**3)**(3/2)/(9*d) - (c + d*x**3)* 
*(5/2)/(15*d))/d, Ne(d, 0)), (sqrt(c)*x**6/48, True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.93 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {2 \, {\left (1620 \, c^{\frac {5}{2}} \log \left (\frac {\sqrt {d x^{3} + c} - 3 \, \sqrt {c}}{\sqrt {d x^{3} + c} + 3 \, \sqrt {c}}\right ) + 3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} + 40 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c + 1080 \, \sqrt {d x^{3} + c} c^{2}\right )}}{45 \, d^{2}} \] Input:

integrate(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c),x, algorithm="maxima")
 

Output:

-2/45*(1620*c^(5/2)*log((sqrt(d*x^3 + c) - 3*sqrt(c))/(sqrt(d*x^3 + c) + 3 
*sqrt(c))) + 3*(d*x^3 + c)^(5/2) + 40*(d*x^3 + c)^(3/2)*c + 1080*sqrt(d*x^ 
3 + c)*c^2)/d^2
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.94 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=-\frac {144 \, c^{3} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{\sqrt {-c} d^{2}} - \frac {2 \, {\left (3 \, {\left (d x^{3} + c\right )}^{\frac {5}{2}} d^{8} + 40 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} c d^{8} + 1080 \, \sqrt {d x^{3} + c} c^{2} d^{8}\right )}}{45 \, d^{10}} \] Input:

integrate(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c),x, algorithm="giac")
 

Output:

-144*c^3*arctan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*d^2) - 2/45*(3*(d* 
x^3 + c)^(5/2)*d^8 + 40*(d*x^3 + c)^(3/2)*c*d^8 + 1080*sqrt(d*x^3 + c)*c^2 
*d^8)/d^10
 

Mupad [B] (verification not implemented)

Time = 1.73 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.08 \[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\frac {72\,c^{5/2}\,\ln \left (\frac {10\,c+d\,x^3+6\,\sqrt {c}\,\sqrt {d\,x^3+c}}{8\,c-d\,x^3}\right )}{d^2}-\frac {2\,x^6\,\sqrt {d\,x^3+c}}{15}-\frac {2246\,c^2\,\sqrt {d\,x^3+c}}{45\,d^2}-\frac {92\,c\,x^3\,\sqrt {d\,x^3+c}}{45\,d} \] Input:

int((x^5*(c + d*x^3)^(3/2))/(8*c - d*x^3),x)
 

Output:

(72*c^(5/2)*log((10*c + d*x^3 + 6*c^(1/2)*(c + d*x^3)^(1/2))/(8*c - d*x^3) 
))/d^2 - (2*x^6*(c + d*x^3)^(1/2))/15 - (2246*c^2*(c + d*x^3)^(1/2))/(45*d 
^2) - (92*c*x^3*(c + d*x^3)^(1/2))/(45*d)
 

Reduce [F]

\[ \int \frac {x^5 \left (c+d x^3\right )^{3/2}}{8 c-d x^3} \, dx=\frac {184 \sqrt {d \,x^{3}+c}\, c^{2}-92 \sqrt {d \,x^{3}+c}\, c d \,x^{3}-6 \sqrt {d \,x^{3}+c}\, d^{2} x^{6}+3645 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{-d^{2} x^{6}+7 c d \,x^{3}+8 c^{2}}d x \right ) c^{2} d^{2}}{45 d^{2}} \] Input:

int(x^5*(d*x^3+c)^(3/2)/(-d*x^3+8*c),x)
 

Output:

(184*sqrt(c + d*x**3)*c**2 - 92*sqrt(c + d*x**3)*c*d*x**3 - 6*sqrt(c + d*x 
**3)*d**2*x**6 + 3645*int((sqrt(c + d*x**3)*x**5)/(8*c**2 + 7*c*d*x**3 - d 
**2*x**6),x)*c**2*d**2)/(45*d**2)