\(\int \frac {\sqrt {c+d x^3}}{x^4 (a+b x^3)} \, dx\) [535]

Optimal result

Integrand size = 24, antiderivative size = 115 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{3 a x^3}+\frac {(2 b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^2 \sqrt {c}}-\frac {2 \sqrt {b} \sqrt {b c-a d} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.47 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )} \, dx=\frac {-\frac {a \sqrt {c+d x^3}}{x^3}-2 \sqrt {b} \sqrt {-b c+a d} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )+\frac {(2 b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{\sqrt {c}}}{3 a^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {948, 110, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 110

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.83

method	result	size
pseudoelliptic	\(\frac {-\frac {a \sqrt {d \,x^{3}+c}}{x^{3}}-\frac {\left (a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{\sqrt {c}}-\frac {2 b \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}}{3 a^{2}}\)	\(96\)
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{3 a \,x^{3}}-\frac {\frac {2 \left (a d -2 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a \sqrt {c}}+\frac {4 b \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a \sqrt {\left (a d -b c \right ) b}}}{2 a}\)	\(105\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}}{a}+\frac {2 b \left (\sqrt {d \,x^{3}+c}-\frac {\left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a^{2}}-\frac {b \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{a^{2}}\)	\(140\)
elliptic	\(\text {Expression too large to display}\)	\(1589\)

Input:

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

Output:

1/3/a^2*(-a*(d*x^3+c)^(1/2)/x^3-(a*d-2*b*c)/c^(1/2)*arctanh((d*x^3+c)^(1/2 
)/c^(1/2))-2*b*(a*d-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 = 0.13 (sec) , antiderivative size = 507, normalized size of antiderivative = 4.41 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )} \, dx=\left [\frac {2 \, \sqrt {b^{2} c - a b d} c x^{3} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) - {\left (2 \, b c - a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} c x^{3}}, \frac {4 \, \sqrt {-b^{2} c + a b d} c x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - {\left (2 \, b c - a d\right )} \sqrt {c} x^{3} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 2 \, \sqrt {d x^{3} + c} a c}{6 \, a^{2} c x^{3}}, -\frac {{\left (2 \, b c - a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) - \sqrt {b^{2} c - a b d} c x^{3} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} \sqrt {b^{2} c - a b d}}{b x^{3} + a}\right ) + \sqrt {d x^{3} + c} a c}{3 \, a^{2} c x^{3}}, \frac {2 \, \sqrt {-b^{2} c + a b d} c x^{3} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-b^{2} c + a b d}}{b d x^{3} + b c}\right ) - {\left (2 \, b c - a d\right )} \sqrt {-c} x^{3} \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) - \sqrt {d x^{3} + c} a c}{3 \, a^{2} c x^{3}}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )} \, dx=\frac {2 \, {\left (b^{2} c - a b d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} a^{2}} - \frac {{\left (2 \, b c - a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{2} \sqrt {-c}} - \frac {\sqrt {d x^{3} + c}}{3 \, a x^{3}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.69 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )} \, dx=\frac {\ln \left (\frac {a\,d-2\,b\,c+2\,\sqrt {d\,x^3+c}\,\sqrt {b^2\,c-a\,b\,d}-b\,d\,x^3}{b\,x^3+a}\right )\,\sqrt {b^2\,c-a\,b\,d}}{3\,a^2}-\frac {\sqrt {d\,x^3+c}}{3\,a\,x^3}+\frac {\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )\,\left (a\,d-2\,b\,c\right )}{6\,a^2\,\sqrt {c}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{b \,x^{7}+a \,x^{4}}d x \] Input:

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

Output:

int(sqrt(c + d*x**3)/(a*x**4 + b*x**7),x)