3.3.88 \(\int \frac {\sqrt {c+d x^3}}{x^7 (8 c-d x^3)} \, dx\) [288]

3.3.88.1 Optimal result

Integrand size = 27, antiderivative size = 107 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{48 c x^6}-\frac {d \sqrt {c+d x^3}}{64 c^2 x^3}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{256 c^{5/2}}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{256 c^{5/2}} \]

output

1/256*d^2*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)+1/256*d^2*arctanh(( 
d*x^3+c)^(1/2)/c^(1/2))/c^(5/2)-1/48*(d*x^3+c)^(1/2)/c/x^6-1/64*d*(d*x^3+c 
)^(1/2)/c^2/x^3
 

3.3.88.2 Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx=\frac {\left (-4 c-3 d x^3\right ) \sqrt {c+d x^3}}{192 c^2 x^6}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{256 c^{5/2}}+\frac {d^2 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{256 c^{5/2}} \]

input

Integrate[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)),x]
 

output

((-4*c - 3*d*x^3)*Sqrt[c + d*x^3])/(192*c^2*x^6) + (d^2*ArcTanh[Sqrt[c + d 
*x^3]/(3*Sqrt[c])])/(256*c^(5/2)) + (d^2*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]) 
/(256*c^(5/2))
 

3.3.88.3 Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 110, 27, 168, 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.

\(\displaystyle \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 110

\(\displaystyle \frac {1}{3} \left (\frac {\int \frac {3 d \left (d x^3+4 c\right )}{2 x^6 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{16 c}-\frac {\sqrt {c+d x^3}}{16 c x^6}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {3 d \int \frac {d x^3+4 c}{x^6 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6}\right )\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {1}{3} \left (\frac {3 d \left (-\frac {\int \frac {2 c d \left (2 c-d x^3\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{8 c^2}-\frac {\sqrt {c+d x^3}}{2 c x^3}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {3 d \left (-\frac {d \int \frac {2 c-d x^3}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{4 c}-\frac {\sqrt {c+d x^3}}{2 c x^3}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6}\right )\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {1}{3} \left (\frac {3 d \left (-\frac {d \left (\frac {1}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3-\frac {3}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3\right )}{4 c}-\frac {\sqrt {c+d x^3}}{2 c x^3}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{3} \left (\frac {3 d \left (-\frac {d \left (\frac {\int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}-\frac {3}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}\right )}{4 c}-\frac {\sqrt {c+d x^3}}{2 c x^3}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6}\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{3} \left (\frac {3 d \left (-\frac {d \left (\frac {\int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}\right )}{4 c}-\frac {\sqrt {c+d x^3}}{2 c x^3}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (\frac {3 d \left (-\frac {d \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )}{4 c}-\frac {\sqrt {c+d x^3}}{2 c x^3}\right )}{32 c}-\frac {\sqrt {c+d x^3}}{16 c x^6}\right )\)

input

Int[Sqrt[c + d*x^3]/(x^7*(8*c - d*x^3)),x]
 

output

(-1/16*Sqrt[c + d*x^3]/(c*x^6) + (3*d*(-1/2*Sqrt[c + d*x^3]/(c*x^3) - (d*( 
-1/2*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])]/Sqrt[c] - ArcTanh[Sqrt[c + d*x^3 
]/Sqrt[c]]/(2*Sqrt[c])))/(4*c)))/(32*c))/3
 

3.3.88.3.1 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 110

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

rule 168

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] && ILtQ[m, -1]
 

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]]
 

3.3.88.4 Maple [A] (verified)

Time = 4.45 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.72

method	result	size
risch	\(-\frac {\sqrt {d \,x^{3}+c}\, \left (3 d \,x^{3}+4 c \right )}{192 c^{2} x^{6}}-\frac {3 d^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{6 \sqrt {c}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{6 \sqrt {c}}\right )}{128 c^{2}}\)	\(77\)
pseudoelliptic	\(\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) d^{2} x^{6}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) d^{2} x^{6}-12 d \,x^{3} \sqrt {d \,x^{3}+c}\, \sqrt {c}-16 \sqrt {d \,x^{3}+c}\, c^{\frac {3}{2}}}{768 c^{\frac {5}{2}} x^{6}}\)	\(87\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{6 x^{6}}-\frac {d \sqrt {d \,x^{3}+c}}{12 c \,x^{3}}+\frac {d^{2} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{12 c^{\frac {3}{2}}}}{8 c}+\frac {d \left (-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}\right )}{64 c^{2}}+\frac {d^{2} \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) \sqrt {c}}{3}\right )}{512 c^{3}}+\frac {d^{2} \left (-2 \sqrt {d \,x^{3}+c}+6 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right ) \sqrt {c}\right )}{1536 c^{3}}\)	\(182\)
elliptic	\(\text {Expression too large to display}\)	\(1551\)

input

int((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x,method=_RETURNVERBOSE)
 

output

-1/192*(d*x^3+c)^(1/2)*(3*d*x^3+4*c)/c^2/x^6-3/128*d^2/c^2*(-1/6*arctanh(( 
d*x^3+c)^(1/2)/c^(1/2))/c^(1/2)-1/6*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c 
^(1/2))
 

3.3.88.5 Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.76 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx=\left [\frac {3 \, \sqrt {c} d^{2} x^{6} \log \left (\frac {d^{2} x^{6} + 24 \, c d x^{3} + 8 \, {\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} \sqrt {c} + 32 \, c^{2}}{d x^{6} - 8 \, c x^{3}}\right ) - 8 \, {\left (3 \, c d x^{3} + 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{1536 \, c^{3} x^{6}}, -\frac {3 \, \sqrt {-c} d^{2} x^{6} \arctan \left (\frac {{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c} \sqrt {-c}}{4 \, {\left (c d x^{3} + c^{2}\right )}}\right ) + 4 \, {\left (3 \, c d x^{3} + 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{768 \, c^{3} x^{6}}\right ] \]

input

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="fricas")
 

output

[1/1536*(3*sqrt(c)*d^2*x^6*log((d^2*x^6 + 24*c*d*x^3 + 8*(d*x^3 + 4*c)*sqr 
t(d*x^3 + c)*sqrt(c) + 32*c^2)/(d*x^6 - 8*c*x^3)) - 8*(3*c*d*x^3 + 4*c^2)* 
sqrt(d*x^3 + c))/(c^3*x^6), -1/768*(3*sqrt(-c)*d^2*x^6*arctan(1/4*(d*x^3 + 
 4*c)*sqrt(d*x^3 + c)*sqrt(-c)/(c*d*x^3 + c^2)) + 4*(3*c*d*x^3 + 4*c^2)*sq 
rt(d*x^3 + c))/(c^3*x^6)]
 

3.3.88.6 Sympy [F]

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

input

integrate((d*x**3+c)**(1/2)/x**7/(-d*x**3+8*c),x)
 

output

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

3.3.88.7 Maxima [F]

\[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx=\int { -\frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )} x^{7}} \,d x } \]

input

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="maxima")
 

output

-integrate(sqrt(d*x^3 + c)/((d*x^3 - 8*c)*x^7), x)
 

3.3.88.8 Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx=-\frac {d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{256 \, \sqrt {-c} c^{2}} - \frac {d^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{256 \, \sqrt {-c} c^{2}} - \frac {3 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d^{2} + \sqrt {d x^{3} + c} c d^{2}}{192 \, c^{2} d^{2} x^{6}} \]

input

integrate((d*x^3+c)^(1/2)/x^7/(-d*x^3+8*c),x, algorithm="giac")
 

output

-1/256*d^2*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/256*d^2*arc 
tan(1/3*sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c^2) - 1/192*(3*(d*x^3 + c)^(3 
/2)*d^2 + sqrt(d*x^3 + c)*c*d^2)/(c^2*d^2*x^6)
 

3.3.88.9 Mupad [B] (verification not implemented)

Time = 7.57 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {c+d x^3}}{x^7 \left (8 c-d x^3\right )} \, dx=\frac {d^2\,\mathrm {atanh}\left (\frac {d^4\,\sqrt {d\,x^3+c}}{2048\,c^{7/2}\,\left (\frac {d^4}{2048\,c^3}+\frac {d^5\,x^3}{8192\,c^4}\right )}\right )}{256\,c^{5/2}}-\frac {\sqrt {d\,x^3+c}}{192\,c\,x^6}-\frac {{\left (d\,x^3+c\right )}^{3/2}}{64\,c^2\,x^6} \]

input

int((c + d*x^3)^(1/2)/(x^7*(8*c - d*x^3)),x)
 

output

(d^2*atanh((d^4*(c + d*x^3)^(1/2))/(2048*c^(7/2)*(d^4/(2048*c^3) + (d^5*x^ 
3)/(8192*c^4)))))/(256*c^(5/2)) - (c + d*x^3)^(1/2)/(192*c*x^6) - (c + d*x 
^3)^(3/2)/(64*c^2*x^6)