\(\int \frac {(e x)^m}{(c+d x)^2 (a+b x^2)^{5/2}} \, dx\) [1752]

Optimal result

Integrand size = 24, antiderivative size = 261 \[ \int \frac {(e x)^m}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {(e x)^{1+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {5}{2},2,\frac {3+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{a^2 c^2 e (1+m) \sqrt {a+b x^2}}-\frac {2 d (e x)^{2+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {2+m}{2},\frac {5}{2},2,\frac {4+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{a^2 c^3 e^2 (2+m) \sqrt {a+b x^2}}+\frac {d^2 (e x)^{3+m} \sqrt {1+\frac {b x^2}{a}} \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {5}{2},2,\frac {5+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{a^2 c^4 e^3 (3+m) \sqrt {a+b x^2}} \] Output:

(e*x)^(1+m)*(1+b*x^2/a)^(1/2)*AppellF1(1/2+1/2*m,2,5/2,3/2+1/2*m,d^2*x^2/c 
^2,-b*x^2/a)/a^2/c^2/e/(1+m)/(b*x^2+a)^(1/2)-2*d*(e*x)^(2+m)*(1+b*x^2/a)^( 
1/2)*AppellF1(1+1/2*m,2,5/2,2+1/2*m,d^2*x^2/c^2,-b*x^2/a)/a^2/c^3/e^2/(2+m 
)/(b*x^2+a)^(1/2)+d^2*(e*x)^(3+m)*(1+b*x^2/a)^(1/2)*AppellF1(3/2+1/2*m,2,5 
/2,5/2+1/2*m,d^2*x^2/c^2,-b*x^2/a)/a^2/c^4/e^3/(3+m)/(b*x^2+a)^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 257, normalized size of antiderivative = 0.98, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {623, 622, 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.

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

\(\Big \downarrow \) 623

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

\(\Big \downarrow \) 622

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

\(\Big \downarrow \) 2009

\(\displaystyle x^{-m} (e x)^m \left (\frac {x^{m+1} \sqrt {\frac {b x^2}{a}+1} \operatorname {AppellF1}\left (\frac {m+1}{2},\frac {5}{2},2,\frac {m+3}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{a^2 c^2 (m+1) \sqrt {a+b x^2}}+\frac {d^2 x^{m+3} \sqrt {\frac {b x^2}{a}+1} \operatorname {AppellF1}\left (\frac {m+3}{2},\frac {5}{2},2,\frac {m+5}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{a^2 c^4 (m+3) \sqrt {a+b x^2}}-\frac {2 d x^{m+2} \sqrt {\frac {b x^2}{a}+1} \operatorname {AppellF1}\left (\frac {m+2}{2},\frac {5}{2},2,\frac {m+4}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{a^2 c^3 (m+2) \sqrt {a+b x^2}}\right )\)

Input:

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

Output:

((e*x)^m*((x^(1 + m)*Sqrt[1 + (b*x^2)/a]*AppellF1[(1 + m)/2, 5/2, 2, (3 + 
m)/2, -((b*x^2)/a), (d^2*x^2)/c^2])/(a^2*c^2*(1 + m)*Sqrt[a + b*x^2]) - (2 
*d*x^(2 + m)*Sqrt[1 + (b*x^2)/a]*AppellF1[(2 + m)/2, 5/2, 2, (4 + m)/2, -( 
(b*x^2)/a), (d^2*x^2)/c^2])/(a^2*c^3*(2 + m)*Sqrt[a + b*x^2]) + (d^2*x^(3 
+ m)*Sqrt[1 + (b*x^2)/a]*AppellF1[(3 + m)/2, 5/2, 2, (5 + m)/2, -((b*x^2)/ 
a), (d^2*x^2)/c^2])/(a^2*c^4*(3 + m)*Sqrt[a + b*x^2])))/x^m
 

Defintions of rubi rules used

rule 622

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Int[ExpandIntegrand[x^m*(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 
 - d^2*x^2)))^(-n), x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[n, -1]
 

rule 623

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

rule 2009

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(sqrt(b*x^2 + a)*(e*x)^m/(b^3*d^2*x^8 + 2*b^3*c*d*x^7 + 6*a*b^2*c* 
d*x^5 + 6*a^2*b*c*d*x^3 + (b^3*c^2 + 3*a*b^2*d^2)*x^6 + 2*a^3*c*d*x + a^3* 
c^2 + 3*(a*b^2*c^2 + a^2*b*d^2)*x^4 + (3*a^2*b*c^2 + a^3*d^2)*x^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=e^{m} \left (\int \frac {x^{m} \sqrt {b \,x^{2}+a}}{b^{3} d^{2} x^{8}+2 b^{3} c d \,x^{7}+3 a \,b^{2} d^{2} x^{6}+b^{3} c^{2} x^{6}+6 a \,b^{2} c d \,x^{5}+3 a^{2} b \,d^{2} x^{4}+3 a \,b^{2} c^{2} x^{4}+6 a^{2} b c d \,x^{3}+a^{3} d^{2} x^{2}+3 a^{2} b \,c^{2} x^{2}+2 a^{3} c d x +a^{3} c^{2}}d x \right ) \] Input:

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

Output:

e**m*int((x**m*sqrt(a + b*x**2))/(a**3*c**2 + 2*a**3*c*d*x + a**3*d**2*x** 
2 + 3*a**2*b*c**2*x**2 + 6*a**2*b*c*d*x**3 + 3*a**2*b*d**2*x**4 + 3*a*b**2 
*c**2*x**4 + 6*a*b**2*c*d*x**5 + 3*a*b**2*d**2*x**6 + b**3*c**2*x**6 + 2*b 
**3*c*d*x**7 + b**3*d**2*x**8),x)