\(\int \frac {(c+d x^3)^{3/2}}{x (a+b x^3)} \, dx\) [544]

Optimal result

Integrand size = 24, antiderivative size = 104 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (a+b x^3\right )} \, dx=\frac {2 d \sqrt {c+d x^3}}{3 b}-\frac {2 c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{3 a}+\frac {2 (b c-a d)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {b c-a d}}\right )}{3 a b^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.02 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (a+b x^3\right )} \, dx=\frac {2 \left (a \sqrt {b} d \sqrt {c+d x^3}-(-b c+a d)^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {c+d x^3}}{\sqrt {-b c+a d}}\right )-b^{3/2} c^{3/2} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )\right )}{3 a b^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {948, 95, 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.

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 95

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 95

Int[((e_.) + (f_.)*(x_))^(p_)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), 
x_] :> Simp[f*((e + f*x)^(p - 1)/(b*d*(p - 1))), x] + Simp[1/(b*d)   Int[(b 
*d*e^2 - a*c*f^2 + f*(2*b*d*e - b*c*f - a*d*f)*x)*((e + f*x)^(p - 2)/((a + 
b*x)*(c + d*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 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 = 1.23 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00

method	result	size
pseudoelliptic	\(\frac {-\frac {2 \left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3}+\frac {2 \left (a d \sqrt {d \,x^{3}+c}-c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) b \right ) \sqrt {\left (a d -b c \right ) b}}{3}}{a b \sqrt {\left (a d -b c \right ) b}}\)	\(104\)
default	\(\frac {\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9}+\frac {8 c \sqrt {d \,x^{3}+c}}{9}-\frac {2 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}}{a}+\frac {-\frac {2 \left (a d -b c \right )^{2} \arctan \left (\frac {b \sqrt {d \,x^{3}+c}}{\sqrt {\left (a d -b c \right ) b}}\right )}{3}+\frac {2 \left (\frac {\left (-d \,x^{3}-4 c \right ) b}{3}+a d \right ) \sqrt {d \,x^{3}+c}\, \sqrt {\left (a d -b c \right ) b}}{3}}{a b \sqrt {\left (a d -b c \right ) b}}\)	\(149\)
elliptic	\(\text {Expression too large to display}\)	\(1613\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 480, normalized size of antiderivative = 4.62 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (a+b x^3\right )} \, dx=\left [\frac {b c^{\frac {3}{2}} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, \sqrt {d x^{3} + c} a d - {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right )}{3 \, a b}, \frac {b c^{\frac {3}{2}} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) + 2 \, \sqrt {d x^{3} + c} a d + 2 \, {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right )}{3 \, a b}, \frac {2 \, b \sqrt {-c} c \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + 2 \, \sqrt {d x^{3} + c} a d - {\left (b c - a d\right )} \sqrt {\frac {b c - a d}{b}} \log \left (\frac {b d x^{3} + 2 \, b c - a d - 2 \, \sqrt {d x^{3} + c} b \sqrt {\frac {b c - a d}{b}}}{b x^{3} + a}\right )}{3 \, a b}, \frac {2 \, {\left (b \sqrt {-c} c \arctan \left (\frac {\sqrt {-c}}{\sqrt {d x^{3} + c}}\right ) + \sqrt {d x^{3} + c} a d + {\left (b c - a d\right )} \sqrt {-\frac {b c - a d}{b}} \arctan \left (-\frac {\sqrt {d x^{3} + c} b \sqrt {-\frac {b c - a d}{b}}}{b c - a d}\right )\right )}}{3 \, a b}\right ] \] Input:

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

Output:

[1/3*(b*c^(3/2)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 2*sqr 
t(d*x^3 + c)*a*d - (b*c - a*d)*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)))/(a*b), 1/3*(b 
*c^(3/2)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) + 2*sqrt(d*x^3 
 + c)*a*d + 2*(b*c - a*d)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x^3 + c)*b*s 
qrt(-(b*c - a*d)/b)/(b*c - a*d)))/(a*b), 1/3*(2*b*sqrt(-c)*c*arctan(sqrt(- 
c)/sqrt(d*x^3 + c)) + 2*sqrt(d*x^3 + c)*a*d - (b*c - a*d)*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)))/(a*b), 2/3*(b*sqrt(-c)*c*arctan(sqrt(-c)/sqrt(d*x^3 + c)) + 
sqrt(d*x^3 + c)*a*d + (b*c - a*d)*sqrt(-(b*c - a*d)/b)*arctan(-sqrt(d*x^3 
+ c)*b*sqrt(-(b*c - a*d)/b)/(b*c - a*d)))/(a*b)]
                                                                                    
                                                                                    
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (90) = 180\).

Time = 7.82 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.82 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (a+b x^3\right )} \, dx=\begin {cases} \frac {2 \left (\frac {d^{2} \sqrt {c + d x^{3}}}{3 b} + \frac {c^{2} d \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {- c}} \right )}}{3 a \sqrt {- c}} - \frac {d \left (a d - b c\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {c + d x^{3}}}{\sqrt {\frac {a d - b c}{b}}} \right )}}{3 a b^{2} \sqrt {\frac {a d - b c}{b}}}\right )}{d} & \text {for}\: d \neq 0 \\c^{\frac {3}{2}} \left (- \frac {2 b \left (\begin {cases} \frac {\frac {a}{2 b} + x^{3}}{a} & \text {for}\: b = 0 \\- \frac {\log {\left (a - 2 b \left (\frac {a}{2 b} + x^{3}\right ) \right )}}{2 b} & \text {otherwise} \end {cases}\right )}{3 a} - \frac {2 b \left (\begin {cases} \frac {\frac {a}{2 b} + x^{3}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + 2 b \left (\frac {a}{2 b} + x^{3}\right ) \right )}}{2 b} & \text {otherwise} \end {cases}\right )}{3 a}\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((2*(d**2*sqrt(c + d*x**3)/(3*b) + c**2*d*atan(sqrt(c + d*x**3)/s 
qrt(-c))/(3*a*sqrt(-c)) - d*(a*d - b*c)**2*atan(sqrt(c + d*x**3)/sqrt((a*d 
 - b*c)/b))/(3*a*b**2*sqrt((a*d - b*c)/b)))/d, Ne(d, 0)), (c**(3/2)*(-2*b* 
Piecewise(((a/(2*b) + x**3)/a, Eq(b, 0)), (-log(a - 2*b*(a/(2*b) + x**3))/ 
(2*b), True))/(3*a) - 2*b*Piecewise(((a/(2*b) + x**3)/a, Eq(b, 0)), (log(a 
 + 2*b*(a/(2*b) + x**3))/(2*b), True))/(3*a)), True))
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (a+b x^3\right )} \, dx=\frac {2 \, c^{2} \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{3 \, a \sqrt {-c}} + \frac {2 \, \sqrt {d x^{3} + c} d}{3 \, b} - \frac {2 \, {\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \arctan \left (\frac {\sqrt {d x^{3} + c} b}{\sqrt {-b^{2} c + a b d}}\right )}{3 \, \sqrt {-b^{2} c + a b d} a b} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 6.39 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.49 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (a+b x^3\right )} \, dx=\frac {c^{3/2}\,\ln \left (\frac {{\left (\sqrt {d\,x^3+c}-\sqrt {c}\right )}^3\,\left (\sqrt {d\,x^3+c}+\sqrt {c}\right )}{x^6}\right )}{3\,a}+\frac {2\,d\,\sqrt {d\,x^3+c}}{3\,b}+\frac {\ln \left (\frac {a^2\,d^2+2\,b^2\,c^2-a\,b\,d^2\,x^3+b^2\,c\,d\,x^3-3\,a\,b\,c\,d+\sqrt {b}\,\sqrt {d\,x^3+c}\,{\left (a\,d-b\,c\right )}^{3/2}\,2{}\mathrm {i}}{b\,x^3+a}\right )\,{\left (a\,d-b\,c\right )}^{3/2}\,1{}\mathrm {i}}{3\,a\,b^{3/2}} \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x \left (a+b x^3\right )} \, dx=\frac {4 \sqrt {d \,x^{3}+c}\, c +3 \left (\int \frac {\sqrt {d \,x^{3}+c}}{b d \,x^{7}+a d \,x^{4}+b c \,x^{4}+a c x}d x \right ) a \,c^{2}+3 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) a \,d^{2}-6 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{5}}{b d \,x^{6}+a d \,x^{3}+b c \,x^{3}+a c}d x \right ) b c d}{3 a} \] Input:

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

Output:

(4*sqrt(c + d*x**3)*c + 3*int(sqrt(c + d*x**3)/(a*c*x + a*d*x**4 + b*c*x** 
4 + b*d*x**7),x)*a*c**2 + 3*int((sqrt(c + d*x**3)*x**5)/(a*c + a*d*x**3 + 
b*c*x**3 + b*d*x**6),x)*a*d**2 - 6*int((sqrt(c + d*x**3)*x**5)/(a*c + a*d* 
x**3 + b*c*x**3 + b*d*x**6),x)*b*c*d)/(3*a)