Integrand size = 21, antiderivative size = 187 \[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx=-\frac {3 d \left (c-\frac {3 a d (1+n)}{b (3+2 n)}\right ) x \left (a+b x^n\right )^{2/3}}{a b n}+\frac {3 (b c-a d) x \left (c+d x^n\right )}{a b n \sqrt [3]{a+b x^n}}-\frac {\left (9 a^2 d^2 (1+n)-6 a b c d (3+2 n)+b^2 c^2 \left (9+3 n-2 n^2\right )\right ) x \sqrt [3]{1+\frac {b x^n}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b^2 n (3+2 n) \sqrt [3]{a+b x^n}} \] Output:
-3*d*(c-3*a*d*(1+n)/b/(3+2*n))*x*(a+b*x^n)^(2/3)/a/b/n+3*(-a*d+b*c)*x*(c+d *x^n)/a/b/n/(a+b*x^n)^(1/3)-(9*a^2*d^2*(1+n)-6*a*b*c*d*(3+2*n)+b^2*c^2*(-2 *n^2+3*n+9))*x*(1+b*x^n/a)^(1/3)*hypergeom([1/3, 1/n],[1+1/n],-b*x^n/a)/a/ b^2/n/(3+2*n)/(a+b*x^n)^(1/3)
Time = 5.31 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.75 \[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx=\frac {3 x \left ((b c-a d)^2 (3+2 n)+a d^2 n \left (a+b x^n\right )\right )-\left (9 a^2 d^2 (1+n)-6 a b c d (3+2 n)+b^2 c^2 \left (9+3 n-2 n^2\right )\right ) x \sqrt [3]{1+\frac {b x^n}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b^2 n (3+2 n) \sqrt [3]{a+b x^n}} \] Input:
Integrate[(c + d*x^n)^2/(a + b*x^n)^(4/3),x]
Output:
(3*x*((b*c - a*d)^2*(3 + 2*n) + a*d^2*n*(a + b*x^n)) - (9*a^2*d^2*(1 + n) - 6*a*b*c*d*(3 + 2*n) + b^2*c^2*(9 + 3*n - 2*n^2))*x*(1 + (b*x^n)/a)^(1/3) *Hypergeometric2F1[1/3, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b^2*n*(3 + 2 *n)*(a + b*x^n)^(1/3))
Time = 0.65 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {930, 27, 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.
\(\displaystyle \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx\) |
\(\Big \downarrow \) 930 |
\(\displaystyle \frac {3 \int \frac {d (3 a d (n+1)-b c (2 n+3)) x^n+c (3 a d-b c (3-n))}{3 \sqrt [3]{b x^n+a}}dx}{a b n}+\frac {3 x (b c-a d) \left (c+d x^n\right )}{a b n \sqrt [3]{a+b x^n}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {d (3 a d (n+1)-b c (2 n+3)) x^n+c (3 a d-b c (3-n))}{\sqrt [3]{b x^n+a}}dx}{a b n}+\frac {3 x (b c-a d) \left (c+d x^n\right )}{a b n \sqrt [3]{a+b x^n}}\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {-\frac {\left (9 a^2 d^2 (n+1)-6 a b c d (2 n+3)+b^2 c^2 \left (-2 n^2+3 n+9\right )\right ) \int \frac {1}{\sqrt [3]{b x^n+a}}dx}{b (2 n+3)}-3 d x \left (a+b x^n\right )^{2/3} \left (c-\frac {3 a d (n+1)}{b (2 n+3)}\right )}{a b n}+\frac {3 x (b c-a d) \left (c+d x^n\right )}{a b n \sqrt [3]{a+b x^n}}\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \frac {-\frac {\sqrt [3]{\frac {b x^n}{a}+1} \left (9 a^2 d^2 (n+1)-6 a b c d (2 n+3)+b^2 c^2 \left (-2 n^2+3 n+9\right )\right ) \int \frac {1}{\sqrt [3]{\frac {b x^n}{a}+1}}dx}{b (2 n+3) \sqrt [3]{a+b x^n}}-3 d x \left (a+b x^n\right )^{2/3} \left (c-\frac {3 a d (n+1)}{b (2 n+3)}\right )}{a b n}+\frac {3 x (b c-a d) \left (c+d x^n\right )}{a b n \sqrt [3]{a+b x^n}}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {-\frac {x \sqrt [3]{\frac {b x^n}{a}+1} \left (9 a^2 d^2 (n+1)-6 a b c d (2 n+3)+b^2 c^2 \left (-2 n^2+3 n+9\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b (2 n+3) \sqrt [3]{a+b x^n}}-3 d x \left (a+b x^n\right )^{2/3} \left (c-\frac {3 a d (n+1)}{b (2 n+3)}\right )}{a b n}+\frac {3 x (b c-a d) \left (c+d x^n\right )}{a b n \sqrt [3]{a+b x^n}}\) |
Input:
Int[(c + d*x^n)^2/(a + b*x^n)^(4/3),x]
Output:
(3*(b*c - a*d)*x*(c + d*x^n))/(a*b*n*(a + b*x^n)^(1/3)) + (-3*d*(c - (3*a* d*(1 + n))/(b*(3 + 2*n)))*x*(a + b*x^n)^(2/3) - ((9*a^2*d^2*(1 + n) - 6*a* b*c*d*(3 + 2*n) + b^2*c^2*(9 + 3*n - 2*n^2))*x*(1 + (b*x^n)/a)^(1/3)*Hyper geometric2F1[1/3, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(b*(3 + 2*n)*(a + b*x ^n)^(1/3)))/(a*b*n)
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[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]
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]
\[\int \frac {\left (c +d \,x^{n}\right )^{2}}{\left (a +b \,x^{n}\right )^{\frac {4}{3}}}d x\]
Input:
int((c+d*x^n)^2/(a+b*x^n)^(4/3),x)
Output:
int((c+d*x^n)^2/(a+b*x^n)^(4/3),x)
Exception generated. \[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((c+d*x^n)^2/(a+b*x^n)^(4/3),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx=\int \frac {\left (c + d x^{n}\right )^{2}}{\left (a + b x^{n}\right )^{\frac {4}{3}}}\, dx \] Input:
integrate((c+d*x**n)**2/(a+b*x**n)**(4/3),x)
Output:
Integral((c + d*x**n)**2/(a + b*x**n)**(4/3), x)
\[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{2}}{{\left (b x^{n} + a\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((c+d*x^n)^2/(a+b*x^n)^(4/3),x, algorithm="maxima")
Output:
integrate((d*x^n + c)^2/(b*x^n + a)^(4/3), x)
\[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{2}}{{\left (b x^{n} + a\right )}^{\frac {4}{3}}} \,d x } \] Input:
integrate((c+d*x^n)^2/(a+b*x^n)^(4/3),x, algorithm="giac")
Output:
integrate((d*x^n + c)^2/(b*x^n + a)^(4/3), x)
Timed out. \[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx=\int \frac {{\left (c+d\,x^n\right )}^2}{{\left (a+b\,x^n\right )}^{4/3}} \,d x \] Input:
int((c + d*x^n)^2/(a + b*x^n)^(4/3),x)
Output:
int((c + d*x^n)^2/(a + b*x^n)^(4/3), x)
\[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{4/3}} \, dx=\left (\int \frac {x^{2 n}}{x^{n} \left (x^{n} b +a \right )^{\frac {1}{3}} b +\left (x^{n} b +a \right )^{\frac {1}{3}} a}d x \right ) d^{2}+2 \left (\int \frac {x^{n}}{x^{n} \left (x^{n} b +a \right )^{\frac {1}{3}} b +\left (x^{n} b +a \right )^{\frac {1}{3}} a}d x \right ) c d +\left (\int \frac {1}{x^{n} \left (x^{n} b +a \right )^{\frac {1}{3}} b +\left (x^{n} b +a \right )^{\frac {1}{3}} a}d x \right ) c^{2} \] Input:
int((c+d*x^n)^2/(a+b*x^n)^(4/3),x)
Output:
int(x**(2*n)/(x**n*(x**n*b + a)**(1/3)*b + (x**n*b + a)**(1/3)*a),x)*d**2 + 2*int(x**n/(x**n*(x**n*b + a)**(1/3)*b + (x**n*b + a)**(1/3)*a),x)*c*d + int(1/(x**n*(x**n*b + a)**(1/3)*b + (x**n*b + a)**(1/3)*a),x)*c**2