\(\int \frac {(c+d x^n)^2}{(a+b x^n)^{3/2}} \, dx\) [103]

Optimal result

Integrand size = 21, antiderivative size = 178 \[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx=-\frac {2 d \left (c-\frac {2 a d (1+n)}{b (2+n)}\right ) x \sqrt {a+b x^n}}{a b n}+\frac {2 (b c-a d) x \left (c+d x^n\right )}{a b n \sqrt {a+b x^n}}-\frac {\left (4 a^2 d^2 (1+n)-4 a b c d (2+n)+b^2 c^2 \left (4-n^2\right )\right ) x \sqrt {1+\frac {b x^n}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b^2 n (2+n) \sqrt {a+b x^n}} \] Output:

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

Mathematica [A] (verified)

Time = 5.26 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.73 \[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx=\frac {2 x \left ((b c-a d)^2 (2+n)+a d^2 n \left (a+b x^n\right )\right )-\left (4 a^2 d^2 (1+n)-4 a b c d (2+n)-b^2 c^2 \left (-4+n^2\right )\right ) x \sqrt {1+\frac {b x^n}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a b^2 n (2+n) \sqrt {a+b x^n}} \] Input:

Integrate[(c + d*x^n)^2/(a + b*x^n)^(3/2),x]
 

Output:

(2*x*((b*c - a*d)^2*(2 + n) + a*d^2*n*(a + b*x^n)) - (4*a^2*d^2*(1 + n) - 
4*a*b*c*d*(2 + n) - b^2*c^2*(-4 + n^2))*x*Sqrt[1 + (b*x^n)/a]*Hypergeometr 
ic2F1[1/2, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b^2*n*(2 + n)*Sqrt[a + b* 
x^n])
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 174, 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.

\(\displaystyle \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 930

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 913

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

\(\Big \downarrow \) 779

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

\(\Big \downarrow \) 778

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

Input:

Int[(c + d*x^n)^2/(a + b*x^n)^(3/2),x]
 

Output:

(2*(b*c - a*d)*x*(c + d*x^n))/(a*b*n*Sqrt[a + b*x^n]) + (-2*d*(c - (2*a*d* 
(1 + n))/(b*(2 + n)))*x*Sqrt[a + b*x^n] - ((4*a^2*d^2*(1 + n) - 4*a*b*c*d* 
(2 + n) + b^2*c^2*(4 - n^2))*x*Sqrt[1 + (b*x^n)/a]*Hypergeometric2F1[1/2, 
n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(b*(2 + n)*Sqrt[a + b*x^n]))/(a*b*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 930

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]
 

Maple [F]

\[\int \frac {\left (c +d \,x^{n}\right )^{2}}{\left (a +b \,x^{n}\right )^{\frac {3}{2}}}d x\]

Input:

int((c+d*x^n)^2/(a+b*x^n)^(3/2),x)
 

Output:

int((c+d*x^n)^2/(a+b*x^n)^(3/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((c+d*x^n)^2/(a+b*x^n)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx=\int \frac {\left (c + d x^{n}\right )^{2}}{\left (a + b x^{n}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((c+d*x**n)**2/(a+b*x**n)**(3/2),x)
 

Output:

Integral((c + d*x**n)**2/(a + b*x**n)**(3/2), x)
 

Maxima [F]

\[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{2}}{{\left (b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c+d*x^n)^2/(a+b*x^n)^(3/2),x, algorithm="maxima")
 

Output:

integrate((d*x^n + c)^2/(b*x^n + a)^(3/2), x)
 

Giac [F]

\[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx=\int { \frac {{\left (d x^{n} + c\right )}^{2}}{{\left (b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c+d*x^n)^2/(a+b*x^n)^(3/2),x, algorithm="giac")
 

Output:

integrate((d*x^n + c)^2/(b*x^n + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx=\int \frac {{\left (c+d\,x^n\right )}^2}{{\left (a+b\,x^n\right )}^{3/2}} \,d x \] Input:

int((c + d*x^n)^2/(a + b*x^n)^(3/2),x)
 

Output:

int((c + d*x^n)^2/(a + b*x^n)^(3/2), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^n\right )^2}{\left (a+b x^n\right )^{3/2}} \, dx=\text {too large to display} \] Input:

int((c+d*x^n)^2/(a+b*x^n)^(3/2),x)
 

Output:

(2*x**n*sqrt(x**n*b + a)*b*d**2*n*x - 4*x**n*sqrt(x**n*b + a)*b*d**2*x + 4 
*sqrt(x**n*b + a)*a*d**2*n*x + 4*sqrt(x**n*b + a)*a*d**2*x - 4*sqrt(x**n*b 
 + a)*b*c*d*n*x - 8*sqrt(x**n*b + a)*b*c*d*x - 4*x**n*int(sqrt(x**n*b + a) 
/(x**(2*n)*b**2*n**2 - 4*x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 8*x**n*a*b + a* 
*2*n**2 - 4*a**2),x)*a**2*b*d**2*n**3 - 4*x**n*int(sqrt(x**n*b + a)/(x**(2 
*n)*b**2*n**2 - 4*x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 8*x**n*a*b + a**2*n**2 
 - 4*a**2),x)*a**2*b*d**2*n**2 + 16*x**n*int(sqrt(x**n*b + a)/(x**(2*n)*b* 
*2*n**2 - 4*x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 8*x**n*a*b + a**2*n**2 - 4*a 
**2),x)*a**2*b*d**2*n + 16*x**n*int(sqrt(x**n*b + a)/(x**(2*n)*b**2*n**2 - 
 4*x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 8*x**n*a*b + a**2*n**2 - 4*a**2),x)*a 
**2*b*d**2 + 4*x**n*int(sqrt(x**n*b + a)/(x**(2*n)*b**2*n**2 - 4*x**(2*n)* 
b**2 + 2*x**n*a*b*n**2 - 8*x**n*a*b + a**2*n**2 - 4*a**2),x)*a*b**2*c*d*n* 
*3 + 8*x**n*int(sqrt(x**n*b + a)/(x**(2*n)*b**2*n**2 - 4*x**(2*n)*b**2 + 2 
*x**n*a*b*n**2 - 8*x**n*a*b + a**2*n**2 - 4*a**2),x)*a*b**2*c*d*n**2 - 16* 
x**n*int(sqrt(x**n*b + a)/(x**(2*n)*b**2*n**2 - 4*x**(2*n)*b**2 + 2*x**n*a 
*b*n**2 - 8*x**n*a*b + a**2*n**2 - 4*a**2),x)*a*b**2*c*d*n - 32*x**n*int(s 
qrt(x**n*b + a)/(x**(2*n)*b**2*n**2 - 4*x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 
8*x**n*a*b + a**2*n**2 - 4*a**2),x)*a*b**2*c*d + x**n*int(sqrt(x**n*b + a) 
/(x**(2*n)*b**2*n**2 - 4*x**(2*n)*b**2 + 2*x**n*a*b*n**2 - 8*x**n*a*b + a* 
*2*n**2 - 4*a**2),x)*b**3*c**2*n**4 - 8*x**n*int(sqrt(x**n*b + a)/(x**(...