Integrand size = 19, antiderivative size = 200 \[ \int \frac {\left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=-\frac {d \left (b^2 c^2 (1+n)+a^2 d^2 \left (1+3 n+2 n^2\right )-a b c d \left (2+4 n+3 n^2\right )\right ) x}{a b^3 n (1+n)}-\frac {d (b c (1+n)-a d (1+2 n)) x \left (c+d x^n\right )}{a b^2 n (1+n)}+\frac {(b c-a d) x \left (c+d x^n\right )^2}{a b n \left (a+b x^n\right )}-\frac {(b c-a d)^2 (b c (1-n)-a d (1+2 n)) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 b^3 n} \]
-d*(b^2*c^2*(1+n)+a^2*d^2*(2*n^2+3*n+1)-a*b*c*d*(3*n^2+4*n+2))*x/a/b^3/n/( 1+n)-d*(b*c*(1+n)-a*d*(1+2*n))*x*(c+d*x^n)/a/b^2/n/(1+n)+(-a*d+b*c)*x*(c+d *x^n)^2/a/b/n/(a+b*x^n)-(-a*d+b*c)^2*(b*c*(1-n)-a*d*(1+2*n))*x*hypergeom([ 1, 1/n],[1+1/n],-b*x^n/a)/a^2/b^3/n
Result contains higher order function than in optimal. Order 9 vs. order 5 in optimal.
Time = 4.83 (sec) , antiderivative size = 2050, normalized size of antiderivative = 10.25 \[ \int \frac {\left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\text {Result too large to show} \]
(x*(3*a*(1 + 10*n + 35*n^2 + 50*n^3 + 24*n^4)*(c^3*(1 + n)^4 + 3*c^2*d*(1 + 4*n + 6*n^2 + 2*n^3 + n^4)*x^n + 3*c*d^2*(1 + n)^4*x^(2*n) + d^3*(1 + n) ^4*x^(3*n))*HurwitzLerchPhi[-((b*x^n)/a), 1, 1 + n^(-1)] - 3*a*(1 + 10*n + 35*n^2 + 50*n^3 + 24*n^4)*(c^3*(1 + 2*n)^4 + 3*c^2*d*(1 + 2*n)^4*x^n + 3* c*d^2*(1 + 8*n + 24*n^2 + 34*n^3 + 18*n^4)*x^(2*n) + d^3*(1 + 2*n)^4*x^(3* n))*HurwitzLerchPhi[-((b*x^n)/a), 1, 2 + n^(-1)] + a*c^3*HurwitzLerchPhi[- ((b*x^n)/a), 1, 3 + n^(-1)] + 22*a*c^3*n*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 209*a*c^3*n^2*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 1118*a*c^3*n^3*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 3675*a*c^3* n^4*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 7578*a*c^3*n^5*HurwitzL erchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 9531*a*c^3*n^6*HurwitzLerchPhi[-((b *x^n)/a), 1, 3 + n^(-1)] + 6642*a*c^3*n^7*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 1944*a*c^3*n^8*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 3*a*c^2*d*x^n*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 66*a*c^2*d *n*x^n*HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 627*a*c^2*d*n^2*x^n* HurwitzLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 3354*a*c^2*d*n^3*x^n*Hurwit zLerchPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 11025*a*c^2*d*n^4*x^n*HurwitzLerc hPhi[-((b*x^n)/a), 1, 3 + n^(-1)] + 22734*a*c^2*d*n^5*x^n*HurwitzLerchPhi[ -((b*x^n)/a), 1, 3 + n^(-1)] + 28593*a*c^2*d*n^6*x^n*HurwitzLerchPhi[-((b* x^n)/a), 1, 3 + n^(-1)] + 19926*a*c^2*d*n^7*x^n*HurwitzLerchPhi[-((b*x^...
Time = 0.45 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {930, 1025, 913, 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.
| \(\displaystyle \int \frac {\left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx\) |
| \(\Big \downarrow \) 930 |
| \(\displaystyle \frac {\int \frac {\left (d x^n+c\right ) \left (c (a d-b c (1-n))-d (b c (n+1)-a d (2 n+1)) x^n\right )}{b x^n+a}dx}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^2}{a b n \left (a+b x^n\right )}\) |
| \(\Big \downarrow \) 1025 |
| \(\displaystyle \frac {\frac {\int \frac {c \left (-b^2 \left (1-n^2\right ) c^2+2 a b d (n+1) c-a^2 d^2 (2 n+1)\right )-d \left (b^2 (n+1) c^2-a b d \left (3 n^2+4 n+2\right ) c+a^2 d^2 \left (2 n^2+3 n+1\right )\right ) x^n}{b x^n+a}dx}{b (n+1)}-\frac {d x \left (c+d x^n\right ) (b c (n+1)-a d (2 n+1))}{b (n+1)}}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^2}{a b n \left (a+b x^n\right )}\) |
| \(\Big \downarrow \) 913 |
| \(\displaystyle \frac {\frac {-\frac {(n+1) (b c-a d)^2 (b c (1-n)-a d (2 n+1)) \int \frac {1}{b x^n+a}dx}{b}-\frac {d x \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (3 n^2+4 n+2\right )+b^2 c^2 (n+1)\right )}{b}}{b (n+1)}-\frac {d x \left (c+d x^n\right ) (b c (n+1)-a d (2 n+1))}{b (n+1)}}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^2}{a b n \left (a+b x^n\right )}\) |
| \(\Big \downarrow \) 778 |
| \(\displaystyle \frac {\frac {-\frac {d x \left (a^2 d^2 \left (2 n^2+3 n+1\right )-a b c d \left (3 n^2+4 n+2\right )+b^2 c^2 (n+1)\right )}{b}-\frac {(n+1) x (b c-a d)^2 (b c (1-n)-a d (2 n+1)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b}}{b (n+1)}-\frac {d x \left (c+d x^n\right ) (b c (n+1)-a d (2 n+1))}{b (n+1)}}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^2}{a b n \left (a+b x^n\right )}\) |
((b*c - a*d)*x*(c + d*x^n)^2)/(a*b*n*(a + b*x^n)) + (-((d*(b*c*(1 + n) - a *d*(1 + 2*n))*x*(c + d*x^n))/(b*(1 + n))) + (-((d*(b^2*c^2*(1 + n) + a^2*d ^2*(1 + 3*n + 2*n^2) - a*b*c*d*(2 + 4*n + 3*n^2))*x)/b) - ((b*c - a*d)^2*( 1 + n)*(b*c*(1 - n) - a*d*(1 + 2*n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^ (-1), -((b*x^n)/a)])/(a*b))/(b*(1 + n)))/(a*b*n)
3.4.6.3.1 Defintions of rubi rules used
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_)*((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[((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]
\[\int \frac {\left (c +d \,x^{n}\right )^{3}}{\left (a +b \,x^{n}\right )^{2}}d x\]
\[ \int \frac {\left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{3}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
integral((d^3*x^(3*n) + 3*c*d^2*x^(2*n) + 3*c^2*d*x^n + c^3)/(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)
\[ \int \frac {\left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int \frac {\left (c + d x^{n}\right )^{3}}{\left (a + b x^{n}\right )^{2}}\, dx \]
\[ \int \frac {\left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{3}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
(a^3*d^3*(2*n + 1) - 3*a^2*b*c*d^2*(n + 1) + b^3*c^3*(n - 1) + 3*a*b^2*c^2 *d)*integrate(1/(a*b^4*n*x^n + a^2*b^3*n), x) + (a*b^2*d^3*n*x*x^(2*n) + ( 3*(n^2 + n)*a*b^2*c*d^2 - (2*n^2 + n)*a^2*b*d^3)*x*x^n + (3*(n^2 + 2*n + 1 )*a^2*b*c*d^2 - (2*n^2 + 3*n + 1)*a^3*d^3 + b^3*c^3*(n + 1) - 3*a*b^2*c^2* d*(n + 1))*x)/((n^2 + n)*a*b^4*x^n + (n^2 + n)*a^2*b^3)
\[ \int \frac {\left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{3}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (c+d x^n\right )^3}{\left (a+b x^n\right )^2} \, dx=\int \frac {{\left (c+d\,x^n\right )}^3}{{\left (a+b\,x^n\right )}^2} \,d x \]