\(\int \frac {(c+d x^3)^{3/2}}{x^4 (a+b x^3)^2} \, dx\) [647]

Optimal result

Integrand size = 24, antiderivative size = 170 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (a+b x^3\right )^2} \, dx=-\frac {(2 b c-a d) \sqrt {c+d x^3}}{3 a^2 \left (a+b x^3\right )}-\frac {c \sqrt {c+d x^3}}{3 a x^3 \left (a+b x^3\right )}+\frac {\sqrt {c} (4 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3}-\frac {\sqrt {b c-a d} (4 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a^3 \sqrt {b}} \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (a+b x^3\right )^2} \, dx=\frac {\frac {a \sqrt {c+d x^3} \left (-a c-2 b c x^3+a d x^3\right )}{x^3 \left (a+b x^3\right )}+\frac {\left (4 b^2 c^2-5 a b c d+a^2 d^2\right ) \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )}{\sqrt {b} \sqrt {-b c+a d}}+\sqrt {c} (4 b c-3 a d) \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a^3} \] Input:

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

Output:

((a*Sqrt[c + d*x^3]*(-(a*c) - 2*b*c*x^3 + a*d*x^3))/(x^3*(a + b*x^3)) + (( 
4*b^2*c^2 - 5*a*b*c*d + a^2*d^2)*ArcTan[(Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[-(b 
*c) + a*d]])/(Sqrt[b]*Sqrt[-(b*c) + a*d]) + Sqrt[c]*(4*b*c - 3*a*d)*ArcTan 
h[Sqrt[c + d*x^3]/Sqrt[c]])/(3*a^3)
 

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.16, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {948, 109, 27, 168, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-((c*Sqrt[c + d*x^3])/(a*x^3*(a + b*x^3))) - ((2*(2*b*c - a*d)*Sqrt[c + d 
*x^3])/(a*(a + b*x^3)) + ((-2*Sqrt[c]*(4*b*c - 3*a*d)*(b*c - a*d)*ArcTanh[ 
Sqrt[c + d*x^3]/Sqrt[c]])/a + (2*(b*c - a*d)^(3/2)*(4*b*c - a*d)*ArcTanh[( 
Sqrt[b]*Sqrt[c + d*x^3])/Sqrt[b*c - a*d]])/(a*Sqrt[b]))/(a*(b*c - a*d)))/( 
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 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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 = 1.42 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {\frac {4 \left (b \,x^{3}+a \right ) \left (-a d +b c \right ) \left (b c -\frac {a d}{4}\right ) \sqrt {c}\, x^{3} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3}-\frac {\sqrt {\left (a d -b c \right ) b}\, \left (3 c \left (a d -\frac {4 b c}{3}\right ) \left (b \,x^{3}+a \right ) x^{3} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )+\left (2 x^{3} b c +a \left (-d \,x^{3}+c \right )\right ) a \sqrt {c}\, \sqrt {d \,x^{3}+c}\right )}{3}}{a^{3} x^{3} \sqrt {c}\, \left (b \,x^{3}+a \right ) \sqrt {\left (a d -b c \right ) b}}\)	\(170\)
risch	\(-\frac {c \sqrt {d \,x^{3}+c}}{3 a^{2} x^{3}}-\frac {\frac {\left (-2 a^{2} d^{2}+4 a b c d -2 b^{2} c^{2}\right ) \left (\frac {\sqrt {d \,x^{3}+c}}{b \,x^{3}+a}+\frac {d \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a d -3 b c}+\frac {2 \sqrt {c}\, \left (3 a d -4 b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 a}+\frac {8 b c \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3 a \sqrt {\left (a d -b c \right ) b}}}{2 a^{2}}\)	\(201\)
default	\(\frac {-\frac {c \sqrt {d \,x^{3}+c}}{3 x^{3}}+\frac {2 d \sqrt {d \,x^{3}+c}}{3}-\sqrt {c}\, d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{a^{2}}+\frac {-d \left (b \,x^{3}+a \right ) \left (a d -b c \right ) \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\sqrt {\left (a d -b c \right ) b}\, \sqrt {d \,x^{3}+c}\, \left (\frac {\left (2 d \,x^{3}-c \right ) b}{3}+a d \right )}{a^{2} \sqrt {\left (a d -b c \right ) b}\, \left (b \,x^{3}+a \right )}-\frac {2 b \left (\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9}+\frac {8 c \sqrt {d \,x^{3}+c}}{9}-\frac {2 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{a^{3}}-\frac {4 \left (-\left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )+\left (\frac {\left (-d \,x^{3}-4 c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}\right )}{3 a^{3} \sqrt {\left (a d -b c \right ) b}}\)	\(307\)
elliptic	\(\text {Expression too large to display}\)	\(1656\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 832, normalized size of antiderivative = 4.89 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (a+b x^3\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.27 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (a+b x^3\right )^2} \, dx=\frac {{\left (4 \, b^{2} c^{2} - 5 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} a^{3}} - \frac {{\left (4 \, b c^{2} - 3 \, a c d\right )} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a^{3} \sqrt {-c}} - \frac {2 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} b c d - 2 \, \sqrt {d x^{3} + c} b c^{2} d - {\left (d x^{3} + c\right )}^{\frac {3}{2}} a d^{2} + 2 \, \sqrt {d x^{3} + c} a c d^{2}}{3 \, {\left ({\left (d x^{3} + c\right )}^{2} b - 2 \, {\left (d x^{3} + c\right )} b c + b c^{2} + {\left (d x^{3} + c\right )} a d - a c d\right )} a^{2}} \] Input:

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

Output:

1/3*(4*b^2*c^2 - 5*a*b*c*d + a^2*d^2)*arctan(sqrt(d*x^3 + c)*b/sqrt(-b^2*c 
 + a*b*d))/(sqrt(-b^2*c + a*b*d)*a^3) - 1/3*(4*b*c^2 - 3*a*c*d)*arctan(sqr 
t(d*x^3 + c)/sqrt(-c))/(a^3*sqrt(-c)) - 1/3*(2*(d*x^3 + c)^(3/2)*b*c*d - 2 
*sqrt(d*x^3 + c)*b*c^2*d - (d*x^3 + c)^(3/2)*a*d^2 + 2*sqrt(d*x^3 + c)*a*c 
*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 = 10.13 (sec) , antiderivative size = 531, normalized size of antiderivative = 3.12 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (a+b x^3\right )^2} \, dx=\frac {\sqrt {c}\,\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^3}-\frac {c\,\sqrt {d\,x^3+c}}{3\,a^2\,x^3}-\frac {\sqrt {d\,x^3+c}\,\left (\frac {3\,a\,d-4\,b\,c}{2\,a^2}-\frac {a\,\left (\frac {a\,\left (\frac {a\,\left (\frac {b\,d^2\,\left (a\,d+b\,c\right )}{a^3\,c^2}-\frac {a\,\left (\frac {b^2\,d^3}{2\,a^3\,c^2}-\frac {b^2\,d^3\,\left (3\,a\,d-4\,b\,c\right )}{6\,a^2\,c^2\,\left (a^2\,d-a\,b\,c\right )}+\frac {b^2\,d^2\,\left (a\,d+b\,c\right )\,\left (3\,a\,d-4\,b\,c\right )}{3\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}+\frac {b\,\left (3\,a\,d-4\,b\,c\right )\,\left (a^2\,d^3+4\,a\,b\,c\,d^2-b^2\,c^2\,d\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {a^2\,d^3+4\,a\,b\,c\,d^2-b^2\,c^2\,d}{2\,a^3\,c^2}+\frac {b\,\left (3\,a\,d-4\,b\,c\right )\,\left (-4\,a^2\,c\,d^2+2\,a\,b\,c^2\,d+2\,b^2\,c^3\right )}{6\,a^3\,c^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}-\frac {-4\,a^2\,c\,d^2+2\,a\,b\,c^2\,d+2\,b^2\,c^3}{2\,a^3\,c^2}+\frac {b\,{\left (3\,a\,d-4\,b\,c\right )}^2}{6\,a^2\,\left (a^2\,d-a\,b\,c\right )}\right )}{b}\right )}{b\,x^3+a}+\frac {\ln \left (\frac {2\,b\,c-a\,d+b\,d\,x^3+\sqrt {b}\,\sqrt {d\,x^3+c}\,\sqrt {a\,d-b\,c}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,\sqrt {a\,d-b\,c}\,\left (a\,d-4\,b\,c\right )\,1{}\mathrm {i}}{6\,a^3\,\sqrt {b}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (a+b x^3\right )^2} \, dx=\text {too large to display} \] Input:

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

Output:

( - 2*sqrt(c + d*x**3)*a*c*d + 4*sqrt(c + d*x**3)*a*d**2*x**3 - 8*sqrt(c + 
 d*x**3)*b*c**2 + 9*int(sqrt(c + d*x**3)/(a**3*c*d*x + a**3*d**2*x**4 + 4* 
a**2*b*c**2*x + 6*a**2*b*c*d*x**4 + 2*a**2*b*d**2*x**7 + 8*a*b**2*c**2*x** 
4 + 9*a*b**2*c*d*x**7 + a*b**2*d**2*x**10 + 4*b**3*c**2*x**7 + 4*b**3*c*d* 
x**10),x)*a**4*c*d**3*x**3 + 60*int(sqrt(c + d*x**3)/(a**3*c*d*x + a**3*d* 
*2*x**4 + 4*a**2*b*c**2*x + 6*a**2*b*c*d*x**4 + 2*a**2*b*d**2*x**7 + 8*a*b 
**2*c**2*x**4 + 9*a*b**2*c*d*x**7 + a*b**2*d**2*x**10 + 4*b**3*c**2*x**7 + 
 4*b**3*c*d*x**10),x)*a**3*b*c**2*d**2*x**3 + 9*int(sqrt(c + d*x**3)/(a**3 
*c*d*x + a**3*d**2*x**4 + 4*a**2*b*c**2*x + 6*a**2*b*c*d*x**4 + 2*a**2*b*d 
**2*x**7 + 8*a*b**2*c**2*x**4 + 9*a*b**2*c*d*x**7 + a*b**2*d**2*x**10 + 4* 
b**3*c**2*x**7 + 4*b**3*c*d*x**10),x)*a**3*b*c*d**3*x**6 + 48*int(sqrt(c + 
 d*x**3)/(a**3*c*d*x + a**3*d**2*x**4 + 4*a**2*b*c**2*x + 6*a**2*b*c*d*x** 
4 + 2*a**2*b*d**2*x**7 + 8*a*b**2*c**2*x**4 + 9*a*b**2*c*d*x**7 + a*b**2*d 
**2*x**10 + 4*b**3*c**2*x**7 + 4*b**3*c*d*x**10),x)*a**2*b**2*c**3*d*x**3 
+ 60*int(sqrt(c + d*x**3)/(a**3*c*d*x + a**3*d**2*x**4 + 4*a**2*b*c**2*x + 
 6*a**2*b*c*d*x**4 + 2*a**2*b*d**2*x**7 + 8*a*b**2*c**2*x**4 + 9*a*b**2*c* 
d*x**7 + a*b**2*d**2*x**10 + 4*b**3*c**2*x**7 + 4*b**3*c*d*x**10),x)*a**2* 
b**2*c**2*d**2*x**6 - 192*int(sqrt(c + d*x**3)/(a**3*c*d*x + a**3*d**2*x** 
4 + 4*a**2*b*c**2*x + 6*a**2*b*c*d*x**4 + 2*a**2*b*d**2*x**7 + 8*a*b**2*c* 
*2*x**4 + 9*a*b**2*c*d*x**7 + a*b**2*d**2*x**10 + 4*b**3*c**2*x**7 + 4*...