\(\int \frac {(e x)^m (a+b x^2)}{(c+d x)^{3/2}} \, dx\) [134]

Optimal result

Integrand size = 22, antiderivative size = 159 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{(c+d x)^{3/2}} \, dx=\frac {2 \left (b c^2+a d^2\right ) (e x)^{1+m}}{c d^2 e \sqrt {c+d x}}+\frac {2 b (e x)^{1+m} \sqrt {c+d x}}{d^2 e (3+2 m)}-\frac {2 \left (4 b c^2 \left (2+3 m+m^2\right )+a d^2 \left (3+8 m+4 m^2\right )\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )}{c d^3 (3+2 m)} \] Output:

2*(a*d^2+b*c^2)*(e*x)^(1+m)/c/d^2/e/(d*x+c)^(1/2)+2*b*(e*x)^(1+m)*(d*x+c)^ 
(1/2)/d^2/e/(3+2*m)-2*(4*b*c^2*(m^2+3*m+2)+a*d^2*(4*m^2+8*m+3))*(e*x)^m*(d 
*x+c)^(1/2)*hypergeom([1/2, -m],[3/2],1+d*x/c)/c/d^3/(3+2*m)/((-d*x/c)^m)
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.75 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{(c+d x)^{3/2}} \, dx=\frac {2 \left (-\frac {d x}{c}\right )^{-m} (e x)^m \left (-3 \left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m,\frac {1}{2},1+\frac {d x}{c}\right )-b (c+d x) \left (6 c \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )-(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )\right )\right )}{3 d^3 \sqrt {c+d x}} \] Input:

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

Output:

(2*(e*x)^m*(-3*(b*c^2 + a*d^2)*Hypergeometric2F1[-1/2, -m, 1/2, 1 + (d*x)/ 
c] - b*(c + d*x)*(6*c*Hypergeometric2F1[1/2, -m, 3/2, 1 + (d*x)/c] - (c + 
d*x)*Hypergeometric2F1[3/2, -m, 5/2, 1 + (d*x)/c])))/(3*d^3*(-((d*x)/c))^m 
*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {519, 27, 90, 77, 75}

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.

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

\(\Big \downarrow \) 519

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {2 (e x)^{m+1} \left (a+\frac {b c^2}{d^2}\right )}{c e \sqrt {c+d x}}-\frac {\frac {\left (a d^2 \left (4 m^2+8 m+3\right )+4 b c^2 \left (m^2+3 m+2\right )\right ) \int \frac {(e x)^m}{\sqrt {c+d x}}dx}{d (2 m+3)}-\frac {2 b c \sqrt {c+d x} (e x)^{m+1}}{d e (2 m+3)}}{c d}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {2 (e x)^{m+1} \left (a+\frac {b c^2}{d^2}\right )}{c e \sqrt {c+d x}}-\frac {\frac {(e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a d^2 \left (4 m^2+8 m+3\right )+4 b c^2 \left (m^2+3 m+2\right )\right ) \int \frac {\left (-\frac {d x}{c}\right )^m}{\sqrt {c+d x}}dx}{d (2 m+3)}-\frac {2 b c \sqrt {c+d x} (e x)^{m+1}}{d e (2 m+3)}}{c d}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {2 (e x)^{m+1} \left (a+\frac {b c^2}{d^2}\right )}{c e \sqrt {c+d x}}-\frac {\frac {2 \sqrt {c+d x} (e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a d^2 \left (4 m^2+8 m+3\right )+4 b c^2 \left (m^2+3 m+2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},\frac {d x}{c}+1\right )}{d^2 (2 m+3)}-\frac {2 b c \sqrt {c+d x} (e x)^{m+1}}{d e (2 m+3)}}{c d}\)

Input:

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

Output:

(2*(a + (b*c^2)/d^2)*(e*x)^(1 + m))/(c*e*Sqrt[c + d*x]) - ((-2*b*c*(e*x)^( 
1 + m)*Sqrt[c + d*x])/(d*e*(3 + 2*m)) + (2*(4*b*c^2*(2 + 3*m + m^2) + a*d^ 
2*(3 + 8*m + 4*m^2))*(e*x)^m*Sqrt[c + d*x]*Hypergeometric2F1[1/2, -m, 3/2, 
 1 + (d*x)/c])/(d^2*(3 + 2*m)*(-((d*x)/c))^m))/(c*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 75

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x 
)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))*Hypergeometric2F1[-m, n + 1, n + 2, 1 + 
 d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (IntegerQ[m] 
 || GtQ[-d/(b*c), 0])
 

rule 77

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((-b)*(c/ 
d))^IntPart[m]*((b*x)^FracPart[m]/((-d)*(x/c))^FracPart[m])   Int[((-d)*(x/ 
c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && 
 !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 519

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2)^p, c + d*x, x], R = 
 PolynomialRemainder[(a + b*x^2)^p, c + d*x, x]}, Simp[(-R)*(e*x)^(m + 1)*( 
(c + d*x)^(n + 1)/(c*e*(n + 1))), x] + Simp[1/(c*(n + 1))   Int[(e*x)^m*(c 
+ d*x)^(n + 1)*ExpandToSum[c*(n + 1)*Qx + R*(m + n + 2), x], x], x]] /; Fre 
eQ[{a, b, c, d, e, m}, x] && IGtQ[p, 0] && LtQ[n, -1] &&  !IntegerQ[m]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.55 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.52 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{(c+d x)^{3/2}} \, dx=\frac {a e^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{c^{\frac {3}{2}} \Gamma \left (m + 2\right )} + \frac {b e^{m} x^{m + 3} \Gamma \left (m + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, m + 3 \\ m + 4 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{c^{\frac {3}{2}} \Gamma \left (m + 4\right )} \] Input:

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

Output:

a*e**m*x**(m + 1)*gamma(m + 1)*hyper((3/2, m + 1), (m + 2,), d*x*exp_polar 
(I*pi)/c)/(c**(3/2)*gamma(m + 2)) + b*e**m*x**(m + 3)*gamma(m + 3)*hyper(( 
3/2, m + 3), (m + 4,), d*x*exp_polar(I*pi)/c)/(c**(3/2)*gamma(m + 4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*e**m*(4*x**m*sqrt(c + d*x)*a*d**2*m**2 + 8*x**m*sqrt(c + d*x)*a*d**2*m 
+ 3*x**m*sqrt(c + d*x)*a*d**2 + 4*x**m*sqrt(c + d*x)*b*c**2*m**2 + 12*x**m 
*sqrt(c + d*x)*b*c**2*m + 8*x**m*sqrt(c + d*x)*b*c**2 - 4*x**m*sqrt(c + d* 
x)*b*c*d*m**2*x - 6*x**m*sqrt(c + d*x)*b*c*d*m*x + 4*x**m*sqrt(c + d*x)*b* 
c*d*x + 4*x**m*sqrt(c + d*x)*b*d**2*m**2*x**2 - x**m*sqrt(c + d*x)*b*d**2* 
x**2 - 32*int((x**m*sqrt(c + d*x))/(8*c**2*m**3*x + 12*c**2*m**2*x - 2*c** 
2*m*x - 3*c**2*x + 16*c*d*m**3*x**2 + 24*c*d*m**2*x**2 - 4*c*d*m*x**2 - 6* 
c*d*x**2 + 8*d**2*m**3*x**3 + 12*d**2*m**2*x**3 - 2*d**2*m*x**3 - 3*d**2*x 
**3),x)*a*c**2*d**2*m**6 - 112*int((x**m*sqrt(c + d*x))/(8*c**2*m**3*x + 1 
2*c**2*m**2*x - 2*c**2*m*x - 3*c**2*x + 16*c*d*m**3*x**2 + 24*c*d*m**2*x** 
2 - 4*c*d*m*x**2 - 6*c*d*x**2 + 8*d**2*m**3*x**3 + 12*d**2*m**2*x**3 - 2*d 
**2*m*x**3 - 3*d**2*x**3),x)*a*c**2*d**2*m**5 - 112*int((x**m*sqrt(c + d*x 
))/(8*c**2*m**3*x + 12*c**2*m**2*x - 2*c**2*m*x - 3*c**2*x + 16*c*d*m**3*x 
**2 + 24*c*d*m**2*x**2 - 4*c*d*m*x**2 - 6*c*d*x**2 + 8*d**2*m**3*x**3 + 12 
*d**2*m**2*x**3 - 2*d**2*m*x**3 - 3*d**2*x**3),x)*a*c**2*d**2*m**4 - 8*int 
((x**m*sqrt(c + d*x))/(8*c**2*m**3*x + 12*c**2*m**2*x - 2*c**2*m*x - 3*c** 
2*x + 16*c*d*m**3*x**2 + 24*c*d*m**2*x**2 - 4*c*d*m*x**2 - 6*c*d*x**2 + 8* 
d**2*m**3*x**3 + 12*d**2*m**2*x**3 - 2*d**2*m*x**3 - 3*d**2*x**3),x)*a*c** 
2*d**2*m**3 + 30*int((x**m*sqrt(c + d*x))/(8*c**2*m**3*x + 12*c**2*m**2*x 
- 2*c**2*m*x - 3*c**2*x + 16*c*d*m**3*x**2 + 24*c*d*m**2*x**2 - 4*c*d*m...