\(\int \frac {(d+e x)^m (2+x+3 x^2-5 x^3+4 x^4)}{3+2 x+5 x^2} \, dx\) [206]

Optimal result

Integrand size = 38, antiderivative size = 254 \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{1+m}}{125 e^3 (1+m)}-\frac {(40 d+33 e) (d+e x)^{2+m}}{25 e^3 (2+m)}+\frac {4 (d+e x)^{3+m}}{5 e^3 (3+m)}-\frac {\left (6412 i+423 \sqrt {14}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d-e-i \sqrt {14} e}\right )}{3500 \left (5 i d-\left (i-\sqrt {14}\right ) e\right ) (1+m)}-\frac {\left (6412 i-423 \sqrt {14}\right ) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d-e+i \sqrt {14} e}\right )}{3500 \left (5 i d-\left (i+\sqrt {14}\right ) e\right ) (1+m)} \] Output:

1/125*(100*d^2+165*d*e+81*e^2)*(e*x+d)^(1+m)/e^3/(1+m)-1/25*(40*d+33*e)*(e 
*x+d)^(2+m)/e^3/(2+m)+4/5*(e*x+d)^(3+m)/e^3/(3+m)-1/3500*(6412*I+423*14^(1 
/2))*(e*x+d)^(1+m)*hypergeom([1, 1+m],[2+m],5*(e*x+d)/(5*d-e-I*14^(1/2)*e) 
)/(5*I*d-(I-14^(1/2))*e)/(1+m)-1/3500*(6412*I-423*14^(1/2))*(e*x+d)^(1+m)* 
hypergeom([1, 1+m],[2+m],5*(e*x+d)/(5*d-e+I*14^(1/2)*e))/(5*I*d-(I+14^(1/2 
))*e)/(1+m)
 

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.87 \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {28 \left (100 d^2+165 d e+81 e^2\right )}{e^3 (1+m)}-\frac {140 (40 d+33 e) (d+e x)}{e^3 (2+m)}+\frac {2800 (d+e x)^2}{e^3 (3+m)}-\frac {\left (6412 i+423 \sqrt {14}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d+\left (-1-i \sqrt {14}\right ) e}\right )}{\left (5 i d+\left (-i+\sqrt {14}\right ) e\right ) (1+m)}-\frac {\left (-6412 i+423 \sqrt {14}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {5 (d+e x)}{5 d+i \left (i+\sqrt {14}\right ) e}\right )}{\left (-5 i d+\left (i+\sqrt {14}\right ) e\right ) (1+m)}\right )}{3500} \] Input:

Integrate[((d + e*x)^m*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(3 + 2*x + 5*x^2), 
x]
 

Output:

((d + e*x)^(1 + m)*((28*(100*d^2 + 165*d*e + 81*e^2))/(e^3*(1 + m)) - (140 
*(40*d + 33*e)*(d + e*x))/(e^3*(2 + m)) + (2800*(d + e*x)^2)/(e^3*(3 + m)) 
 - ((6412*I + 423*Sqrt[14])*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x 
))/(5*d + (-1 - I*Sqrt[14])*e)])/(((5*I)*d + (-I + Sqrt[14])*e)*(1 + m)) - 
 ((-6412*I + 423*Sqrt[14])*Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x) 
)/(5*d + I*(I + Sqrt[14])*e)])/(((-5*I)*d + (I + Sqrt[14])*e)*(1 + m))))/3 
500
 

Rubi [A] (verified)

Time = 1.09 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2159, 2009}

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.

\(\displaystyle \int \frac {\left (4 x^4-5 x^3+3 x^2+x+2\right ) (d+e x)^m}{5 x^2+2 x+3} \, dx\)

\(\Big \downarrow \) 2159

\(\displaystyle \int \left (\frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^m}{125 e^2}+\frac {(-40 d-33 e) (d+e x)^{m+1}}{25 e^2}+\frac {4 (d+e x)^{m+2}}{5 e^2}+\frac {\left (\frac {458}{125}+\frac {423 i}{125 \sqrt {14}}\right ) (d+e x)^m}{10 x-2 i \sqrt {14}+2}+\frac {\left (\frac {458}{125}-\frac {423 i}{125 \sqrt {14}}\right ) (d+e x)^m}{10 x+2 i \sqrt {14}+2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (100 d^2+165 d e+81 e^2\right ) (d+e x)^{m+1}}{125 e^3 (m+1)}-\frac {(40 d+33 e) (d+e x)^{m+2}}{25 e^3 (m+2)}+\frac {4 (d+e x)^{m+3}}{5 e^3 (m+3)}-\frac {\left (-423 \sqrt {14}+6412 i\right ) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {5 (d+e x)}{5 d+i \sqrt {14} e-e}\right )}{3500 (m+1) \left (5 i d-\left (\sqrt {14}+i\right ) e\right )}-\frac {\left (423 \sqrt {14}+6412 i\right ) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {5 (d+e x)}{5 d-\left (1+i \sqrt {14}\right ) e}\right )}{3500 (m+1) \left (5 i d-\left (-\sqrt {14}+i\right ) e\right )}\)

Input:

Int[((d + e*x)^m*(2 + x + 3*x^2 - 5*x^3 + 4*x^4))/(3 + 2*x + 5*x^2),x]
 

Output:

((100*d^2 + 165*d*e + 81*e^2)*(d + e*x)^(1 + m))/(125*e^3*(1 + m)) - ((40* 
d + 33*e)*(d + e*x)^(2 + m))/(25*e^3*(2 + m)) + (4*(d + e*x)^(3 + m))/(5*e 
^3*(3 + m)) - ((6412*I - 423*Sqrt[14])*(d + e*x)^(1 + m)*Hypergeometric2F1 
[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d - e + I*Sqrt[14]*e)])/(3500*((5*I)*d 
- (I + Sqrt[14])*e)*(1 + m)) - ((6412*I + 423*Sqrt[14])*(d + e*x)^(1 + m)* 
Hypergeometric2F1[1, 1 + m, 2 + m, (5*(d + e*x))/(5*d - (1 + I*Sqrt[14])*e 
)])/(3500*((5*I)*d - (I - Sqrt[14])*e)*(1 + m))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2159

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x 
], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
 

Maple [F]

\[\int \frac {\left (e x +d \right )^{m} \left (4 x^{4}-5 x^{3}+3 x^{2}+x +2\right )}{5 x^{2}+2 x +3}d x\]

Input:

int((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x)
 

Output:

int((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x)
 

Fricas [F]

\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3} \,d x } \] Input:

integrate((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="fr 
icas")
 

Output:

integral((4*x^4 - 5*x^3 + 3*x^2 + x + 2)*(e*x + d)^m/(5*x^2 + 2*x + 3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\text {Timed out} \] Input:

integrate((e*x+d)**m*(4*x**4-5*x**3+3*x**2+x+2)/(5*x**2+2*x+3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3} \,d x } \] Input:

integrate((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="ma 
xima")
 

Output:

integrate((4*x^4 - 5*x^3 + 3*x^2 + x + 2)*(e*x + d)^m/(5*x^2 + 2*x + 3), x 
)
 

Giac [F]

\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\int { \frac {{\left (4 \, x^{4} - 5 \, x^{3} + 3 \, x^{2} + x + 2\right )} {\left (e x + d\right )}^{m}}{5 \, x^{2} + 2 \, x + 3} \,d x } \] Input:

integrate((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x, algorithm="gi 
ac")
 

Output:

integrate((4*x^4 - 5*x^3 + 3*x^2 + x + 2)*(e*x + d)^m/(5*x^2 + 2*x + 3), x 
)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx=\int \frac {{\left (d+e\,x\right )}^m\,\left (4\,x^4-5\,x^3+3\,x^2+x+2\right )}{5\,x^2+2\,x+3} \,d x \] Input:

int(((d + e*x)^m*(x + 3*x^2 - 5*x^3 + 4*x^4 + 2))/(2*x + 5*x^2 + 3),x)
 

Output:

int(((d + e*x)^m*(x + 3*x^2 - 5*x^3 + 4*x^4 + 2))/(2*x + 5*x^2 + 3), x)
 

Reduce [F]

\[ \int \frac {(d+e x)^m \left (2+x+3 x^2-5 x^3+4 x^4\right )}{3+2 x+5 x^2} \, dx =\text {Too large to display} \] Input:

int((e*x+d)^m*(4*x^4-5*x^3+3*x^2+x+2)/(5*x^2+2*x+3),x)
 

Output:

(400*(d + e*x)**m*d**3*m - 400*(d + e*x)**m*d**2*e*m**2*x + 330*(d + e*x)* 
*m*d**2*e*m**2 + 990*(d + e*x)**m*d**2*e*m + 200*(d + e*x)**m*d*e**2*m**3* 
x**2 - 330*(d + e*x)**m*d*e**2*m**3*x + 620*(d + e*x)**m*d*e**2*m**3 + 200 
*(d + e*x)**m*d*e**2*m**2*x**2 - 990*(d + e*x)**m*d*e**2*m**2*x + 3558*(d 
+ e*x)**m*d*e**2*m**2 + 6010*(d + e*x)**m*d*e**2*m + 2748*(d + e*x)**m*d*e 
**2 + 200*(d + e*x)**m*e**3*m**3*x**3 - 330*(d + e*x)**m*e**3*m**3*x**2 + 
162*(d + e*x)**m*e**3*m**3*x + 7*(d + e*x)**m*e**3*m**3 + 600*(d + e*x)**m 
*e**3*m**2*x**3 - 1320*(d + e*x)**m*e**3*m**2*x**2 + 810*(d + e*x)**m*e**3 
*m**2*x + 42*(d + e*x)**m*e**3*m**2 + 400*(d + e*x)**m*e**3*m*x**3 - 990*( 
d + e*x)**m*e**3*m*x**2 + 972*(d + e*x)**m*e**3*m*x + 77*(d + e*x)**m*e**3 
*m + 42*(d + e*x)**m*e**3 - 1360*int((d + e*x)**m/(5*d*x**2 + 2*d*x + 3*d 
+ 5*e*x**3 + 2*e*x**2 + 3*e*x),x)*d*e**3*m**4 - 8160*int((d + e*x)**m/(5*d 
*x**2 + 2*d*x + 3*d + 5*e*x**3 + 2*e*x**2 + 3*e*x),x)*d*e**3*m**3 - 14960* 
int((d + e*x)**m/(5*d*x**2 + 2*d*x + 3*d + 5*e*x**3 + 2*e*x**2 + 3*e*x),x) 
*d*e**3*m**2 - 8160*int((d + e*x)**m/(5*d*x**2 + 2*d*x + 3*d + 5*e*x**3 + 
2*e*x**2 + 3*e*x),x)*d*e**3*m - 21*int((d + e*x)**m/(5*d*x**2 + 2*d*x + 3* 
d + 5*e*x**3 + 2*e*x**2 + 3*e*x),x)*e**4*m**4 - 126*int((d + e*x)**m/(5*d* 
x**2 + 2*d*x + 3*d + 5*e*x**3 + 2*e*x**2 + 3*e*x),x)*e**4*m**3 - 231*int(( 
d + e*x)**m/(5*d*x**2 + 2*d*x + 3*d + 5*e*x**3 + 2*e*x**2 + 3*e*x),x)*e**4 
*m**2 - 126*int((d + e*x)**m/(5*d*x**2 + 2*d*x + 3*d + 5*e*x**3 + 2*e*x...