\(\int (e x)^m \sqrt {c+d x} (a+b x^2) \, dx\) [132]

Optimal result

Integrand size = 22, antiderivative size = 168 \[ \int (e x)^m \sqrt {c+d x} \left (a+b x^2\right ) \, dx=-\frac {4 b c (2+m) (e x)^{1+m} (c+d x)^{3/2}}{d^2 e (5+2 m) (7+2 m)}+\frac {2 b (e x)^{2+m} (c+d x)^{3/2}}{d e^2 (7+2 m)}+\frac {2 \left (4 b c^2 (1+m) (2+m)+a d^2 (5+2 m) (7+2 m)\right ) \left (-\frac {d x}{c}\right )^{-m} (e x)^m (c+d x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )}{3 d^3 (5+2 m) (7+2 m)} \] Output:

-4*b*c*(2+m)*(e*x)^(1+m)*(d*x+c)^(3/2)/d^2/e/(5+2*m)/(7+2*m)+2*b*(e*x)^(2+ 
m)*(d*x+c)^(3/2)/d/e^2/(7+2*m)+2/3*(4*b*c^2*(1+m)*(2+m)+a*d^2*(5+2*m)*(7+2 
*m))*(e*x)^m*(d*x+c)^(3/2)*hypergeom([3/2, -m],[5/2],1+d*x/c)/d^3/(5+2*m)/ 
(7+2*m)/((-d*x/c)^m)
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.71 \[ \int (e x)^m \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {2 \left (-\frac {d x}{c}\right )^{-m} (e x)^m (c+d x)^{3/2} \left (35 \left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},1+\frac {d x}{c}\right )-3 b (c+d x) \left (14 c \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-m,\frac {7}{2},1+\frac {d x}{c}\right )-5 (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {7}{2},-m,\frac {9}{2},1+\frac {d x}{c}\right )\right )\right )}{105 d^3} \] Input:

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

Output:

(2*(e*x)^m*(c + d*x)^(3/2)*(35*(b*c^2 + a*d^2)*Hypergeometric2F1[3/2, -m, 
5/2, 1 + (d*x)/c] - 3*b*(c + d*x)*(14*c*Hypergeometric2F1[5/2, -m, 7/2, 1 
+ (d*x)/c] - 5*(c + d*x)*Hypergeometric2F1[7/2, -m, 9/2, 1 + (d*x)/c])))/( 
105*d^3*(-((d*x)/c))^m)
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 166, 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.

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

\(\Big \downarrow \) 521

\(\displaystyle \frac {2 \int \frac {1}{2} e^2 (e x)^m \sqrt {c+d x} (a d (2 m+7)-2 b c (m+2) x)dx}{d e^2 (2 m+7)}+\frac {2 b (c+d x)^{3/2} (e x)^{m+2}}{d e^2 (2 m+7)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (e x)^m \sqrt {c+d x} (a d (2 m+7)-2 b c (m+2) x)dx}{d (2 m+7)}+\frac {2 b (c+d x)^{3/2} (e x)^{m+2}}{d e^2 (2 m+7)}\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {\frac {\left (a d^2 (2 m+5) (2 m+7)+4 b c^2 (m+1) (m+2)\right ) \int (e x)^m \sqrt {c+d x}dx}{d (2 m+5)}-\frac {4 b c (m+2) (c+d x)^{3/2} (e x)^{m+1}}{d e (2 m+5)}}{d (2 m+7)}+\frac {2 b (c+d x)^{3/2} (e x)^{m+2}}{d e^2 (2 m+7)}\)

\(\Big \downarrow \) 77

\(\displaystyle \frac {\frac {(e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a d^2 (2 m+5) (2 m+7)+4 b c^2 (m+1) (m+2)\right ) \int \left (-\frac {d x}{c}\right )^m \sqrt {c+d x}dx}{d (2 m+5)}-\frac {4 b c (m+2) (c+d x)^{3/2} (e x)^{m+1}}{d e (2 m+5)}}{d (2 m+7)}+\frac {2 b (c+d x)^{3/2} (e x)^{m+2}}{d e^2 (2 m+7)}\)

\(\Big \downarrow \) 75

\(\displaystyle \frac {\frac {2 (c+d x)^{3/2} (e x)^m \left (-\frac {d x}{c}\right )^{-m} \left (a d^2 (2 m+5) (2 m+7)+4 b c^2 (m+1) (m+2)\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-m,\frac {5}{2},\frac {d x}{c}+1\right )}{3 d^2 (2 m+5)}-\frac {4 b c (m+2) (c+d x)^{3/2} (e x)^{m+1}}{d e (2 m+5)}}{d (2 m+7)}+\frac {2 b (c+d x)^{3/2} (e x)^{m+2}}{d e^2 (2 m+7)}\)

Input:

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

Output:

(2*b*(e*x)^(2 + m)*(c + d*x)^(3/2))/(d*e^2*(7 + 2*m)) + ((-4*b*c*(2 + m)*( 
e*x)^(1 + m)*(c + d*x)^(3/2))/(d*e*(5 + 2*m)) + (2*(4*b*c^2*(1 + m)*(2 + m 
) + a*d^2*(5 + 2*m)*(7 + 2*m))*(e*x)^m*(c + d*x)^(3/2)*Hypergeometric2F1[3 
/2, -m, 5/2, 1 + (d*x)/c])/(3*d^2*(5 + 2*m)*(-((d*x)/c))^m))/(d*(7 + 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.76 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.52 \[ \int (e x)^m \sqrt {c+d x} \left (a+b x^2\right ) \, dx=\frac {a \sqrt {c} e^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 2\right )} + \frac {b \sqrt {c} e^{m} x^{m + 3} \Gamma \left (m + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, m + 3 \\ m + 4 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 4\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

(2*e**m*(4*x**m*sqrt(c + d*x)*a*c*d**2*m**2 + 24*x**m*sqrt(c + d*x)*a*c*d* 
*2*m + 35*x**m*sqrt(c + d*x)*a*c*d**2 + 8*x**m*sqrt(c + d*x)*a*d**3*m**3*x 
 + 52*x**m*sqrt(c + d*x)*a*d**3*m**2*x + 94*x**m*sqrt(c + d*x)*a*d**3*m*x 
+ 35*x**m*sqrt(c + d*x)*a*d**3*x + 4*x**m*sqrt(c + d*x)*b*c**3*m**2 + 12*x 
**m*sqrt(c + d*x)*b*c**3*m + 8*x**m*sqrt(c + d*x)*b*c**3 - 4*x**m*sqrt(c + 
 d*x)*b*c**2*d*m**2*x - 10*x**m*sqrt(c + d*x)*b*c**2*d*m*x - 4*x**m*sqrt(c 
 + d*x)*b*c**2*d*x + 4*x**m*sqrt(c + d*x)*b*c*d**2*m**2*x**2 + 8*x**m*sqrt 
(c + d*x)*b*c*d**2*m*x**2 + 3*x**m*sqrt(c + d*x)*b*c*d**2*x**2 + 8*x**m*sq 
rt(c + d*x)*b*d**3*m**3*x**3 + 36*x**m*sqrt(c + d*x)*b*d**3*m**2*x**3 + 46 
*x**m*sqrt(c + d*x)*b*d**3*m*x**3 + 15*x**m*sqrt(c + d*x)*b*d**3*x**3 - 64 
*int((x**m*sqrt(c + d*x))/(16*c*m**4*x + 128*c*m**3*x + 344*c*m**2*x + 352 
*c*m*x + 105*c*x + 16*d*m**4*x**2 + 128*d*m**3*x**2 + 344*d*m**2*x**2 + 35 
2*d*m*x**2 + 105*d*x**2),x)*a*c**2*d**2*m**7 - 896*int((x**m*sqrt(c + d*x) 
)/(16*c*m**4*x + 128*c*m**3*x + 344*c*m**2*x + 352*c*m*x + 105*c*x + 16*d* 
m**4*x**2 + 128*d*m**3*x**2 + 344*d*m**2*x**2 + 352*d*m*x**2 + 105*d*x**2) 
,x)*a*c**2*d**2*m**6 - 5008*int((x**m*sqrt(c + d*x))/(16*c*m**4*x + 128*c* 
m**3*x + 344*c*m**2*x + 352*c*m*x + 105*c*x + 16*d*m**4*x**2 + 128*d*m**3* 
x**2 + 344*d*m**2*x**2 + 352*d*m*x**2 + 105*d*x**2),x)*a*c**2*d**2*m**5 - 
14144*int((x**m*sqrt(c + d*x))/(16*c*m**4*x + 128*c*m**3*x + 344*c*m**2*x 
+ 352*c*m*x + 105*c*x + 16*d*m**4*x**2 + 128*d*m**3*x**2 + 344*d*m**2*x...