\(\int \frac {(c+d x^3)^3}{(a+b x^3)^{4/3}} \, dx\) [103]

Optimal result

Integrand size = 21, antiderivative size = 205 \[ \int \frac {\left (c+d x^3\right )^3}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {(b c-a d)^3 x}{a b^3 \sqrt [3]{a+b x^3}}+\frac {d^2 (9 b c-5 a d) x \left (a+b x^3\right )^{2/3}}{9 b^3}+\frac {d^3 x^4 \left (a+b x^3\right )^{2/3}}{6 b^2}+\frac {d \left (27 b^2 c^2-36 a b c d+14 a^2 d^2\right ) \arctan \left (\frac {1+\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}}{\sqrt {3}}\right )}{9 \sqrt {3} b^{10/3}}-\frac {d \left (27 b^2 c^2-36 a b c d+14 a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )}{18 b^{10/3}} \] Output:

(-a*d+b*c)^3*x/a/b^3/(b*x^3+a)^(1/3)+1/9*d^2*(-5*a*d+9*b*c)*x*(b*x^3+a)^(2 
/3)/b^3+1/6*d^3*x^4*(b*x^3+a)^(2/3)/b^2+1/27*d*(14*a^2*d^2-36*a*b*c*d+27*b 
^2*c^2)*arctan(1/3*(1+2*b^(1/3)*x/(b*x^3+a)^(1/3))*3^(1/2))*3^(1/2)/b^(10/ 
3)-1/18*d*(14*a^2*d^2-36*a*b*c*d+27*b^2*c^2)*ln(-b^(1/3)*x+(b*x^3+a)^(1/3) 
)/b^(10/3)
 

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.33 \[ \int \frac {\left (c+d x^3\right )^3}{\left (a+b x^3\right )^{4/3}} \, dx=\frac {\frac {3 \sqrt [3]{b} x \left (18 b^3 c^3-28 a^3 d^3+a^2 b d^2 \left (72 c-7 d x^3\right )+3 a b^2 d \left (-18 c^2+6 c d x^3+d^2 x^6\right )\right )}{a \sqrt [3]{a+b x^3}}+2 \sqrt {3} d \left (27 b^2 c^2-36 a b c d+14 a^2 d^2\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{a+b x^3}}\right )-2 d \left (27 b^2 c^2-36 a b c d+14 a^2 d^2\right ) \log \left (-\sqrt [3]{b} x+\sqrt [3]{a+b x^3}\right )+d \left (27 b^2 c^2-36 a b c d+14 a^2 d^2\right ) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{a+b x^3}+\left (a+b x^3\right )^{2/3}\right )}{54 b^{10/3}} \] Input:

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

Output:

((3*b^(1/3)*x*(18*b^3*c^3 - 28*a^3*d^3 + a^2*b*d^2*(72*c - 7*d*x^3) + 3*a* 
b^2*d*(-18*c^2 + 6*c*d*x^3 + d^2*x^6)))/(a*(a + b*x^3)^(1/3)) + 2*Sqrt[3]* 
d*(27*b^2*c^2 - 36*a*b*c*d + 14*a^2*d^2)*ArcTan[(Sqrt[3]*b^(1/3)*x)/(b^(1/ 
3)*x + 2*(a + b*x^3)^(1/3))] - 2*d*(27*b^2*c^2 - 36*a*b*c*d + 14*a^2*d^2)* 
Log[-(b^(1/3)*x) + (a + b*x^3)^(1/3)] + d*(27*b^2*c^2 - 36*a*b*c*d + 14*a^ 
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)]) 
/(54*b^(10/3))
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {930, 27, 1025, 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.

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

\(\Big \downarrow \) 930

\(\displaystyle \frac {\int \frac {d \left (d x^3+c\right ) \left (a c-(6 b c-7 a d) x^3\right )}{\sqrt [3]{b x^3+a}}dx}{a b}+\frac {x \left (c+d x^3\right )^2 (b c-a d)}{a b \sqrt [3]{a+b x^3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d \int \frac {\left (d x^3+c\right ) \left (a c-(6 b c-7 a d) x^3\right )}{\sqrt [3]{b x^3+a}}dx}{a b}+\frac {x \left (c+d x^3\right )^2 (b c-a d)}{a b \sqrt [3]{a+b x^3}}\)

\(\Big \downarrow \) 1025

\(\displaystyle \frac {d \left (\frac {\int \frac {a c (12 b c-7 a d)-\left (18 b^2 c^2-51 a b d c+28 a^2 d^2\right ) x^3}{\sqrt [3]{b x^3+a}}dx}{6 b}-\frac {x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) (6 b c-7 a d)}{6 b}\right )}{a b}+\frac {x \left (c+d x^3\right )^2 (b c-a d)}{a b \sqrt [3]{a+b x^3}}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {d \left (\frac {\frac {2 a \left (14 a^2 d^2-36 a b c d+27 b^2 c^2\right ) \int \frac {1}{\sqrt [3]{b x^3+a}}dx}{3 b}-\frac {x \left (a+b x^3\right )^{2/3} \left (28 a^2 d^2-51 a b c d+18 b^2 c^2\right )}{3 b}}{6 b}-\frac {x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) (6 b c-7 a d)}{6 b}\right )}{a b}+\frac {x \left (c+d x^3\right )^2 (b c-a d)}{a b \sqrt [3]{a+b x^3}}\)

\(\Big \downarrow \) 769

\(\displaystyle \frac {d \left (\frac {\frac {2 a \left (14 a^2 d^2-36 a b c d+27 b^2 c^2\right ) \left (\frac {\arctan \left (\frac {\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a+b x^3}}+1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a+b x^3}-\sqrt [3]{b} x\right )}{2 \sqrt [3]{b}}\right )}{3 b}-\frac {x \left (a+b x^3\right )^{2/3} \left (28 a^2 d^2-51 a b c d+18 b^2 c^2\right )}{3 b}}{6 b}-\frac {x \left (a+b x^3\right )^{2/3} \left (c+d x^3\right ) (6 b c-7 a d)}{6 b}\right )}{a b}+\frac {x \left (c+d x^3\right )^2 (b c-a d)}{a b \sqrt [3]{a+b x^3}}\)

Input:

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

Output:

((b*c - a*d)*x*(c + d*x^3)^2)/(a*b*(a + b*x^3)^(1/3)) + (d*(-1/6*((6*b*c - 
 7*a*d)*x*(a + b*x^3)^(2/3)*(c + d*x^3))/b + (-1/3*((18*b^2*c^2 - 51*a*b*c 
*d + 28*a^2*d^2)*x*(a + b*x^3)^(2/3))/b + (2*a*(27*b^2*c^2 - 36*a*b*c*d + 
14*a^2*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))))/(3*b))/(6 
*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]
 

rule 1025

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

Maple [A] (verified)

Time = 1.73 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.04

method	result	size
pseudoelliptic	\(-\frac {14 \left (a d \left (\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (b \,x^{3}+a \right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right )+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (b \,x^{3}+a \right )^{\frac {1}{3}} x +\left (b \,x^{3}+a \right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) \left (a^{2} d^{2}-\frac {18}{7} a b c d +\frac {27}{14} b^{2} c^{2}\right ) \left (b \,x^{3}+a \right )^{\frac {1}{3}}+3 x \left (\frac {27 a d \left (-\frac {1}{18} d^{2} x^{6}-\frac {1}{3} c d \,x^{3}+c^{2}\right ) b^{\frac {7}{3}}}{14}-\frac {18 a^{2} d^{2} \left (-\frac {7 d \,x^{3}}{72}+c \right ) b^{\frac {4}{3}}}{7}+a^{3} d^{3} b^{\frac {1}{3}}-\frac {9 b^{\frac {10}{3}} c^{3}}{14}\right )\right )}{27 \left (b \,x^{3}+a \right )^{\frac {1}{3}} b^{\frac {10}{3}} a}\)	\(213\)

Input:

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

Output:

-14/27/(b*x^3+a)^(1/3)/b^(10/3)*(a*d*(3^(1/2)*arctan(1/3*3^(1/2)*(b^(1/3)* 
x+2*(b*x^3+a)^(1/3))/b^(1/3)/x)+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^2*d^2-18/ 
7*a*b*c*d+27/14*b^2*c^2)*(b*x^3+a)^(1/3)+3*x*(27/14*a*d*(-1/18*d^2*x^6-1/3 
*c*d*x^3+c^2)*b^(7/3)-18/7*a^2*d^2*(-7/72*d*x^3+c)*b^(4/3)+a^3*d^3*b^(1/3) 
-9/14*b^(10/3)*c^3))/a
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (176) = 352\).

Time = 0.14 (sec) , antiderivative size = 956, normalized size of antiderivative = 4.66 \[ \int \frac {\left (c+d x^3\right )^3}{\left (a+b x^3\right )^{4/3}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/54*(3*sqrt(1/3)*(27*a^2*b^3*c^2*d - 36*a^3*b^2*c*d^2 + 14*a^4*b*d^3 + ( 
27*a*b^4*c^2*d - 36*a^2*b^3*c*d^2 + 14*a^3*b^2*d^3)*x^3)*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)*sqr 
t((-b)^(1/3)/b) + 2*a) - 2*(27*a^2*b^2*c^2*d - 36*a^3*b*c*d^2 + 14*a^4*d^3 
 + (27*a*b^3*c^2*d - 36*a^2*b^2*c*d^2 + 14*a^3*b*d^3)*x^3)*(-b)^(2/3)*log( 
((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) + (27*a^2*b^2*c^2*d - 36*a^3*b*c*d^2 
 + 14*a^4*d^3 + (27*a*b^3*c^2*d - 36*a^2*b^2*c*d^2 + 14*a^3*b*d^3)*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*(3*a*b^3*d^3*x^7 + (18*a*b^3*c*d^2 - 7*a^2*b^2*d^3)*x^4 
+ 2*(9*b^4*c^3 - 27*a*b^3*c^2*d + 36*a^2*b^2*c*d^2 - 14*a^3*b*d^3)*x)*(b*x 
^3 + a)^(2/3))/(a*b^5*x^3 + a^2*b^4), -1/54*(6*sqrt(1/3)*(27*a^2*b^3*c^2*d 
 - 36*a^3*b^2*c*d^2 + 14*a^4*b*d^3 + (27*a*b^4*c^2*d - 36*a^2*b^3*c*d^2 + 
14*a^3*b^2*d^3)*x^3)*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) + 2*(27*a^2*b^2*c^2*d - 36*a^ 
3*b*c*d^2 + 14*a^4*d^3 + (27*a*b^3*c^2*d - 36*a^2*b^2*c*d^2 + 14*a^3*b*d^3 
)*x^3)*(-b)^(2/3)*log(((-b)^(1/3)*x + (b*x^3 + a)^(1/3))/x) - (27*a^2*b^2* 
c^2*d - 36*a^3*b*c*d^2 + 14*a^4*d^3 + (27*a*b^3*c^2*d - 36*a^2*b^2*c*d^2 + 
 14*a^3*b*d^3)*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*(3*a*b^3*d^3*x^7 + (18*a*b^3*c*...
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

Integral((c + d*x**3)**3/(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. 514 vs. \(2 (176) = 352\).

Time = 0.12 (sec) , antiderivative size = 514, normalized size of antiderivative = 2.51 \[ \int \frac {\left (c+d x^3\right )^3}{\left (a+b x^3\right )^{4/3}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/54*d^3*(28*sqrt(3)*a^2*arctan(1/3*sqrt(3)*(b^(1/3) + 2*(b*x^3 + a)^(1/3 
)/x)/b^(1/3))/b^(10/3) + 3*(18*a^2*b^2 - 49*(b*x^3 + a)*a^2*b/x^3 + 28*(b* 
x^3 + a)^2*a^2/x^6)/((b*x^3 + a)^(1/3)*b^5/x - 2*(b*x^3 + a)^(4/3)*b^4/x^4 
 + (b*x^3 + a)^(7/3)*b^3/x^7) - 14*a^2*log(b^(2/3) + (b*x^3 + a)^(1/3)*b^( 
1/3)/x + (b*x^3 + a)^(2/3)/x^2)/b^(10/3) + 28*a^2*log(-b^(1/3) + (b*x^3 + 
a)^(1/3)/x)/b^(10/3)) + 1/3*c*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)) - 1/2*c^2*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^3*x/ 
((b*x^3 + a)^(1/3)*a)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^3}{\left (a+b x^3\right )^{4/3}} \, dx=\left (\int \frac {x^{9}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a +\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}}d x \right ) d^{3}+3 \left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a +\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}}d x \right ) c \,d^{2}+3 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a +\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}}d x \right ) c^{2} d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {1}{3}} a +\left (b \,x^{3}+a \right )^{\frac {1}{3}} b \,x^{3}}d x \right ) c^{3} \] Input:

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

Output:

int(x**9/((a + b*x**3)**(1/3)*a + (a + b*x**3)**(1/3)*b*x**3),x)*d**3 + 3* 
int(x**6/((a + b*x**3)**(1/3)*a + (a + b*x**3)**(1/3)*b*x**3),x)*c*d**2 + 
3*int(x**3/((a + b*x**3)**(1/3)*a + (a + b*x**3)**(1/3)*b*x**3),x)*c**2*d 
+ int(1/((a + b*x**3)**(1/3)*a + (a + b*x**3)**(1/3)*b*x**3),x)*c**3