3.4.82 \(\int \frac {1}{x^4 (a+b x^3) \sqrt {c+d x^3}} \, dx\) [382]

3.4.82.1 Optimal result

Integrand size = 24, antiderivative size = 117 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {c+d x^3}}{3 a c x^3}+\frac {(2 b c+a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 c^{3/2}}-\frac {2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2 \sqrt {b c-a d}} \]

output

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

3.4.82.2 Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {-\frac {a \sqrt {c+d x^3}}{c x^3}+\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{\sqrt {-b c+a d}}+\frac {(2 b c+a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{c^{3/2}}}{3 a^2} \]

input

Integrate[1/(x^4*(a + b*x^3)*Sqrt[c + d*x^3]),x]
 

output

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

3.4.82.3 Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {948, 114, 27, 174, 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.

\(\displaystyle \int \frac {1}{x^4 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx\)

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{3} \left (-\frac {\frac {4 b^{3/2} c \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{a \sqrt {b c-a d}}-\frac {2 (a d+2 b c) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{a \sqrt {c}}}{2 a c}-\frac {\sqrt {c+d x^3}}{a c x^3}\right )\)

input

Int[1/(x^4*(a + b*x^3)*Sqrt[c + d*x^3]),x]
 

output

(-(Sqrt[c + d*x^3]/(a*c*x^3)) - ((-2*(2*b*c + a*d)*ArcTanh[Sqrt[c + d*x^3] 
/Sqrt[c]])/(a*Sqrt[c]) + (4*b^(3/2)*c*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sq 
rt[b*c - a*d]])/(a*Sqrt[b*c - a*d]))/(2*a*c))/3
 

3.4.82.3.1 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 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 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 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]]
 

3.4.82.4 Maple [A] (verified)

Time = 4.61 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79

method	result	size
pseudoelliptic	\(\frac {\frac {2 b^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}-\frac {a \sqrt {d \,x^{3}+c}}{c \,x^{3}}+\frac {\left (a d +2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{c^{\frac {3}{2}}}}{3 a^{2}}\)	\(92\)
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{3 a c \,x^{3}}-\frac {-\frac {2 \left (a d +2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a \sqrt {c}}-\frac {4 b^{2} c \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a \sqrt {\left (a d -b c \right ) b}}}{2 a c}\)	\(106\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 c \,x^{3}}+\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 c^{\frac {3}{2}}}}{a}+\frac {2 b \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a^{2} \sqrt {c}}+\frac {2 b^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a^{2} \sqrt {\left (a d -b c \right ) b}}\)	\(111\)
elliptic	\(\text {Expression too large to display}\)	\(1652\)

input

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

output

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

3.4.82.5 Fricas [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 565, normalized size of antiderivative = 4.83 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\left [\frac {2 \, b c^{2} x^{3} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) + {\left (2 \, b c + a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} c^{2} x^{3}}, -\frac {4 \, b c^{2} x^{3} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) - {\left (2 \, b c + a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} + 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} c^{2} x^{3}}, \frac {b c^{2} x^{3} \sqrt {\frac {b}{b c - a d}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {\frac {b}{b c - a d}}}{b x^{3} + a}\right ) - {\left (2 \, b c + a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - \sqrt {d x^{3} + c} a c}{3 \, a^{2} c^{2} x^{3}}, -\frac {2 \, b c^{2} x^{3} \sqrt {-\frac {b}{b c - a d}} \arctan \left (-\frac {\sqrt {d x^{3} + c} {\left (b c - a d\right )} \sqrt {-\frac {b}{b c - a d}}}{b d x^{3} + b c}\right ) + {\left (2 \, b c + a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) + \sqrt {d x^{3} + c} a c}{3 \, a^{2} c^{2} x^{3}}\right ] \]

input

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

output

[1/6*(2*b*c^2*x^3*sqrt(b/(b*c - a*d))*log((b*d*x^3 + 2*b*c - a*d - 2*sqrt( 
d*x^3 + c)*(b*c - a*d)*sqrt(b/(b*c - a*d)))/(b*x^3 + a)) + (2*b*c + a*d)*s 
qrt(c)*x^3*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 2*sqrt(d*x 
^3 + c)*a*c)/(a^2*c^2*x^3), -1/6*(4*b*c^2*x^3*sqrt(-b/(b*c - a*d))*arctan( 
-sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(-b/(b*c - a*d))/(b*d*x^3 + b*c)) - (2*b* 
c + a*d)*sqrt(c)*x^3*log((d*x^3 + 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 
2*sqrt(d*x^3 + c)*a*c)/(a^2*c^2*x^3), 1/3*(b*c^2*x^3*sqrt(b/(b*c - a*d))*l 
og((b*d*x^3 + 2*b*c - a*d - 2*sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(b/(b*c - a* 
d)))/(b*x^3 + a)) - (2*b*c + a*d)*sqrt(-c)*x^3*arctan(sqrt(d*x^3 + c)*sqrt 
(-c)/c) - sqrt(d*x^3 + c)*a*c)/(a^2*c^2*x^3), -1/3*(2*b*c^2*x^3*sqrt(-b/(b 
*c - a*d))*arctan(-sqrt(d*x^3 + c)*(b*c - a*d)*sqrt(-b/(b*c - a*d))/(b*d*x 
^3 + b*c)) + (2*b*c + a*d)*sqrt(-c)*x^3*arctan(sqrt(d*x^3 + c)*sqrt(-c)/c) 
 + sqrt(d*x^3 + c)*a*c)/(a^2*c^2*x^3)]
 

3.4.82.6 Sympy [F]

\[ \int \frac {1}{x^4 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^{4} \left (a + b x^{3}\right ) \sqrt {c + d x^{3}}}\, dx \]

input

integrate(1/x**4/(b*x**3+a)/(d*x**3+c)**(1/2),x)
 

output

Integral(1/(x**4*(a + b*x**3)*sqrt(c + d*x**3)), x)
 

3.4.82.7 Maxima [F]

\[ \int \frac {1}{x^4 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\int { \frac {1}{{\left (b x^{3} + a\right )} \sqrt {d x^{3} + c} x^{4}} \,d x } \]

input

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

output

integrate(1/((b*x^3 + a)*sqrt(d*x^3 + c)*x^4), x)
 

3.4.82.8 Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {2 \, b^{2} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} a^{2}} - \frac {{\left (2 \, b c + a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c} c} - \frac {\sqrt {d x^{3} + c}}{3 \, a c x^{3}} \]

input

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

output

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

3.4.82.9 Mupad [B] (verification not implemented)

Time = 12.83 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^4 \left (a+b x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {\ln \left (\frac {\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )\,{\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}^3}{x^6}\right )\,\left (a\,d+2\,b\,c\right )}{6\,a^2\,c^{3/2}}-\frac {\sqrt {d\,x^3+c}}{3\,a\,c\,x^3}+\frac {b^{3/2}\,\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,1{}\mathrm {i}}{3\,a^2\,\sqrt {a\,d-b\,c}} \]

input

int(1/(x^4*(a + b*x^3)*(c + d*x^3)^(1/2)),x)
 

output

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