\(\int \sqrt {a+b x^n} (c+d x^n)^2 \, dx\) [95]

Optimal result

Integrand size = 21, antiderivative size = 187 \[ \int \sqrt {a+b x^n} \left (c+d x^n\right )^2 \, dx=-\frac {2 d (2 a d (1+n)-b c (2+7 n)) x \left (a+b x^n\right )^{3/2}}{b^2 (2+3 n) (2+5 n)}+\frac {2 d x \left (a+b x^n\right )^{3/2} \left (c+d x^n\right )}{b (2+5 n)}+\frac {\left (4 a^2 d^2 (1+n)-4 a b c d (2+5 n)+b^2 c^2 \left (4+16 n+15 n^2\right )\right ) x \sqrt {a+b x^n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b^2 (2+3 n) (2+5 n) \sqrt {1+\frac {b x^n}{a}}} \] Output:

-2*d*(2*a*d*(1+n)-b*c*(2+7*n))*x*(a+b*x^n)^(3/2)/b^2/(2+3*n)/(2+5*n)+2*d*x 
*(a+b*x^n)^(3/2)*(c+d*x^n)/b/(2+5*n)+(4*a^2*d^2*(1+n)-4*a*b*c*d*(2+5*n)+b^ 
2*c^2*(15*n^2+16*n+4))*x*(a+b*x^n)^(1/2)*hypergeom([-1/2, 1/n],[1+1/n],-b* 
x^n/a)/b^2/(2+3*n)/(2+5*n)/(1+b*x^n/a)^(1/2)
 

Mathematica [A] (verified)

Time = 5.40 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.14 \[ \int \sqrt {a+b x^n} \left (c+d x^n\right )^2 \, dx=\frac {2 x \left (a+b x^n\right ) \left (-2 a^2 d^2 n (1+n)+a b d n \left (2 c (2+5 n)+d (2+n) x^n\right )+b^2 \left (c^2 \left (4+16 n+15 n^2\right )+2 c d \left (4+12 n+5 n^2\right ) x^n+d^2 \left (4+8 n+3 n^2\right ) x^{2 n}\right )\right )+a n \left (4 a^2 d^2 (1+n)-4 a b c d (2+5 n)+b^2 c^2 \left (4+16 n+15 n^2\right )\right ) x \sqrt {1+\frac {b x^n}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b^2 (2+n) (2+3 n) (2+5 n) \sqrt {a+b x^n}} \] Input:

Integrate[Sqrt[a + b*x^n]*(c + d*x^n)^2,x]
 

Output:

(2*x*(a + b*x^n)*(-2*a^2*d^2*n*(1 + n) + a*b*d*n*(2*c*(2 + 5*n) + d*(2 + n 
)*x^n) + b^2*(c^2*(4 + 16*n + 15*n^2) + 2*c*d*(4 + 12*n + 5*n^2)*x^n + d^2 
*(4 + 8*n + 3*n^2)*x^(2*n))) + a*n*(4*a^2*d^2*(1 + n) - 4*a*b*c*d*(2 + 5*n 
) + b^2*c^2*(4 + 16*n + 15*n^2))*x*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[1 
/2, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(b^2*(2 + n)*(2 + 3*n)*(2 + 5*n)*Sq 
rt[a + b*x^n])
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {933, 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.

\(\displaystyle \int \sqrt {a+b x^n} \left (c+d x^n\right )^2 \, dx\)

\(\Big \downarrow \) 933

\(\displaystyle \frac {2 \int -\frac {1}{2} \sqrt {b x^n+a} \left (d (2 a d (n+1)-b c (7 n+2)) x^n+c (2 a d-b c (5 n+2))\right )dx}{b (5 n+2)}+\frac {2 d x \left (a+b x^n\right )^{3/2} \left (c+d x^n\right )}{b (5 n+2)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 d x \left (a+b x^n\right )^{3/2} \left (c+d x^n\right )}{b (5 n+2)}-\frac {\int \sqrt {b x^n+a} \left (d (2 a d (n+1)-b c (7 n+2)) x^n+c (2 a d-b c (5 n+2))\right )dx}{b (5 n+2)}\)

\(\Big \downarrow \) 913

\(\displaystyle \frac {2 d x \left (a+b x^n\right )^{3/2} \left (c+d x^n\right )}{b (5 n+2)}-\frac {\frac {2 d x \left (a+b x^n\right )^{3/2} (2 a d (n+1)-b c (7 n+2))}{b (3 n+2)}-\frac {\left (4 a^2 d^2 (n+1)-4 a b c d (5 n+2)+b^2 c^2 \left (15 n^2+16 n+4\right )\right ) \int \sqrt {b x^n+a}dx}{b (3 n+2)}}{b (5 n+2)}\)

\(\Big \downarrow \) 779

\(\displaystyle \frac {2 d x \left (a+b x^n\right )^{3/2} \left (c+d x^n\right )}{b (5 n+2)}-\frac {\frac {2 d x \left (a+b x^n\right )^{3/2} (2 a d (n+1)-b c (7 n+2))}{b (3 n+2)}-\frac {\sqrt {a+b x^n} \left (4 a^2 d^2 (n+1)-4 a b c d (5 n+2)+b^2 c^2 \left (15 n^2+16 n+4\right )\right ) \int \sqrt {\frac {b x^n}{a}+1}dx}{b (3 n+2) \sqrt {\frac {b x^n}{a}+1}}}{b (5 n+2)}\)

\(\Big \downarrow \) 778

\(\displaystyle \frac {2 d x \left (a+b x^n\right )^{3/2} \left (c+d x^n\right )}{b (5 n+2)}-\frac {\frac {2 d x \left (a+b x^n\right )^{3/2} (2 a d (n+1)-b c (7 n+2))}{b (3 n+2)}-\frac {x \sqrt {a+b x^n} \left (4 a^2 d^2 (n+1)-4 a b c d (5 n+2)+b^2 c^2 \left (15 n^2+16 n+4\right )\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{b (3 n+2) \sqrt {\frac {b x^n}{a}+1}}}{b (5 n+2)}\)

Input:

Int[Sqrt[a + b*x^n]*(c + d*x^n)^2,x]
 

Output:

(2*d*x*(a + b*x^n)^(3/2)*(c + d*x^n))/(b*(2 + 5*n)) - ((2*d*(2*a*d*(1 + n) 
 - b*c*(2 + 7*n))*x*(a + b*x^n)^(3/2))/(b*(2 + 3*n)) - ((4*a^2*d^2*(1 + n) 
 - 4*a*b*c*d*(2 + 5*n) + b^2*c^2*(4 + 16*n + 15*n^2))*x*Sqrt[a + b*x^n]*Hy 
pergeometric2F1[-1/2, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(b*(2 + 3*n)*Sqrt 
[1 + (b*x^n)/a]))/(b*(2 + 5*n))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 933

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[d*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q - 1)/(b*(n*(p + q) + 1))), 
x] + Simp[1/(b*(n*(p + q) + 1))   Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Sim 
p[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q 
- 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d 
, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntBinomialQ[ 
a, b, c, d, n, p, q, x]
 

Maple [F]

\[\int \sqrt {a +b \,x^{n}}\, \left (c +d \,x^{n}\right )^{2}d x\]

Input:

int((a+b*x^n)^(1/2)*(c+d*x^n)^2,x)
 

Output:

int((a+b*x^n)^(1/2)*(c+d*x^n)^2,x)
 

Fricas [F(-2)]

Exception generated. \[ \int \sqrt {a+b x^n} \left (c+d x^n\right )^2 \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*x^n)^(1/2)*(c+d*x^n)^2,x, algorithm="fricas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.82 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.99 \[ \int \sqrt {a+b x^n} \left (c+d x^n\right )^2 \, dx=\frac {a^{\frac {1}{n}} a^{\frac {1}{2} - \frac {1}{n}} c^{2} x \Gamma \left (\frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {1}{n} \\ 1 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (1 + \frac {1}{n}\right )} + \frac {a^{- \frac {3}{2} - \frac {1}{n}} a^{2 + \frac {1}{n}} d^{2} x^{2 n + 1} \Gamma \left (2 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, 2 + \frac {1}{n} \\ 3 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (3 + \frac {1}{n}\right )} + \frac {2 a^{- \frac {1}{2} - \frac {1}{n}} a^{1 + \frac {1}{n}} c d x^{n + 1} \Gamma \left (1 + \frac {1}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, 1 + \frac {1}{n} \\ 2 + \frac {1}{n} \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (2 + \frac {1}{n}\right )} \] Input:

integrate((a+b*x**n)**(1/2)*(c+d*x**n)**2,x)
 

Output:

a**(1/n)*a**(1/2 - 1/n)*c**2*x*gamma(1/n)*hyper((-1/2, 1/n), (1 + 1/n,), b 
*x**n*exp_polar(I*pi)/a)/(n*gamma(1 + 1/n)) + a**(-3/2 - 1/n)*a**(2 + 1/n) 
*d**2*x**(2*n + 1)*gamma(2 + 1/n)*hyper((-1/2, 2 + 1/n), (3 + 1/n,), b*x** 
n*exp_polar(I*pi)/a)/(n*gamma(3 + 1/n)) + 2*a**(-1/2 - 1/n)*a**(1 + 1/n)*c 
*d*x**(n + 1)*gamma(1 + 1/n)*hyper((-1/2, 1 + 1/n), (2 + 1/n,), b*x**n*exp 
_polar(I*pi)/a)/(n*gamma(2 + 1/n))
 

Maxima [F]

\[ \int \sqrt {a+b x^n} \left (c+d x^n\right )^2 \, dx=\int { \sqrt {b x^{n} + a} {\left (d x^{n} + c\right )}^{2} \,d x } \] Input:

integrate((a+b*x^n)^(1/2)*(c+d*x^n)^2,x, algorithm="maxima")
 

Output:

integrate(sqrt(b*x^n + a)*(d*x^n + c)^2, x)
 

Giac [F]

\[ \int \sqrt {a+b x^n} \left (c+d x^n\right )^2 \, dx=\int { \sqrt {b x^{n} + a} {\left (d x^{n} + c\right )}^{2} \,d x } \] Input:

integrate((a+b*x^n)^(1/2)*(c+d*x^n)^2,x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate(sqrt(b*x^n + a)*(d*x^n + c)^2, x)
 

Mupad [F(-1)]

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

int((a + b*x^n)^(1/2)*(c + d*x^n)^2,x)
 

Output:

int((a + b*x^n)^(1/2)*(c + d*x^n)^2, x)
 

Reduce [F]

\[ \int \sqrt {a+b x^n} \left (c+d x^n\right )^2 \, dx =\text {Too large to display} \] Input:

int((a+b*x^n)^(1/2)*(c+d*x^n)^2,x)
 

Output:

(6*x**(2*n)*sqrt(x**n*b + a)*b**2*d**2*n**2*x + 16*x**(2*n)*sqrt(x**n*b + 
a)*b**2*d**2*n*x + 8*x**(2*n)*sqrt(x**n*b + a)*b**2*d**2*x + 2*x**n*sqrt(x 
**n*b + a)*a*b*d**2*n**2*x + 4*x**n*sqrt(x**n*b + a)*a*b*d**2*n*x + 20*x** 
n*sqrt(x**n*b + a)*b**2*c*d*n**2*x + 48*x**n*sqrt(x**n*b + a)*b**2*c*d*n*x 
 + 16*x**n*sqrt(x**n*b + a)*b**2*c*d*x - 4*sqrt(x**n*b + a)*a**2*d**2*n**2 
*x - 4*sqrt(x**n*b + a)*a**2*d**2*n*x + 20*sqrt(x**n*b + a)*a*b*c*d*n**2*x 
 + 8*sqrt(x**n*b + a)*a*b*c*d*n*x + 30*sqrt(x**n*b + a)*b**2*c**2*n**2*x + 
 32*sqrt(x**n*b + a)*b**2*c**2*n*x + 8*sqrt(x**n*b + a)*b**2*c**2*x + 60*i 
nt(sqrt(x**n*b + a)/(15*x**n*b*n**3 + 46*x**n*b*n**2 + 36*x**n*b*n + 8*x** 
n*b + 15*a*n**3 + 46*a*n**2 + 36*a*n + 8*a),x)*a**3*d**2*n**5 + 244*int(sq 
rt(x**n*b + a)/(15*x**n*b*n**3 + 46*x**n*b*n**2 + 36*x**n*b*n + 8*x**n*b + 
 15*a*n**3 + 46*a*n**2 + 36*a*n + 8*a),x)*a**3*d**2*n**4 + 328*int(sqrt(x* 
*n*b + a)/(15*x**n*b*n**3 + 46*x**n*b*n**2 + 36*x**n*b*n + 8*x**n*b + 15*a 
*n**3 + 46*a*n**2 + 36*a*n + 8*a),x)*a**3*d**2*n**3 + 176*int(sqrt(x**n*b 
+ a)/(15*x**n*b*n**3 + 46*x**n*b*n**2 + 36*x**n*b*n + 8*x**n*b + 15*a*n**3 
 + 46*a*n**2 + 36*a*n + 8*a),x)*a**3*d**2*n**2 + 32*int(sqrt(x**n*b + a)/( 
15*x**n*b*n**3 + 46*x**n*b*n**2 + 36*x**n*b*n + 8*x**n*b + 15*a*n**3 + 46* 
a*n**2 + 36*a*n + 8*a),x)*a**3*d**2*n - 300*int(sqrt(x**n*b + a)/(15*x**n* 
b*n**3 + 46*x**n*b*n**2 + 36*x**n*b*n + 8*x**n*b + 15*a*n**3 + 46*a*n**2 + 
 36*a*n + 8*a),x)*a**2*b*c*d*n**5 - 1040*int(sqrt(x**n*b + a)/(15*x**n*...