\(\int \frac {(e x)^m (a+b x^2)}{(c+d x)^{5/2}} \, dx\) [135]

Optimal result

Integrand size = 22, antiderivative size = 160 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \left (b c^2+a d^2\right ) (e x)^{1+m}}{3 c d^2 e (c+d x)^{3/2}}+\frac {2 b (e x)^{1+m}}{d^2 e (1+2 m) \sqrt {c+d x}}-\frac {2 \left (a d^2 \left (1-4 m^2\right )-4 b c^2 \left (2+3 m+m^2\right )\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m,\frac {1}{2},1+\frac {d x}{c}\right )}{3 c d^3 (1+2 m) \sqrt {c+d x}} \] Output:

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

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.74 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {2 \left (-\frac {d x}{c}\right )^{-m} (e x)^m \left (-\left (\left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-m,-\frac {1}{2},1+\frac {d x}{c}\right )\right )+3 b (c+d x) \left (2 c \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m,\frac {1}{2},1+\frac {d x}{c}\right )+(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )\right )\right )}{3 d^3 (c+d x)^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {519, 27, 87, 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.

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

\(\Big \downarrow \) 519

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 77

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

\(\Big \downarrow \) 75

\(\displaystyle \frac {\frac {2 (e x)^{m+1} \left (a d^2 (1-2 m)-b c^2 (2 m+5)\right )}{c d e \sqrt {c+d x}}-\frac {2 \sqrt {c+d x} (e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (\frac {a d \left (1-4 m^2\right )}{c}-\frac {4 b c \left (m^2+3 m+2\right )}{d}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},\frac {d x}{c}+1\right )}{d}}{3 c d}+\frac {2 (e x)^{m+1} \left (a+\frac {b c^2}{d^2}\right )}{3 c e (c+d x)^{3/2}}\)

Input:

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

Output:

(2*(a + (b*c^2)/d^2)*(e*x)^(1 + m))/(3*c*e*(c + d*x)^(3/2)) + ((2*(a*d^2*( 
1 - 2*m) - b*c^2*(5 + 2*m))*(e*x)^(1 + m))/(c*d*e*Sqrt[c + d*x]) - (2*((a* 
d*(1 - 4*m^2))/c - (4*b*c*(2 + 3*m + m^2))/d)*(e*x)^m*Sqrt[c + d*x]*Hyperg 
eometric2F1[1/2, -m, 3/2, 1 + (d*x)/c])/(d*(-((d*x)/c))^m))/(3*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 87

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.85 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.52 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{(c+d x)^{5/2}} \, dx=\frac {a e^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{c^{\frac {5}{2}} \Gamma \left (m + 2\right )} + \frac {b e^{m} x^{m + 3} \Gamma \left (m + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{2}, m + 3 \\ m + 4 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{c^{\frac {5}{2}} \Gamma \left (m + 4\right )} \] Input:

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

Output:

a*e**m*x**(m + 1)*gamma(m + 1)*hyper((5/2, m + 1), (m + 2,), d*x*exp_polar 
(I*pi)/c)/(c**(5/2)*gamma(m + 2)) + b*e**m*x**(m + 3)*gamma(m + 3)*hyper(( 
5/2, m + 3), (m + 4,), d*x*exp_polar(I*pi)/c)/(c**(5/2)*gamma(m + 4))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

(2*e**m*(4*x**m*sqrt(c + d*x)*a*d**2*m**2 - 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*s 
qrt(c + d*x)*b*c**2 - 4*x**m*sqrt(c + d*x)*b*c*d*m**2*x - 2*x**m*sqrt(c + 
d*x)*b*c*d*m*x + 12*x**m*sqrt(c + d*x)*b*c*d*x + 4*x**m*sqrt(c + d*x)*b*d* 
*2*m**2*x**2 - 8*x**m*sqrt(c + d*x)*b*d**2*m*x**2 + 3*x**m*sqrt(c + d*x)*b 
*d**2*x**2 - 32*int((x**m*sqrt(c + d*x))/(8*c**3*m**3*x - 12*c**3*m**2*x - 
 2*c**3*m*x + 3*c**3*x + 24*c**2*d*m**3*x**2 - 36*c**2*d*m**2*x**2 - 6*c** 
2*d*m*x**2 + 9*c**2*d*x**2 + 24*c*d**2*m**3*x**3 - 36*c*d**2*m**2*x**3 - 6 
*c*d**2*m*x**3 + 9*c*d**2*x**3 + 8*d**3*m**3*x**4 - 12*d**3*m**2*x**4 - 2* 
d**3*m*x**4 + 3*d**3*x**4),x)*a*c**3*d**2*m**6 + 48*int((x**m*sqrt(c + d*x 
))/(8*c**3*m**3*x - 12*c**3*m**2*x - 2*c**3*m*x + 3*c**3*x + 24*c**2*d*m** 
3*x**2 - 36*c**2*d*m**2*x**2 - 6*c**2*d*m*x**2 + 9*c**2*d*x**2 + 24*c*d**2 
*m**3*x**3 - 36*c*d**2*m**2*x**3 - 6*c*d**2*m*x**3 + 9*c*d**2*x**3 + 8*d** 
3*m**3*x**4 - 12*d**3*m**2*x**4 - 2*d**3*m*x**4 + 3*d**3*x**4),x)*a*c**3*d 
**2*m**5 + 16*int((x**m*sqrt(c + d*x))/(8*c**3*m**3*x - 12*c**3*m**2*x - 2 
*c**3*m*x + 3*c**3*x + 24*c**2*d*m**3*x**2 - 36*c**2*d*m**2*x**2 - 6*c**2* 
d*m*x**2 + 9*c**2*d*x**2 + 24*c*d**2*m**3*x**3 - 36*c*d**2*m**2*x**3 - 6*c 
*d**2*m*x**3 + 9*c*d**2*x**3 + 8*d**3*m**3*x**4 - 12*d**3*m**2*x**4 - 2*d* 
*3*m*x**4 + 3*d**3*x**4),x)*a*c**3*d**2*m**4 - 24*int((x**m*sqrt(c + d*x)) 
/(8*c**3*m**3*x - 12*c**3*m**2*x - 2*c**3*m*x + 3*c**3*x + 24*c**2*d*m*...