\(\int (e x)^m (c+d x)^3 (a+b x^2)^p \, dx\) [287]

Optimal result

Integrand size = 22, antiderivative size = 254 \[ \int (e x)^m (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {3 c d^2 (e x)^{1+m} \left (a+b x^2\right )^{1+p}}{b e (3+m+2 p)}+\frac {d^3 (e x)^{2+m} \left (a+b x^2\right )^{1+p}}{b e^2 (4+m+2 p)}+\frac {c \left (\frac {c^2}{1+m}-\frac {3 a d^2}{b (3+m+2 p)}\right ) (e x)^{1+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )}{e}+\frac {d \left (\frac {3 c^2}{2+m}-\frac {a d^2}{b (4+m+2 p)}\right ) (e x)^{2+m} \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},-p,\frac {4+m}{2},-\frac {b x^2}{a}\right )}{e^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.72 \[ \int (e x)^m (c+d x)^3 \left (a+b x^2\right )^p \, dx=x (e x)^m \left (a+b x^2\right )^p \left (1+\frac {b x^2}{a}\right )^{-p} \left (\frac {c^3 \operatorname {Hypergeometric2F1}\left (\frac {1+m}{2},-p,\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+d x \left (\frac {3 c^2 \operatorname {Hypergeometric2F1}\left (\frac {2+m}{2},-p,\frac {4+m}{2},-\frac {b x^2}{a}\right )}{2+m}+d x \left (\frac {3 c \operatorname {Hypergeometric2F1}\left (\frac {3+m}{2},-p,\frac {5+m}{2},-\frac {b x^2}{a}\right )}{3+m}+\frac {d x \operatorname {Hypergeometric2F1}\left (\frac {4+m}{2},-p,\frac {6+m}{2},-\frac {b x^2}{a}\right )}{4+m}\right )\right )\right ) \] Input:

Integrate[(e*x)^m*(c + d*x)^3*(a + b*x^2)^p,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.15, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {559, 2340, 25, 27, 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.

\(\displaystyle \int (c+d x)^3 (e x)^m \left (a+b x^2\right )^p \, dx\)

\(\Big \downarrow \) 559

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 279

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

\(\Big \downarrow \) 278

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

Input:

Int[(e*x)^m*(c + d*x)^3*(a + b*x^2)^p,x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

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 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]
 

rule 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

Maple [F]

\[\int \left (e x \right )^{m} \left (d x +c \right )^{3} \left (b \,x^{2}+a \right )^{p}d x\]

Input:

int((e*x)^m*(d*x+c)^3*(b*x^2+a)^p,x)
 

Output:

int((e*x)^m*(d*x+c)^3*(b*x^2+a)^p,x)
 

Fricas [F]

\[ \int (e x)^m (c+d x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m} \,d x } \] Input:

integrate((e*x)^m*(d*x+c)^3*(b*x^2+a)^p,x, algorithm="fricas")
 

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 116.29 (sec) , antiderivative size = 230, normalized size of antiderivative = 0.91 \[ \int (e x)^m (c+d x)^3 \left (a+b x^2\right )^p \, dx=\frac {a^{p} c^{3} e^{m} x^{m + 1} \Gamma \left (\frac {m}{2} + \frac {1}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {1}{2} \\ \frac {m}{2} + \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {3 a^{p} c^{2} d e^{m} x^{m + 2} \Gamma \left (\frac {m}{2} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + 1 \\ \frac {m}{2} + 2 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 2\right )} + \frac {3 a^{p} c d^{2} e^{m} x^{m + 3} \Gamma \left (\frac {m}{2} + \frac {3}{2}\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + \frac {3}{2} \\ \frac {m}{2} + \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {a^{p} d^{3} e^{m} x^{m + 4} \Gamma \left (\frac {m}{2} + 2\right ) {{}_{2}F_{1}\left (\begin {matrix} - p, \frac {m}{2} + 2 \\ \frac {m}{2} + 3 \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {m}{2} + 3\right )} \] Input:

integrate((e*x)**m*(d*x+c)**3*(b*x**2+a)**p,x)
 

Output:

a**p*c**3*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)) + 3*a**p*c**2*d*e**m* 
x**(m + 2)*gamma(m/2 + 1)*hyper((-p, m/2 + 1), (m/2 + 2,), b*x**2*exp_pola 
r(I*pi)/a)/(2*gamma(m/2 + 2)) + 3*a**p*c*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*gam 
ma(m/2 + 5/2)) + a**p*d**3*e**m*x**(m + 4)*gamma(m/2 + 2)*hyper((-p, m/2 + 
 2), (m/2 + 3,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(m/2 + 3))
 

Maxima [F]

\[ \int (e x)^m (c+d x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m} \,d x } \] Input:

integrate((e*x)^m*(d*x+c)^3*(b*x^2+a)^p,x, algorithm="maxima")
 

Output:

integrate((d*x + c)^3*(b*x^2 + a)^p*(e*x)^m, x)
 

Giac [F]

\[ \int (e x)^m (c+d x)^3 \left (a+b x^2\right )^p \, dx=\int { {\left (d x + c\right )}^{3} {\left (b x^{2} + a\right )}^{p} \left (e x\right )^{m} \,d x } \] Input:

integrate((e*x)^m*(d*x+c)^3*(b*x^2+a)^p,x, algorithm="giac")
 

Output:

integrate((d*x + c)^3*(b*x^2 + a)^p*(e*x)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (e x)^m (c+d x)^3 \left (a+b x^2\right )^p \, dx=\int {\left (e\,x\right )}^m\,{\left (b\,x^2+a\right )}^p\,{\left (c+d\,x\right )}^3 \,d x \] Input:

int((e*x)^m*(a + b*x^2)^p*(c + d*x)^3,x)
 

Output:

int((e*x)^m*(a + b*x^2)^p*(c + d*x)^3, x)
 

Reduce [F]

\[ \int (e x)^m (c+d x)^3 \left (a+b x^2\right )^p \, dx=\text {too large to display} \] Input:

int((e*x)^m*(d*x+c)^3*(b*x^2+a)^p,x)
 

Output:

(e**m*( - 2*x**m*(a + b*x**2)**p*a**2*d**3*m**3*p - 8*x**m*(a + b*x**2)**p 
*a**2*d**3*m**2*p**2 - 12*x**m*(a + b*x**2)**p*a**2*d**3*m**2*p - 8*x**m*( 
a + b*x**2)**p*a**2*d**3*m*p**3 - 32*x**m*(a + b*x**2)**p*a**2*d**3*m*p**2 
 - 22*x**m*(a + b*x**2)**p*a**2*d**3*m*p - 16*x**m*(a + b*x**2)**p*a**2*d* 
*3*p**3 - 32*x**m*(a + b*x**2)**p*a**2*d**3*p**2 - 12*x**m*(a + b*x**2)**p 
*a**2*d**3*p + 6*x**m*(a + b*x**2)**p*a*b*c**2*d*m**3*p + 36*x**m*(a + b*x 
**2)**p*a*b*c**2*d*m**2*p**2 + 48*x**m*(a + b*x**2)**p*a*b*c**2*d*m**2*p + 
 72*x**m*(a + b*x**2)**p*a*b*c**2*d*m*p**3 + 192*x**m*(a + b*x**2)**p*a*b* 
c**2*d*m*p**2 + 114*x**m*(a + b*x**2)**p*a*b*c**2*d*m*p + 48*x**m*(a + b*x 
**2)**p*a*b*c**2*d*p**4 + 192*x**m*(a + b*x**2)**p*a*b*c**2*d*p**3 + 228*x 
**m*(a + b*x**2)**p*a*b*c**2*d*p**2 + 72*x**m*(a + b*x**2)**p*a*b*c**2*d*p 
 + 6*x**m*(a + b*x**2)**p*a*b*c*d**2*m**3*p*x + 36*x**m*(a + b*x**2)**p*a* 
b*c*d**2*m**2*p**2*x + 36*x**m*(a + b*x**2)**p*a*b*c*d**2*m**2*p*x + 72*x* 
*m*(a + b*x**2)**p*a*b*c*d**2*m*p**3*x + 144*x**m*(a + b*x**2)**p*a*b*c*d* 
*2*m*p**2*x + 48*x**m*(a + b*x**2)**p*a*b*c*d**2*m*p*x + 48*x**m*(a + b*x* 
*2)**p*a*b*c*d**2*p**4*x + 144*x**m*(a + b*x**2)**p*a*b*c*d**2*p**3*x + 96 
*x**m*(a + b*x**2)**p*a*b*c*d**2*p**2*x + 2*x**m*(a + b*x**2)**p*a*b*d**3* 
m**3*p*x**2 + 12*x**m*(a + b*x**2)**p*a*b*d**3*m**2*p**2*x**2 + 8*x**m*(a 
+ b*x**2)**p*a*b*d**3*m**2*p*x**2 + 24*x**m*(a + b*x**2)**p*a*b*d**3*m*p** 
3*x**2 + 32*x**m*(a + b*x**2)**p*a*b*d**3*m*p**2*x**2 + 6*x**m*(a + b*x...