\(\int \frac {(e x)^m (c+d x)^2}{a+b x^2} \, dx\) [1710]

Optimal result

Integrand size = 22, antiderivative size = 129 \[ \int \frac {(e x)^m (c+d x)^2}{a+b x^2} \, dx=\frac {d^2 (e x)^{1+m}}{b e (1+m)}+\frac {\left (b c^2-a d^2\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a b e (1+m)}+\frac {2 c d (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a e^2 (2+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.83 \[ \int \frac {(e x)^m (c+d x)^2}{a+b x^2} \, dx=\frac {x (e x)^m \left (\left (b c^2-a d^2\right ) (2+m) \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )+d \left (a d (2+m)+2 b c (1+m) x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )\right )\right )}{a b (1+m) (2+m)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {559, 27, 557, 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.

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

\(\Big \downarrow \) 559

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

(d^2*(e*x)^(1 + m))/(b*e*(1 + m)) + (((b*c^2 - a*d^2)*(e*x)^(1 + m)*Hyperg 
eometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*e*(1 + m)) + (2*b*c 
*d*(e*x)^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)]) 
/(a*e^2*(2 + m)))/b
 

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 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 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.40 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.32 \[ \int \frac {(e x)^m (c+d x)^2}{a+b x^2} \, dx=\frac {c^{2} e^{m} m x^{m + 1} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {c^{2} e^{m} x^{m + 1} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {1}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {1}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )} + \frac {c d e^{m} m x^{m + 2} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{2 a \Gamma \left (\frac {m}{2} + 2\right )} + \frac {c d e^{m} x^{m + 2} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + 1\right ) \Gamma \left (\frac {m}{2} + 1\right )}{a \Gamma \left (\frac {m}{2} + 2\right )} + \frac {d^{2} e^{m} m x^{m + 3} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} + \frac {3 d^{2} e^{m} x^{m + 3} \Phi \left (\frac {b x^{2} e^{i \pi }}{a}, 1, \frac {m}{2} + \frac {3}{2}\right ) \Gamma \left (\frac {m}{2} + \frac {3}{2}\right )}{4 a \Gamma \left (\frac {m}{2} + \frac {5}{2}\right )} \] Input:

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

Output:

c**2*e**m*m*x**(m + 1)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*ga 
mma(m/2 + 1/2)/(4*a*gamma(m/2 + 3/2)) + c**2*e**m*x**(m + 1)*lerchphi(b*x* 
*2*exp_polar(I*pi)/a, 1, m/2 + 1/2)*gamma(m/2 + 1/2)/(4*a*gamma(m/2 + 3/2) 
) + c*d*e**m*m*x**(m + 2)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 1)*g 
amma(m/2 + 1)/(2*a*gamma(m/2 + 2)) + c*d*e**m*x**(m + 2)*lerchphi(b*x**2*e 
xp_polar(I*pi)/a, 1, m/2 + 1)*gamma(m/2 + 1)/(a*gamma(m/2 + 2)) + d**2*e** 
m*m*x**(m + 3)*lerchphi(b*x**2*exp_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 
+ 3/2)/(4*a*gamma(m/2 + 5/2)) + 3*d**2*e**m*x**(m + 3)*lerchphi(b*x**2*exp 
_polar(I*pi)/a, 1, m/2 + 3/2)*gamma(m/2 + 3/2)/(4*a*gamma(m/2 + 5/2))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e x)^m (c+d x)^2}{a+b x^2} \, dx=\frac {e^{m} \left (2 x^{m} c d m +2 x^{m} c d +x^{m} d^{2} m x -2 \left (\int \frac {x^{m}}{b \,x^{3}+a x}d x \right ) a c d \,m^{2}-2 \left (\int \frac {x^{m}}{b \,x^{3}+a x}d x \right ) a c d m -\left (\int \frac {x^{m}}{b \,x^{2}+a}d x \right ) a \,d^{2} m^{2}-\left (\int \frac {x^{m}}{b \,x^{2}+a}d x \right ) a \,d^{2} m +\left (\int \frac {x^{m}}{b \,x^{2}+a}d x \right ) b \,c^{2} m^{2}+\left (\int \frac {x^{m}}{b \,x^{2}+a}d x \right ) b \,c^{2} m \right )}{b m \left (m +1\right )} \] Input:

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

Output:

(e**m*(2*x**m*c*d*m + 2*x**m*c*d + x**m*d**2*m*x - 2*int(x**m/(a*x + b*x** 
3),x)*a*c*d*m**2 - 2*int(x**m/(a*x + b*x**3),x)*a*c*d*m - int(x**m/(a + b* 
x**2),x)*a*d**2*m**2 - int(x**m/(a + b*x**2),x)*a*d**2*m + int(x**m/(a + b 
*x**2),x)*b*c**2*m**2 + int(x**m/(a + b*x**2),x)*b*c**2*m))/(b*m*(m + 1))