\(\int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx\) [309]

Optimal result

Integrand size = 28, antiderivative size = 222 \[ \int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {e (b d-a e)^3 x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(b d-a e)^2 (a+b x) (d+e x)^2}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(b d-a e) (a+b x) (d+e x)^3}{3 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^4}{4 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(b d-a e)^4 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.59 \[ \int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {(a+b x) \left (b e x \left (-12 a^3 e^3+6 a^2 b e^2 (8 d+e x)-4 a b^2 e \left (18 d^2+6 d e x+e^2 x^2\right )+b^3 \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )+12 (b d-a e)^4 \log (a+b x)\right )}{12 b^5 \sqrt {(a+b x)^2}} \] Input:

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

Output:

((a + b*x)*(b*e*x*(-12*a^3*e^3 + 6*a^2*b*e^2*(8*d + e*x) - 4*a*b^2*e*(18*d 
^2 + 6*d*e*x + e^2*x^2) + b^3*(48*d^3 + 36*d^2*e*x + 16*d*e^2*x^2 + 3*e^3* 
x^3)) + 12*(b*d - a*e)^4*Log[a + b*x]))/(12*b^5*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.55, 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.

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

\(\Big \downarrow \) 1102

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 49

\(\displaystyle \frac {(a+b x) \int \left (\frac {(b d-a e)^4}{b^4 (a+b x)}+\frac {e (b d-a e)^3}{b^4}+\frac {e (d+e x) (b d-a e)^2}{b^3}+\frac {e (d+e x)^2 (b d-a e)}{b^2}+\frac {e (d+e x)^3}{b}\right )dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {(a+b x) \left (\frac {(b d-a e)^4 \log (a+b x)}{b^5}+\frac {e x (b d-a e)^3}{b^4}+\frac {(d+e x)^2 (b d-a e)^2}{2 b^3}+\frac {(d+e x)^3 (b d-a e)}{3 b^2}+\frac {(d+e x)^4}{4 b}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

Input:

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

Output:

((a + b*x)*((e*(b*d - a*e)^3*x)/b^4 + ((b*d - a*e)^2*(d + e*x)^2)/(2*b^3) 
+ ((b*d - a*e)*(d + e*x)^3)/(3*b^2) + (d + e*x)^4/(4*b) + ((b*d - a*e)^4*L 
og[a + b*x])/b^5))/Sqrt[a^2 + 2*a*b*x + b^2*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 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 = 1.16 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.99

method	result	size
risch	\(-\frac {\sqrt {\left (b x +a \right )^{2}}\, e \left (-\frac {b^{3} x^{4} e^{3}}{4}+\frac {\left (\left (a e -2 b d \right ) b^{2} e^{2}-2 b^{3} e^{2} d \right ) x^{3}}{3}+\frac {\left (2 \left (a e -2 b d \right ) d e \,b^{2}-b e \left (e^{2} a^{2}-2 a b d e +2 b^{2} d^{2}\right )\right ) x^{2}}{2}+\left (a e -2 b d \right ) \left (e^{2} a^{2}-2 a b d e +2 b^{2} d^{2}\right ) x \right )}{\left (b x +a \right ) b^{4}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{5}}\)	\(220\)
default	\(\frac {\left (b x +a \right ) \left (3 b^{4} x^{4} e^{4}-4 x^{3} a \,b^{3} e^{4}+16 x^{3} b^{4} d \,e^{3}+6 x^{2} a^{2} b^{2} e^{4}-24 x^{2} a \,b^{3} d \,e^{3}+36 x^{2} b^{4} d^{2} e^{2}+12 \ln \left (b x +a \right ) a^{4} e^{4}-48 \ln \left (b x +a \right ) a^{3} b d \,e^{3}+72 \ln \left (b x +a \right ) a^{2} b^{2} d^{2} e^{2}-48 \ln \left (b x +a \right ) a \,b^{3} d^{3} e +12 \ln \left (b x +a \right ) b^{4} d^{4}-12 x \,a^{3} b \,e^{4}+48 x \,a^{2} b^{2} d \,e^{3}-72 x a \,b^{3} d^{2} e^{2}+48 x \,b^{4} d^{3} e \right )}{12 \sqrt {\left (b x +a \right )^{2}}\, b^{5}}\)	\(223\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.82 \[ \int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, b^{4} e^{4} x^{4} + 4 \, {\left (4 \, b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (6 \, b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + 12 \, {\left (4 \, b^{4} d^{3} e - 6 \, a b^{3} d^{2} e^{2} + 4 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x + 12 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \log \left (b x + a\right )}{12 \, b^{5}} \] Input:

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

Output:

1/12*(3*b^4*e^4*x^4 + 4*(4*b^4*d*e^3 - a*b^3*e^4)*x^3 + 6*(6*b^4*d^2*e^2 - 
 4*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + 12*(4*b^4*d^3*e - 6*a*b^3*d^2*e^2 + 4* 
a^2*b^2*d*e^3 - a^3*b*e^4)*x + 12*(b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2 
*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*log(b*x + a))/b^5
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 746 vs. \(2 (162) = 324\).

Time = 1.50 (sec) , antiderivative size = 746, normalized size of antiderivative = 3.36 \[ \int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((sqrt(a**2 + 2*a*b*x + b**2*x**2)*(e**4*x**3/(4*b**2) + x**2*(-7 
*a*e**4/(4*b) + 4*d*e**3)/(3*b**2) + x*(-3*a**2*e**4/(4*b**2) - 5*a*(-7*a* 
e**4/(4*b) + 4*d*e**3)/(3*b) + 6*d**2*e**2)/(2*b**2) + (-2*a**2*(-7*a*e**4 
/(4*b) + 4*d*e**3)/(3*b**2) - 3*a*(-3*a**2*e**4/(4*b**2) - 5*a*(-7*a*e**4/ 
(4*b) + 4*d*e**3)/(3*b) + 6*d**2*e**2)/(2*b) + 4*d**3*e)/b**2) + (a/b + x) 
*(-a**2*(-3*a**2*e**4/(4*b**2) - 5*a*(-7*a*e**4/(4*b) + 4*d*e**3)/(3*b) + 
6*d**2*e**2)/(2*b**2) - a*(-2*a**2*(-7*a*e**4/(4*b) + 4*d*e**3)/(3*b**2) - 
 3*a*(-3*a**2*e**4/(4*b**2) - 5*a*(-7*a*e**4/(4*b) + 4*d*e**3)/(3*b) + 6*d 
**2*e**2)/(2*b) + 4*d**3*e)/b + d**4)*log(a/b + x)/sqrt(b**2*(a/b + x)**2) 
, Ne(b**2, 0)), ((2*d**4*sqrt(a**2 + 2*a*b*x) + 4*d**3*e*(-a**2*sqrt(a**2 
+ 2*a*b*x) + (a**2 + 2*a*b*x)**(3/2)/3)/(a*b) + 3*d**2*e**2*(a**4*sqrt(a** 
2 + 2*a*b*x) - 2*a**2*(a**2 + 2*a*b*x)**(3/2)/3 + (a**2 + 2*a*b*x)**(5/2)/ 
5)/(a**2*b**2) + d*e**3*(-a**6*sqrt(a**2 + 2*a*b*x) + a**4*(a**2 + 2*a*b*x 
)**(3/2) - 3*a**2*(a**2 + 2*a*b*x)**(5/2)/5 + (a**2 + 2*a*b*x)**(7/2)/7)/( 
a**3*b**3) + e**4*(a**8*sqrt(a**2 + 2*a*b*x) - 4*a**6*(a**2 + 2*a*b*x)**(3 
/2)/3 + 6*a**4*(a**2 + 2*a*b*x)**(5/2)/5 - 4*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))/(2*a*b), Ne(a*b, 0)), ((d**4 
*x + 2*d**3*e*x**2 + 2*d**2*e**2*x**3 + d*e**3*x**4 + e**4*x**5/5)/sqrt(a* 
*2), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (161) = 322\).

Time = 0.04 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.57 \[ \int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} e^{4} x^{3}}{4 \, b^{2}} + \frac {3 \, d^{2} e^{2} x^{2}}{b} - \frac {10 \, a d e^{3} x^{2}}{3 \, b^{2}} + \frac {13 \, a^{2} e^{4} x^{2}}{12 \, b^{3}} + \frac {4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d e^{3} x^{2}}{3 \, b^{2}} - \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a e^{4} x^{2}}{12 \, b^{3}} - \frac {6 \, a d^{2} e^{2} x}{b^{2}} + \frac {20 \, a^{2} d e^{3} x}{3 \, b^{3}} - \frac {13 \, a^{3} e^{4} x}{6 \, b^{4}} + \frac {d^{4} \log \left (x + \frac {a}{b}\right )}{b} - \frac {4 \, a d^{3} e \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {6 \, a^{2} d^{2} e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {4 \, a^{3} d e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {a^{4} e^{4} \log \left (x + \frac {a}{b}\right )}{b^{5}} + \frac {4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d^{3} e}{b^{2}} - \frac {8 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} d e^{3}}{3 \, b^{4}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3} e^{4}}{6 \, b^{5}} \] Input:

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

Output:

1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*e^4*x^3/b^2 + 3*d^2*e^2*x^2/b - 10/3*a*d 
*e^3*x^2/b^2 + 13/12*a^2*e^4*x^2/b^3 + 4/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*d 
*e^3*x^2/b^2 - 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a*e^4*x^2/b^3 - 6*a*d^2* 
e^2*x/b^2 + 20/3*a^2*d*e^3*x/b^3 - 13/6*a^3*e^4*x/b^4 + d^4*log(x + a/b)/b 
 - 4*a*d^3*e*log(x + a/b)/b^2 + 6*a^2*d^2*e^2*log(x + a/b)/b^3 - 4*a^3*d*e 
^3*log(x + a/b)/b^4 + a^4*e^4*log(x + a/b)/b^5 + 4*sqrt(b^2*x^2 + 2*a*b*x 
+ a^2)*d^3*e/b^2 - 8/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*d*e^3/b^4 + 7/6*s 
qrt(b^2*x^2 + 2*a*b*x + a^2)*a^3*e^4/b^5
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.23 \[ \int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {3 \, b^{3} e^{4} x^{4} \mathrm {sgn}\left (b x + a\right ) + 16 \, b^{3} d e^{3} x^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{2} e^{4} x^{3} \mathrm {sgn}\left (b x + a\right ) + 36 \, b^{3} d^{2} e^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) - 24 \, a b^{2} d e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b e^{4} x^{2} \mathrm {sgn}\left (b x + a\right ) + 48 \, b^{3} d^{3} e x \mathrm {sgn}\left (b x + a\right ) - 72 \, a b^{2} d^{2} e^{2} x \mathrm {sgn}\left (b x + a\right ) + 48 \, a^{2} b d e^{3} x \mathrm {sgn}\left (b x + a\right ) - 12 \, a^{3} e^{4} x \mathrm {sgn}\left (b x + a\right )}{12 \, b^{4}} + \frac {{\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{5}} \] Input:

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

Output:

1/12*(3*b^3*e^4*x^4*sgn(b*x + a) + 16*b^3*d*e^3*x^3*sgn(b*x + a) - 4*a*b^2 
*e^4*x^3*sgn(b*x + a) + 36*b^3*d^2*e^2*x^2*sgn(b*x + a) - 24*a*b^2*d*e^3*x 
^2*sgn(b*x + a) + 6*a^2*b*e^4*x^2*sgn(b*x + a) + 48*b^3*d^3*e*x*sgn(b*x + 
a) - 72*a*b^2*d^2*e^2*x*sgn(b*x + a) + 48*a^2*b*d*e^3*x*sgn(b*x + a) - 12* 
a^3*e^4*x*sgn(b*x + a))/b^4 + (b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b* 
x + a) + 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 4*a^3*b*d*e^3*sgn(b*x + a) + a^4 
*e^4*sgn(b*x + a))*log(abs(b*x + a))/b^5
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^4}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx=\frac {12 \,\mathrm {log}\left (b x +a \right ) a^{4} e^{4}-48 \,\mathrm {log}\left (b x +a \right ) a^{3} b d \,e^{3}+72 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} d^{2} e^{2}-48 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} d^{3} e +12 \,\mathrm {log}\left (b x +a \right ) b^{4} d^{4}-12 a^{3} b \,e^{4} x +48 a^{2} b^{2} d \,e^{3} x +6 a^{2} b^{2} e^{4} x^{2}-72 a \,b^{3} d^{2} e^{2} x -24 a \,b^{3} d \,e^{3} x^{2}-4 a \,b^{3} e^{4} x^{3}+48 b^{4} d^{3} e x +36 b^{4} d^{2} e^{2} x^{2}+16 b^{4} d \,e^{3} x^{3}+3 b^{4} e^{4} x^{4}}{12 b^{5}} \] Input:

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

Output:

(12*log(a + b*x)*a**4*e**4 - 48*log(a + b*x)*a**3*b*d*e**3 + 72*log(a + b* 
x)*a**2*b**2*d**2*e**2 - 48*log(a + b*x)*a*b**3*d**3*e + 12*log(a + b*x)*b 
**4*d**4 - 12*a**3*b*e**4*x + 48*a**2*b**2*d*e**3*x + 6*a**2*b**2*e**4*x** 
2 - 72*a*b**3*d**2*e**2*x - 24*a*b**3*d*e**3*x**2 - 4*a*b**3*e**4*x**3 + 4 
8*b**4*d**3*e*x + 36*b**4*d**2*e**2*x**2 + 16*b**4*d*e**3*x**3 + 3*b**4*e* 
*4*x**4)/(12*b**5)