\(\int \frac {(e x)^m (c+d x)^3}{(a+b x^2)^2} \, dx\) [123]

Optimal result

Integrand size = 22, antiderivative size = 205 \[ \int \frac {(e x)^m (c+d x)^3}{\left (a+b x^2\right )^2} \, dx=-\frac {3 c d^2 (e x)^{1+m}}{b e (1-m) \left (a+b x^2\right )}+\frac {d^3 (e x)^{2+m}}{b e^2 m \left (a+b x^2\right )}+\frac {c \left (b c^2 (1-m)+3 a d^2 (1+m)\right ) (e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{a^2 b e \left (1-m^2\right )}+\frac {d \left (3 b c^2 m-a d^2 (2+m)\right ) (e x)^{2+m} \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{a^2 b e^2 m (2+m)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.86 \[ \int \frac {(e x)^m (c+d x)^3}{\left (a+b x^2\right )^2} \, dx=\frac {x (e x)^m \left (\frac {3 a c d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+\frac {a d^3 x \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{2+m}+\frac {c \left (b c^2-3 a d^2\right ) \operatorname {Hypergeometric2F1}\left (2,\frac {1+m}{2},\frac {3+m}{2},-\frac {b x^2}{a}\right )}{1+m}+\frac {d \left (3 b c^2-a d^2\right ) x \operatorname {Hypergeometric2F1}\left (2,\frac {2+m}{2},\frac {4+m}{2},-\frac {b x^2}{a}\right )}{2+m}\right )}{a^2 b} \] Input:

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

Output:

(x*(e*x)^m*((3*a*c*d^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2 
)/a)])/(1 + m) + (a*d^3*x*Hypergeometric2F1[1, (2 + m)/2, (4 + m)/2, -((b* 
x^2)/a)])/(2 + m) + (c*(b*c^2 - 3*a*d^2)*Hypergeometric2F1[2, (1 + m)/2, ( 
3 + m)/2, -((b*x^2)/a)])/(1 + m) + (d*(3*b*c^2 - a*d^2)*x*Hypergeometric2F 
1[2, (2 + m)/2, (4 + m)/2, -((b*x^2)/a)])/(2 + m)))/(a^2*b)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.98, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {558, 25, 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.

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

\(\Big \downarrow \) 558

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 557

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

\(\Big \downarrow \) 278

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

Input:

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

Output:

((e*x)^(1 + m)*(c*(c^2 - (3*a*d^2)/b) + d*(3*c^2 - (a*d^2)/b)*x))/(2*a*e*( 
a + b*x^2)) + ((c*(b*c^2*(1 - m) + 3*a*d^2*(1 + m))*(e*x)^(1 + m)*Hypergeo 
metric2F1[1, (1 + m)/2, (3 + m)/2, -((b*x^2)/a)])/(a*e*(1 + m)) - (d*(3*b* 
c^2*m - a*d^2*(2 + m))*(e*x)^(2 + m)*Hypergeometric2F1[1, (2 + m)/2, (4 + 
m)/2, -((b*x^2)/a)])/(a*e^2*(2 + m)))/(2*a*b)
 

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 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 558

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), 
 x_Symbol] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, a + b*x^2, x], f = 
 Coeff[PolynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 0], g = Coeff[Pol 
ynomialRemainder[(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(-(e*x)^(m + 1))* 
(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*e*(p + 1))), x] + Simp[1/(2*a*(p + 1)) 
  Int[(e*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + f*(m + 2*p + 
 3) + g*(m + 2*p + 4)*x, x], x], x]] /; FreeQ[{a, b, c, d, e, m}, x] && IGt 
Q[n, 1] &&  !IntegerQ[m] && LtQ[p, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*(e*x)^m/(b^2*x^4 + 2*a* 
b*x^2 + a^2), x)
 

Sympy [F]

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

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

Output:

Integral((e*x)**m*(c + d*x)**3/(a + b*x**2)**2, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(e**m*( - x**m*a*d**3*m**2 - x**m*a*d**3*m + 2*x**m*a*d**3 + 3*x**m*b*c**2 
*d*m**2 - 3*x**m*b*c**2*d*m + 3*x**m*b*c*d**2*m**2*x - 6*x**m*b*c*d**2*m*x 
 + x**m*b*d**3*m**2*x**2 - 3*x**m*b*d**3*m*x**2 + 2*x**m*b*d**3*x**2 + int 
(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x** 
3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x)*a**3*d** 
3*m**5 - 2*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 
 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x** 
5),x)*a**3*d**3*m**4 - 3*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2 
*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x** 
5 + 2*b**2*x**5),x)*a**3*d**3*m**3 + 8*int(x**m/(a**2*m**2*x - 3*a**2*m*x 
+ 2*a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 
- 3*b**2*m*x**5 + 2*b**2*x**5),x)*a**3*d**3*m**2 - 4*int(x**m/(a**2*m**2*x 
 - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b 
**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x)*a**3*d**3*m - 3*int(x**m/( 
a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a 
*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5),x)*a**2*b*c**2*d*m 
**5 + 12*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x + 2*a*b*m**2*x**3 - 
 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2*m*x**5 + 2*b**2*x**5) 
,x)*a**2*b*c**2*d*m**4 - 15*int(x**m/(a**2*m**2*x - 3*a**2*m*x + 2*a**2*x 
+ 2*a*b*m**2*x**3 - 6*a*b*m*x**3 + 4*a*b*x**3 + b**2*m**2*x**5 - 3*b**2...