\(\int \frac {(e x)^m (a+b x^2)^{3/2}}{(c+d x)^2} \, dx\) [166]

Optimal result

Integrand size = 24, antiderivative size = 255 \[ \int \frac {(e x)^m \left (a+b x^2\right )^{3/2}}{(c+d x)^2} \, dx=\frac {a (e x)^{1+m} \sqrt {a+b x^2} \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {3}{2},2,\frac {3+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^2 e (1+m) \sqrt {1+\frac {b x^2}{a}}}-\frac {2 a d (e x)^{2+m} \sqrt {a+b x^2} \operatorname {AppellF1}\left (\frac {2+m}{2},-\frac {3}{2},2,\frac {4+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^3 e^2 (2+m) \sqrt {1+\frac {b x^2}{a}}}+\frac {a d^2 (e x)^{3+m} \sqrt {a+b x^2} \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {3}{2},2,\frac {5+m}{2},-\frac {b x^2}{a},\frac {d^2 x^2}{c^2}\right )}{c^4 e^3 (3+m) \sqrt {1+\frac {b x^2}{a}}} \] Output:

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

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 251, 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.

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

\(\Big \downarrow \) 623

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

\(\Big \downarrow \) 622

\(\displaystyle x^{-m} (e x)^m \int \left (\frac {c^2 \left (b x^2+a\right )^{3/2} x^m}{\left (c^2-d^2 x^2\right )^2}-\frac {2 c d \left (b x^2+a\right )^{3/2} x^{m+1}}{\left (c^2-d^2 x^2\right )^2}+\frac {d^2 \left (b x^2+a\right )^{3/2} x^{m+2}}{\left (d^2 x^2-c^2\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*( - 2*x**m*sqrt(a + b*x**2)*a*d - x**m*sqrt(a + b*x**2)*b*c*m**2*x - 
 4*x**m*sqrt(a + b*x**2)*b*c*m*x - 3*x**m*sqrt(a + b*x**2)*b*c*x + x**m*sq 
rt(a + b*x**2)*b*d*m**2*x**2 + 2*x**m*sqrt(a + b*x**2)*b*d*m*x**2 + x**m*s 
qrt(a + b*x**2)*b*d*x**2 + int((x**m*sqrt(a + b*x**2)*x**2)/(a*c**2*m**3 + 
 4*a*c**2*m**2 + 5*a*c**2*m + 2*a*c**2 + 2*a*c*d*m**3*x + 8*a*c*d*m**2*x + 
 10*a*c*d*m*x + 4*a*c*d*x + a*d**2*m**3*x**2 + 4*a*d**2*m**2*x**2 + 5*a*d* 
*2*m*x**2 + 2*a*d**2*x**2 + b*c**2*m**3*x**2 + 4*b*c**2*m**2*x**2 + 5*b*c* 
*2*m*x**2 + 2*b*c**2*x**2 + 2*b*c*d*m**3*x**3 + 8*b*c*d*m**2*x**3 + 10*b*c 
*d*m*x**3 + 4*b*c*d*x**3 + b*d**2*m**3*x**4 + 4*b*d**2*m**2*x**4 + 5*b*d** 
2*m*x**4 + 2*b*d**2*x**4),x)*a*b*c*d**2*m**6 + 9*int((x**m*sqrt(a + b*x**2 
)*x**2)/(a*c**2*m**3 + 4*a*c**2*m**2 + 5*a*c**2*m + 2*a*c**2 + 2*a*c*d*m** 
3*x + 8*a*c*d*m**2*x + 10*a*c*d*m*x + 4*a*c*d*x + a*d**2*m**3*x**2 + 4*a*d 
**2*m**2*x**2 + 5*a*d**2*m*x**2 + 2*a*d**2*x**2 + b*c**2*m**3*x**2 + 4*b*c 
**2*m**2*x**2 + 5*b*c**2*m*x**2 + 2*b*c**2*x**2 + 2*b*c*d*m**3*x**3 + 8*b* 
c*d*m**2*x**3 + 10*b*c*d*m*x**3 + 4*b*c*d*x**3 + b*d**2*m**3*x**4 + 4*b*d* 
*2*m**2*x**4 + 5*b*d**2*m*x**4 + 2*b*d**2*x**4),x)*a*b*c*d**2*m**5 + 34*in 
t((x**m*sqrt(a + b*x**2)*x**2)/(a*c**2*m**3 + 4*a*c**2*m**2 + 5*a*c**2*m + 
 2*a*c**2 + 2*a*c*d*m**3*x + 8*a*c*d*m**2*x + 10*a*c*d*m*x + 4*a*c*d*x + a 
*d**2*m**3*x**2 + 4*a*d**2*m**2*x**2 + 5*a*d**2*m*x**2 + 2*a*d**2*x**2 + b 
*c**2*m**3*x**2 + 4*b*c**2*m**2*x**2 + 5*b*c**2*m*x**2 + 2*b*c**2*x**2 ...