\(\int \frac {(e x)^m (c+d x)^{5/2}}{(a+b x^2)^2} \, dx\) [149]

Optimal result

Integrand size = 24, antiderivative size = 437 \[ \int \frac {(e x)^m (c+d x)^{5/2}}{\left (a+b x^2\right )^2} \, dx=-\frac {d^2 (e x)^{1+m} \sqrt {c+d x}}{2 a b e}+\frac {(e x)^{1+m} (c+d x)^{5/2}}{2 a e \left (a+b x^2\right )}+\frac {\left (a b c^2 d (1-6 m)+2 \sqrt {-a} b^{3/2} c^3 (1-m)+a^2 d^3 (3+2 m)+2 \sqrt {-a} a \sqrt {b} c d^2 (2+3 m)\right ) (e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{8 (-a)^{5/2} b^{3/2} e (1+m) \sqrt {c+d x}}-\frac {\left (a b c^2 d (1-6 m)-2 \sqrt {-a} b^{3/2} c^3 (1-m)+a^2 d^3 (3+2 m)-2 \sqrt {-a} a \sqrt {b} c d^2 (2+3 m)\right ) (e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{8 (-a)^{5/2} b^{3/2} e (1+m) \sqrt {c+d x}}-\frac {c d (1+2 m) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )}{a b} \] Output:

-1/2*d^2*(e*x)^(1+m)*(d*x+c)^(1/2)/a/b/e+1/2*(e*x)^(1+m)*(d*x+c)^(5/2)/a/e 
/(b*x^2+a)+1/8*(a*b*c^2*d*(1-6*m)+2*(-a)^(1/2)*b^(3/2)*c^3*(1-m)+a^2*d^3*( 
3+2*m)+2*(-a)^(1/2)*a*b^(1/2)*c*d^2*(2+3*m))*(e*x)^(1+m)*(1+d*x/c)^(1/2)*A 
ppellF1(1+m,1,1/2,2+m,-b^(1/2)*x/(-a)^(1/2),-d*x/c)/(-a)^(5/2)/b^(3/2)/e/( 
1+m)/(d*x+c)^(1/2)-1/8*(a*b*c^2*d*(1-6*m)-2*(-a)^(1/2)*b^(3/2)*c^3*(1-m)+a 
^2*d^3*(3+2*m)-2*(-a)^(1/2)*a*b^(1/2)*c*d^2*(2+3*m))*(e*x)^(1+m)*(1+d*x/c) 
^(1/2)*AppellF1(1+m,1,1/2,2+m,b^(1/2)*x/(-a)^(1/2),-d*x/c)/(-a)^(5/2)/b^(3 
/2)/e/(1+m)/(d*x+c)^(1/2)-c*d*(1+2*m)*(e*x)^m*(d*x+c)^(1/2)*hypergeom([1/2 
, -m],[3/2],1+d*x/c)/a/b/((-d*x/c)^m)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 315, normalized size of antiderivative = 0.72, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {615, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(c^2*(e*x)^(1 + m)*Sqrt[c + d*x]*AppellF1[1 + m, -5/2, 1, 2 + m, -((d*x)/c 
), -((Sqrt[b]*x)/Sqrt[-a])])/(4*a^2*e*(1 + m)*Sqrt[1 + (d*x)/c]) + (c^2*(e 
*x)^(1 + m)*Sqrt[c + d*x]*AppellF1[1 + m, -5/2, 1, 2 + m, -((d*x)/c), (Sqr 
t[b]*x)/Sqrt[-a]])/(4*a^2*e*(1 + m)*Sqrt[1 + (d*x)/c]) + (c^2*(e*x)^(1 + m 
)*Sqrt[c + d*x]*AppellF1[1 + m, -5/2, 2, 2 + m, -((d*x)/c), -((Sqrt[b]*x)/ 
Sqrt[-a])])/(4*a^2*e*(1 + m)*Sqrt[1 + (d*x)/c]) + (c^2*(e*x)^(1 + m)*Sqrt[ 
c + d*x]*AppellF1[1 + m, -5/2, 2, 2 + m, -((d*x)/c), (Sqrt[b]*x)/Sqrt[-a]] 
)/(4*a^2*e*(1 + m)*Sqrt[1 + (d*x)/c])
 

Defintions of rubi rules used

rule 615

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] 
 /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
 

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*( - 4*x**m*sqrt(c + d*x)*a*d**3*m - 6*x**m*sqrt(c + d*x)*a*d**3 + 12 
*x**m*sqrt(c + d*x)*b*c**2*d*m - 6*x**m*sqrt(c + d*x)*b*c**2*d + 4*x**m*sq 
rt(c + d*x)*b*c*d**2*m*x - 8*x**m*sqrt(c + d*x)*b*c*d**2*x + 8*int((x**m*s 
qrt(c + d*x)*x**2)/(2*a**2*c*m**2 - 5*a**2*c*m + 2*a**2*c + 2*a**2*d*m**2* 
x - 5*a**2*d*m*x + 2*a**2*d*x + 4*a*b*c*m**2*x**2 - 10*a*b*c*m*x**2 + 4*a* 
b*c*x**2 + 4*a*b*d*m**2*x**3 - 10*a*b*d*m*x**3 + 4*a*b*d*x**3 + 2*b**2*c*m 
**2*x**4 - 5*b**2*c*m*x**4 + 2*b**2*c*x**4 + 2*b**2*d*m**2*x**5 - 5*b**2*d 
*m*x**5 + 2*b**2*d*x**5),x)*a**2*b*d**4*m**4 - 20*int((x**m*sqrt(c + d*x)* 
x**2)/(2*a**2*c*m**2 - 5*a**2*c*m + 2*a**2*c + 2*a**2*d*m**2*x - 5*a**2*d* 
m*x + 2*a**2*d*x + 4*a*b*c*m**2*x**2 - 10*a*b*c*m*x**2 + 4*a*b*c*x**2 + 4* 
a*b*d*m**2*x**3 - 10*a*b*d*m*x**3 + 4*a*b*d*x**3 + 2*b**2*c*m**2*x**4 - 5* 
b**2*c*m*x**4 + 2*b**2*c*x**4 + 2*b**2*d*m**2*x**5 - 5*b**2*d*m*x**5 + 2*b 
**2*d*x**5),x)*a**2*b*d**4*m**3 - 10*int((x**m*sqrt(c + d*x)*x**2)/(2*a**2 
*c*m**2 - 5*a**2*c*m + 2*a**2*c + 2*a**2*d*m**2*x - 5*a**2*d*m*x + 2*a**2* 
d*x + 4*a*b*c*m**2*x**2 - 10*a*b*c*m*x**2 + 4*a*b*c*x**2 + 4*a*b*d*m**2*x* 
*3 - 10*a*b*d*m*x**3 + 4*a*b*d*x**3 + 2*b**2*c*m**2*x**4 - 5*b**2*c*m*x**4 
 + 2*b**2*c*x**4 + 2*b**2*d*m**2*x**5 - 5*b**2*d*m*x**5 + 2*b**2*d*x**5),x 
)*a**2*b*d**4*m**2 + 45*int((x**m*sqrt(c + d*x)*x**2)/(2*a**2*c*m**2 - 5*a 
**2*c*m + 2*a**2*c + 2*a**2*d*m**2*x - 5*a**2*d*m*x + 2*a**2*d*x + 4*a*b*c 
*m**2*x**2 - 10*a*b*c*m*x**2 + 4*a*b*c*x**2 + 4*a*b*d*m**2*x**3 - 10*a*...