\(\int \frac {(e x)^m (c+d x)}{(a+b x^2)^{5/2}} \, dx\) [181]

Optimal result

Integrand size = 22, antiderivative size = 145 \[ \int \frac {(e x)^m (c+d x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {c (e x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2 e (1+m) \sqrt {a+b x^2}}+\frac {d (e x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a^2 e^2 (2+m) \sqrt {a+b x^2}} \] Output:

c*(e*x)^(1+m)*(1+b*x^2/a)^(1/2)*hypergeom([5/2, 1/2+1/2*m],[3/2+1/2*m],-b* 
x^2/a)/a^2/e/(1+m)/(b*x^2+a)^(1/2)+d*(e*x)^(2+m)*(1+b*x^2/a)^(1/2)*hyperge 
om([5/2, 1+1/2*m],[2+1/2*m],-b*x^2/a)/a^2/e^2/(2+m)/(b*x^2+a)^(1/2)
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.77 \[ \int \frac {(e x)^m (c+d x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {b x^2}{a}} \left (c (2+m) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+d (1+m) x \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )\right )}{a^2 (1+m) (2+m) \sqrt {a+b x^2}} \] Input:

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

Output:

(x*(e*x)^m*Sqrt[1 + (b*x^2)/a]*(c*(2 + m)*Hypergeometric2F1[5/2, (1 + m)/2 
, (3 + m)/2, -((b*x^2)/a)] + d*(1 + m)*x*Hypergeometric2F1[5/2, (2 + m)/2, 
 (4 + m)/2, -((b*x^2)/a)]))/(a^2*(1 + m)*(2 + m)*Sqrt[a + b*x^2])
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {557, 279, 278}

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.

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

\(\Big \downarrow \) 557

\(\displaystyle c \int \frac {(e x)^m}{\left (b x^2+a\right )^{5/2}}dx+\frac {d \int \frac {(e x)^{m+1}}{\left (b x^2+a\right )^{5/2}}dx}{e}\)

\(\Big \downarrow \) 279

\(\displaystyle \frac {c \sqrt {\frac {b x^2}{a}+1} \int \frac {(e x)^m}{\left (\frac {b x^2}{a}+1\right )^{5/2}}dx}{a^2 \sqrt {a+b x^2}}+\frac {d \sqrt {\frac {b x^2}{a}+1} \int \frac {(e x)^{m+1}}{\left (\frac {b x^2}{a}+1\right )^{5/2}}dx}{a^2 e \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 278

\(\displaystyle \frac {c \sqrt {\frac {b x^2}{a}+1} (e x)^{m+1} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+1}{2},\frac {m+3}{2},-\frac {b x^2}{a}\right )}{a^2 e (m+1) \sqrt {a+b x^2}}+\frac {d \sqrt {\frac {b x^2}{a}+1} (e x)^{m+2} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {m+2}{2},\frac {m+4}{2},-\frac {b x^2}{a}\right )}{a^2 e^2 (m+2) \sqrt {a+b x^2}}\)

Input:

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

Output:

(c*(e*x)^(1 + m)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[5/2, (1 + m)/2, (3 
+ m)/2, -((b*x^2)/a)])/(a^2*e*(1 + m)*Sqrt[a + b*x^2]) + (d*(e*x)^(2 + m)* 
Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[5/2, (2 + m)/2, (4 + m)/2, -((b*x^2) 
/a)])/(a^2*e^2*(2 + m)*Sqrt[a + b*x^2])
 

Defintions of rubi rules used

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

Maple [F]

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

Input:

int((e*x)^m*(d*x+c)/(b*x^2+a)^(5/2),x)
 

Output:

int((e*x)^m*(d*x+c)/(b*x^2+a)^(5/2),x)
 

Fricas [F]

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

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

Output:

integral(sqrt(b*x^2 + a)*(d*x + c)*(e*x)^m/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2* 
b*x^2 + a^3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 66.23 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.76 \[ \int \frac {(e x)^m (c+d x)}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {c e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {d e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} \Gamma \left (\frac {m}{2} + 2\right )} \] Input:

integrate((e*x)**m*(d*x+c)/(b*x**2+a)**(5/2),x)
 

Output:

c*e**m*x**(m + 1)*gamma(m/2 + 1/2)*hyper((5/2, m/2 + 1/2), (m/2 + 3/2,), b 
*x**2*exp_polar(I*pi)/a)/(2*a**(5/2)*gamma(m/2 + 3/2)) + d*e**m*x**(m + 2) 
*gamma(m/2 + 1)*hyper((5/2, m/2 + 1), (m/2 + 2,), b*x**2*exp_polar(I*pi)/a 
)/(2*a**(5/2)*gamma(m/2 + 2))
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

integrate((d*x + c)*(e*x)^m/(b*x^2 + a)^(5/2), x)
 

Giac [F]

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

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

Output:

integrate((d*x + c)*(e*x)^m/(b*x^2 + a)^(5/2), x)
 

Mupad [F(-1)]

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

int(((e*x)^m*(c + d*x))/(a + b*x^2)^(5/2),x)
 

Output:

int(((e*x)^m*(c + d*x))/(a + b*x^2)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(e x)^m (c+d x)}{\left (a+b x^2\right )^{5/2}} \, dx=e^{m} \left (\left (\int \frac {x^{m}}{\sqrt {b \,x^{2}+a}\, a^{2}+2 \sqrt {b \,x^{2}+a}\, a b \,x^{2}+\sqrt {b \,x^{2}+a}\, b^{2} x^{4}}d x \right ) c +\left (\int \frac {x^{m} x}{\sqrt {b \,x^{2}+a}\, a^{2}+2 \sqrt {b \,x^{2}+a}\, a b \,x^{2}+\sqrt {b \,x^{2}+a}\, b^{2} x^{4}}d x \right ) d \right ) \] Input:

int((e*x)^m*(d*x+c)/(b*x^2+a)^(5/2),x)
 

Output:

e**m*(int(x**m/(sqrt(a + b*x**2)*a**2 + 2*sqrt(a + b*x**2)*a*b*x**2 + sqrt 
(a + b*x**2)*b**2*x**4),x)*c + int((x**m*x)/(sqrt(a + b*x**2)*a**2 + 2*sqr 
t(a + b*x**2)*a*b*x**2 + sqrt(a + b*x**2)*b**2*x**4),x)*d)