\(\int (a+b x^n+c x^{2 n})^{5/2} \, dx\) [100]

Optimal result

Integrand size = 18, antiderivative size = 142 \[ \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx=\frac {a^2 x \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (\frac {1}{n},-\frac {5}{2},-\frac {5}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}}} \] Output:

a^2*x*(a+b*x^n+c*x^(2*n))^(1/2)*AppellF1(1/n,-5/2,-5/2,1+1/n,-2*c*x^n/(b-( 
-4*a*c+b^2)^(1/2)),-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(1+2*c*x^n/(b-(-4*a*c+ 
b^2)^(1/2)))^(1/2)/(1+2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))^(1/2)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(706\) vs. \(2(142)=284\).

Time = 3.83 (sec) , antiderivative size = 706, normalized size of antiderivative = 4.97 \[ \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx=\frac {x \left (15 b n^3 \left (b^4 \left (4+8 n+3 n^2\right )+48 a^2 c^2 \left (1+5 n+5 n^2\right )-4 a b^2 c \left (7+23 n+10 n^2\right )\right ) x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (1+\frac {1}{n},\frac {1}{2},\frac {1}{2},2+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+2 (1+n) \left (\left (a+x^n \left (b+c x^n\right )\right ) \left (-15 b^4 n^3 (2+3 n)+16 a^2 c^2 \left (1+15 n+85 n^2+210 n^3+184 n^4\right )+30 b^3 c n^3 (1+n) x^n+4 b^2 c^2 \left (4+50 n+215 n^2+355 n^3+186 n^4\right ) x^{2 n}+8 b c^3 \left (4+45 n+170 n^2+255 n^3+126 n^4\right ) x^{3 n}+16 c^4 \left (1+10 n+35 n^2+50 n^3+24 n^4\right ) x^{4 n}+4 a c \left (45 b^2 n^3 (1+3 n)+2 b c \left (4+55 n+270 n^2+530 n^3+311 n^4\right ) x^n+4 c^2 \left (2+25 n+105 n^2+170 n^3+88 n^4\right ) x^{2 n}\right )\right )+15 a n^3 \left (-12 a b^2 c (1+3 n)+b^4 (2+3 n)+16 a^2 c^2 \left (1+6 n+8 n^2\right )\right ) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {1}{n},\frac {1}{2},\frac {1}{2},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )\right )}{32 c^2 (1+n)^2 (1+2 n) (1+3 n) (1+4 n) (1+5 n) \sqrt {a+x^n \left (b+c x^n\right )}} \] Input:

Integrate[(a + b*x^n + c*x^(2*n))^(5/2),x]
 

Output:

(x*(15*b*n^3*(b^4*(4 + 8*n + 3*n^2) + 48*a^2*c^2*(1 + 5*n + 5*n^2) - 4*a*b 
^2*c*(7 + 23*n + 10*n^2))*x^n*Sqrt[(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - 
Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 
 4*a*c])]*AppellF1[1 + n^(-1), 1/2, 1/2, 2 + n^(-1), (-2*c*x^n)/(b + Sqrt[ 
b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + 2*(1 + n)*((a + x^n*( 
b + c*x^n))*(-15*b^4*n^3*(2 + 3*n) + 16*a^2*c^2*(1 + 15*n + 85*n^2 + 210*n 
^3 + 184*n^4) + 30*b^3*c*n^3*(1 + n)*x^n + 4*b^2*c^2*(4 + 50*n + 215*n^2 + 
 355*n^3 + 186*n^4)*x^(2*n) + 8*b*c^3*(4 + 45*n + 170*n^2 + 255*n^3 + 126* 
n^4)*x^(3*n) + 16*c^4*(1 + 10*n + 35*n^2 + 50*n^3 + 24*n^4)*x^(4*n) + 4*a* 
c*(45*b^2*n^3*(1 + 3*n) + 2*b*c*(4 + 55*n + 270*n^2 + 530*n^3 + 311*n^4)*x 
^n + 4*c^2*(2 + 25*n + 105*n^2 + 170*n^3 + 88*n^4)*x^(2*n))) + 15*a*n^3*(- 
12*a*b^2*c*(1 + 3*n) + b^4*(2 + 3*n) + 16*a^2*c^2*(1 + 6*n + 8*n^2))*Sqrt[ 
(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[(b + Sqrt[ 
b^2 - 4*a*c] + 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]*AppellF1[n^(-1), 1/2, 1/2 
, 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 
 - 4*a*c])])))/(32*c^2*(1 + n)^2*(1 + 2*n)*(1 + 3*n)*(1 + 4*n)*(1 + 5*n)*S 
qrt[a + x^n*(b + c*x^n)])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {1686, 936}

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.

\(\displaystyle \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx\)

\(\Big \downarrow \) 1686

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

\(\Big \downarrow \) 936

\(\displaystyle \frac {a^2 x \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (\frac {1}{n},-\frac {5}{2},-\frac {5}{2},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}}\)

Input:

Int[(a + b*x^n + c*x^(2*n))^(5/2),x]
 

Output:

(a^2*x*Sqrt[a + b*x^n + c*x^(2*n)]*AppellF1[n^(-1), -5/2, -5/2, 1 + n^(-1) 
, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])]) 
/(Sqrt[1 + (2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^n)/(b + Sqrt 
[b^2 - 4*a*c])])
 

Defintions of rubi rules used

rule 936

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] 
:> Simp[a^p*c^q*x*AppellF1[1/n, -p, -q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c) 
], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] 
 && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 1686

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^ 
IntPart[p]*((a + b*x^n + c*x^(2*n))^FracPart[p]/((1 + 2*c*(x^n/(b + Rt[b^2 
- 4*a*c, 2])))^FracPart[p]*(1 + 2*c*(x^n/(b - Rt[b^2 - 4*a*c, 2])))^FracPar 
t[p]))   Int[(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c])))^p*(1 + 2*c*(x^n/(b - S 
qrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] 
&& NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]
 

Maple [F]

\[\int \left (a +b \,x^{n}+c \,x^{2 n}\right )^{\frac {5}{2}}d x\]

Input:

int((a+b*x^n+c*x^(2*n))^(5/2),x)
 

Output:

int((a+b*x^n+c*x^(2*n))^(5/2),x)
 

Fricas [F(-2)]

Exception generated. \[ \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*x^n+c*x^(2*n))^(5/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx=\int \left (a + b x^{n} + c x^{2 n}\right )^{\frac {5}{2}}\, dx \] Input:

integrate((a+b*x**n+c*x**(2*n))**(5/2),x)
 

Output:

Integral((a + b*x**n + c*x**(2*n))**(5/2), x)
 

Maxima [F]

\[ \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx=\int { {\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+b*x^n+c*x^(2*n))^(5/2),x, algorithm="maxima")
 

Output:

integrate((c*x^(2*n) + b*x^n + a)^(5/2), x)
 

Giac [F]

\[ \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx=\int { {\left (c x^{2 \, n} + b x^{n} + a\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+b*x^n+c*x^(2*n))^(5/2),x, algorithm="giac")
 

Output:

integrate((c*x^(2*n) + b*x^n + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx=\int {\left (a+b\,x^n+c\,x^{2\,n}\right )}^{5/2} \,d x \] Input:

int((a + b*x^n + c*x^(2*n))^(5/2),x)
 

Output:

int((a + b*x^n + c*x^(2*n))^(5/2), x)
 

Reduce [F]

\[ \int \left (a+b x^n+c x^{2 n}\right )^{5/2} \, dx=\text {too large to display} \] Input:

int((a+b*x^n+c*x^(2*n))^(5/2),x)
 

Output:

(384*x**(4*n)*sqrt(x**(2*n)*c + x**n*b + a)*c**3*n**4*x + 1184*x**(4*n)*sq 
rt(x**(2*n)*c + x**n*b + a)*c**3*n**3*x + 976*x**(4*n)*sqrt(x**(2*n)*c + x 
**n*b + a)*c**3*n**2*x + 304*x**(4*n)*sqrt(x**(2*n)*c + x**n*b + a)*c**3*n 
*x + 32*x**(4*n)*sqrt(x**(2*n)*c + x**n*b + a)*c**3*x + 1008*x**(3*n)*sqrt 
(x**(2*n)*c + x**n*b + a)*b*c**2*n**4*x + 3048*x**(3*n)*sqrt(x**(2*n)*c + 
x**n*b + a)*b*c**2*n**3*x + 2392*x**(3*n)*sqrt(x**(2*n)*c + x**n*b + a)*b* 
c**2*n**2*x + 688*x**(3*n)*sqrt(x**(2*n)*c + x**n*b + a)*b*c**2*n*x + 64*x 
**(3*n)*sqrt(x**(2*n)*c + x**n*b + a)*b*c**2*x + 1408*x**(2*n)*sqrt(x**(2* 
n)*c + x**n*b + a)*a*c**2*n**4*x + 4128*x**(2*n)*sqrt(x**(2*n)*c + x**n*b 
+ a)*a*c**2*n**3*x + 2992*x**(2*n)*sqrt(x**(2*n)*c + x**n*b + a)*a*c**2*n* 
*2*x + 768*x**(2*n)*sqrt(x**(2*n)*c + x**n*b + a)*a*c**2*n*x + 64*x**(2*n) 
*sqrt(x**(2*n)*c + x**n*b + a)*a*c**2*x + 744*x**(2*n)*sqrt(x**(2*n)*c + x 
**n*b + a)*b**2*c*n**4*x + 2164*x**(2*n)*sqrt(x**(2*n)*c + x**n*b + a)*b** 
2*c*n**3*x + 1536*x**(2*n)*sqrt(x**(2*n)*c + x**n*b + a)*b**2*c*n**2*x + 3 
84*x**(2*n)*sqrt(x**(2*n)*c + x**n*b + a)*b**2*c*n*x + 32*x**(2*n)*sqrt(x* 
*(2*n)*c + x**n*b + a)*b**2*c*x + 2488*x**n*sqrt(x**(2*n)*c + x**n*b + a)* 
a*b*c*n**4*x + 6728*x**n*sqrt(x**(2*n)*c + x**n*b + a)*a*b*c*n**3*x + 3912 
*x**n*sqrt(x**(2*n)*c + x**n*b + a)*a*b*c*n**2*x + 848*x**n*sqrt(x**(2*n)* 
c + x**n*b + a)*a*b*c*n*x + 64*x**n*sqrt(x**(2*n)*c + x**n*b + a)*a*b*c*x 
+ 30*x**n*sqrt(x**(2*n)*c + x**n*b + a)*b**3*n**4*x + 60*x**n*sqrt(x**(...