\(\int \frac {(a+b x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [185]

Optimal result

Integrand size = 33, antiderivative size = 163 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {(b d-a e)^3}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (3 b d-2 a e) x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 x^2 (a+b x)}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (b d-a e)^2 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.74 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {2 a^3 e^3-2 a^2 b e^2 (3 d+2 e x)+3 a b^2 e \left (2 d^2+2 d e x-e^2 x^2\right )+b^3 \left (-2 d^3+6 d e^2 x^2+e^3 x^3\right )+6 e (b d-a e)^2 (a+b x) \log (a+b x)}{2 b^4 \sqrt {(a+b x)^2}} \] Input:

Integrate[((a + b*x)*(d + e*x)^3)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 

Output:

(2*a^3*e^3 - 2*a^2*b*e^2*(3*d + 2*e*x) + 3*a*b^2*e*(2*d^2 + 2*d*e*x - e^2* 
x^2) + b^3*(-2*d^3 + 6*d*e^2*x^2 + e^3*x^3) + 6*e*(b*d - a*e)^2*(a + b*x)* 
Log[a + b*x])/(2*b^4*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.62, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1187, 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.

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

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

Int[((a + b*x)*(d + e*x)^3)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
 

Output:

((a + b*x)*((e^2*(3*b*d - 2*a*e)*x)/b^3 + (e^3*x^2)/(2*b^2) - (b*d - a*e)^ 
3/(b^4*(a + b*x)) + (3*e*(b*d - a*e)^2*Log[a + b*x])/b^4))/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 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 = 1.42 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.91

method	result	size
risch	\(-\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{2} \left (-\frac {1}{2} b e \,x^{2}+2 a e x -3 b d x \right )}{\left (b x +a \right ) b^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}{\left (b x +a \right )^{2} b^{4}}+\frac {3 \sqrt {\left (b x +a \right )^{2}}\, e \left (e^{2} a^{2}-2 a b d e +b^{2} d^{2}\right ) \ln \left (b x +a \right )}{\left (b x +a \right ) b^{4}}\)	\(149\)
default	\(\frac {\left (e^{3} x^{3} b^{3}+6 \ln \left (b x +a \right ) x \,a^{2} b \,e^{3}-12 \ln \left (b x +a \right ) x a \,b^{2} d \,e^{2}+6 \ln \left (b x +a \right ) x \,b^{3} d^{2} e -3 x^{2} a \,b^{2} e^{3}+6 x^{2} b^{3} d \,e^{2}+6 \ln \left (b x +a \right ) a^{3} e^{3}-12 \ln \left (b x +a \right ) a^{2} b d \,e^{2}+6 \ln \left (b x +a \right ) a \,b^{2} d^{2} e -4 a^{2} b \,e^{3} x +6 x a \,b^{2} d \,e^{2}+2 e^{3} a^{3}-6 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{2 b^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(209\)

Input:

int((b*x+a)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {b^{3} e^{3} x^{3} - 2 \, b^{3} d^{3} + 6 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} + 2 \, a^{3} e^{3} + 3 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (3 \, a b^{2} d e^{2} - 2 \, a^{2} b e^{3}\right )} x + 6 \, {\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a b^{4}\right )}} \] Input:

integrate((b*x+a)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fric 
as")
 

Output:

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

Sympy [F]

\[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (a + b x\right ) \left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((b*x+a)*(e*x+d)**3/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
 

Output:

Integral((a + b*x)*(d + e*x)**3/((a + b*x)**2)**(3/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (117) = 234\).

Time = 0.03 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.95 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^{3} x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b} - \frac {5 \, a e^{3} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {6 \, a^{2} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {5 \, a^{3} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {12 \, a^{3} e^{3} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {a d^{3}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, a^{4} e^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {3 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {3 \, {\left (b d^{2} e + a d e^{2}\right )} \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {b d^{3} + 3 \, a d^{2} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {6 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {6 \, {\left (b d^{2} e + a d e^{2}\right )} a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, {\left (b d^{2} e + a d e^{2}\right )} a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \] Input:

integrate((b*x+a)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxi 
ma")
 

Output:

1/2*e^3*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b) - 5/2*a*e^3*x^2/(sqrt(b^2*x^ 
2 + 2*a*b*x + a^2)*b^2) + 6*a^2*e^3*log(x + a/b)/b^4 - 5*a^3*e^3/(sqrt(b^2 
*x^2 + 2*a*b*x + a^2)*b^4) + (3*b*d*e^2 + a*e^3)*x^2/(sqrt(b^2*x^2 + 2*a*b 
*x + a^2)*b^2) + 12*a^3*e^3*x/(b^5*(x + a/b)^2) - 1/2*a*d^3/(b^3*(x + a/b) 
^2) + 23/2*a^4*e^3/(b^6*(x + a/b)^2) - 3*(3*b*d*e^2 + a*e^3)*a*log(x + a/b 
)/b^4 + 3*(b*d^2*e + a*d*e^2)*log(x + a/b)/b^3 + 2*(3*b*d*e^2 + a*e^3)*a^2 
/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - (b*d^3 + 3*a*d^2*e)/(sqrt(b^2*x^2 + 
 2*a*b*x + a^2)*b^2) - 6*(3*b*d*e^2 + a*e^3)*a^2*x/(b^5*(x + a/b)^2) + 6*( 
b*d^2*e + a*d*e^2)*a*x/(b^4*(x + a/b)^2) - 11/2*(3*b*d*e^2 + a*e^3)*a^3/(b 
^6*(x + a/b)^2) + 9/2*(b*d^2*e + a*d*e^2)*a^2/(b^5*(x + a/b)^2) + 1/2*(b*d 
^3 + 3*a*d^2*e)*a/(b^4*(x + a/b)^2)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {b^{2} e^{3} x^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{2} d e^{2} x \mathrm {sgn}\left (b x + a\right ) - 4 \, a b e^{3} x \mathrm {sgn}\left (b x + a\right )}{2 \, b^{4}} + \frac {3 \, {\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}}{{\left (b x + a\right )} b^{4} \mathrm {sgn}\left (b x + a\right )} \] Input:

integrate((b*x+a)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac 
")
 

Output:

1/2*(b^2*e^3*x^2*sgn(b*x + a) + 6*b^2*d*e^2*x*sgn(b*x + a) - 4*a*b*e^3*x*s 
gn(b*x + a))/b^4 + 3*(b^2*d^2*e - 2*a*b*d*e^2 + a^2*e^3)*log(abs(b*x + a)) 
/(b^4*sgn(b*x + a)) - (b^3*d^3 - 3*a*b^2*d^2*e + 3*a^2*b*d*e^2 - a^3*e^3)/ 
((b*x + a)*b^4*sgn(b*x + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^3}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \] Input:

int(((a + b*x)*(d + e*x)^3)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
 

Output:

int(((a + b*x)*(d + e*x)^3)/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.21 \[ \int \frac {(a+b x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {6 \,\mathrm {log}\left (b x +a \right ) a^{4} e^{3}-12 \,\mathrm {log}\left (b x +a \right ) a^{3} b d \,e^{2}+6 \,\mathrm {log}\left (b x +a \right ) a^{3} b \,e^{3} x +6 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} d^{2} e -12 \,\mathrm {log}\left (b x +a \right ) a^{2} b^{2} d \,e^{2} x +6 \,\mathrm {log}\left (b x +a \right ) a \,b^{3} d^{2} e x -6 a^{3} b \,e^{3} x +12 a^{2} b^{2} d \,e^{2} x -3 a^{2} b^{2} e^{3} x^{2}-6 a \,b^{3} d^{2} e x +6 a \,b^{3} d \,e^{2} x^{2}+a \,b^{3} e^{3} x^{3}+2 b^{4} d^{3} x}{2 a \,b^{4} \left (b x +a \right )} \] Input:

int((b*x+a)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
 

Output:

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