\(\int \frac {(e x)^m (c+d x)^{3/2}}{a+b x^2} \, dx\) [143]

Optimal result

Integrand size = 24, antiderivative size = 154 \[ \int \frac {(e x)^m (c+d x)^{3/2}}{a+b x^2} \, dx=\frac {c (e x)^{1+m} \sqrt {c+d x} \operatorname {AppellF1}\left (1+m,-\frac {3}{2},1,2+m,-\frac {d x}{c},-\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{2 a e (1+m) \sqrt {1+\frac {d x}{c}}}+\frac {c (e x)^{1+m} \sqrt {c+d x} \operatorname {AppellF1}\left (1+m,-\frac {3}{2},1,2+m,-\frac {d x}{c},\frac {\sqrt {b} x}{\sqrt {-a}}\right )}{2 a e (1+m) \sqrt {1+\frac {d x}{c}}} \] Output:

1/2*c*(e*x)^(1+m)*(d*x+c)^(1/2)*AppellF1(1+m,1,-3/2,2+m,-b^(1/2)*x/(-a)^(1 
/2),-d*x/c)/a/e/(1+m)/(1+d*x/c)^(1/2)+1/2*c*(e*x)^(1+m)*(d*x+c)^(1/2)*Appe 
llF1(1+m,1,-3/2,2+m,b^(1/2)*x/(-a)^(1/2),-d*x/c)/a/e/(1+m)/(1+d*x/c)^(1/2)
 

Mathematica [F]

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

Integrate[((e*x)^m*(c + d*x)^(3/2))/(a + b*x^2),x]
 

Output:

Integrate[((e*x)^m*(c + d*x)^(3/2))/(a + b*x^2), x]
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 154, 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.

\(\displaystyle \int \frac {(c+d x)^{3/2} (e x)^m}{a+b x^2} \, dx\)

\(\Big \downarrow \) 615

\(\displaystyle \int \left (\frac {\sqrt {-a} (c+d x)^{3/2} (e x)^m}{2 a \left (\sqrt {-a}-\sqrt {b} x\right )}+\frac {\sqrt {-a} (c+d x)^{3/2} (e x)^m}{2 a \left (\sqrt {-a}+\sqrt {b} x\right )}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[((e*x)^m*(c + d*x)^(3/2))/(a + b*x^2),x]
 

Output:

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

\[\int \frac {\left (e x \right )^{m} \left (d x +c \right )^{\frac {3}{2}}}{b \,x^{2}+a}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate((e*x)^m*(d*x+c)^(3/2)/(b*x^2+a),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((e*x)**m*(d*x+c)**(3/2)/(b*x**2+a),x)
 

Output:

Integral((e*x)**m*(c + d*x)**(3/2)/(a + b*x**2), x)
 

Maxima [F]

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

integrate((e*x)^m*(d*x+c)^(3/2)/(b*x^2+a),x, algorithm="maxima")
 

Output:

integrate((d*x + c)^(3/2)*(e*x)^m/(b*x^2 + a), x)
 

Giac [F]

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

integrate((e*x)^m*(d*x+c)^(3/2)/(b*x^2+a),x, algorithm="giac")
 

Output:

integrate((d*x + c)^(3/2)*(e*x)^m/(b*x^2 + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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