\(\int \frac {(c+d x^3)^2}{(a+b x^3)^{5/3}} \, dx\) [135]

Optimal result

Integrand size = 21, antiderivative size = 132 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\frac {(b c-a d)^2 x}{2 a b^2 \left (a+b x^3\right )^{2/3}}+\frac {d^2 x \sqrt [3]{a+b x^3}}{2 b^2}+\frac {\left (b^2 c^2+2 a b c d-2 a^2 d^2\right ) x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{2 a b^2 \left (a+b x^3\right )^{2/3}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 13.14 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.30 \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\frac {x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Gamma}\left (\frac {2}{3}\right ) \left (4 a \left (14 c^2+7 c d x^3+2 d^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{3},\frac {10}{3},-\frac {b x^3}{a}\right )-b x^3 \left (11 c^2+16 c d x^3+5 d^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {8}{3},\frac {13}{3},-\frac {b x^3}{a}\right )-3 b x^3 \left (c+d x^3\right )^2 \, _3F_2\left (\frac {4}{3},2,\frac {8}{3};1,\frac {13}{3};-\frac {b x^3}{a}\right )\right )}{84 a^2 \left (a+b x^3\right )^{2/3} \operatorname {Gamma}\left (\frac {5}{3}\right )} \] Input:

Integrate[(c + d*x^3)^2/(a + b*x^3)^(5/3),x]
 

Output:

(x*(1 + (b*x^3)/a)^(2/3)*Gamma[2/3]*(4*a*(14*c^2 + 7*c*d*x^3 + 2*d^2*x^6)* 
Hypergeometric2F1[1/3, 5/3, 10/3, -((b*x^3)/a)] - b*x^3*(11*c^2 + 16*c*d*x 
^3 + 5*d^2*x^6)*Hypergeometric2F1[4/3, 8/3, 13/3, -((b*x^3)/a)] - 3*b*x^3* 
(c + d*x^3)^2*HypergeometricPFQ[{4/3, 2, 8/3}, {1, 13/3}, -((b*x^3)/a)]))/ 
(84*a^2*(a + b*x^3)^(2/3)*Gamma[5/3])
 

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.11, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {930, 913, 779, 778}

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.

\(\displaystyle \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx\)

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 913

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

\(\Big \downarrow \) 779

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

\(\Big \downarrow \) 778

\(\displaystyle \frac {\frac {x \left (\frac {b x^3}{a}+1\right )^{2/3} \left (-2 a^2 d^2+2 a b c d+b^2 c^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{b \left (a+b x^3\right )^{2/3}}-\frac {d x \sqrt [3]{a+b x^3} (b c-2 a d)}{b}}{2 a b}+\frac {x \left (c+d x^3\right ) (b c-a d)}{2 a b \left (a+b x^3\right )^{2/3}}\)

Input:

Int[(c + d*x^3)^2/(a + b*x^3)^(5/3),x]
 

Output:

((b*c - a*d)*x*(c + d*x^3))/(2*a*b*(a + b*x^3)^(2/3)) + (-((d*(b*c - 2*a*d 
)*x*(a + b*x^3)^(1/3))/b) + ((b^2*c^2 + 2*a*b*c*d - 2*a^2*d^2)*x*(1 + (b*x 
^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/3, 4/3, -((b*x^3)/a)])/(b*(a + b*x^3 
)^(2/3)))/(2*a*b)
 

Defintions of rubi rules used

rule 778

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 
1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p 
, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || 
GtQ[a, 0])
 

rule 779

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x 
^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p])   Int[(1 + b*(x^n/a))^p, x], x 
] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Si 
mplify[1/n + p], 0] &&  !(IntegerQ[p] || GtQ[a, 0])
 

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 [F]

\[\int \frac {\left (d \,x^{3}+c \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {5}{3}}}d x\]

Input:

int((d*x^3+c)^2/(b*x^3+a)^(5/3),x)
 

Output:

int((d*x^3+c)^2/(b*x^3+a)^(5/3),x)
 

Fricas [F]

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

integrate((d*x^3+c)^2/(b*x^3+a)^(5/3),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((d*x**3+c)**2/(b*x**3+a)**(5/3),x)
 

Output:

Integral((c + d*x**3)**2/(a + b*x**3)**(5/3), x)
 

Maxima [F]

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

integrate((d*x^3+c)^2/(b*x^3+a)^(5/3),x, algorithm="maxima")
 

Output:

integrate((d*x^3 + c)^2/(b*x^3 + a)^(5/3), x)
 

Giac [F]

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

integrate((d*x^3+c)^2/(b*x^3+a)^(5/3),x, algorithm="giac")
 

Output:

integrate((d*x^3 + c)^2/(b*x^3 + a)^(5/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\int \frac {{\left (d\,x^3+c\right )}^2}{{\left (b\,x^3+a\right )}^{5/3}} \,d x \] Input:

int((c + d*x^3)^2/(a + b*x^3)^(5/3),x)
 

Output:

int((c + d*x^3)^2/(a + b*x^3)^(5/3), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^3\right )^2}{\left (a+b x^3\right )^{5/3}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a +\left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}}d x \right ) d^{2}+2 \left (\int \frac {x^{3}}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a +\left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}}d x \right ) c d +\left (\int \frac {1}{\left (b \,x^{3}+a \right )^{\frac {2}{3}} a +\left (b \,x^{3}+a \right )^{\frac {2}{3}} b \,x^{3}}d x \right ) c^{2} \] Input:

int((d*x^3+c)^2/(b*x^3+a)^(5/3),x)
 

Output:

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