\(\int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [509]

Optimal result

Integrand size = 33, antiderivative size = 171 \[ \int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {(b d-a e) (B d-A e) (d+e x)^{1+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (1+m) (a+b x)}-\frac {(2 b B d-A b e-a B e) (d+e x)^{2+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (2+m) (a+b x)}+\frac {b B (d+e x)^{3+m} \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (3+m) (a+b x)} \] Output:

(-a*e+b*d)*(-A*e+B*d)*(e*x+d)^(1+m)*((b*x+a)^2)^(1/2)/e^3/(1+m)/(b*x+a)-(- 
A*b*e-B*a*e+2*B*b*d)*(e*x+d)^(2+m)*((b*x+a)^2)^(1/2)/e^3/(2+m)/(b*x+a)+b*B 
*(e*x+d)^(3+m)*((b*x+a)^2)^(1/2)/e^3/(3+m)/(b*x+a)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.71 \[ \int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {\sqrt {(a+b x)^2} (d+e x)^{1+m} \left (a e (3+m) (-B d+A e (2+m)+B e (1+m) x)+b \left (A e (3+m) (-d+e (1+m) x)+B \left (2 d^2-2 d e (1+m) x+e^2 \left (2+3 m+m^2\right ) x^2\right )\right )\right )}{e^3 (1+m) (2+m) (3+m) (a+b x)} \] Input:

Integrate[(A + B*x)*(d + e*x)^m*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
 

Output:

(Sqrt[(a + b*x)^2]*(d + e*x)^(1 + m)*(a*e*(3 + m)*(-(B*d) + A*e*(2 + m) + 
B*e*(1 + m)*x) + b*(A*e*(3 + m)*(-d + e*(1 + m)*x) + B*(2*d^2 - 2*d*e*(1 + 
 m)*x + e^2*(2 + 3*m + m^2)*x^2))))/(e^3*(1 + m)*(2 + m)*(3 + m)*(a + b*x) 
)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.69, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 27, 86, 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.

\(\displaystyle \int \sqrt {a^2+2 a b x+b^2 x^2} (A+B x) (d+e x)^m \, dx\)

\(\Big \downarrow \) 1187

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int b (a+b x) (A+B x) (d+e x)^mdx}{b (a+b x)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x) (A+B x) (d+e x)^mdx}{a+b x}\)

\(\Big \downarrow \) 86

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {(b d-a e) (B d-A e) (d+e x)^{m+1}}{e^3 (m+1)}-\frac {(d+e x)^{m+2} (-a B e-A b e+2 b B d)}{e^3 (m+2)}+\frac {b B (d+e x)^{m+3}}{e^3 (m+3)}\right )}{a+b x}\)

Input:

Int[(A + B*x)*(d + e*x)^m*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]
 

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(((b*d - a*e)*(B*d - A*e)*(d + e*x)^(1 + m) 
)/(e^3*(1 + m)) - ((2*b*B*d - A*b*e - a*B*e)*(d + e*x)^(2 + m))/(e^3*(2 + 
m)) + (b*B*(d + e*x)^(3 + m))/(e^3*(3 + m))))/(a + b*x)
 

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 86

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ 
.), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; 
 FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 
] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p 
+ 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
 

rule 1187

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

rule 2009

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

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.20

method	result	size
gosper	\(\frac {\left (e x +d \right )^{1+m} \sqrt {\left (b x +a \right )^{2}}\, \left (B b \,e^{2} m^{2} x^{2}+A b \,e^{2} m^{2} x +B a \,e^{2} m^{2} x +3 B b \,e^{2} m \,x^{2}+A a \,e^{2} m^{2}+4 A b \,e^{2} m x +4 B a \,e^{2} m x -2 B b d e m x +2 B b \,e^{2} x^{2}+5 A a \,e^{2} m -A b d e m +3 A b \,e^{2} x -a B d e m +3 B a \,e^{2} x -2 B b d e x +6 A a \,e^{2}-3 A b d e -3 B a d e +2 B b \,d^{2}\right )}{e^{3} \left (b x +a \right ) \left (m^{3}+6 m^{2}+11 m +6\right )}\)	\(205\)
orering	\(\frac {\left (e x +d \right ) \left (B b \,e^{2} m^{2} x^{2}+A b \,e^{2} m^{2} x +B a \,e^{2} m^{2} x +3 B b \,e^{2} m \,x^{2}+A a \,e^{2} m^{2}+4 A b \,e^{2} m x +4 B a \,e^{2} m x -2 B b d e m x +2 B b \,e^{2} x^{2}+5 A a \,e^{2} m -A b d e m +3 A b \,e^{2} x -a B d e m +3 B a \,e^{2} x -2 B b d e x +6 A a \,e^{2}-3 A b d e -3 B a d e +2 B b \,d^{2}\right ) \left (e x +d \right )^{m} \sqrt {\left (b x +a \right )^{2}}}{\left (b x +a \right ) e^{3} \left (m^{3}+6 m^{2}+11 m +6\right )}\)	\(208\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (B b \,e^{3} m^{2} x^{3}+A b \,e^{3} m^{2} x^{2}+B a \,e^{3} m^{2} x^{2}+B b d \,e^{2} m^{2} x^{2}+3 B b \,e^{3} m \,x^{3}+A a \,e^{3} m^{2} x +A b d \,e^{2} m^{2} x +4 A b \,e^{3} m \,x^{2}+B a d \,e^{2} m^{2} x +4 B a \,e^{3} m \,x^{2}+B b d \,e^{2} m \,x^{2}+2 B b \,x^{3} e^{3}+A a d \,e^{2} m^{2}+5 A a \,e^{3} m x +3 A b d \,e^{2} m x +3 A b \,e^{3} x^{2}+3 B a d \,e^{2} m x +3 B \,x^{2} a \,e^{3}-2 B b \,d^{2} e m x +5 A a d \,e^{2} m +6 A a \,e^{3} x -A b \,d^{2} e m -B a \,d^{2} e m +6 A a d \,e^{2}-3 A b \,d^{2} e -3 B a \,d^{2} e +2 B b \,d^{3}\right ) \left (e x +d \right )^{m}}{\left (b x +a \right ) \left (2+m \right ) \left (3+m \right ) \left (1+m \right ) e^{3}}\)	\(314\)

Input:

int((B*x+A)*(e*x+d)^m*((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/e^3*(e*x+d)^(1+m)/(b*x+a)*((b*x+a)^2)^(1/2)/(m^3+6*m^2+11*m+6)*(B*b*e^2* 
m^2*x^2+A*b*e^2*m^2*x+B*a*e^2*m^2*x+3*B*b*e^2*m*x^2+A*a*e^2*m^2+4*A*b*e^2* 
m*x+4*B*a*e^2*m*x-2*B*b*d*e*m*x+2*B*b*e^2*x^2+5*A*a*e^2*m-A*b*d*e*m+3*A*b* 
e^2*x-B*a*d*e*m+3*B*a*e^2*x-2*B*b*d*e*x+6*A*a*e^2-3*A*b*d*e-3*B*a*d*e+2*B* 
b*d^2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.50 \[ \int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (A a d e^{2} m^{2} + 2 \, B b d^{3} + 6 \, A a d e^{2} - 3 \, {\left (B a + A b\right )} d^{2} e + {\left (B b e^{3} m^{2} + 3 \, B b e^{3} m + 2 \, B b e^{3}\right )} x^{3} + {\left (3 \, {\left (B a + A b\right )} e^{3} + {\left (B b d e^{2} + {\left (B a + A b\right )} e^{3}\right )} m^{2} + {\left (B b d e^{2} + 4 \, {\left (B a + A b\right )} e^{3}\right )} m\right )} x^{2} + {\left (5 \, A a d e^{2} - {\left (B a + A b\right )} d^{2} e\right )} m + {\left (6 \, A a e^{3} + {\left (A a e^{3} + {\left (B a + A b\right )} d e^{2}\right )} m^{2} - {\left (2 \, B b d^{2} e - 5 \, A a e^{3} - 3 \, {\left (B a + A b\right )} d e^{2}\right )} m\right )} x\right )} {\left (e x + d\right )}^{m}}{e^{3} m^{3} + 6 \, e^{3} m^{2} + 11 \, e^{3} m + 6 \, e^{3}} \] Input:

integrate((B*x+A)*(e*x+d)^m*((b*x+a)^2)^(1/2),x, algorithm="fricas")
 

Output:

(A*a*d*e^2*m^2 + 2*B*b*d^3 + 6*A*a*d*e^2 - 3*(B*a + A*b)*d^2*e + (B*b*e^3* 
m^2 + 3*B*b*e^3*m + 2*B*b*e^3)*x^3 + (3*(B*a + A*b)*e^3 + (B*b*d*e^2 + (B* 
a + A*b)*e^3)*m^2 + (B*b*d*e^2 + 4*(B*a + A*b)*e^3)*m)*x^2 + (5*A*a*d*e^2 
- (B*a + A*b)*d^2*e)*m + (6*A*a*e^3 + (A*a*e^3 + (B*a + A*b)*d*e^2)*m^2 - 
(2*B*b*d^2*e - 5*A*a*e^3 - 3*(B*a + A*b)*d*e^2)*m)*x)*(e*x + d)^m/(e^3*m^3 
 + 6*e^3*m^2 + 11*e^3*m + 6*e^3)
 

Sympy [F(-2)]

Exception generated. \[ \int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((B*x+A)*(e*x+d)**m*((b*x+a)**2)**(1/2),x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.04 \[ \int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (b e^{2} {\left (m + 1\right )} x^{2} + a d e {\left (m + 2\right )} - b d^{2} + {\left (a e^{2} {\left (m + 2\right )} + b d e m\right )} x\right )} {\left (e x + d\right )}^{m} A}{{\left (m^{2} + 3 \, m + 2\right )} e^{2}} + \frac {{\left ({\left (m^{2} + 3 \, m + 2\right )} b e^{3} x^{3} - a d^{2} e {\left (m + 3\right )} + 2 \, b d^{3} + {\left ({\left (m^{2} + m\right )} b d e^{2} + {\left (m^{2} + 4 \, m + 3\right )} a e^{3}\right )} x^{2} + {\left ({\left (m^{2} + 3 \, m\right )} a d e^{2} - 2 \, b d^{2} e m\right )} x\right )} {\left (e x + d\right )}^{m} B}{{\left (m^{3} + 6 \, m^{2} + 11 \, m + 6\right )} e^{3}} \] Input:

integrate((B*x+A)*(e*x+d)^m*((b*x+a)^2)^(1/2),x, algorithm="maxima")
 

Output:

(b*e^2*(m + 1)*x^2 + a*d*e*(m + 2) - b*d^2 + (a*e^2*(m + 2) + b*d*e*m)*x)* 
(e*x + d)^m*A/((m^2 + 3*m + 2)*e^2) + ((m^2 + 3*m + 2)*b*e^3*x^3 - a*d^2*e 
*(m + 3) + 2*b*d^3 + ((m^2 + m)*b*d*e^2 + (m^2 + 4*m + 3)*a*e^3)*x^2 + ((m 
^2 + 3*m)*a*d*e^2 - 2*b*d^2*e*m)*x)*(e*x + d)^m*B/((m^3 + 6*m^2 + 11*m + 6 
)*e^3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 652 vs. \(2 (138) = 276\).

Time = 0.19 (sec) , antiderivative size = 652, normalized size of antiderivative = 3.81 \[ \int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx =\text {Too large to display} \] Input:

integrate((B*x+A)*(e*x+d)^m*((b*x+a)^2)^(1/2),x, algorithm="giac")
 

Output:

((e*x + d)^m*B*b*e^3*m^2*x^3*sgn(b*x + a) + (e*x + d)^m*B*b*d*e^2*m^2*x^2* 
sgn(b*x + a) + (e*x + d)^m*B*a*e^3*m^2*x^2*sgn(b*x + a) + (e*x + d)^m*A*b* 
e^3*m^2*x^2*sgn(b*x + a) + 3*(e*x + d)^m*B*b*e^3*m*x^3*sgn(b*x + a) + (e*x 
 + d)^m*B*a*d*e^2*m^2*x*sgn(b*x + a) + (e*x + d)^m*A*b*d*e^2*m^2*x*sgn(b*x 
 + a) + (e*x + d)^m*A*a*e^3*m^2*x*sgn(b*x + a) + (e*x + d)^m*B*b*d*e^2*m*x 
^2*sgn(b*x + a) + 4*(e*x + d)^m*B*a*e^3*m*x^2*sgn(b*x + a) + 4*(e*x + d)^m 
*A*b*e^3*m*x^2*sgn(b*x + a) + 2*(e*x + d)^m*B*b*e^3*x^3*sgn(b*x + a) + (e* 
x + d)^m*A*a*d*e^2*m^2*sgn(b*x + a) - 2*(e*x + d)^m*B*b*d^2*e*m*x*sgn(b*x 
+ a) + 3*(e*x + d)^m*B*a*d*e^2*m*x*sgn(b*x + a) + 3*(e*x + d)^m*A*b*d*e^2* 
m*x*sgn(b*x + a) + 5*(e*x + d)^m*A*a*e^3*m*x*sgn(b*x + a) + 3*(e*x + d)^m* 
B*a*e^3*x^2*sgn(b*x + a) + 3*(e*x + d)^m*A*b*e^3*x^2*sgn(b*x + a) - (e*x + 
 d)^m*B*a*d^2*e*m*sgn(b*x + a) - (e*x + d)^m*A*b*d^2*e*m*sgn(b*x + a) + 5* 
(e*x + d)^m*A*a*d*e^2*m*sgn(b*x + a) + 6*(e*x + d)^m*A*a*e^3*x*sgn(b*x + a 
) + 2*(e*x + d)^m*B*b*d^3*sgn(b*x + a) - 3*(e*x + d)^m*B*a*d^2*e*sgn(b*x + 
 a) - 3*(e*x + d)^m*A*b*d^2*e*sgn(b*x + a) + 6*(e*x + d)^m*A*a*d*e^2*sgn(b 
*x + a))/(e^3*m^3 + 6*e^3*m^2 + 11*e^3*m + 6*e^3)
 

Mupad [B] (verification not implemented)

Time = 11.86 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.82 \[ \int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (d+e\,x\right )}^m\,\left (\frac {B\,x^3\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (m^2+3\,m+2\right )}{m^3+6\,m^2+11\,m+6}+\frac {x\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (6\,A\,a\,e^3+5\,A\,a\,e^3\,m+A\,a\,e^3\,m^2+3\,A\,b\,d\,e^2\,m+3\,B\,a\,d\,e^2\,m-2\,B\,b\,d^2\,e\,m+A\,b\,d\,e^2\,m^2+B\,a\,d\,e^2\,m^2\right )}{b\,e^3\,\left (m^3+6\,m^2+11\,m+6\right )}+\frac {d\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (6\,A\,a\,e^2+2\,B\,b\,d^2+5\,A\,a\,e^2\,m+A\,a\,e^2\,m^2-3\,A\,b\,d\,e-3\,B\,a\,d\,e-A\,b\,d\,e\,m-B\,a\,d\,e\,m\right )}{b\,e^3\,\left (m^3+6\,m^2+11\,m+6\right )}+\frac {x^2\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (m+1\right )\,\left (3\,A\,b\,e+3\,B\,a\,e+A\,b\,e\,m+B\,a\,e\,m+B\,b\,d\,m\right )}{b\,e\,\left (m^3+6\,m^2+11\,m+6\right )}\right )}{x+\frac {a}{b}} \] Input:

int(((a + b*x)^2)^(1/2)*(A + B*x)*(d + e*x)^m,x)
 

Output:

((d + e*x)^m*((B*x^3*((a + b*x)^2)^(1/2)*(3*m + m^2 + 2))/(11*m + 6*m^2 + 
m^3 + 6) + (x*((a + b*x)^2)^(1/2)*(6*A*a*e^3 + 5*A*a*e^3*m + A*a*e^3*m^2 + 
 3*A*b*d*e^2*m + 3*B*a*d*e^2*m - 2*B*b*d^2*e*m + A*b*d*e^2*m^2 + B*a*d*e^2 
*m^2))/(b*e^3*(11*m + 6*m^2 + m^3 + 6)) + (d*((a + b*x)^2)^(1/2)*(6*A*a*e^ 
2 + 2*B*b*d^2 + 5*A*a*e^2*m + A*a*e^2*m^2 - 3*A*b*d*e - 3*B*a*d*e - A*b*d* 
e*m - B*a*d*e*m))/(b*e^3*(11*m + 6*m^2 + m^3 + 6)) + (x^2*((a + b*x)^2)^(1 
/2)*(m + 1)*(3*A*b*e + 3*B*a*e + A*b*e*m + B*a*e*m + B*b*d*m))/(b*e*(11*m 
+ 6*m^2 + m^3 + 6))))/(x + a/b)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.41 \[ \int (A+B x) (d+e x)^m \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {\left (e x +d \right )^{m} \left (b^{2} e^{3} m^{2} x^{3}+2 a b \,e^{3} m^{2} x^{2}+b^{2} d \,e^{2} m^{2} x^{2}+3 b^{2} e^{3} m \,x^{3}+a^{2} e^{3} m^{2} x +2 a b d \,e^{2} m^{2} x +8 a b \,e^{3} m \,x^{2}+b^{2} d \,e^{2} m \,x^{2}+2 b^{2} e^{3} x^{3}+a^{2} d \,e^{2} m^{2}+5 a^{2} e^{3} m x +6 a b d \,e^{2} m x +6 a b \,e^{3} x^{2}-2 b^{2} d^{2} e m x +5 a^{2} d \,e^{2} m +6 a^{2} e^{3} x -2 a b \,d^{2} e m +6 a^{2} d \,e^{2}-6 a b \,d^{2} e +2 b^{2} d^{3}\right )}{e^{3} \left (m^{3}+6 m^{2}+11 m +6\right )} \] Input:

int((B*x+A)*(e*x+d)^m*((b*x+a)^2)^(1/2),x)
 

Output:

((d + e*x)**m*(a**2*d*e**2*m**2 + 5*a**2*d*e**2*m + 6*a**2*d*e**2 + a**2*e 
**3*m**2*x + 5*a**2*e**3*m*x + 6*a**2*e**3*x - 2*a*b*d**2*e*m - 6*a*b*d**2 
*e + 2*a*b*d*e**2*m**2*x + 6*a*b*d*e**2*m*x + 2*a*b*e**3*m**2*x**2 + 8*a*b 
*e**3*m*x**2 + 6*a*b*e**3*x**2 + 2*b**2*d**3 - 2*b**2*d**2*e*m*x + b**2*d* 
e**2*m**2*x**2 + b**2*d*e**2*m*x**2 + b**2*e**3*m**2*x**3 + 3*b**2*e**3*m* 
x**3 + 2*b**2*e**3*x**3))/(e**3*(m**3 + 6*m**2 + 11*m + 6))