\(\int (-\frac {b x^{1+m}}{(a+b x^2)^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}) \, dx\) [713]

Optimal result

Integrand size = 38, antiderivative size = 15 \[ \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx=\frac {x^m}{\sqrt {a+b x^2}} \] Output:

x^m/(b*x^2+a)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.00 (sec) , antiderivative size = 103, normalized size of antiderivative = 6.87 \[ \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx=\frac {x^m \sqrt {1+\frac {b x^2}{a}} \left (a (2+m) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m}{2},\frac {2+m}{2},-\frac {b x^2}{a}\right )+b (-1+m) x^2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )\right )}{a (2+m) \sqrt {a+b x^2}} \] Input:

Integrate[-((b*x^(1 + m))/(a + b*x^2)^(3/2)) + (m*x^(-1 + m))/Sqrt[a + b*x 
^2],x]
 

Output:

(x^m*Sqrt[1 + (b*x^2)/a]*(a*(2 + m)*Hypergeometric2F1[3/2, m/2, (2 + m)/2, 
 -((b*x^2)/a)] + b*(-1 + m)*x^2*Hypergeometric2F1[3/2, (2 + m)/2, (4 + m)/ 
2, -((b*x^2)/a)]))/(a*(2 + m)*Sqrt[a + b*x^2])
 

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.23 (sec) , antiderivative size = 123, normalized size of antiderivative = 8.20, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {2009}

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.

\(\displaystyle \int \left (\frac {m x^{m-1}}{\sqrt {a+b x^2}}-\frac {b x^{m+1}}{\left (a+b x^2\right )^{3/2}}\right ) \, dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^m \sqrt {\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {m+2}{2},-\frac {b x^2}{a}\right )}{\sqrt {a+b x^2}}-\frac {b x^{m+2} \sqrt {\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {b x^2}{a}\right )}{a (m+2) \sqrt {a+b x^2}}\)

Input:

Int[-((b*x^(1 + m))/(a + b*x^2)^(3/2)) + (m*x^(-1 + m))/Sqrt[a + b*x^2],x]
 

Output:

(x^m*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, m/2, (2 + m)/2, -((b*x^2)/ 
a)])/Sqrt[a + b*x^2] - (b*x^(2 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[ 
3/2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/(a*(2 + m)*Sqrt[a + b*x^2])
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(61\) vs. \(2(13)=26\).

Time = 0.46 (sec) , antiderivative size = 62, normalized size of antiderivative = 4.13

method	result	size
orering	\(\frac {x \left (b \,x^{2}+a \right ) \left (-\frac {b \,x^{1+m}}{\left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {m \,x^{m -1}}{\sqrt {b \,x^{2}+a}}\right )}{b m \,x^{2}-b \,x^{2}+a m}\)	\(62\)

Input:

int(-b*x^(1+m)/(b*x^2+a)^(3/2)+m*x^(m-1)/(b*x^2+a)^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

x*(b*x^2+a)/(b*m*x^2-b*x^2+a*m)*(-b*x^(1+m)/(b*x^2+a)^(3/2)+m*x^(m-1)/(b*x 
^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.73 \[ \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx=\frac {\sqrt {b x^{2} + a} x^{m + 1}}{b x^{3} + a x} \] Input:

integrate(-b*x^(1+m)/(b*x^2+a)^(3/2)+m*x^(-1+m)/(b*x^2+a)^(1/2),x, algorit 
hm="fricas")
 

Output:

sqrt(b*x^2 + a)*x^(m + 1)/(b*x^3 + a*x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 6.80 \[ \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx=\frac {a^{\frac {m}{2}} a^{- \frac {m}{2} - \frac {1}{2}} m x^{m} \Gamma \left (\frac {m}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {m}{2} \\ \frac {m}{2} + 1 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 1\right )} - \frac {b x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (\frac {m}{2} + 2\right )} \] Input:

integrate(-b*x**(1+m)/(b*x**2+a)**(3/2)+m*x**(-1+m)/(b*x**2+a)**(1/2),x)
 

Output:

a**(m/2)*a**(-m/2 - 1/2)*m*x**m*gamma(m/2)*hyper((1/2, m/2), (m/2 + 1,), b 
*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 1)) - b*x**(m + 2)*gamma(m/2 + 1)* 
hyper((3/2, m/2 + 1), (m/2 + 2,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*ga 
mma(m/2 + 2))
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx=\frac {x^{m}}{\sqrt {b x^{2} + a}} \] Input:

integrate(-b*x^(1+m)/(b*x^2+a)^(3/2)+m*x^(-1+m)/(b*x^2+a)^(1/2),x, algorit 
hm="maxima")
 

Output:

x^m/sqrt(b*x^2 + a)
 

Giac [F]

\[ \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx=\int { \frac {m x^{m - 1}}{\sqrt {b x^{2} + a}} - \frac {b x^{m + 1}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(-b*x^(1+m)/(b*x^2+a)^(3/2)+m*x^(-1+m)/(b*x^2+a)^(1/2),x, algorit 
hm="giac")
 

Output:

integrate(m*x^(m - 1)/sqrt(b*x^2 + a) - b*x^(m + 1)/(b*x^2 + a)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx=-\int \frac {b\,x^{m+1}}{{\left (b\,x^2+a\right )}^{3/2}}-\frac {m\,x^{m-1}}{\sqrt {b\,x^2+a}} \,d x \] Input:

int((m*x^(m - 1))/(a + b*x^2)^(1/2) - (b*x^(m + 1))/(a + b*x^2)^(3/2),x)
 

Output:

-int((b*x^(m + 1))/(a + b*x^2)^(3/2) - (m*x^(m - 1))/(a + b*x^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 206, normalized size of antiderivative = 13.73 \[ \int \left (-\frac {b x^{1+m}}{\left (a+b x^2\right )^{3/2}}+\frac {m x^{-1+m}}{\sqrt {a+b x^2}}\right ) \, dx=\frac {\sqrt {b \,x^{2}+a}\, \left (a^{\frac {m}{2}} \left (\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +b \,x^{2}\right )^{m} \left (\sqrt {b \,x^{2}+a}+\sqrt {b}\, x \right )^{m} 2^{m} m -a^{\frac {m}{2}} \left (\sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +b \,x^{2}\right )^{m} \left (\sqrt {b \,x^{2}+a}+\sqrt {b}\, x \right )^{m} 2^{m}+\left (2 \sqrt {b}\, \sqrt {b \,x^{2}+a}\, x +2 b \,x^{2}\right )^{m} \left (\sqrt {a}\, \sqrt {b \,x^{2}+a}+\sqrt {b}\, \sqrt {a}\, x \right )^{m}\right )}{b^{\frac {m}{2}} \left (\sqrt {a}\, \sqrt {b \,x^{2}+a}+\sqrt {b}\, \sqrt {a}\, x \right )^{m} \left (\sqrt {b \,x^{2}+a}+\sqrt {b}\, x \right )^{m} 2^{m} m \left (b \,x^{2}+a \right )} \] Input:

int(-b*x^(1+m)/(b*x^2+a)^(3/2)+m*x^(-1+m)/(b*x^2+a)^(1/2),x)
 

Output:

(sqrt(a + b*x**2)*(a**(m/2)*(sqrt(b)*sqrt(a + b*x**2)*x + b*x**2)**m*(sqrt 
(a + b*x**2) + sqrt(b)*x)**m*2**m*m - a**(m/2)*(sqrt(b)*sqrt(a + b*x**2)*x 
 + b*x**2)**m*(sqrt(a + b*x**2) + sqrt(b)*x)**m*2**m + (2*sqrt(b)*sqrt(a + 
 b*x**2)*x + 2*b*x**2)**m*(sqrt(a)*sqrt(a + b*x**2) + sqrt(b)*sqrt(a)*x)** 
m))/(b**(m/2)*(sqrt(a)*sqrt(a + b*x**2) + sqrt(b)*sqrt(a)*x)**m*(sqrt(a + 
b*x**2) + sqrt(b)*x)**m*2**m*m*(a + b*x**2))