\(\int \frac {(a+b x)^m (2+d x)}{\sqrt {4-d^2 x^2}} \, dx\) [130]

Optimal result

Integrand size = 27, antiderivative size = 82 \[ \int \frac {(a+b x)^m (2+d x)}{\sqrt {4-d^2 x^2}} \, dx=-\frac {4 (a+b x)^m \left (\frac {d (a+b x)}{2 b+a d}\right )^{-m} \sqrt {2-d x} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{2},-m,\frac {3}{2},\frac {1}{4} (2-d x),\frac {b (2-d x)}{2 b+a d}\right )}{d} \] Output:

-4*(b*x+a)^m*(-d*x+2)^(1/2)*AppellF1(1/2,-m,-1/2,3/2,b*(-d*x+2)/(a*d+2*b), 
-1/4*d*x+1/2)/d/((d*(b*x+a)/(a*d+2*b))^m)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(233\) vs. \(2(82)=164\).

Time = 3.94 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.84 \[ \int \frac {(a+b x)^m (2+d x)}{\sqrt {4-d^2 x^2}} \, dx=\frac {(a+b x)^m \left (-4+d^2 x^2-\frac {\left (-4+d^2 x^2\right ) \operatorname {AppellF1}\left (m,-\frac {1}{2},-\frac {1}{2},1+m,\frac {d (a+b x)}{-2 b+a d},\frac {d (a+b x)}{2 b+a d}\right )}{\sqrt {-\frac {b (-2+d x)}{2 b+a d}} \sqrt {\frac {b (2+d x)}{2 b-a d}}}+\frac {2 d (a+b x) \sqrt {-\frac {b (-2+d x)}{2 b+a d}} \sqrt {\frac {b (2+d x)}{2 b-a d}} \operatorname {AppellF1}\left (1+m,\frac {1}{2},\frac {1}{2},2+m,\frac {d (a+b x)}{-2 b+a d},\frac {d (a+b x)}{2 b+a d}\right )}{b (1+m)}\right )}{d \sqrt {4-d^2 x^2}} \] Input:

Integrate[((a + b*x)^m*(2 + d*x))/Sqrt[4 - d^2*x^2],x]
 

Output:

((a + b*x)^m*(-4 + d^2*x^2 - ((-4 + d^2*x^2)*AppellF1[m, -1/2, -1/2, 1 + m 
, (d*(a + b*x))/(-2*b + a*d), (d*(a + b*x))/(2*b + a*d)])/(Sqrt[-((b*(-2 + 
 d*x))/(2*b + a*d))]*Sqrt[(b*(2 + d*x))/(2*b - a*d)]) + (2*d*(a + b*x)*Sqr 
t[-((b*(-2 + d*x))/(2*b + a*d))]*Sqrt[(b*(2 + d*x))/(2*b - a*d)]*AppellF1[ 
1 + m, 1/2, 1/2, 2 + m, (d*(a + b*x))/(-2*b + a*d), (d*(a + b*x))/(2*b + a 
*d)])/(b*(1 + m))))/(d*Sqrt[4 - d^2*x^2])
 

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {717, 156, 155}

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.

\(\displaystyle \int \frac {(d x+2) (a+b x)^m}{\sqrt {4-d^2 x^2}} \, dx\)

\(\Big \downarrow \) 717

\(\displaystyle \int \frac {\sqrt {d x+2} (a+b x)^m}{\sqrt {2-d x}}dx\)

\(\Big \downarrow \) 156

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

\(\Big \downarrow \) 155

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

Input:

Int[((a + b*x)^m*(2 + d*x))/Sqrt[4 - d^2*x^2],x]
 

Output:

(-4*(a + b*x)^m*Sqrt[2 - d*x]*AppellF1[1/2, -1/2, -m, 3/2, (2 - d*x)/4, (b 
*(2 - d*x))/(2*b + a*d)])/(d*((d*(a + b*x))/(2*b + a*d))^m)
 

Defintions of rubi rules used

rule 155

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

rule 156

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[(e + f*x)^FracPart[p]/(Simplify[b/(b*e - a*f)]^IntPart[p 
]*(b*((e + f*x)/(b*e - a*f)))^FracPart[p])   Int[(a + b*x)^m*(c + d*x)^n*Si 
mp[b*(e/(b*e - a*f)) + b*f*(x/(b*e - a*f)), x]^p, x], x] /; FreeQ[{a, b, c, 
 d, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !IntegerQ[p] & 
& GtQ[Simplify[b/(b*c - a*d)], 0] &&  !GtQ[Simplify[b/(b*e - a*f)], 0]
 

rule 717

Int[((d_) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_) 
^2)^(p_), x_Symbol] :> Int[(d + e*x)^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, 
 x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a 
, 0] && GtQ[d, 0]
 

Maple [F]

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

Input:

int((b*x+a)^m*(d*x+2)/(-d^2*x^2+4)^(1/2),x)
 

Output:

int((b*x+a)^m*(d*x+2)/(-d^2*x^2+4)^(1/2),x)
 

Fricas [F]

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

integrate((b*x+a)^m*(d*x+2)/(-d^2*x^2+4)^(1/2),x, algorithm="fricas")
 

Output:

integral(-sqrt(-d^2*x^2 + 4)*(b*x + a)^m/(d*x - 2), x)
 

Sympy [F]

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

integrate((b*x+a)**m*(d*x+2)/(-d**2*x**2+4)**(1/2),x)
 

Output:

Integral((a + b*x)**m*(d*x + 2)/sqrt(-(d*x - 2)*(d*x + 2)), x)
 

Maxima [F]

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

integrate((b*x+a)^m*(d*x+2)/(-d^2*x^2+4)^(1/2),x, algorithm="maxima")
 

Output:

integrate((d*x + 2)*(b*x + a)^m/sqrt(-d^2*x^2 + 4), x)
 

Giac [F]

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

integrate((b*x+a)^m*(d*x+2)/(-d^2*x^2+4)^(1/2),x, algorithm="giac")
 

Output:

integrate((d*x + 2)*(b*x + a)^m/sqrt(-d^2*x^2 + 4), x)
 

Mupad [F(-1)]

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

int(((d*x + 2)*(a + b*x)^m)/(4 - d^2*x^2)^(1/2),x)
 

Output:

int(((d*x + 2)*(a + b*x)^m)/(4 - d^2*x^2)^(1/2), x)
 

Reduce [F]

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

int((b*x+a)^m*(d*x+2)/(-d^2*x^2+4)^(1/2),x)
 

Output:

2*int((a + b*x)**m/sqrt( - d**2*x**2 + 4),x) + int(((a + b*x)**m*x)/sqrt( 
- d**2*x**2 + 4),x)*d