\(\int (e x)^m (c+d x)^2 (a+b x^2)^p \, dx\) [288]

Optimal result

Integrand size = 22, antiderivative size = 194 \[ \int (e x)^m (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {d^2 (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{b e (3+m+2 p)}+\frac {\left (\frac {c^2}{1+m}-\frac {a d^2}{b (3+m+2 p)}\right ) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )}{e}+\frac {2 c d (e x)^{2+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},-p,\frac {4+m}{2},-\frac {b x^2}{a}\right )}{e^2 (2+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.81 \[ \int (e x)^m (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {x (e x)^m \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (c^2 \left (6+5 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )+d (1+m) x \left (2 c (3+m) \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},-p,\frac {4+m}{2},-\frac {b x^2}{a}\right )+d (2+m) x \operatorname {Hypergeometric2F1}\left (\frac {3+m}{2},-p,\frac {5+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{(1+m) (2+m) (3+m)} \] Input:

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

Output:

(x*(e*x)^m*(a + b*x^2)^p*(c^2*(6 + 5*m + m^2)*Hypergeometric2F1[(1 + m)/2, 
 -p, (3 + m)/2, -((b*x^2)/a)] + d*(1 + m)*x*(2*c*(3 + m)*Hypergeometric2F1 
[(2 + m)/2, -p, (4 + m)/2, -((b*x^2)/a)] + d*(2 + m)*x*Hypergeometric2F1[( 
3 + m)/2, -p, (5 + m)/2, -((b*x^2)/a)])))/((1 + m)*(2 + m)*(3 + m)*(1 + (b 
*x^2)/a)^p)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {559, 25, 557, 279, 278}

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.

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

\(\Big \downarrow \) 559

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 278

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^p*(( 
c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/2, (m + 1)/2 + 1, ( 
-b)*(x^2/a)], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && (ILtQ[p, 0 
] || GtQ[a, 0])
 

rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[a^IntP 
art[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p])   Int[(c*x)^m* 
(1 + b*(x^2/a))^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IGtQ[p, 0] && 
!(ILtQ[p, 0] || GtQ[a, 0])
 

rule 557

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Sym 
bol] :> Simp[c   Int[(e*x)^m*(a + b*x^2)^p, x], x] + Simp[d/e   Int[(e*x)^( 
m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x]
 

rule 559

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

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 78.52 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.87 \[ \int (e x)^m (c+d x)^2 \left (a+b x^2\right )^p \, dx=\frac {a^{p} c^{2} e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {a^{p} c d e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{\Gamma \left (\frac {m}{2} + 2\right )} + \frac {a^{p} d^{2} e^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*(4*x**m*(a + b*x**2)**p*a*c*d*m**2*p + 16*x**m*(a + b*x**2)**p*a*c*d 
*m*p**2 + 16*x**m*(a + b*x**2)**p*a*c*d*m*p + 16*x**m*(a + b*x**2)**p*a*c* 
d*p**3 + 32*x**m*(a + b*x**2)**p*a*c*d*p**2 + 12*x**m*(a + b*x**2)**p*a*c* 
d*p + 2*x**m*(a + b*x**2)**p*a*d**2*m**2*p*x + 8*x**m*(a + b*x**2)**p*a*d* 
*2*m*p**2*x + 4*x**m*(a + b*x**2)**p*a*d**2*m*p*x + 8*x**m*(a + b*x**2)**p 
*a*d**2*p**3*x + 8*x**m*(a + b*x**2)**p*a*d**2*p**2*x + x**m*(a + b*x**2)* 
*p*b*c**2*m**3*x + 6*x**m*(a + b*x**2)**p*b*c**2*m**2*p*x + 5*x**m*(a + b* 
x**2)**p*b*c**2*m**2*x + 12*x**m*(a + b*x**2)**p*b*c**2*m*p**2*x + 20*x**m 
*(a + b*x**2)**p*b*c**2*m*p*x + 6*x**m*(a + b*x**2)**p*b*c**2*m*x + 8*x**m 
*(a + b*x**2)**p*b*c**2*p**3*x + 20*x**m*(a + b*x**2)**p*b*c**2*p**2*x + 1 
2*x**m*(a + b*x**2)**p*b*c**2*p*x + 2*x**m*(a + b*x**2)**p*b*c*d*m**3*x**2 
 + 12*x**m*(a + b*x**2)**p*b*c*d*m**2*p*x**2 + 8*x**m*(a + b*x**2)**p*b*c* 
d*m**2*x**2 + 24*x**m*(a + b*x**2)**p*b*c*d*m*p**2*x**2 + 32*x**m*(a + b*x 
**2)**p*b*c*d*m*p*x**2 + 6*x**m*(a + b*x**2)**p*b*c*d*m*x**2 + 16*x**m*(a 
+ b*x**2)**p*b*c*d*p**3*x**2 + 32*x**m*(a + b*x**2)**p*b*c*d*p**2*x**2 + 1 
2*x**m*(a + b*x**2)**p*b*c*d*p*x**2 + x**m*(a + b*x**2)**p*b*d**2*m**3*x** 
3 + 6*x**m*(a + b*x**2)**p*b*d**2*m**2*p*x**3 + 3*x**m*(a + b*x**2)**p*b*d 
**2*m**2*x**3 + 12*x**m*(a + b*x**2)**p*b*d**2*m*p**2*x**3 + 12*x**m*(a + 
b*x**2)**p*b*d**2*m*p*x**3 + 2*x**m*(a + b*x**2)**p*b*d**2*m*x**3 + 8*x**m 
*(a + b*x**2)**p*b*d**2*p**3*x**3 + 12*x**m*(a + b*x**2)**p*b*d**2*p**2...