\(\int (e x)^m (c+d x)^n (a+b x^2) \, dx\) [1821]

Optimal result

Integrand size = 20, antiderivative size = 164 \[ \int (e x)^m (c+d x)^n \left (a+b x^2\right ) \, dx=-\frac {b c (2+m) (e x)^{1+m} (c+d x)^{1+n}}{d^2 e (2+m+n) (3+m+n)}+\frac {b (e x)^{2+m} (c+d x)^{1+n}}{d e^2 (3+m+n)}+\frac {\left (b c^2 (1+m) (2+m)+a d^2 (2+m+n) (3+m+n)\right ) (e x)^{1+m} (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {d x}{c}\right )}{d^2 e (1+m) (2+m+n) (3+m+n)} \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.69 \[ \int (e x)^m (c+d x)^n \left (a+b x^2\right ) \, dx=\frac {x (e x)^m (c+d x)^n \left (1+\frac {d x}{c}\right )^{-n} \left (b c^2 \operatorname {Hypergeometric2F1}\left (1+m,-2-n,2+m,-\frac {d x}{c}\right )-2 b c^2 \operatorname {Hypergeometric2F1}\left (1+m,-1-n,2+m,-\frac {d x}{c}\right )+\left (b c^2+a d^2\right ) \operatorname {Hypergeometric2F1}\left (1+m,-n,2+m,-\frac {d x}{c}\right )\right )}{d^2 (1+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {521, 27, 90, 76, 74}

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.

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

\(\Big \downarrow \) 521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 90

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

\(\Big \downarrow \) 76

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

\(\Big \downarrow \) 74

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

Input:

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

Output:

(b*(e*x)^(2 + m)*(c + d*x)^(1 + n))/(d*e^2*(3 + m + n)) + (-((b*c*(2 + m)* 
(e*x)^(1 + m)*(c + d*x)^(1 + n))/(d*e*(2 + m + n))) + (((b*c^2*(1 + m)*(2 
+ m))/(d*(2 + m + n)) + a*d*(3 + m + n))*(e*x)^(1 + m)*(c + d*x)^n*Hyperge 
ometric2F1[1 + m, -n, 2 + m, -((d*x)/c)])/(e*(1 + m)*(1 + (d*x)/c)^n))/(d* 
(3 + m + n))
 

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 74

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

rule 76

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.46 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.49 \[ \int (e x)^m (c+d x)^n \left (a+b x^2\right ) \, dx=\frac {a c^{n} e^{m} x^{m + 1} \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - n, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 2\right )} + \frac {b c^{n} e^{m} x^{m + 3} \Gamma \left (m + 3\right ) {{}_{2}F_{1}\left (\begin {matrix} - n, m + 3 \\ m + 4 \end {matrix}\middle | {\frac {d x e^{i \pi }}{c}} \right )}}{\Gamma \left (m + 4\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*(x**m*(c + d*x)**n*a*c*d**2*m**2*n + 2*x**m*(c + d*x)**n*a*c*d**2*m* 
n**2 + 5*x**m*(c + d*x)**n*a*c*d**2*m*n + x**m*(c + d*x)**n*a*c*d**2*n**3 
+ 5*x**m*(c + d*x)**n*a*c*d**2*n**2 + 6*x**m*(c + d*x)**n*a*c*d**2*n + x** 
m*(c + d*x)**n*a*d**3*m**3*x + 3*x**m*(c + d*x)**n*a*d**3*m**2*n*x + 5*x** 
m*(c + d*x)**n*a*d**3*m**2*x + 3*x**m*(c + d*x)**n*a*d**3*m*n**2*x + 10*x* 
*m*(c + d*x)**n*a*d**3*m*n*x + 6*x**m*(c + d*x)**n*a*d**3*m*x + x**m*(c + 
d*x)**n*a*d**3*n**3*x + 5*x**m*(c + d*x)**n*a*d**3*n**2*x + 6*x**m*(c + d* 
x)**n*a*d**3*n*x + x**m*(c + d*x)**n*b*c**3*m**2*n + 3*x**m*(c + d*x)**n*b 
*c**3*m*n + 2*x**m*(c + d*x)**n*b*c**3*n - x**m*(c + d*x)**n*b*c**2*d*m**2 
*n*x - x**m*(c + d*x)**n*b*c**2*d*m*n**2*x - 2*x**m*(c + d*x)**n*b*c**2*d* 
m*n*x - 2*x**m*(c + d*x)**n*b*c**2*d*n**2*x + x**m*(c + d*x)**n*b*c*d**2*m 
**2*n*x**2 + 2*x**m*(c + d*x)**n*b*c*d**2*m*n**2*x**2 + x**m*(c + d*x)**n* 
b*c*d**2*m*n*x**2 + x**m*(c + d*x)**n*b*c*d**2*n**3*x**2 + x**m*(c + d*x)* 
*n*b*c*d**2*n**2*x**2 + x**m*(c + d*x)**n*b*d**3*m**3*x**3 + 3*x**m*(c + d 
*x)**n*b*d**3*m**2*n*x**3 + 3*x**m*(c + d*x)**n*b*d**3*m**2*x**3 + 3*x**m* 
(c + d*x)**n*b*d**3*m*n**2*x**3 + 6*x**m*(c + d*x)**n*b*d**3*m*n*x**3 + 2* 
x**m*(c + d*x)**n*b*d**3*m*x**3 + x**m*(c + d*x)**n*b*d**3*n**3*x**3 + 3*x 
**m*(c + d*x)**n*b*d**3*n**2*x**3 + 2*x**m*(c + d*x)**n*b*d**3*n*x**3 - in 
t((x**m*(c + d*x)**n)/(c*m**4*x + 4*c*m**3*n*x + 6*c*m**3*x + 6*c*m**2*n** 
2*x + 18*c*m**2*n*x + 11*c*m**2*x + 4*c*m*n**3*x + 18*c*m*n**2*x + 22*c...