\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{4/3}} \, dx\) [125]

Optimal result

Integrand size = 21, antiderivative size = 144 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {(b c-a d)^2 x}{a b^2 \sqrt [3]{a+b x^3}}+\frac {d^2 x \left (a+b x^3\right )^{2/3}}{3 b^2}+\frac {2 d (3 b c-2 a d) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{3 \sqrt {3} b^{7/3}}-\frac {d (3 b c-2 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{3 b^{7/3}} \] Output:

(-a*d+b*c)^2*x/a/b^2/(b*x^3+a)^(1/3)+1/3*d^2*x*(b*x^3+a)^(2/3)/b^2+2/9*d*( 
-2*a*d+3*b*c)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/ 
b^(7/3)-1/3*d*(-2*a*d+3*b*c)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(7/3)
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.38 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {\frac {3 \sqrt [3]{b} x \left (3 b^2 c^2+4 a^2 d^2+a b d \left (-6 c+d x^3\right )\right )}{a \sqrt [3]{a+b x^3}}+2 \sqrt {3} d (3 b c-2 a d) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )+2 d (-3 b c+2 a d) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+d (3 b c-2 a d) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{9 b^{7/3}} \] Input:

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

Output:

((3*b^(1/3)*x*(3*b^2*c^2 + 4*a^2*d^2 + a*b*d*(-6*c + d*x^3)))/(a*(a + b*x^ 
3)^(1/3)) + 2*Sqrt[3]*d*(3*b*c - 2*a*d)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3 
)*x + 2*(a + b*x^3)^(1/3))] + 2*d*(-3*b*c + 2*a*d)*Log[-(b^(1/3)*x) + (a + 
 b*x^3)^(1/3)] + d*(3*b*c - 2*a*d)*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3) 
^(1/3) + (a + b*x^3)^(2/3)])/(9*b^(7/3))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.10, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {930, 27, 913, 769}

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.

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

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 913

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

\(\Big \downarrow \) 769

\(\displaystyle \frac {d \left (\frac {2 a (3 b c-2 a d) \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{3 b}-\frac {x \left (a+b x^3\right )^{2/3} (3 b c-4 a d)}{3 b}\right )}{a b}+\frac {x \left (c+d x^3\right ) (b c-a d)}{a b \sqrt [3]{a+b x^3}}\)

Input:

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

Output:

((b*c - a*d)*x*(c + d*x^3))/(a*b*(a + b*x^3)^(1/3)) + (d*(-1/3*((3*b*c - 4 
*a*d)*x*(a + b*x^3)^(2/3))/b + (2*a*(3*b*c - 2*a*d)*(ArcTan[(1 + (2*b^(1/3 
)*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*b^(1/3)) - Log[-(b^(1/3)*x) + (a 
 + b*x^3)^(1/3)]/(2*b^(1/3))))/(3*b)))/(a*b)
 

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 769

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]* 
(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sqrt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^ 
3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]
 

rule 913

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[d*x*((a + b*x^n)^(p + 1)/(b*(n*(p + 1) + 1))), x] - Simp[(a*d - b*c*(n*( 
p + 1) + 1))/(b*(n*(p + 1) + 1))   Int[(a + b*x^n)^p, x], x] /; FreeQ[{a, b 
, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]
 

rule 930

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[(a*d - c*b)*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 
1))), x] - Simp[1/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q 
- 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*( 
p + q) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q, x]
 

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.15

method	result	size
pseudoelliptic	\(\frac {\frac {4 a d \,b^{2} \left (\arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) \sqrt {3}+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) \left (a d -\frac {3 b c}{2}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{9}+\left (b^{2} c^{2}-2 a d \left (-\frac {d \,x^{3}}{6}+c \right ) b +\frac {4 a^{2} d^{2}}{3}\right ) x \,b^{\frac {7}{3}}}{b^{\frac {13}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} a}\)	\(166\)

Input:

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

Output:

4/9*(a*d*b^2*(arctan(1/3*3^(1/2)*(2*(b*x^3+a)^(1/3)/b^(1/3)+x)/x)*3^(1/2)+ 
ln((-b^(1/3)*x+(b*x^3+a)^(1/3))/x)-1/2*ln((b^(2/3)*x^2+b^(1/3)*(b*x^3+a)^( 
1/3)*x+(b*x^3+a)^(2/3))/x^2))*(a*d-3/2*b*c)*(b*x^3+a)^(1/3)+9/4*(b^2*c^2-2 
*a*d*(-1/6*d*x^3+c)*b+4/3*a^2*d^2)*x*b^(7/3))/(b*x^3+a)^(1/3)/b^(13/3)/a
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 299 vs. \(2 (119) = 238\).

Time = 0.12 (sec) , antiderivative size = 652, normalized size of antiderivative = 4.53 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{4/3}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/9*(3*sqrt(1/3)*(3*a^2*b^2*c*d - 2*a^3*b*d^2 + (3*a*b^3*c*d - 2*a^2*b^2 
*d^2)*x^3)*sqrt(-1/b^(2/3))*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*b^(2/3)*x^2 
- 3*sqrt(1/3)*(b^(4/3)*x^3 + (b*x^3 + a)^(1/3)*b*x^2 - 2*(b*x^3 + a)^(2/3) 
*b^(2/3)*x)*sqrt(-1/b^(2/3)) + 2*a) + 2*(3*a^2*b*c*d - 2*a^3*d^2 + (3*a*b^ 
2*c*d - 2*a^2*b*d^2)*x^3)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) 
- (3*a^2*b*c*d - 2*a^3*d^2 + (3*a*b^2*c*d - 2*a^2*b*d^2)*x^3)*b^(2/3)*log( 
(b^(2/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) - 3*( 
a*b^2*d^2*x^4 + (3*b^3*c^2 - 6*a*b^2*c*d + 4*a^2*b*d^2)*x)*(b*x^3 + a)^(2/ 
3))/(a*b^4*x^3 + a^2*b^3), -1/9*(2*(3*a^2*b*c*d - 2*a^3*d^2 + (3*a*b^2*c*d 
 - 2*a^2*b*d^2)*x^3)*b^(2/3)*log(-(b^(1/3)*x - (b*x^3 + a)^(1/3))/x) - (3* 
a^2*b*c*d - 2*a^3*d^2 + (3*a*b^2*c*d - 2*a^2*b*d^2)*x^3)*b^(2/3)*log((b^(2 
/3)*x^2 + (b*x^3 + a)^(1/3)*b^(1/3)*x + (b*x^3 + a)^(2/3))/x^2) + 6*sqrt(1 
/3)*(3*a^2*b^2*c*d - 2*a^3*b*d^2 + (3*a*b^3*c*d - 2*a^2*b^2*d^2)*x^3)*arct 
an(sqrt(1/3)*(b^(1/3)*x + 2*(b*x^3 + a)^(1/3))/(b^(1/3)*x))/b^(1/3) - 3*(a 
*b^2*d^2*x^4 + (3*b^3*c^2 - 6*a*b^2*c*d + 4*a^2*b*d^2)*x)*(b*x^3 + a)^(2/3 
))/(a*b^4*x^3 + a^2*b^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (119) = 238\).

Time = 0.11 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.09 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {1}{9} \, d^{2} {\left (\frac {4 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {7}{3}}} + \frac {3 \, {\left (3 \, a b - \frac {4 \, {\left (b x^{3} + a\right )} a}{x^{3}}\right )}}{\frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{3}}{x} - \frac {{\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{2}}{x^{4}}} - \frac {2 \, a \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {7}{3}}} + \frac {4 \, a \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {7}{3}}}\right )} - \frac {1}{3} \, c d {\left (\frac {2 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {4}{3}}} + \frac {6 \, x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} b} - \frac {\log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {4}{3}}} + \frac {2 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {4}{3}}}\right )} + \frac {c^{2} x}{{\left (b x^{3} + a\right )}^{\frac {1}{3}} a} \] Input:

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

Output:

1/9*d^2*(4*sqrt(3)*a*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x)/ 
b^(1/3))/b^(7/3) + 3*(3*a*b - 4*(b*x^3 + a)*a/x^3)/((b*x^3 + a)^(1/3)*b^3/ 
x - (b*x^3 + a)^(4/3)*b^2/x^4) - 2*a*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^(1/ 
3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(7/3) + 4*a*log(-b^(1/3) + (b*x^3 + a)^(1/ 
3)/x)/b^(7/3)) - 1/3*c*d*(2*sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 
 + a)^(1/3)/x)/b^(1/3))/b^(4/3) + 6*x/((b*x^3 + a)^(1/3)*b) - log(b^(2/3) 
+ (b*x^3 + a)^(1/3)*b^(1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(4/3) + 2*log(-b^ 
(1/3) + (b*x^3 + a)^(1/3)/x)/b^(4/3)) + c^2*x/((b*x^3 + a)^(1/3)*a)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{4/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a +\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}}d x \right ) d^{2}+2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a +\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a +\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}}d x \right ) c^{2} \] Input:

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

Output:

int(x**6/((a + b*x**3)**(1/3)*a + (a + b*x**3)**(1/3)*b*x**3),x)*d**2 + 2* 
int(x**3/((a + b*x**3)**(1/3)*a + (a + b*x**3)**(1/3)*b*x**3),x)*c*d + int 
(1/((a + b*x**3)**(1/3)*a + (a + b*x**3)**(1/3)*b*x**3),x)*c**2