Integrand size = 20, antiderivative size = 148 \[ \int x \sqrt {a+b x^n+c x^{2 n}} \, dx=\frac {x^2 \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (\frac {2}{n},-\frac {1}{2},-\frac {1}{2},\frac {2+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {1+\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}}} \] Output:
1/2*x^2*(a+b*x^n+c*x^(2*n))^(1/2)*AppellF1(2/n,-1/2,-1/2,(2+n)/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)
Leaf count is larger than twice the leaf count of optimal. \(364\) vs. \(2(148)=296\).
Time = 0.83 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.46 \[ \int x \sqrt {a+b x^n+c x^{2 n}} \, dx=\frac {x^2 \left (2 (2+n) \left (a+x^n \left (b+c x^n\right )\right )+a n (2+n) \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {2}{n},\frac {1}{2},\frac {1}{2},\frac {2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )+b n x^n \sqrt {\frac {b-\sqrt {b^2-4 a c}+2 c x^n}{b-\sqrt {b^2-4 a c}}} \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^n}{b+\sqrt {b^2-4 a c}}} \operatorname {AppellF1}\left (\frac {2+n}{n},\frac {1}{2},\frac {1}{2},2+\frac {2}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )\right )}{2 (2+n)^2 \sqrt {a+x^n \left (b+c x^n\right )}} \] Input:
Integrate[x*Sqrt[a + b*x^n + c*x^(2*n)],x]
Output:
(x^2*(2*(2 + n)*(a + x^n*(b + c*x^n)) + a*n*(2 + 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[2/n, 1/2, 1/2, (2 + n)/n, (-2*c*x ^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])] + b*n*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[(2 + n)/n, 1/2, 1/2, 2 + 2/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x^n)/(-b + Sq rt[b^2 - 4*a*c])]))/(2*(2 + n)^2*Sqrt[a + x^n*(b + c*x^n)])
Time = 0.30 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {1721, 1012}
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.
\(\displaystyle \int x \sqrt {a+b x^n+c x^{2 n}} \, dx\) |
\(\Big \downarrow \) 1721 |
\(\displaystyle \frac {\sqrt {a+b x^n+c x^{2 n}} \int x \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}+1}dx}{\sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {x^2 \sqrt {a+b x^n+c x^{2 n}} \operatorname {AppellF1}\left (\frac {2}{n},-\frac {1}{2},-\frac {1}{2},\frac {n+2}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 \sqrt {\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}+1} \sqrt {\frac {2 c x^n}{\sqrt {b^2-4 a c}+b}+1}}\) |
Input:
Int[x*Sqrt[a + b*x^n + c*x^(2*n)],x]
Output:
(x^2*Sqrt[a + b*x^n + c*x^(2*n)]*AppellF1[2/n, -1/2, -1/2, (2 + n)/n, (-2* c*x^n)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(2*Sq rt[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])])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((d_.)*(x_))^(m_.)*((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])))^FracPart[p])) Int[(d*x)^m*(1 + 2*c*(x^n/(b + Sqrt[b^2 - 4*a*c ])))^p*(1 + 2*c*(x^n/(b - Sqrt[b^2 - 4*a*c])))^p, x], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n]
\[\int x \sqrt {a +b \,x^{n}+c \,x^{2 n}}d x\]
Input:
int(x*(a+b*x^n+c*x^(2*n))^(1/2),x)
Output:
int(x*(a+b*x^n+c*x^(2*n))^(1/2),x)
Exception generated. \[ \int x \sqrt {a+b x^n+c x^{2 n}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x*(a+b*x^n+c*x^(2*n))^(1/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int x \sqrt {a+b x^n+c x^{2 n}} \, dx=\int x \sqrt {a + b x^{n} + c x^{2 n}}\, dx \] Input:
integrate(x*(a+b*x**n+c*x**(2*n))**(1/2),x)
Output:
Integral(x*sqrt(a + b*x**n + c*x**(2*n)), x)
\[ \int x \sqrt {a+b x^n+c x^{2 n}} \, dx=\int { \sqrt {c x^{2 \, n} + b x^{n} + a} x \,d x } \] Input:
integrate(x*(a+b*x^n+c*x^(2*n))^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(c*x^(2*n) + b*x^n + a)*x, x)
\[ \int x \sqrt {a+b x^n+c x^{2 n}} \, dx=\int { \sqrt {c x^{2 \, n} + b x^{n} + a} x \,d x } \] Input:
integrate(x*(a+b*x^n+c*x^(2*n))^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(c*x^(2*n) + b*x^n + a)*x, x)
Timed out. \[ \int x \sqrt {a+b x^n+c x^{2 n}} \, dx=\int x\,\sqrt {a+b\,x^n+c\,x^{2\,n}} \,d x \] Input:
int(x*(a + b*x^n + c*x^(2*n))^(1/2),x)
Output:
int(x*(a + b*x^n + c*x^(2*n))^(1/2), x)
\[ \int x \sqrt {a+b x^n+c x^{2 n}} \, dx=\frac {2 \sqrt {x^{2 n} c +x^{n} b +a}\, x^{2}-\left (\int \frac {x^{2 n} \sqrt {x^{2 n} c +x^{n} b +a}\, x}{x^{2 n} c n +4 x^{2 n} c +x^{n} b n +4 x^{n} b +a n +4 a}d x \right ) c \,n^{2}-4 \left (\int \frac {x^{2 n} \sqrt {x^{2 n} c +x^{n} b +a}\, x}{x^{2 n} c n +4 x^{2 n} c +x^{n} b n +4 x^{n} b +a n +4 a}d x \right ) c n +\left (\int \frac {\sqrt {x^{2 n} c +x^{n} b +a}\, x}{x^{2 n} c n +4 x^{2 n} c +x^{n} b n +4 x^{n} b +a n +4 a}d x \right ) a \,n^{2}+4 \left (\int \frac {\sqrt {x^{2 n} c +x^{n} b +a}\, x}{x^{2 n} c n +4 x^{2 n} c +x^{n} b n +4 x^{n} b +a n +4 a}d x \right ) a n}{n +4} \] Input:
int(x*(a+b*x^n+c*x^(2*n))^(1/2),x)
Output:
(2*sqrt(x**(2*n)*c + x**n*b + a)*x**2 - int((x**(2*n)*sqrt(x**(2*n)*c + x* *n*b + a)*x)/(x**(2*n)*c*n + 4*x**(2*n)*c + x**n*b*n + 4*x**n*b + a*n + 4* a),x)*c*n**2 - 4*int((x**(2*n)*sqrt(x**(2*n)*c + x**n*b + a)*x)/(x**(2*n)* c*n + 4*x**(2*n)*c + x**n*b*n + 4*x**n*b + a*n + 4*a),x)*c*n + int((sqrt(x **(2*n)*c + x**n*b + a)*x)/(x**(2*n)*c*n + 4*x**(2*n)*c + x**n*b*n + 4*x** n*b + a*n + 4*a),x)*a*n**2 + 4*int((sqrt(x**(2*n)*c + x**n*b + a)*x)/(x**( 2*n)*c*n + 4*x**(2*n)*c + x**n*b*n + 4*x**n*b + a*n + 4*a),x)*a*n)/(n + 4)