\(\int \frac {(e x)^m (a+b x^2)}{\sqrt {c+d x}} \, dx\) [133]

Optimal result

Integrand size = 22, antiderivative size = 166 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=-\frac {4 b c (2+m) (e x)^{1+m} \sqrt {c+d x}}{d^2 e (3+2 m) (5+2 m)}+\frac {2 b (e x)^{2+m} \sqrt {c+d x}}{d e^2 (5+2 m)}+\frac {2 \left (4 b c^2 (1+m) (2+m)+a d^2 (3+2 m) (5+2 m)\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )}{d^3 (3+2 m) (5+2 m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.72 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {2 \left (-\frac {d x}{c}\right )^{-m} (e x)^m \sqrt {c+d x} \left (15 \left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+\frac {d x}{c}\right )-b (c+d x) \left (10 c \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )-3 (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-m,\frac {7}{2},1+\frac {d x}{c}\right )\right )\right )}{15 d^3} \] Input:

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

Output:

(2*(e*x)^m*Sqrt[c + d*x]*(15*(b*c^2 + a*d^2)*Hypergeometric2F1[1/2, -m, 3/ 
2, 1 + (d*x)/c] - b*(c + d*x)*(10*c*Hypergeometric2F1[3/2, -m, 5/2, 1 + (d 
*x)/c] - 3*(c + d*x)*Hypergeometric2F1[5/2, -m, 7/2, 1 + (d*x)/c])))/(15*d 
^3*(-((d*x)/c))^m)
 

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {521, 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.

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

\(\Big \downarrow \) 521

\(\displaystyle \frac {2 \int \frac {e^2 (e x)^m (a d (2 m+5)-2 b c (m+2) x)}{2 \sqrt {c+d x}}dx}{d e^2 (2 m+5)}+\frac {2 b \sqrt {c+d x} (e x)^{m+2}}{d e^2 (2 m+5)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(e x)^m (a d (2 m+5)-2 b c (m+2) x)}{\sqrt {c+d x}}dx}{d (2 m+5)}+\frac {2 b \sqrt {c+d x} (e x)^{m+2}}{d e^2 (2 m+5)}\)

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 77

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

\(\Big \downarrow \) 75

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

Input:

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

Output:

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

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 521

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), 
 x_Symbol] :> Simp[b^p*(e*x)^(m + 2*p)*((c + d*x)^(n + 1)/(d*e^(2*p)*(m + n 
 + 2*p + 1))), x] + Simp[1/(d*e^(2*p)*(m + n + 2*p + 1))   Int[(e*x)^m*(c + 
 d*x)^n*ExpandToSum[d*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x^2)^p - b^p*(e*x)^ 
(2*p)) - b^p*(e*c)*(m + 2*p)*(e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, 
 d, e}, x] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] &&  !IntegerQ[m] &&  !I 
ntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.59 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.50 \[ \int \frac {(e x)^m \left (a+b x^2\right )}{\sqrt {c+d x}} \, dx=\frac {a e^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\sqrt {c} \Gamma \left (m + 2\right )} + \frac {b e^{m} x^{m + 3} \Gamma \left (m + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, m + 3 \\ m + 4 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\sqrt {c} \Gamma \left (m + 4\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(2*e**m*(4*x**m*sqrt(c + d*x)*a*d**2*m**2 + 16*x**m*sqrt(c + d*x)*a*d**2*m 
 + 15*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 - 10*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 + 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*m**3*x + 36*c*m**2*x + 46*c*m*x + 15*c*x + 8*d*m**3*x**2 + 36*d*m* 
*2*x**2 + 46*d*m*x**2 + 15*d*x**2),x)*a*c*d**2*m**6 - 272*int((x**m*sqrt(c 
 + d*x))/(8*c*m**3*x + 36*c*m**2*x + 46*c*m*x + 15*c*x + 8*d*m**3*x**2 + 3 
6*d*m**2*x**2 + 46*d*m*x**2 + 15*d*x**2),x)*a*c*d**2*m**5 - 880*int((x**m* 
sqrt(c + d*x))/(8*c*m**3*x + 36*c*m**2*x + 46*c*m*x + 15*c*x + 8*d*m**3*x* 
*2 + 36*d*m**2*x**2 + 46*d*m*x**2 + 15*d*x**2),x)*a*c*d**2*m**4 - 1336*int 
((x**m*sqrt(c + d*x))/(8*c*m**3*x + 36*c*m**2*x + 46*c*m*x + 15*c*x + 8*d* 
m**3*x**2 + 36*d*m**2*x**2 + 46*d*m*x**2 + 15*d*x**2),x)*a*c*d**2*m**3 - 9 
30*int((x**m*sqrt(c + d*x))/(8*c*m**3*x + 36*c*m**2*x + 46*c*m*x + 15*c*x 
+ 8*d*m**3*x**2 + 36*d*m**2*x**2 + 46*d*m*x**2 + 15*d*x**2),x)*a*c*d**2*m* 
*2 - 225*int((x**m*sqrt(c + d*x))/(8*c*m**3*x + 36*c*m**2*x + 46*c*m*x + 1 
5*c*x + 8*d*m**3*x**2 + 36*d*m**2*x**2 + 46*d*m*x**2 + 15*d*x**2),x)*a*c*d 
**2*m - 32*int((x**m*sqrt(c + d*x))/(8*c*m**3*x + 36*c*m**2*x + 46*c*m*x + 
 15*c*x + 8*d*m**3*x**2 + 36*d*m**2*x**2 + 46*d*m*x**2 + 15*d*x**2),x)*...