\(\int \frac {(e x)^m}{(c+d x)^{5/2} (a+b x^2)} \, dx\) [147]

Optimal result

Integrand size = 24, antiderivative size = 158 \[ \int \frac {(e x)^m}{(c+d x)^{5/2} \left (a+b x^2\right )} \, dx=\frac {(e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {5}{2},1,2+m,-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{2 a c^2 e (1+m) \sqrt {c+d x}}+\frac {(e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {5}{2},1,2+m,-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{2 a c^2 e (1+m) \sqrt {c+d x}} \] Output:

1/2*(e*x)^(1+m)*(1+d*x/c)^(1/2)*AppellF1(1+m,1,5/2,2+m,-b^(1/2)*x/(-a)^(1/ 
2),-d*x/c)/a/c^2/e/(1+m)/(d*x+c)^(1/2)+1/2*(e*x)^(1+m)*(1+d*x/c)^(1/2)*App 
ellF1(1+m,1,5/2,2+m,b^(1/2)*x/(-a)^(1/2),-d*x/c)/a/c^2/e/(1+m)/(d*x+c)^(1/ 
2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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