\(\int \frac {\sqrt {c+d x^3}}{x^4 (a+b x^3)^2} \, dx\) [637]

Optimal result

Integrand size = 24, antiderivative size = 161 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )^2} \, dx=-\frac {2 b \sqrt {c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac {\sqrt {c+d x^3}}{3 a x^3 \left (a+b x^3\right )}+\frac {(4 b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3 \sqrt {c}}-\frac {\sqrt {b} (4 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 \sqrt {b c-a d}} \] Output:

-2/3*b*(d*x^3+c)^(1/2)/a^2/(b*x^3+a)-1/3*(d*x^3+c)^(1/2)/a/x^3/(b*x^3+a)+1 
/3*(-a*d+4*b*c)*arctanh((d*x^3+c)^(1/2)/c^(1/2))/a^3/c^(1/2)-1/3*b^(1/2)*( 
-3*a*d+4*b*c)*arctanh(b^(1/2)*(d*x^3+c)^(1/2)/(-a*d+b*c)^(1/2))/a^3/(-a*d+ 
b*c)^(1/2)
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )^2} \, dx=\frac {-\frac {a \left (a+2 b x^3\right ) \sqrt {c+d x^3}}{x^3 \left (a+b x^3\right )}+\frac {\sqrt {b} (4 b c-3 a d) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{\sqrt {-b c+a d}}+\frac {(4 b c-a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{\sqrt {c}}}{3 a^3} \] Input:

Integrate[Sqrt[c + d*x^3]/(x^4*(a + b*x^3)^2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 110, 27, 168, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 110

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Sqrt[c + d*x^3]/(x^4*(a + b*x^3)^2),x]
 

Output:

(-(Sqrt[c + d*x^3]/(a*x^3*(a + b*x^3))) - ((4*b*Sqrt[c + d*x^3])/(a*(a + b 
*x^3)) + ((-2*(4*b*c - a*d)*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(a*Sqrt[c]) 
+ (2*Sqrt[b]*(4*b*c - 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - 
a*d]])/(a*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 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 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 = 1.43 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.94

method	result	size
pseudoelliptic	\(\frac {\frac {4 \sqrt {c}\, \left (b \,x^{3}+a \right ) b \,x^{3} \left (b c -\frac {3 a d}{4}\right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3}-\frac {\sqrt {\left (a d -b c \right ) b}\, \left (x^{3} \left (b \,x^{3}+a \right ) \left (a d -4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )+\sqrt {c}\, a \left (2 b \,x^{3}+a \right ) \sqrt {d \,x^{3}+c}\right )}{3}}{x^{3} \sqrt {c}\, a^{3} \left (b \,x^{3}+a \right ) \sqrt {\left (a d -b c \right ) b}}\)	\(151\)
risch	\(-\frac {\sqrt {d \,x^{3}+c}}{3 a^{2} x^{3}}-\frac {\frac {2 \left (a d -4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a \sqrt {c}}+\frac {2 b \left (\frac {\sqrt {d \,x^{3}+c}}{b \,x^{3}+a}+\frac {d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3}+\frac {4 b \left (a d -2 b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a \sqrt {\left (a d -b c \right ) b}}}{2 a^{2}}\)	\(166\)
default	\(\frac {-\frac {\sqrt {d \,x^{3}+c}}{3 x^{3}}-\frac {d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}}{a^{2}}+\frac {b \left (-\frac {\sqrt {d \,x^{3}+c}}{b \,x^{3}+a}+\frac {d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a^{2}}-\frac {2 b \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-\frac {2 \sqrt {c}\, \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{a^{3}}+\frac {4 b \left (\sqrt {d \,x^{3}+c}-\frac {\left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a^{3}}\)	\(205\)
elliptic	\(\text {Expression too large to display}\)	\(1708\)

Input:

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

Output:

4/3/((a*d-b*c)*b)^(1/2)/c^(1/2)*(c^(1/2)*(b*x^3+a)*b*x^3*(b*c-3/4*a*d)*arc 
tan(b*(d*x^3+c)^(1/2)/((a*d-b*c)*b)^(1/2))-1/4*((a*d-b*c)*b)^(1/2)*(x^3*(b 
*x^3+a)*(a*d-4*b*c)*arctanh((d*x^3+c)^(1/2)/c^(1/2))+c^(1/2)*a*(2*b*x^3+a) 
*(d*x^3+c)^(1/2)))/x^3/a^3/(b*x^3+a)
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 825, normalized size of antiderivative = 5.12 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/6*(((4*b^2*c^2 - 3*a*b*c*d)*x^6 + (4*a*b*c^2 - 3*a^2*c*d)*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)*sq 
rt(b/(b*c - a*d)))/(b*x^3 + a)) + ((4*b^2*c - a*b*d)*x^6 + (4*a*b*c - a^2* 
d)*x^3)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 2*(2* 
a*b*c*x^3 + a^2*c)*sqrt(d*x^3 + c))/(a^3*b*c*x^6 + a^4*c*x^3), 1/6*(2*((4* 
b^2*c^2 - 3*a*b*c*d)*x^6 + (4*a*b*c^2 - 3*a^2*c*d)*x^3)*sqrt(-b/(b*c - a*d 
))*arctan(sqrt(d*x^3 + c)*sqrt(-b/(b*c - a*d))) - ((4*b^2*c - a*b*d)*x^6 + 
 (4*a*b*c - a^2*d)*x^3)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2 
*c)/x^3) - 2*(2*a*b*c*x^3 + a^2*c)*sqrt(d*x^3 + c))/(a^3*b*c*x^6 + a^4*c*x 
^3), -1/6*(2*((4*b^2*c - a*b*d)*x^6 + (4*a*b*c - a^2*d)*x^3)*sqrt(-c)*arct 
an(sqrt(-c)/sqrt(d*x^3 + c)) + ((4*b^2*c^2 - 3*a*b*c*d)*x^6 + (4*a*b*c^2 - 
 3*a^2*c*d)*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*(2*a*b*c*x^3 + 
 a^2*c)*sqrt(d*x^3 + c))/(a^3*b*c*x^6 + a^4*c*x^3), 1/3*(((4*b^2*c^2 - 3*a 
*b*c*d)*x^6 + (4*a*b*c^2 - 3*a^2*c*d)*x^3)*sqrt(-b/(b*c - a*d))*arctan(sqr 
t(d*x^3 + c)*sqrt(-b/(b*c - a*d))) - ((4*b^2*c - a*b*d)*x^6 + (4*a*b*c - a 
^2*d)*x^3)*sqrt(-c)*arctan(sqrt(-c)/sqrt(d*x^3 + c)) - (2*a*b*c*x^3 + a^2* 
c)*sqrt(d*x^3 + c))/(a^3*b*c*x^6 + a^4*c*x^3)]
 

Sympy [F]

\[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )^2} \, dx=\int \frac {\sqrt {c + d x^{3}}}{x^{4} \left (a + b x^{3}\right )^{2}}\, dx \] Input:

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

Output:

Integral(sqrt(c + d*x**3)/(x**4*(a + b*x**3)**2), x)
 

Maxima [F]

\[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )^2} \, dx=\int { \frac {\sqrt {d x^{3} + c}}{{\left (b x^{3} + a\right )}^{2} x^{4}} \,d x } \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )^2} \, dx=\frac {{\left (4 \, b^{2} c - 3 \, a b d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} a^{3}} - \frac {{\left (4 \, b c - a d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{3} \sqrt {-c}} - \frac {2 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} b d - 2 \, \sqrt {d x^{3} + c} b c d + \sqrt {d x^{3} + c} a d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )}^{2} b - 2 \, {\left (d x^{3} + c\right )} b c + b c^{2} + {\left (d x^{3} + c\right )} a d - a c d\right )} a^{2}} \] Input:

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

Output:

1/3*(4*b^2*c - 3*a*b*d)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c + a*b*d))/(sq 
rt(-b^2*c + a*b*d)*a^3) - 1/3*(4*b*c - a*d)*arctan(sqrt(d*x^3 + c)/sqrt(-c 
))/(a^3*sqrt(-c)) - 1/3*(2*(d*x^3 + c)^(3/2)*b*d - 2*sqrt(d*x^3 + c)*b*c*d 
 + sqrt(d*x^3 + c)*a*d^2)/(((d*x^3 + c)^2*b - 2*(d*x^3 + c)*b*c + b*c^2 + 
(d*x^3 + c)*a*d - a*c*d)*a^2)
 

Mupad [B] (verification not implemented)

Time = 9.20 (sec) , antiderivative size = 438, normalized size of antiderivative = 2.72 \[ \int \frac {\sqrt {c+d x^3}}{x^4 \left (a+b x^3\right )^2} \, dx=\frac {\left (\frac {a\,\left (\frac {a\,\left (\frac {a\,\left (\frac {b^2\,d^2}{2\,a^3\,c^2}-\frac {b^2\,d^2\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^2\,c^2\,\left (a^2\,d-a\,b\,c\right )}+\frac {b^2\,d\,\left (2\,a\,d-b\,c\right )\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {b\,d\,\left (2\,a\,d-b\,c\right )}{2\,a^3\,c^2}+\frac {b\,\left (3\,a\,d-4\,b\,c\right )\,\left (-a^2\,d^2+2\,a\,b\,c\,d+2\,b^2\,c^2\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {-a^2\,d^2+2\,a\,b\,c\,d+2\,b^2\,c^2}{2\,a^3\,c^2}+\frac {b\,\left (a\,d-4\,b\,c\right )\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^2\,c\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {a\,d-4\,b\,c}{2\,a^2\,c}\right )\,\sqrt {d\,x^3+c}}{b\,x^3+a}-\frac {\sqrt {d\,x^3+c}}{3\,a^2\,x^3}+\frac {\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )\,\left (a\,d-4\,b\,c\right )}{6\,a^3\,\sqrt {c}}+\frac {\sqrt {b}\,\ln \left (\frac {a\,d-2\,b\,c-b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\left (3\,a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,a^3\,\sqrt {a\,d-b\,c}} \] Input:

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

Output:

(((a*((a*((a*((b^2*d^2)/(2*a^3*c^2) - (b^2*d^2*(3*a*d - 4*b*c))/(6*a^2*c^2 
*(a^2*d - a*b*c)) + (b^2*d*(2*a*d - b*c)*(3*a*d - 4*b*c))/(6*a^3*c^2*(a^2* 
d - a*b*c))))/b - (b*d*(2*a*d - b*c))/(2*a^3*c^2) + (b*(3*a*d - 4*b*c)*(2* 
b^2*c^2 - a^2*d^2 + 2*a*b*c*d))/(6*a^3*c^2*(a^2*d - a*b*c))))/b - (2*b^2*c 
^2 - a^2*d^2 + 2*a*b*c*d)/(2*a^3*c^2) + (b*(a*d - 4*b*c)*(3*a*d - 4*b*c))/ 
(6*a^2*c*(a^2*d - a*b*c))))/b - (a*d - 4*b*c)/(2*a^2*c))*(c + d*x^3)^(1/2) 
)/(a + b*x^3) - (c + d*x^3)^(1/2)/(3*a^2*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 - 4*b*c))/(6*a^3*c^(1 
/2)) + (b^(1/2)*log((a*d - 2*b*c + b^(1/2)*(c + d*x^3)^(1/2)*(a*d - b*c)^( 
1/2)*2i - b*d*x^3)/(a + b*x^3))*(3*a*d - 4*b*c)*1i)/(6*a^3*(a*d - b*c)^(1/ 
2))
 

Reduce [F]

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

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

Output:

int(sqrt(c + d*x**3)/(a**2*x**4 + 2*a*b*x**7 + b**2*x**10),x)