Integrand size = 19, antiderivative size = 264 \[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=-\frac {d \left (3 b^3 c^3-6 a b^2 c^2 d (1+n)+4 a^2 b c d^2 (1+2 n)-a^3 d^3 (1+3 n)\right ) x}{a b^4 n}-\frac {d^2 \left (3 b^2 c^2 (1+n)-4 a b c d (1+2 n)+a^2 d^2 (1+3 n)\right ) x^{1+n}}{a b^3 n (1+n)}+\frac {d^3 (a d (1+3 n)-b (c+2 c n)) x^{1+2 n}}{a b^2 n (1+2 n)}+\frac {(b c-a d) x \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}-\frac {(b c-a d)^3 (b c (1-n)-a d (1+3 n)) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 b^4 n} \] Output:
-d*(3*b^3*c^3-6*a*b^2*c^2*d*(1+n)+4*a^2*b*c*d^2*(1+2*n)-a^3*d^3*(1+3*n))*x /a/b^4/n-d^2*(3*b^2*c^2*(1+n)-4*a*b*c*d*(1+2*n)+a^2*d^2*(1+3*n))*x^(1+n)/a /b^3/n/(1+n)+d^3*(a*d*(1+3*n)-b*(2*c*n+c))*x^(1+2*n)/a/b^2/n/(1+2*n)+(-a*d +b*c)*x*(c+d*x^n)^3/a/b/n/(a+b*x^n)-(-a*d+b*c)^3*(b*c*(1-n)-a*d*(1+3*n))*x *hypergeom([1, 1/n],[1+1/n],-b*x^n/a)/a^2/b^4/n
Time = 6.46 (sec) , antiderivative size = 217, normalized size of antiderivative = 0.82 \[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\frac {x \left (\frac {4 a b^3 c^3 d-6 a^2 b^2 c^2 d^2+4 a^3 b c d^3-a^4 d^4+b^4 c^4 (-1+n)}{a^2 n}+\frac {(-b c+a d)^3 (b c (-1+n)+a d (1+3 n))}{a^2 n}+\frac {2 b d^3 (2 b c-a d) x^n}{1+n}+\frac {b^2 d^4 x^{2 n}}{1+2 n}+\frac {(b c-a d)^4}{a n \left (a+b x^n\right )}+\frac {(b c-a d)^3 (b c (-1+n)+a d (1+3 n)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 n}\right )}{b^4} \] Input:
Integrate[(c + d*x^n)^4/(a + b*x^n)^2,x]
Output:
(x*((4*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - a^4*d^4 + b^4*c^4 *(-1 + n))/(a^2*n) + ((-(b*c) + a*d)^3*(b*c*(-1 + n) + a*d*(1 + 3*n)))/(a^ 2*n) + (2*b*d^3*(2*b*c - a*d)*x^n)/(1 + n) + (b^2*d^4*x^(2*n))/(1 + 2*n) + (b*c - a*d)^4/(a*n*(a + b*x^n)) + ((b*c - a*d)^3*(b*c*(-1 + n) + a*d*(1 + 3*n))*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^2*n)))/b ^4
Time = 1.22 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.30, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {930, 1025, 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.
\(\displaystyle \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx\) |
\(\Big \downarrow \) 930 |
\(\displaystyle \frac {\int \frac {\left (d x^n+c\right )^2 \left (d (a d (3 n+1)-b (2 n c+c)) x^n+c (a d-b c (1-n))\right )}{b x^n+a}dx}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}\) |
\(\Big \downarrow \) 1025 |
\(\displaystyle \frac {\frac {\int \frac {\left (d x^n+c\right ) \left (c \left (-b^2 \left (-2 n^2+n+1\right ) c^2+2 a b d (2 n+1) c-a^2 d^2 (3 n+1)\right )-d \left (b^2 \left (2 n^2+3 n+1\right ) c^2-2 a b d \left (5 n^2+4 n+1\right ) c+a^2 d^2 \left (6 n^2+5 n+1\right )\right ) x^n\right )}{b x^n+a}dx}{b (2 n+1)}+\frac {d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (2 c n+c))}{b (2 n+1)}}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}\) |
\(\Big \downarrow \) 1025 |
\(\displaystyle \frac {\frac {\frac {\int \frac {c \left (-b^3 \left (-2 n^3-n^2+2 n+1\right ) c^3+3 a b^2 d \left (2 n^2+3 n+1\right ) c^2-a^2 b d^2 \left (13 n^2+12 n+3\right ) c+a^3 d^3 \left (6 n^2+5 n+1\right )\right )-d \left (b^3 \left (2 n^2+3 n+1\right ) c^3-a b^2 d \left (12 n^3+17 n^2+12 n+3\right ) c^2+a^2 b d^2 \left (16 n^3+26 n^2+15 n+3\right ) c-a^3 d^3 \left (6 n^3+11 n^2+6 n+1\right )\right ) x^n}{b x^n+a}dx}{b (n+1)}-\frac {d x \left (c+d x^n\right ) \left (a^2 d^2 \left (6 n^2+5 n+1\right )-2 a b c d \left (5 n^2+4 n+1\right )+b^2 c^2 \left (2 n^2+3 n+1\right )\right )}{b (n+1)}}{b (2 n+1)}+\frac {d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (2 c n+c))}{b (2 n+1)}}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}\) |
\(\Big \downarrow \) 913 |
\(\displaystyle \frac {\frac {\frac {-\frac {\left (2 n^2+3 n+1\right ) (b c-a d)^3 (b c (1-n)-a d (3 n+1)) \int \frac {1}{b x^n+a}dx}{b}-\frac {d x \left (-a^3 d^3 \left (6 n^3+11 n^2+6 n+1\right )+a^2 b c d^2 \left (16 n^3+26 n^2+15 n+3\right )-a b^2 c^2 d \left (12 n^3+17 n^2+12 n+3\right )+b^3 c^3 \left (2 n^2+3 n+1\right )\right )}{b}}{b (n+1)}-\frac {d x \left (c+d x^n\right ) \left (a^2 d^2 \left (6 n^2+5 n+1\right )-2 a b c d \left (5 n^2+4 n+1\right )+b^2 c^2 \left (2 n^2+3 n+1\right )\right )}{b (n+1)}}{b (2 n+1)}+\frac {d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (2 c n+c))}{b (2 n+1)}}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {\frac {\frac {-\frac {d x \left (-a^3 d^3 \left (6 n^3+11 n^2+6 n+1\right )+a^2 b c d^2 \left (16 n^3+26 n^2+15 n+3\right )-a b^2 c^2 d \left (12 n^3+17 n^2+12 n+3\right )+b^3 c^3 \left (2 n^2+3 n+1\right )\right )}{b}-\frac {\left (2 n^2+3 n+1\right ) x (b c-a d)^3 (b c (1-n)-a d (3 n+1)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b}}{b (n+1)}-\frac {d x \left (c+d x^n\right ) \left (a^2 d^2 \left (6 n^2+5 n+1\right )-2 a b c d \left (5 n^2+4 n+1\right )+b^2 c^2 \left (2 n^2+3 n+1\right )\right )}{b (n+1)}}{b (2 n+1)}+\frac {d x \left (c+d x^n\right )^2 (a d (3 n+1)-b (2 c n+c))}{b (2 n+1)}}{a b n}+\frac {x (b c-a d) \left (c+d x^n\right )^3}{a b n \left (a+b x^n\right )}\) |
Input:
Int[(c + d*x^n)^4/(a + b*x^n)^2,x]
Output:
((b*c - a*d)*x*(c + d*x^n)^3)/(a*b*n*(a + b*x^n)) + ((d*(a*d*(1 + 3*n) - b *(c + 2*c*n))*x*(c + d*x^n)^2)/(b*(1 + 2*n)) + (-((d*(b^2*c^2*(1 + 3*n + 2 *n^2) - 2*a*b*c*d*(1 + 4*n + 5*n^2) + a^2*d^2*(1 + 5*n + 6*n^2))*x*(c + d* x^n))/(b*(1 + n))) + (-((d*(b^3*c^3*(1 + 3*n + 2*n^2) - a^3*d^3*(1 + 6*n + 11*n^2 + 6*n^3) - a*b^2*c^2*d*(3 + 12*n + 17*n^2 + 12*n^3) + a^2*b*c*d^2* (3 + 15*n + 26*n^2 + 16*n^3))*x)/b) - ((b*c - a*d)^3*(1 + 3*n + 2*n^2)*(b* c*(1 - n) - a*d*(1 + 3*n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b *x^n)/a)])/(a*b))/(b*(1 + n)))/(b*(1 + 2*n)))/(a*b*n)
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]
\[\int \frac {\left (c +d \,x^{n}\right )^{4}}{\left (a +b \,x^{n}\right )^{2}}d x\]
Input:
int((c+d*x^n)^4/(a+b*x^n)^2,x)
Output:
int((c+d*x^n)^4/(a+b*x^n)^2,x)
\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{4}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \] Input:
integrate((c+d*x^n)^4/(a+b*x^n)^2,x, algorithm="fricas")
Output:
integral((d^4*x^(4*n) + 4*c*d^3*x^(3*n) + 6*c^2*d^2*x^(2*n) + 4*c^3*d*x^n + c^4)/(b^2*x^(2*n) + 2*a*b*x^n + a^2), x)
\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int \frac {\left (c + d x^{n}\right )^{4}}{\left (a + b x^{n}\right )^{2}}\, dx \] Input:
integrate((c+d*x**n)**4/(a+b*x**n)**2,x)
Output:
Integral((c + d*x**n)**4/(a + b*x**n)**2, x)
\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{4}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \] Input:
integrate((c+d*x^n)^4/(a+b*x^n)^2,x, algorithm="maxima")
Output:
-(a^4*d^4*(3*n + 1) - 4*a^3*b*c*d^3*(2*n + 1) + 6*a^2*b^2*c^2*d^2*(n + 1) - b^4*c^4*(n - 1) - 4*a*b^3*c^3*d)*integrate(1/(a*b^5*n*x^n + a^2*b^4*n), x) + ((n^2 + n)*a*b^3*d^4*x*x^(3*n) + (4*(2*n^2 + n)*a*b^3*c*d^3 - (3*n^2 + n)*a^2*b^2*d^4)*x*x^(2*n) + (6*(2*n^3 + 3*n^2 + n)*a*b^3*c^2*d^2 - 4*(4* n^3 + 4*n^2 + n)*a^2*b^2*c*d^3 + (6*n^3 + 5*n^2 + n)*a^3*b*d^4)*x*x^n + (( 2*n^2 + 3*n + 1)*b^4*c^4 - 4*(2*n^2 + 3*n + 1)*a*b^3*c^3*d + 6*(2*n^3 + 5* n^2 + 4*n + 1)*a^2*b^2*c^2*d^2 - 4*(4*n^3 + 8*n^2 + 5*n + 1)*a^3*b*c*d^3 + (6*n^3 + 11*n^2 + 6*n + 1)*a^4*d^4)*x)/((2*n^3 + 3*n^2 + n)*a*b^5*x^n + ( 2*n^3 + 3*n^2 + n)*a^2*b^4)
\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{4}}{{\left (b x^{n} + a\right )}^{2}} \,d x } \] Input:
integrate((c+d*x^n)^4/(a+b*x^n)^2,x, algorithm="giac")
Output:
integrate((d*x^n + c)^4/(b*x^n + a)^2, x)
Timed out. \[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\int \frac {{\left (c+d\,x^n\right )}^4}{{\left (a+b\,x^n\right )}^2} \,d x \] Input:
int((c + d*x^n)^4/(a + b*x^n)^2,x)
Output:
int((c + d*x^n)^4/(a + b*x^n)^2, x)
\[ \int \frac {\left (c+d x^n\right )^4}{\left (a+b x^n\right )^2} \, dx=\text {too large to display} \] Input:
int((c+d*x^n)^4/(a+b*x^n)^2,x)
Output:
(x**(3*n)*a**2*b*d**4*n*x + x**(3*n)*a**2*b*d**4*x - 3*x**(2*n)*a**3*d**4* n*x - x**(2*n)*a**3*d**4*x + 8*x**(2*n)*a**2*b*c*d**3*n*x + 4*x**(2*n)*a** 2*b*c*d**3*x + 6*x**n*int(x**(2*n)/(x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x* *n*a*b*n + 2*x**n*a*b + a**2*n + a**2),x)*a**4*b*d**4*n**3 + 11*x**n*int(x **(2*n)/(x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x**n*a*b + a** 2*n + a**2),x)*a**4*b*d**4*n**2 + 6*x**n*int(x**(2*n)/(x**(2*n)*b**2*n + x **(2*n)*b**2 + 2*x**n*a*b*n + 2*x**n*a*b + a**2*n + a**2),x)*a**4*b*d**4*n + x**n*int(x**(2*n)/(x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x **n*a*b + a**2*n + a**2),x)*a**4*b*d**4 - 16*x**n*int(x**(2*n)/(x**(2*n)*b **2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x**n*a*b + a**2*n + a**2),x)*a**3 *b**2*c*d**3*n**3 - 32*x**n*int(x**(2*n)/(x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x**n*a*b + a**2*n + a**2),x)*a**3*b**2*c*d**3*n**2 - 20 *x**n*int(x**(2*n)/(x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x** n*a*b + a**2*n + a**2),x)*a**3*b**2*c*d**3*n - 4*x**n*int(x**(2*n)/(x**(2* n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x**n*a*b + a**2*n + a**2),x)* a**3*b**2*c*d**3 + 12*x**n*int(x**(2*n)/(x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x**n*a*b + a**2*n + a**2),x)*a**2*b**3*c**2*d**2*n**3 + 30*x**n*int(x**(2*n)/(x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x **n*a*b + a**2*n + a**2),x)*a**2*b**3*c**2*d**2*n**2 + 24*x**n*int(x**(2*n )/(x**(2*n)*b**2*n + x**(2*n)*b**2 + 2*x**n*a*b*n + 2*x**n*a*b + a**2*n...