\(\int \frac {1}{x^4 (8 c-d x^3) (c+d x^3)^{3/2}} \, dx\) [504]

Optimal result

Integrand size = 27, antiderivative size = 100 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {25 d}{216 c^3 \sqrt {c+d x^3}}-\frac {1}{24 c^2 x^3 \sqrt {c+d x^3}}+\frac {d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2592 c^{7/2}}+\frac {11 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{96 c^{7/2}} \] Output:

-25/216*d/c^3/(d*x^3+c)^(1/2)-1/24/c^2/x^3/(d*x^3+c)^(1/2)+1/2592*d*arctan 
h(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(7/2)+11/96*d*arctanh((d*x^3+c)^(1/2)/c^( 
1/2))/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-\frac {12 \sqrt {c} \left (9 c+25 d x^3\right )}{x^3 \sqrt {c+d x^3}}+d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )+297 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2592 c^{7/2}} \] Input:

Integrate[1/(x^4*(8*c - d*x^3)*(c + d*x^3)^(3/2)),x]
 

Output:

((-12*Sqrt[c]*(9*c + 25*d*x^3))/(x^3*Sqrt[c + d*x^3]) + d*ArcTanh[Sqrt[c + 
 d*x^3]/(3*Sqrt[c])] + 297*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2592*c^(7/ 
2))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.18, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 114, 27, 169, 27, 174, 73, 219, 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.

\(\displaystyle \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 114

\(\displaystyle \frac {1}{3} \left (-\frac {\int \frac {d \left (22 c-3 d x^3\right )}{2 x^3 \left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx^3}{8 c^2}-\frac {1}{8 c^2 x^3 \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-\frac {d \int \frac {22 c-3 d x^3}{x^3 \left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx^3}{16 c^2}-\frac {1}{8 c^2 x^3 \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 169

\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {2 \int \frac {c d \left (198 c-25 d x^3\right )}{2 x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c^2 d}+\frac {50}{9 c \sqrt {c+d x^3}}\right )}{16 c^2}-\frac {1}{8 c^2 x^3 \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {\int \frac {198 c-25 d x^3}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c}+\frac {50}{9 c \sqrt {c+d x^3}}\right )}{16 c^2}-\frac {1}{8 c^2 x^3 \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {\frac {99}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3-\frac {1}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{9 c}+\frac {50}{9 c \sqrt {c+d x^3}}\right )}{16 c^2}-\frac {1}{8 c^2 x^3 \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {\frac {99 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}-\frac {1}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}}{9 c}+\frac {50}{9 c \sqrt {c+d x^3}}\right )}{16 c^2}-\frac {1}{8 c^2 x^3 \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {\frac {99 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}}{9 c}+\frac {50}{9 c \sqrt {c+d x^3}}\right )}{16 c^2}-\frac {1}{8 c^2 x^3 \sqrt {c+d x^3}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (-\frac {d \left (\frac {-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}-\frac {99 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}}{9 c}+\frac {50}{9 c \sqrt {c+d x^3}}\right )}{16 c^2}-\frac {1}{8 c^2 x^3 \sqrt {c+d x^3}}\right )\)

Input:

Int[1/(x^4*(8*c - d*x^3)*(c + d*x^3)^(3/2)),x]
 

Output:

(-1/8*1/(c^2*x^3*Sqrt[c + d*x^3]) - (d*(50/(9*c*Sqrt[c + d*x^3]) + (-1/6*A 
rcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/Sqrt[c] - (99*ArcTanh[Sqrt[c + d*x^3]/ 
Sqrt[c]])/(2*Sqrt[c]))/(9*c)))/(16*c^2))/3
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 114

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[1/((m + 1)*(b*c - a*d)*(b*e 
 - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) 
 - b*(d*e*(m + n + 2) + c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] && (IntegerQ[n] || 
 IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])
 

rule 169

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n, 2*p]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
 

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 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 = 1.24 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.76

method	result	size
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{24 c^{3} x^{3}}-\frac {d \left (-\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{6 \sqrt {c}}+\frac {32}{27 \sqrt {d \,x^{3}+c}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{162 \sqrt {c}}\right )}{16 c^{3}}\)	\(76\)
pseudoelliptic	\(-\frac {d \left (\frac {\frac {\sqrt {d \,x^{3}+c}}{d \,x^{3}}-\frac {11 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{4 \sqrt {c}}}{8 c^{3}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{864 c^{\frac {7}{2}}}+\frac {2}{9 c^{3} \sqrt {d \,x^{3}+c}}\right )}{3}\)	\(80\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 c^{2} x^{3}}-\frac {2 d}{3 c^{2} \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{c^{\frac {5}{2}}}}{8 c}+\frac {d \left (\frac {2}{3 c \sqrt {\left (x^{3}+\frac {c}{d}\right ) d}}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{64 c^{2}}+\frac {d \left (-\frac {1}{c \sqrt {d \,x^{3}+c}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )}{864 c^{2}}\)	\(148\)
elliptic	\(\text {Expression too large to display}\)	\(1542\)

Input:

int(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24/c^3*(d*x^3+c)^(1/2)/x^3-1/16/c^3*d*(-11/6*arctanh((d*x^3+c)^(1/2)/c^ 
(1/2))/c^(1/2)+32/27/(d*x^3+c)^(1/2)-1/162*arctanh(1/3*(d*x^3+c)^(1/2)/c^( 
1/2))/c^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.66 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\left [\frac {{\left (d^{2} x^{6} + c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 297 \, {\left (d^{2} x^{6} + c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 24 \, {\left (25 \, c d x^{3} + 9 \, c^{2}\right )} \sqrt {d x^{3} + c}}{5184 \, {\left (c^{4} d x^{6} + c^{5} x^{3}\right )}}, -\frac {{\left (d^{2} x^{6} + c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {3 \, \sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 297 \, {\left (d^{2} x^{6} + c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 12 \, {\left (25 \, c d x^{3} + 9 \, c^{2}\right )} \sqrt {d x^{3} + c}}{2592 \, {\left (c^{4} d x^{6} + c^{5} x^{3}\right )}}\right ] \] Input:

integrate(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="fricas")
 

Output:

[1/5184*((d^2*x^6 + c*d*x^3)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqrt(c 
) + 10*c)/(d*x^3 - 8*c)) + 297*(d^2*x^6 + c*d*x^3)*sqrt(c)*log((d*x^3 + 2* 
sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 24*(25*c*d*x^3 + 9*c^2)*sqrt(d*x^3 + 
 c))/(c^4*d*x^6 + c^5*x^3), -1/2592*((d^2*x^6 + c*d*x^3)*sqrt(-c)*arctan(3 
*sqrt(-c)/sqrt(d*x^3 + c)) + 297*(d^2*x^6 + c*d*x^3)*sqrt(-c)*arctan(sqrt( 
-c)/sqrt(d*x^3 + c)) + 12*(25*c*d*x^3 + 9*c^2)*sqrt(d*x^3 + c))/(c^4*d*x^6 
 + c^5*x^3)]
 

Sympy [F]

\[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=- \int \frac {1}{- 8 c^{2} x^{4} \sqrt {c + d x^{3}} - 7 c d x^{7} \sqrt {c + d x^{3}} + d^{2} x^{10} \sqrt {c + d x^{3}}}\, dx \] Input:

integrate(1/x**4/(-d*x**3+8*c)/(d*x**3+c)**(3/2),x)
 

Output:

-Integral(1/(-8*c**2*x**4*sqrt(c + d*x**3) - 7*c*d*x**7*sqrt(c + d*x**3) + 
 d**2*x**10*sqrt(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { -\frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )} x^{4}} \,d x } \] Input:

integrate(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="maxima")
 

Output:

-integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)*x^4), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=-\frac {11 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{96 \, \sqrt {-c} c^{3}} - \frac {d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{2592 \, \sqrt {-c} c^{3}} - \frac {25 \, {\left (d x^{3} + c\right )} d - 16 \, c d}{216 \, {\left ({\left (d x^{3} + c\right )}^{\frac {3}{2}} - \sqrt {d x^{3} + c} c\right )} c^{3}} \] Input:

integrate(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="giac")
 

Output:

-11/96*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 1/2592*d*arctan 
(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^3) - 1/216*(25*(d*x^3 + c)*d - 
16*c*d)/(((d*x^3 + c)^(3/2) - sqrt(d*x^3 + c)*c)*c^3)
 

Mupad [B] (verification not implemented)

Time = 2.26 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {11\,d\,\mathrm {atanh}\left (\frac {c^3\,\sqrt {d\,x^3+c}}{\sqrt {c^7}}\right )}{96\,\sqrt {c^7}}-\frac {25\,d}{216\,c^3\,\sqrt {d\,x^3+c}}+\frac {d\,\mathrm {atanh}\left (\frac {c^3\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^7}}\right )}{2592\,\sqrt {c^7}}-\frac {1}{24\,c^2\,x^3\,\sqrt {d\,x^3+c}} \] Input:

int(1/(x^4*(c + d*x^3)^(3/2)*(8*c - d*x^3)),x)
 

Output:

(11*d*atanh((c^3*(c + d*x^3)^(1/2))/(c^7)^(1/2)))/(96*(c^7)^(1/2)) - (25*d 
)/(216*c^3*(c + d*x^3)^(1/2)) + (d*atanh((c^3*(c + d*x^3)^(1/2))/(3*(c^7)^ 
(1/2))))/(2592*(c^7)^(1/2)) - 1/(24*c^2*x^3*(c + d*x^3)^(1/2))
 

Reduce [F]

\[ \int \frac {1}{x^4 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}-66 \left (\int \frac {\sqrt {d \,x^{3}+c}}{-d^{3} x^{10}+6 c \,d^{2} x^{7}+15 c^{2} d \,x^{4}+8 c^{3} x}d x \right ) c^{2} d \,x^{3}-66 \left (\int \frac {\sqrt {d \,x^{3}+c}}{-d^{3} x^{10}+6 c \,d^{2} x^{7}+15 c^{2} d \,x^{4}+8 c^{3} x}d x \right ) c \,d^{2} x^{6}+9 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c \,d^{2} x^{3}+9 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{2}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) d^{3} x^{6}}{48 c^{2} x^{3} \left (d \,x^{3}+c \right )} \] Input:

int(1/x^4/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x)
 

Output:

( - 2*sqrt(c + d*x**3) - 66*int(sqrt(c + d*x**3)/(8*c**3*x + 15*c**2*d*x** 
4 + 6*c*d**2*x**7 - d**3*x**10),x)*c**2*d*x**3 - 66*int(sqrt(c + d*x**3)/( 
8*c**3*x + 15*c**2*d*x**4 + 6*c*d**2*x**7 - d**3*x**10),x)*c*d**2*x**6 + 9 
*int((sqrt(c + d*x**3)*x**2)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d* 
*3*x**9),x)*c*d**2*x**3 + 9*int((sqrt(c + d*x**3)*x**2)/(8*c**3 + 15*c**2* 
d*x**3 + 6*c*d**2*x**6 - d**3*x**9),x)*d**3*x**6)/(48*c**2*x**3*(c + d*x** 
3))