\(\int \frac {(5-2 x) (4+3 x)^m}{\sqrt {2+3 x+x^2}} \, dx\) [654]

Optimal result

Integrand size = 25, antiderivative size = 157 \[ \int \frac {(5-2 x) (4+3 x)^m}{\sqrt {2+3 x+x^2}} \, dx=\frac {23 \sqrt {-1-x} \sqrt {2+x} (4+3 x)^{1+m} \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,4+3 x,\frac {1}{2} (-4-3 x)\right )}{3 \sqrt {2} (1+m) \sqrt {2+3 x+x^2}}-\frac {\sqrt {2} \sqrt {-1-x} \sqrt {2+x} (4+3 x)^{2+m} \operatorname {AppellF1}\left (2+m,\frac {1}{2},\frac {1}{2},3+m,4+3 x,\frac {1}{2} (-4-3 x)\right )}{3 (2+m) \sqrt {2+3 x+x^2}} \] Output:

23/6*(-1-x)^(1/2)*(2+x)^(1/2)*(4+3*x)^(1+m)*AppellF1(1+m,1/2,1/2,2+m,-2-3/ 
2*x,4+3*x)*2^(1/2)/(1+m)/(x^2+3*x+2)^(1/2)-1/3*2^(1/2)*(-1-x)^(1/2)*(2+x)^ 
(1/2)*(4+3*x)^(2+m)*AppellF1(2+m,1/2,1/2,3+m,-2-3/2*x,4+3*x)/(2+m)/(x^2+3* 
x+2)^(1/2)
 

Mathematica [F]

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

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {1269, 1179, 27, 150}

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.

\(\displaystyle \int \frac {(5-2 x) (3 x+4)^m}{\sqrt {x^2+3 x+2}} \, dx\)

\(\Big \downarrow \) 1269

\(\displaystyle \frac {23}{3} \int \frac {(3 x+4)^m}{\sqrt {x^2+3 x+2}}dx-\frac {2}{3} \int \frac {(3 x+4)^{m+1}}{\sqrt {x^2+3 x+2}}dx\)

\(\Big \downarrow \) 1179

\(\displaystyle \frac {23 \sqrt {-x-1} \sqrt {3 x+6} \int \frac {\sqrt {2} (3 x+4)^m}{\sqrt {-3 x-3} \sqrt {3 x+6}}d(3 x+4)}{3 \sqrt {6} \sqrt {x^2+3 x+2}}-\frac {\sqrt {\frac {2}{3}} \sqrt {-x-1} \sqrt {3 x+6} \int \frac {\sqrt {2} (3 x+4)^{m+1}}{\sqrt {-3 x-3} \sqrt {3 x+6}}d(3 x+4)}{3 \sqrt {x^2+3 x+2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {23 \sqrt {-x-1} \sqrt {3 x+6} \int \frac {(3 x+4)^m}{\sqrt {-3 x-3} \sqrt {3 x+6}}d(3 x+4)}{3 \sqrt {3} \sqrt {x^2+3 x+2}}-\frac {2 \sqrt {-x-1} \sqrt {3 x+6} \int \frac {(3 x+4)^{m+1}}{\sqrt {-3 x-3} \sqrt {3 x+6}}d(3 x+4)}{3 \sqrt {3} \sqrt {x^2+3 x+2}}\)

\(\Big \downarrow \) 150

\(\displaystyle \frac {23 \sqrt {-x-1} \sqrt {3 x+6} (3 x+4)^{m+1} \operatorname {AppellF1}\left (m+1,\frac {1}{2},\frac {1}{2},m+2,3 x+4,\frac {1}{2} (-3 x-4)\right )}{3 \sqrt {6} (m+1) \sqrt {x^2+3 x+2}}-\frac {\sqrt {\frac {2}{3}} \sqrt {-x-1} \sqrt {3 x+6} (3 x+4)^{m+2} \operatorname {AppellF1}\left (m+2,\frac {1}{2},\frac {1}{2},m+3,3 x+4,\frac {1}{2} (-3 x-4)\right )}{3 (m+2) \sqrt {x^2+3 x+2}}\)

Input:

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

Output:

(23*Sqrt[-1 - x]*(4 + 3*x)^(1 + m)*Sqrt[6 + 3*x]*AppellF1[1 + m, 1/2, 1/2, 
 2 + m, 4 + 3*x, (-4 - 3*x)/2])/(3*Sqrt[6]*(1 + m)*Sqrt[2 + 3*x + x^2]) - 
(Sqrt[2/3]*Sqrt[-1 - x]*(4 + 3*x)^(2 + m)*Sqrt[6 + 3*x]*AppellF1[2 + m, 1/ 
2, 1/2, 3 + m, 4 + 3*x, (-4 - 3*x)/2])/(3*(2 + m)*Sqrt[2 + 3*x + x^2])
 

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 150

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ 
] :> Simp[c^n*e^p*((b*x)^(m + 1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2 
, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p}, x] &&  !In 
tegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
 

rule 1179

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(a + b*x + c*x^2)^p/(e*(1 - ( 
d + e*x)/(d - e*((b - q)/(2*c))))^p*(1 - (d + e*x)/(d - e*((b + q)/(2*c)))) 
^p)   Subst[Int[x^m*Simp[1 - x/(d - e*((b - q)/(2*c))), x]^p*Simp[1 - x/(d 
- e*((b + q)/(2*c))), x]^p, x], x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m 
, p}, x]
 

rule 1269

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e   Int[(d + e*x)^(m + 1)*(a + b*x + 
 c*x^2)^p, x], x] + Simp[(e*f - d*g)/e   Int[(d + e*x)^m*(a + b*x + c*x^2)^ 
p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] &&  !IGtQ[m, 0]
 

Maple [F]

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

Input:

int((5-2*x)*(3*x+4)^m/(x^2+3*x+2)^(1/2),x)
 

Output:

int((5-2*x)*(3*x+4)^m/(x^2+3*x+2)^(1/2),x)
 

Fricas [F]

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

integrate((5-2*x)*(4+3*x)^m/(x^2+3*x+2)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

integrate((5-2*x)*(4+3*x)**m/(x**2+3*x+2)**(1/2),x)
 

Output:

-Integral(-5*(3*x + 4)**m/sqrt(x**2 + 3*x + 2), x) - Integral(2*x*(3*x + 4 
)**m/sqrt(x**2 + 3*x + 2), x)
 

Maxima [F]

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

integrate((5-2*x)*(4+3*x)^m/(x^2+3*x+2)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((5-2*x)*(4+3*x)^m/(x^2+3*x+2)^(1/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int(-((2*x - 5)*(3*x + 4)^m)/(3*x + x^2 + 2)^(1/2),x)
 

Output:

-int(((2*x - 5)*(3*x + 4)^m)/(3*x + x^2 + 2)^(1/2), x)
 

Reduce [F]

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

int((5-2*x)*(4+3*x)^m/(x^2+3*x+2)^(1/2),x)
 

Output:

5*int((3*x + 4)**m/sqrt(x**2 + 3*x + 2),x) - 2*int(((3*x + 4)**m*x)/sqrt(x 
**2 + 3*x + 2),x)