Integrand size = 22, antiderivative size = 93 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}+\frac {9 x}{77 \left (a+b x^3\right )^{11/3}}+\frac {57 x \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {11}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{77 a^3 \left (a+b x^3\right )^{2/3}} \]
1/7*x*(-b*x^3+a)/(b*x^3+a)^(14/3)+9/77*x/(b*x^3+a)^(11/3)+57/77*x*(1+b*x^3 /a)^(2/3)*hypergeom([1/3, 11/3],[4/3],-b*x^3/a)/a^3/(b*x^3+a)^(2/3)
Time = 10.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.14 \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\frac {x \left (2282 a^4+4879 a^3 b x^3+6270 a^2 b^2 x^6+3591 a b^3 x^9+798 b^4 x^{12}+798 \left (a+b x^3\right )^4 \left (1+\frac {b x^3}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {4}{3},-\frac {b x^3}{a}\right )\right )}{3080 a^3 \left (a+b x^3\right )^{14/3}} \]
(x*(2282*a^4 + 4879*a^3*b*x^3 + 6270*a^2*b^2*x^6 + 3591*a*b^3*x^9 + 798*b^ 4*x^12 + 798*(a + b*x^3)^4*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 2/ 3, 4/3, -((b*x^3)/a)]))/(3080*a^3*(a + b*x^3)^(14/3))
Time = 0.22 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {930, 27, 910, 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.
\(\displaystyle \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx\) |
\(\Big \downarrow \) 930 |
\(\displaystyle \frac {\int \frac {6 a b \left (2 a-b x^3\right )}{\left (b x^3+a\right )^{14/3}}dx}{14 a b}+\frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{7} \int \frac {2 a-b x^3}{\left (b x^3+a\right )^{14/3}}dx+\frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}\) |
\(\Big \downarrow \) 910 |
\(\displaystyle \frac {3}{7} \left (\frac {19}{11} \int \frac {1}{\left (b x^3+a\right )^{11/3}}dx+\frac {3 x}{11 \left (a+b x^3\right )^{11/3}}\right )+\frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \frac {3}{7} \left (\frac {19 \left (\frac {b x^3}{a}+1\right )^{2/3} \int \frac {1}{\left (\frac {b x^3}{a}+1\right )^{11/3}}dx}{11 a^3 \left (a+b x^3\right )^{2/3}}+\frac {3 x}{11 \left (a+b x^3\right )^{11/3}}\right )+\frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {3}{7} \left (\frac {19 x \left (\frac {b x^3}{a}+1\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {11}{3},\frac {4}{3},-\frac {b x^3}{a}\right )}{11 a^3 \left (a+b x^3\right )^{2/3}}+\frac {3 x}{11 \left (a+b x^3\right )^{11/3}}\right )+\frac {x \left (a-b x^3\right )}{7 \left (a+b x^3\right )^{14/3}}\) |
(x*(a - b*x^3))/(7*(a + b*x^3)^(14/3)) + (3*((3*x)/(11*(a + b*x^3)^(11/3)) + (19*x*(1 + (b*x^3)/a)^(2/3)*Hypergeometric2F1[1/3, 11/3, 4/3, -((b*x^3) /a)])/(11*a^3*(a + b*x^3)^(2/3))))/7
3.1.55.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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])
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])
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]
\[\int \frac {\left (-b \,x^{3}+a \right )^{2}}{\left (b \,x^{3}+a \right )^{\frac {17}{3}}}d x\]
\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {17}{3}}} \,d x } \]
integral((b^2*x^6 - 2*a*b*x^3 + a^2)*(b*x^3 + a)^(1/3)/(b^6*x^18 + 6*a*b^5 *x^15 + 15*a^2*b^4*x^12 + 20*a^3*b^3*x^9 + 15*a^4*b^2*x^6 + 6*a^5*b*x^3 + a^6), x)
Timed out. \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {17}{3}}} \,d x } \]
\[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\int { \frac {{\left (b x^{3} - a\right )}^{2}}{{\left (b x^{3} + a\right )}^{\frac {17}{3}}} \,d x } \]
Timed out. \[ \int \frac {\left (a-b x^3\right )^2}{\left (a+b x^3\right )^{17/3}} \, dx=\int \frac {{\left (a-b\,x^3\right )}^2}{{\left (b\,x^3+a\right )}^{17/3}} \,d x \]