Integrand size = 19, antiderivative size = 136 \[ \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx=\frac {5 a^2 x^{1+m} \sqrt {a+b x^{2 (1+m)}}}{16 (1+m)}+\frac {5 a x^{1+m} \left (a+b x^{2 (1+m)}\right )^{3/2}}{24 (1+m)}+\frac {x^{1+m} \left (a+b x^{2 (1+m)}\right )^{5/2}}{6 (1+m)}+\frac {5 a^3 \text {arctanh}\left (\frac {\sqrt {b} x^{1+m}}{\sqrt {a+b x^{2 (1+m)}}}\right )}{16 \sqrt {b} (1+m)} \] Output:
5*a^2*x^(1+m)*(a+b*x^(2+2*m))^(1/2)/(16+16*m)+5*a*x^(1+m)*(a+b*x^(2+2*m))^ (3/2)/(24+24*m)+x^(1+m)*(a+b*x^(2+2*m))^(5/2)/(6+6*m)+5/16*a^3*arctanh(b^( 1/2)*x^(1+m)/(a+b*x^(2+2*m))^(1/2))/b^(1/2)/(1+m)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.65 \[ \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx=\frac {a^2 x^{1+m} \sqrt {a+b x^{2+2 m}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},\frac {1+m}{2+2 m},1+\frac {1+m}{2+2 m},-\frac {b x^{2+2 m}}{a}\right )}{(1+m) \sqrt {1+\frac {b x^{2+2 m}}{a}}} \] Input:
Integrate[x^m*(a + b*x^(2 + 2*m))^(5/2),x]
Output:
(a^2*x^(1 + m)*Sqrt[a + b*x^(2 + 2*m)]*Hypergeometric2F1[-5/2, (1 + m)/(2 + 2*m), 1 + (1 + m)/(2 + 2*m), -((b*x^(2 + 2*m))/a)])/((1 + m)*Sqrt[1 + (b *x^(2 + 2*m))/a])
Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.94, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {868, 211, 211, 211, 224, 219}
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.
\(\displaystyle \int x^m \left (a+b x^{2 m+2}\right )^{5/2} \, dx\) |
\(\Big \downarrow \) 868 |
\(\displaystyle \frac {\int \left (b x^{2 m+2}+a\right )^{5/2}dx^{m+1}}{m+1}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\frac {5}{6} a \int \left (b x^{2 m+2}+a\right )^{3/2}dx^{m+1}+\frac {1}{6} x^{m+1} \left (a+b x^{2 m+2}\right )^{5/2}}{m+1}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {b x^{2 m+2}+a}dx^{m+1}+\frac {1}{4} x^{m+1} \left (a+b x^{2 m+2}\right )^{3/2}\right )+\frac {1}{6} x^{m+1} \left (a+b x^{2 m+2}\right )^{5/2}}{m+1}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^{2 m+2}+a}}dx^{m+1}+\frac {1}{2} x^{m+1} \sqrt {a+b x^{2 m+2}}\right )+\frac {1}{4} x^{m+1} \left (a+b x^{2 m+2}\right )^{3/2}\right )+\frac {1}{6} x^{m+1} \left (a+b x^{2 m+2}\right )^{5/2}}{m+1}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-b x^{2 m+2}}d\frac {x^{m+1}}{\sqrt {b x^{2 m+2}+a}}+\frac {1}{2} x^{m+1} \sqrt {a+b x^{2 m+2}}\right )+\frac {1}{4} x^{m+1} \left (a+b x^{2 m+2}\right )^{3/2}\right )+\frac {1}{6} x^{m+1} \left (a+b x^{2 m+2}\right )^{5/2}}{m+1}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x^{m+1}}{\sqrt {a+b x^{2 m+2}}}\right )}{2 \sqrt {b}}+\frac {1}{2} x^{m+1} \sqrt {a+b x^{2 m+2}}\right )+\frac {1}{4} x^{m+1} \left (a+b x^{2 m+2}\right )^{3/2}\right )+\frac {1}{6} x^{m+1} \left (a+b x^{2 m+2}\right )^{5/2}}{m+1}\) |
Input:
Int[x^m*(a + b*x^(2 + 2*m))^(5/2),x]
Output:
((x^(1 + m)*(a + b*x^(2 + 2*m))^(5/2))/6 + (5*a*((x^(1 + m)*(a + b*x^(2 + 2*m))^(3/2))/4 + (3*a*((x^(1 + m)*Sqrt[a + b*x^(2 + 2*m)])/2 + (a*ArcTanh[ (Sqrt[b]*x^(1 + m))/Sqrt[a + b*x^(2 + 2*m)]])/(2*Sqrt[b])))/4))/6)/(1 + m)
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/(m + 1) Subst[Int[(a + b*x^Simplify[n/(m + 1)])^p, x], x, x^(m + 1)], x] /; FreeQ[ {a, b, m, n, p}, x] && IntegerQ[Simplify[n/(m + 1)]] && !IntegerQ[n]
\[\int x^{m} \left (a +b \,x^{2+2 m}\right )^{\frac {5}{2}}d x\]
Input:
int(x^m*(a+b*x^(2+2*m))^(5/2),x)
Output:
int(x^m*(a+b*x^(2+2*m))^(5/2),x)
Exception generated. \[ \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(x^m*(a+b*x^(2+2*m))^(5/2),x, algorithm="fricas")
Output:
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
Result contains complex when optimal does not.
Time = 53.46 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.79 \[ \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx=\frac {\sqrt {\pi } \sqrt {a} a^{- \frac {m}{2 m + 2} + \frac {5}{2} - \frac {1}{2 m + 2}} x^{m + 1} {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {1}{2} \\ \frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2} \end {matrix}\middle | {\frac {b x^{2 m + 2} e^{i \pi }}{a}} \right )}}{2 m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) + 2 \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )} \] Input:
integrate(x**m*(a+b*x**(2+2*m))**(5/2),x)
Output:
sqrt(pi)*sqrt(a)*a**(-m/(2*m + 2) + 5/2 - 1/(2*m + 2))*x**(m + 1)*hyper((- 5/2, 1/2), (m/(2*m + 2) + 1 + 1/(2*m + 2),), b*x**(2*m + 2)*exp_polar(I*pi )/a)/(2*m*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) + 2*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)))
\[ \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx=\int { {\left (b x^{2 \, m + 2} + a\right )}^{\frac {5}{2}} x^{m} \,d x } \] Input:
integrate(x^m*(a+b*x^(2+2*m))^(5/2),x, algorithm="maxima")
Output:
integrate((b*x^(2*m + 2) + a)^(5/2)*x^m, x)
\[ \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx=\int { {\left (b x^{2 \, m + 2} + a\right )}^{\frac {5}{2}} x^{m} \,d x } \] Input:
integrate(x^m*(a+b*x^(2+2*m))^(5/2),x, algorithm="giac")
Output:
integrate((b*x^(2*m + 2) + a)^(5/2)*x^m, x)
Timed out. \[ \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx=\int x^m\,{\left (a+b\,x^{2\,m+2}\right )}^{5/2} \,d x \] Input:
int(x^m*(a + b*x^(2*m + 2))^(5/2),x)
Output:
int(x^m*(a + b*x^(2*m + 2))^(5/2), x)
\[ \int x^m \left (a+b x^{2+2 m}\right )^{5/2} \, dx=\frac {8 x^{5 m} \sqrt {x^{2 m} b \,x^{2}+a}\, b^{2} x^{5}+26 x^{3 m} \sqrt {x^{2 m} b \,x^{2}+a}\, a b \,x^{3}+33 x^{m} \sqrt {x^{2 m} b \,x^{2}+a}\, a^{2} x +15 \left (\int \frac {x^{m} \sqrt {x^{2 m} b \,x^{2}+a}}{x^{2 m} b \,x^{2}+a}d x \right ) a^{3} m +15 \left (\int \frac {x^{m} \sqrt {x^{2 m} b \,x^{2}+a}}{x^{2 m} b \,x^{2}+a}d x \right ) a^{3}}{48 m +48} \] Input:
int(x^m*(a+b*x^(2+2*m))^(5/2),x)
Output:
(8*x**(5*m)*sqrt(x**(2*m)*b*x**2 + a)*b**2*x**5 + 26*x**(3*m)*sqrt(x**(2*m )*b*x**2 + a)*a*b*x**3 + 33*x**m*sqrt(x**(2*m)*b*x**2 + a)*a**2*x + 15*int ((x**m*sqrt(x**(2*m)*b*x**2 + a))/(x**(2*m)*b*x**2 + a),x)*a**3*m + 15*int ((x**m*sqrt(x**(2*m)*b*x**2 + a))/(x**(2*m)*b*x**2 + a),x)*a**3)/(48*(m + 1))