\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{7/3}} \, dx\) [126]

Optimal result

Integrand size = 21, antiderivative size = 147 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{7/3}} \, dx=\frac {(b c-a d)^2 x}{4 a b^2 \left (a+b x^3\right )^{4/3}}+\frac {(b c-a d) (3 b c+5 a d) x}{4 a^2 b^2 \sqrt [3]{a+b x^3}}+\frac {d^2 \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{\sqrt {3} b^{7/3}}-\frac {d^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{2 b^{7/3}} \] Output:

1/4*(-a*d+b*c)^2*x/a/b^2/(b*x^3+a)^(4/3)+1/4*(-a*d+b*c)*(5*a*d+3*b*c)*x/a^ 
2/b^2/(b*x^3+a)^(1/3)+1/3*d^2*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3 
^(1/2))*3^(1/2)/b^(7/3)-1/2*d^2*ln(-b^(1/3)*x+(b*x^3+a)^(1/3))/b^(7/3)
 

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.27 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{7/3}} \, dx=\frac {\frac {3 \sqrt [3]{b} (b c-a d) x \left (4 a^2 d+3 b^2 c x^3+a b \left (4 c+5 d x^3\right )\right )}{a^2 \left (a+b x^3\right )^{4/3}}+4 \sqrt {3} d^2 \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-4 d^2 \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+2 d^2 \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{12 b^{7/3}} \] Input:

Integrate[(c + d*x^3)^2/(a + b*x^3)^(7/3),x]
 

Output:

((3*b^(1/3)*(b*c - a*d)*x*(4*a^2*d + 3*b^2*c*x^3 + a*b*(4*c + 5*d*x^3)))/( 
a^2*(a + b*x^3)^(4/3)) + 4*Sqrt[3]*d^2*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/3) 
*x + 2*(a + b*x^3)^(1/3))] - 4*d^2*Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + 
 2*d^2*Log[b^(2/3)*x^2 + b^(1/3)*x*(a + b*x^3)^(1/3) + (a + b*x^3)^(2/3)]) 
/(12*b^(7/3))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {930, 910, 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.

\(\displaystyle \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{7/3}} \, dx\)

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 910

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

\(\Big \downarrow \) 769

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

Input:

Int[(c + d*x^3)^2/(a + b*x^3)^(7/3),x]
 

Output:

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

Defintions of rubi rules used

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 910

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si 
mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - 
 b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1))   Int[(a + b*x^n)^(p + 1), x], x] /; 
FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ 
n + p, 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 = 1.24 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.50

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right ) a^{2} d^{2} b^{2} \left (b \,x^{3}+a \right )^{\frac {4}{3}}-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}+x \right )}{3 x}\right ) a^{2} d^{2} b^{2} \left (b \,x^{3}+a \right )^{\frac {4}{3}}-2 \ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right ) a^{2} d^{2} b^{2} \left (b \,x^{3}+a \right )^{\frac {4}{3}}-\frac {3 x \,b^{\frac {7}{3}} \left (5 a^{2} b \,d^{2} x^{3}-2 c \,x^{3} d \,b^{2} a -3 x^{3} b^{3} c^{2}+4 a^{3} d^{2}-4 a \,b^{2} c^{2}\right )}{2}}{6 \left (b \,x^{3}+a \right )^{\frac {4}{3}} b^{\frac {13}{3}} a^{2}}\)	\(221\)

Input:

int((d*x^3+c)^2/(b*x^3+a)^(7/3),x,method=_RETURNVERBOSE)
 

Output:

1/6/(b*x^3+a)^(4/3)*(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^2*d^2*b^2*(b*x^3+a)^(4/3)-2*3^(1/2)*arctan(1/3*3^(1/2)*(2*(b* 
x^3+a)^(1/3)/b^(1/3)+x)/x)*a^2*d^2*b^2*(b*x^3+a)^(4/3)-2*ln((-b^(1/3)*x+(b 
*x^3+a)^(1/3))/x)*a^2*d^2*b^2*(b*x^3+a)^(4/3)-3/2*x*b^(7/3)*(5*a^2*b*d^2*x 
^3-2*a*b^2*c*d*x^3-3*b^3*c^2*x^3+4*a^3*d^2-4*a*b^2*c^2))/b^(13/3)/a^2
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (123) = 246\).

Time = 0.11 (sec) , antiderivative size = 719, normalized size of antiderivative = 4.89 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{7/3}} \, dx =\text {Too large to display} \] Input:

integrate((d*x^3+c)^2/(b*x^3+a)^(7/3),x, algorithm="fricas")
 

Output:

[1/12*(6*sqrt(1/3)*(a^2*b^3*d^2*x^6 + 2*a^3*b^2*d^2*x^3 + a^4*b*d^2)*sqrt( 
(-b)^(1/3)/b)*log(3*b*x^3 - 3*(b*x^3 + a)^(1/3)*(-b)^(2/3)*x^2 - 3*sqrt(1/ 
3)*((-b)^(1/3)*b*x^3 - (b*x^3 + a)^(1/3)*b*x^2 + 2*(b*x^3 + a)^(2/3)*(-b)^ 
(2/3)*x)*sqrt((-b)^(1/3)/b) + 2*a) - 4*(a^2*b^2*d^2*x^6 + 2*a^3*b*d^2*x^3 
+ a^4*d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + 2*(a^2*b 
^2*d^2*x^6 + 2*a^3*b*d^2*x^3 + a^4*d^2)*(-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*((3*b^4*c^2 + 
2*a*b^3*c*d - 5*a^2*b^2*d^2)*x^4 + 4*(a*b^3*c^2 - a^3*b*d^2)*x)*(b*x^3 + a 
)^(2/3))/(a^2*b^5*x^6 + 2*a^3*b^4*x^3 + a^4*b^3), -1/12*(12*sqrt(1/3)*(a^2 
*b^3*d^2*x^6 + 2*a^3*b^2*d^2*x^3 + a^4*b*d^2)*sqrt(-(-b)^(1/3)/b)*arctan(- 
sqrt(1/3)*((-b)^(1/3)*x - 2*(b*x^3 + a)^(1/3))*sqrt(-(-b)^(1/3)/b)/x) + 4* 
(a^2*b^2*d^2*x^6 + 2*a^3*b*d^2*x^3 + a^4*d^2)*(-b)^(2/3)*log(((-b)^(1/3)*x 
 + (b*x^3 + a)^(1/3))/x) - 2*(a^2*b^2*d^2*x^6 + 2*a^3*b*d^2*x^3 + a^4*d^2) 
*(-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*((3*b^4*c^2 + 2*a*b^3*c*d - 5*a^2*b^2*d^2)*x^4 + 4*(a 
*b^3*c^2 - a^3*b*d^2)*x)*(b*x^3 + a)^(2/3))/(a^2*b^5*x^6 + 2*a^3*b^4*x^3 + 
 a^4*b^3)]
 

Sympy [F]

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

integrate((d*x**3+c)**2/(b*x**3+a)**(7/3),x)
 

Output:

Integral((c + d*x**3)**2/(a + b*x**3)**(7/3), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.29 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{7/3}} \, dx=-\frac {{\left (b - \frac {4 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} c^{2} x^{4}}{4 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a^{2}} + \frac {c d x^{4}}{2 \, {\left (b x^{3} + a\right )}^{\frac {4}{3}} a} - \frac {1}{12} \, {\left (\frac {3 \, {\left (b + \frac {4 \, {\left (b x^{3} + a\right )}}{x^{3}}\right )} x^{4}}{{\left (b x^{3} + a\right )}^{\frac {4}{3}} b^{2}} + \frac {4 \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (b^{\frac {1}{3}} + \frac {2 \, {\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{b^{\frac {7}{3}}} - \frac {2 \, \log \left (b^{\frac {2}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}} b^{\frac {1}{3}}}{x} + \frac {{\left (b x^{3} + a\right )}^{\frac {2}{3}}}{x^{2}}\right )}{b^{\frac {7}{3}}} + \frac {4 \, \log \left (-b^{\frac {1}{3}} + \frac {{\left (b x^{3} + a\right )}^{\frac {1}{3}}}{x}\right )}{b^{\frac {7}{3}}}\right )} d^{2} \] Input:

integrate((d*x^3+c)^2/(b*x^3+a)^(7/3),x, algorithm="maxima")
 

Output:

-1/4*(b - 4*(b*x^3 + a)/x^3)*c^2*x^4/((b*x^3 + a)^(4/3)*a^2) + 1/2*c*d*x^4 
/((b*x^3 + a)^(4/3)*a) - 1/12*(3*(b + 4*(b*x^3 + a)/x^3)*x^4/((b*x^3 + a)^ 
(4/3)*b^2) + 4*sqrt(3)*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3)/x 
)/b^(1/3))/b^(7/3) - 2*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*log(-b^(1/3) + (b*x^3 + a)^(1/3)/x)/b^(7/3))*d 
^2
 

Giac [F]

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

integrate((d*x^3+c)^2/(b*x^3+a)^(7/3),x, algorithm="giac")
 

Output:

integrate((d*x^3 + c)^2/(b*x^3 + a)^(7/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{7/3}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^2}{{\left (b\,x^3+a\right )}^{7/3}} \,d x \] Input:

int((c + d*x^3)^2/(a + b*x^3)^(7/3),x)
 

Output:

int((c + d*x^3)^2/(a + b*x^3)^(7/3), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{7/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{6}}d x \right ) d^{2}+2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{6}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a^{2}+2 \left (b \,x^{3}+a \right )^{\frac {1}{3}} a b \,x^{3}+\left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{2} x^{6}}d x \right ) c^{2} \] Input:

int((d*x^3+c)^2/(b*x^3+a)^(7/3),x)
                                                                                    
                                                                                    
 

Output:

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