\(\int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx\) [265]

Optimal result

Integrand size = 28, antiderivative size = 92 \[ \int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=-\frac {(b d-a e) (d+e x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{5 e^2 (a+b x)}+\frac {b (d+e x)^6 \sqrt {a^2+2 a b x+b^2 x^2}}{6 e^2 (a+b x)} \] Output:

-1/5*(-a*e+b*d)*(e*x+d)^5*((b*x+a)^2)^(1/2)/e^2/(b*x+a)+1/6*b*(e*x+d)^6*(( 
b*x+a)^2)^(1/2)/e^2/(b*x+a)
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.21 \[ \int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x \sqrt {(a+b x)^2} \left (6 a \left (5 d^4+10 d^3 e x+10 d^2 e^2 x^2+5 d e^3 x^3+e^4 x^4\right )+b x \left (15 d^4+40 d^3 e x+45 d^2 e^2 x^2+24 d e^3 x^3+5 e^4 x^4\right )\right )}{30 (a+b x)} \] Input:

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

Output:

(x*Sqrt[(a + b*x)^2]*(6*a*(5*d^4 + 10*d^3*e*x + 10*d^2*e^2*x^2 + 5*d*e^3*x 
^3 + e^4*x^4) + b*x*(15*d^4 + 40*d^3*e*x + 45*d^2*e^2*x^2 + 24*d*e^3*x^3 + 
 5*e^4*x^4)))/(30*(a + b*x))
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.72, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1102, 27, 49, 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.

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

\(\Big \downarrow \) 1102

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 49

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b (d+e x)^5}{e}+\frac {(a e-b d) (d+e x)^4}{e}\right )dx}{a+b x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b (d+e x)^6}{6 e^2}-\frac {(d+e x)^5 (b d-a e)}{5 e^2}\right )}{a+b x}\)

Input:

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

Output:

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-1/5*((b*d - a*e)*(d + e*x)^5)/e^2 + (b*(d 
 + e*x)^6)/(6*e^2)))/(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 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 1102

Int[((d_.) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*F 
racPart[p]))   Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, 
 d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0]
 

rule 2009

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

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.24

method	result	size
gosper	\(\frac {x \left (5 b \,e^{4} x^{5}+6 x^{4} e^{4} a +24 x^{4} b d \,e^{3}+30 x^{3} a d \,e^{3}+45 x^{3} b \,d^{2} e^{2}+60 x^{2} a \,d^{2} e^{2}+40 x^{2} b \,d^{3} e +60 x \,d^{3} e a +15 b \,d^{4} x +30 a \,d^{4}\right ) \sqrt {\left (b x +a \right )^{2}}}{30 b x +30 a}\)	\(114\)
orering	\(\frac {x \left (5 b \,e^{4} x^{5}+6 x^{4} e^{4} a +24 x^{4} b d \,e^{3}+30 x^{3} a d \,e^{3}+45 x^{3} b \,d^{2} e^{2}+60 x^{2} a \,d^{2} e^{2}+40 x^{2} b \,d^{3} e +60 x \,d^{3} e a +15 b \,d^{4} x +30 a \,d^{4}\right ) \sqrt {\left (b x +a \right )^{2}}}{30 b x +30 a}\)	\(114\)
default	\(\frac {\operatorname {csgn}\left (b x +a \right ) \left (b x +a \right )^{2} \left (5 b^{4} x^{4} e^{4}-4 x^{3} a \,b^{3} e^{4}+24 x^{3} b^{4} d \,e^{3}+3 x^{2} a^{2} b^{2} e^{4}-18 x^{2} a \,b^{3} d \,e^{3}+45 x^{2} b^{4} d^{2} e^{2}-2 x \,a^{3} b \,e^{4}+12 x \,a^{2} b^{2} d \,e^{3}-30 x a \,b^{3} d^{2} e^{2}+40 x \,b^{4} d^{3} e +a^{4} e^{4}-6 a^{3} b d \,e^{3}+15 a^{2} b^{2} d^{2} e^{2}-20 a \,b^{3} d^{3} e +15 b^{4} d^{4}\right )}{30 b^{5}}\)	\(191\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, b \,e^{4} x^{6}}{6 b x +6 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e^{4} a +4 b d \,e^{3}\right ) x^{5}}{5 b x +5 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (4 a d \,e^{3}+6 b \,d^{2} e^{2}\right ) x^{4}}{4 b x +4 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (6 a \,d^{2} e^{2}+4 b \,d^{3} e \right ) x^{3}}{3 b x +3 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (4 d^{3} e a +b \,d^{4}\right ) x^{2}}{2 b x +2 a}+\frac {\sqrt {\left (b x +a \right )^{2}}\, x a \,d^{4}}{b x +a}\)	\(193\)

Input:

int((e*x+d)^4*((b*x+a)^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/30*x*(5*b*e^4*x^5+6*a*e^4*x^4+24*b*d*e^3*x^4+30*a*d*e^3*x^3+45*b*d^2*e^2 
*x^3+60*a*d^2*e^2*x^2+40*b*d^3*e*x^2+60*a*d^3*e*x+15*b*d^4*x+30*a*d^4)*((b 
*x+a)^2)^(1/2)/(b*x+a)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.04 \[ \int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{6} \, b e^{4} x^{6} + a d^{4} x + \frac {1}{5} \, {\left (4 \, b d e^{3} + a e^{4}\right )} x^{5} + \frac {1}{2} \, {\left (3 \, b d^{2} e^{2} + 2 \, a d e^{3}\right )} x^{4} + \frac {2}{3} \, {\left (2 \, b d^{3} e + 3 \, a d^{2} e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (b d^{4} + 4 \, a d^{3} e\right )} x^{2} \] Input:

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

Output:

1/6*b*e^4*x^6 + a*d^4*x + 1/5*(4*b*d*e^3 + a*e^4)*x^5 + 1/2*(3*b*d^2*e^2 + 
 2*a*d*e^3)*x^4 + 2/3*(2*b*d^3*e + 3*a*d^2*e^2)*x^3 + 1/2*(b*d^4 + 4*a*d^3 
*e)*x^2
 

Sympy [A] (verification not implemented)

Time = 2.00 (sec) , antiderivative size = 666, normalized size of antiderivative = 7.24 \[ \int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx =\text {Too large to display} \] Input:

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

Output:

d**4*Piecewise(((a/(2*b) + x/2)*sqrt(a**2 + 2*a*b*x + b**2*x**2), Ne(b**2, 
 0)), ((a**2 + 2*a*b*x)**(3/2)/(3*a*b), Ne(a*b, 0)), (x*sqrt(a**2), True)) 
 + 4*d**3*e*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**2/(6*b**2) + 
a*x/(6*b) + x**2/3), Ne(b**2, 0)), ((-a**2*(a**2 + 2*a*b*x)**(3/2)/3 + (a* 
*2 + 2*a*b*x)**(5/2)/5)/(2*a**2*b**2), Ne(a*b, 0)), (x**2*sqrt(a**2)/2, Tr 
ue)) + 6*d**2*e**2*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(a**3/(12*b 
**3) - a**2*x/(12*b**2) + a*x**2/(12*b) + x**3/4), Ne(b**2, 0)), ((a**4*(a 
**2 + 2*a*b*x)**(3/2)/3 - 2*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b 
*x)**(7/2)/7)/(4*a**3*b**3), Ne(a*b, 0)), (x**3*sqrt(a**2)/3, True)) + 4*d 
*e**3*Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(-a**4/(20*b**4) + a**3* 
x/(20*b**3) - a**2*x**2/(20*b**2) + a*x**3/(20*b) + x**4/5), Ne(b**2, 0)), 
 ((-a**6*(a**2 + 2*a*b*x)**(3/2)/3 + 3*a**4*(a**2 + 2*a*b*x)**(5/2)/5 - 3* 
a**2*(a**2 + 2*a*b*x)**(7/2)/7 + (a**2 + 2*a*b*x)**(9/2)/9)/(8*a**4*b**4), 
 Ne(a*b, 0)), (x**4*sqrt(a**2)/4, True)) + e**4*Piecewise((sqrt(a**2 + 2*a 
*b*x + b**2*x**2)*(a**5/(30*b**5) - a**4*x/(30*b**4) + a**3*x**2/(30*b**3) 
 - a**2*x**3/(30*b**2) + a*x**4/(30*b) + x**5/6), Ne(b**2, 0)), ((a**8*(a* 
*2 + 2*a*b*x)**(3/2)/3 - 4*a**6*(a**2 + 2*a*b*x)**(5/2)/5 + 6*a**4*(a**2 + 
 2*a*b*x)**(7/2)/7 - 4*a**2*(a**2 + 2*a*b*x)**(9/2)/9 + (a**2 + 2*a*b*x)** 
(11/2)/11)/(16*a**5*b**5), Ne(a*b, 0)), (x**5*sqrt(a**2)/5, True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (66) = 132\).

Time = 0.04 (sec) , antiderivative size = 590, normalized size of antiderivative = 6.41 \[ \int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e^{4} x^{3}}{6 \, b^{2}} + \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d^{4} x - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{3} e x}{b} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d^{2} e^{2} x}{b^{2}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} d e^{3} x}{b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} e^{4} x}{2 \, b^{4}} + \frac {4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d e^{3} x^{2}}{5 \, b^{2}} - \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e^{4} x^{2}}{10 \, b^{3}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d^{4}}{2 \, b} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d^{3} e}{b^{2}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} d^{2} e^{2}}{b^{3}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4} d e^{3}}{b^{4}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{5} e^{4}}{2 \, b^{5}} + \frac {3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d^{2} e^{2} x}{2 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d e^{3} x}{5 \, b^{3}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e^{4} x}{5 \, b^{4}} + \frac {4 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d^{3} e}{3 \, b^{2}} - \frac {5 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d^{2} e^{2}}{2 \, b^{3}} + \frac {9 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} d e^{3}}{5 \, b^{4}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} e^{4}}{15 \, b^{5}} \] Input:

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

Output:

1/6*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*e^4*x^3/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b 
*x + a^2)*d^4*x - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*d^3*e*x/b + 3*sqrt(b^2 
*x^2 + 2*a*b*x + a^2)*a^2*d^2*e^2*x/b^2 - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2)* 
a^3*d*e^3*x/b^3 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^4*e^4*x/b^4 + 4/5*(b 
^2*x^2 + 2*a*b*x + a^2)^(3/2)*d*e^3*x^2/b^2 - 3/10*(b^2*x^2 + 2*a*b*x + a^ 
2)^(3/2)*a*e^4*x^2/b^3 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*d^4/b - 2*sqr 
t(b^2*x^2 + 2*a*b*x + a^2)*a^2*d^3*e/b^2 + 3*sqrt(b^2*x^2 + 2*a*b*x + a^2) 
*a^3*d^2*e^2/b^3 - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^4*d*e^3/b^4 + 1/2*sqr 
t(b^2*x^2 + 2*a*b*x + a^2)*a^5*e^4/b^5 + 3/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/ 
2)*d^2*e^2*x/b^2 - 7/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d*e^3*x/b^3 + 2/5 
*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*e^4*x/b^4 + 4/3*(b^2*x^2 + 2*a*b*x + 
a^2)^(3/2)*d^3*e/b^2 - 5/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d^2*e^2/b^3 + 
 9/5*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2*d*e^3/b^4 - 7/15*(b^2*x^2 + 2*a*b 
*x + a^2)^(3/2)*a^3*e^4/b^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (66) = 132\).

Time = 0.16 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.42 \[ \int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {1}{6} \, b e^{4} x^{6} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{5} \, b d e^{3} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{5} \, a e^{4} x^{5} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, b d^{2} e^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + a d e^{3} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {4}{3} \, b d^{3} e x^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, a d^{2} e^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b d^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, a d^{3} e x^{2} \mathrm {sgn}\left (b x + a\right ) + a d^{4} x \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (15 \, a^{2} b^{4} d^{4} - 20 \, a^{3} b^{3} d^{3} e + 15 \, a^{4} b^{2} d^{2} e^{2} - 6 \, a^{5} b d e^{3} + a^{6} e^{4}\right )} \mathrm {sgn}\left (b x + a\right )}{30 \, b^{5}} \] Input:

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

Output:

1/6*b*e^4*x^6*sgn(b*x + a) + 4/5*b*d*e^3*x^5*sgn(b*x + a) + 1/5*a*e^4*x^5* 
sgn(b*x + a) + 3/2*b*d^2*e^2*x^4*sgn(b*x + a) + a*d*e^3*x^4*sgn(b*x + a) + 
 4/3*b*d^3*e*x^3*sgn(b*x + a) + 2*a*d^2*e^2*x^3*sgn(b*x + a) + 1/2*b*d^4*x 
^2*sgn(b*x + a) + 2*a*d^3*e*x^2*sgn(b*x + a) + a*d^4*x*sgn(b*x + a) + 1/30 
*(15*a^2*b^4*d^4 - 20*a^3*b^3*d^3*e + 15*a^4*b^2*d^2*e^2 - 6*a^5*b*d*e^3 + 
 a^6*e^4)*sgn(b*x + a)/b^5
 

Mupad [B] (verification not implemented)

Time = 9.77 (sec) , antiderivative size = 580, normalized size of antiderivative = 6.30 \[ \int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=d^4\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}+\frac {e^4\,x^3\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{6\,b^2}-\frac {a^2\,e^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{24\,b^5}+\frac {3\,d^2\,e^2\,x\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{2\,b^2}+\frac {4\,d\,e^3\,x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{5\,b^2}-\frac {3\,a\,e^4\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (4\,b^2\,x^2\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-a^4+9\,a^2\,b^2\,x^2+8\,a^3\,b\,x-7\,a\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )\right )}{40\,b^5}+\frac {d^3\,e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{6\,b^4}-\frac {7\,a\,d\,e^3\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^3-5\,a\,b^2\,x^2+3\,b\,x\,\left (a^2+2\,a\,b\,x+b^2\,x^2\right )-4\,a^2\,b\,x\right )}{15\,b^4}-\frac {3\,a^2\,d^2\,e^2\,\left (\frac {x}{2}+\frac {a}{2\,b}\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{2\,b^2}-\frac {5\,a\,d^2\,e^2\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{16\,b^5}-\frac {a^2\,d\,e^3\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{15\,b^6} \] Input:

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

Output:

d^4*(x/2 + a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2) + (e^4*x^3*(a^2 + b^2* 
x^2 + 2*a*b*x)^(3/2))/(6*b^2) - (a^2*e^4*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*( 
a^3 - 5*a*b^2*x^2 + 3*b*x*(a^2 + b^2*x^2 + 2*a*b*x) - 4*a^2*b*x))/(24*b^5) 
 + (3*d^2*e^2*x*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(2*b^2) + (4*d*e^3*x^2*(a 
^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(5*b^2) - (3*a*e^4*(a^2 + b^2*x^2 + 2*a*b*x 
)^(1/2)*(4*b^2*x^2*(a^2 + b^2*x^2 + 2*a*b*x) - a^4 + 9*a^2*b^2*x^2 + 8*a^3 
*b*x - 7*a*b*x*(a^2 + b^2*x^2 + 2*a*b*x)))/(40*b^5) + (d^3*e*(8*b^2*(a^2 + 
 b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(6*b^ 
4) - (7*a*d*e^3*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(a^3 - 5*a*b^2*x^2 + 3*b*x 
*(a^2 + b^2*x^2 + 2*a*b*x) - 4*a^2*b*x))/(15*b^4) - (3*a^2*d^2*e^2*(x/2 + 
a/(2*b))*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(2*b^2) - (5*a*d^2*e^2*(8*b^2*(a 
^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/( 
16*b^5) - (a^2*d*e^3*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 
 + b^2*x^2 + 2*a*b*x)^(1/2))/(15*b^6)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.05 \[ \int (d+e x)^4 \sqrt {a^2+2 a b x+b^2 x^2} \, dx=\frac {x \left (5 b \,e^{4} x^{5}+6 a \,e^{4} x^{4}+24 b d \,e^{3} x^{4}+30 a d \,e^{3} x^{3}+45 b \,d^{2} e^{2} x^{3}+60 a \,d^{2} e^{2} x^{2}+40 b \,d^{3} e \,x^{2}+60 a \,d^{3} e x +15 b \,d^{4} x +30 a \,d^{4}\right )}{30} \] Input:

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

Output:

(x*(30*a*d**4 + 60*a*d**3*e*x + 60*a*d**2*e**2*x**2 + 30*a*d*e**3*x**3 + 6 
*a*e**4*x**4 + 15*b*d**4*x + 40*b*d**3*e*x**2 + 45*b*d**2*e**2*x**3 + 24*b 
*d*e**3*x**4 + 5*b*e**4*x**5))/30