\(\int \frac {(e x)^m \sqrt {c+d x}}{(a+b x^2)^2} \, dx\) [151]

Optimal result

Integrand size = 24, antiderivative size = 243 \[ \int \frac {(e x)^m \sqrt {c+d x}}{\left (a+b x^2\right )^2} \, dx=\frac {(e x)^{1+m} \sqrt {c+d x}}{2 a e \left (a+b x^2\right )}-\frac {\left (\frac {\sqrt {-a} d (1-2 m)}{\sqrt {b}}-2 c (1-m)\right ) (e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{8 a^2 e (1+m) \sqrt {c+d x}}+\frac {\left (\frac {\sqrt {-a} d (1-2 m)}{\sqrt {b}}+2 c (1-m)\right ) (e x)^{1+m} \sqrt {1+\frac {d x}{c}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{8 a^2 e (1+m) \sqrt {c+d x}} \] Output:

1/2*(e*x)^(1+m)*(d*x+c)^(1/2)/a/e/(b*x^2+a)-1/8*((-a)^(1/2)*d*(1-2*m)/b^(1 
/2)-2*c*(1-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^2/e/(1+m)/(d*x+c)^(1/2)+1/8*((-a)^(1/2)*d*(1-2*m)/ 
b^(1/2)+2*c*(1-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^2/e/(1+m)/(d*x+c)^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.25, 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.

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

\(\Big \downarrow \) 615

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

int((e*x)^m*(d*x+c)^(1/2)/(b*x^2+a)^2,x)
 

Output:

int((e*x)^m*(d*x+c)^(1/2)/(b*x^2+a)^2,x)
 

Fricas [F]

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

integrate((e*x)^m*(d*x+c)^(1/2)/(b*x^2+a)^2,x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

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

integrate((e*x)**m*(d*x+c)**(1/2)/(b*x**2+a)**2,x)
 

Output:

Timed out
 

Maxima [F]

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

integrate((e*x)^m*(d*x+c)^(1/2)/(b*x^2+a)^2,x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((e*x)^m*(d*x+c)^(1/2)/(b*x^2+a)^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(((e*x)^m*(c + d*x)^(1/2))/(a + b*x^2)^2,x)
 

Output:

int(((e*x)^m*(c + d*x)^(1/2))/(a + b*x^2)^2, x)
 

Reduce [F]

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

int((e*x)^m*(d*x+c)^(1/2)/(b*x^2+a)^2,x)
 

Output:

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