\(\int \frac {(e x)^m}{(c+d x)^{5/2} (a+b x^2)^2} \, dx\) [154]

Optimal result

Integrand size = 24, antiderivative size = 670 \[ \int \frac {(e x)^m}{(c+d x)^{5/2} \left (a+b x^2\right )^2} \, dx=-\frac {d^2 \left (3 b c^2-2 a d^2\right ) (e x)^{1+m}}{3 a c \left (b c^2+a d^2\right )^2 e (c+d x)^{3/2}}-\frac {d^2 \left (9 b^2 c^4-a b c^2 d^2 (55-8 m)-4 a^2 d^4 (1-2 m)\right ) (e x)^{1+m}}{6 a c^2 \left (b c^2+a d^2\right )^3 e \sqrt {c+d x}}+\frac {b (e x)^{1+m} (c-d x)}{2 a \left (b c^2+a d^2\right ) e (c+d x)^{3/2} \left (a+b x^2\right )}+\frac {b \left (15 \sqrt {-a} a b c^2 d^2-a^2 \sqrt {b} c d^3 (19-4 m)-(-a)^{5/2} d^4 (7-2 m)+2 \sqrt {-a} b^2 c^4 (1-m)+a b^{3/2} c^3 d (1+4 m)\right ) (e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{8 (-a)^{5/2} \left (b c^2+a d^2\right )^3 e (1+m) \sqrt {c+d x}}-\frac {b \left (15 (-a)^{3/2} b c^2 d^2-a^2 \sqrt {b} c d^3 (19-4 m)+(-a)^{5/2} d^4 (7-2 m)-2 \sqrt {-a} b^2 c^4 (1-m)+a b^{3/2} c^3 d (1+4 m)\right ) (e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{8 (-a)^{5/2} \left (b c^2+a d^2\right )^3 e (1+m) \sqrt {c+d x}}+\frac {d \left (9 b^2 c^4-a b c^2 d^2 (55-8 m)-4 a^2 d^4 (1-2 m)\right ) (1+2 m) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )}{6 a c^2 \left (b c^2+a d^2\right )^3} \] Output:

-1/3*d^2*(-2*a*d^2+3*b*c^2)*(e*x)^(1+m)/a/c/(a*d^2+b*c^2)^2/e/(d*x+c)^(3/2 
)-1/6*d^2*(9*b^2*c^4-a*b*c^2*d^2*(55-8*m)-4*a^2*d^4*(1-2*m))*(e*x)^(1+m)/a 
/c^2/(a*d^2+b*c^2)^3/e/(d*x+c)^(1/2)+1/2*b*(e*x)^(1+m)*(-d*x+c)/a/(a*d^2+b 
*c^2)/e/(d*x+c)^(3/2)/(b*x^2+a)+1/8*b*(15*(-a)^(1/2)*a*b*c^2*d^2-a^2*b^(1/ 
2)*c*d^3*(19-4*m)-(-a)^(5/2)*d^4*(7-2*m)+2*(-a)^(1/2)*b^2*c^4*(1-m)+a*b^(3 
/2)*c^3*d*(1+4*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)/(a*d^2+b*c^2)^3/e/(1+m)/(d*x+c)^(1/2) 
-1/8*b*(15*(-a)^(3/2)*b*c^2*d^2-a^2*b^(1/2)*c*d^3*(19-4*m)+(-a)^(5/2)*d^4* 
(7-2*m)-2*(-a)^(1/2)*b^2*c^4*(1-m)+a*b^(3/2)*c^3*d*(1+4*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)/(a*d^2+b*c^2)^3/e/(1+m)/(d*x+c)^(1/2)+1/6*d*(9*b^2*c^4-a*b*c^2*d^2*(55- 
8*m)-4*a^2*d^4*(1-2*m))*(1+2*m)*(e*x)^m*(d*x+c)^(1/2)*hypergeom([1/2, -m], 
[3/2],1+d*x/c)/a/c^2/(a*d^2+b*c^2)^3/((-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] (warning: unable to verify)

Time = 0.63 (sec) , antiderivative size = 315, normalized size of antiderivative = 0.47, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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]

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

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

Output:

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

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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