\(\int \frac {(a+b x) (d+e x)^m}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [265]

Optimal result

Integrand size = 33, antiderivative size = 78 \[ \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3 (a+b x) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^4 (1+m) \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^3 (a+b x) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (4,1+m,2+m,-\frac {b (d+e x)}{-b d+a e}\right )}{(-b d+a e)^4 (1+m) \sqrt {(a+b x)^2}} \] Input:

Integrate[((a + b*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(e^3*(a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[4, 1 + m, 2 + m, -((b*( 
d + e*x))/(-(b*d) + a*e))])/((-(b*d) + a*e)^4*(1 + m)*Sqrt[(a + b*x)^2])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {1187, 27, 78}

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.

\(\displaystyle \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1187

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 78

\(\displaystyle \frac {e^3 (a+b x) (d+e x)^{m+1} \operatorname {Hypergeometric2F1}\left (4,m+1,m+2,\frac {b (d+e x)}{b d-a e}\right )}{(m+1) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}\)

Input:

Int[((a + b*x)*(d + e*x)^m)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
 

Output:

(e^3*(a + b*x)*(d + e*x)^(1 + m)*Hypergeometric2F1[4, 1 + m, 2 + m, (b*(d 
+ e*x))/(b*d - a*e)])/((b*d - a*e)^4*(1 + m)*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 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

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]
 

Maple [F]

\[\int \frac {\left (b x +a \right ) \left (e x +d \right )^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {5}{2}}}d x\]

Input:

int((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Output:

int((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Fricas [F]

\[ \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fric 
as")
 

Output:

integral(sqrt(b^2*x^2 + 2*a*b*x + a^2)*(e*x + d)^m/(b^5*x^5 + 5*a*b^4*x^4 
+ 10*a^2*b^3*x^3 + 10*a^3*b^2*x^2 + 5*a^4*b*x + a^5), x)
 

Sympy [F(-2)]

Exception generated. \[ \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Exception raised: HeuristicGCDFailed} \] Input:

integrate((b*x+a)*(e*x+d)**m/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
 

Output:

Exception raised: HeuristicGCDFailed >> no luck
 

Maxima [F]

\[ \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxi 
ma")
 

Output:

integrate((b*x + a)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)
 

Giac [F]

\[ \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac 
")
 

Output:

integrate((b*x + a)*(e*x + d)^m/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(((a + b*x)*(d + e*x)^m)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
 

Output:

int(((a + b*x)*(d + e*x)^m)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(a+b x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {too large to display} \] Input:

int((b*x+a)*(e*x+d)^m/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
 

Output:

((d + e*x)**m*d + int(((d + e*x)**m*x)/(a**5*d*e*m + a**5*e**2*m*x - 3*a** 
4*b*d**2 + 4*a**4*b*d*e*m*x - 3*a**4*b*d*e*x + 4*a**4*b*e**2*m*x**2 - 12*a 
**3*b**2*d**2*x + 6*a**3*b**2*d*e*m*x**2 - 12*a**3*b**2*d*e*x**2 + 6*a**3* 
b**2*e**2*m*x**3 - 18*a**2*b**3*d**2*x**2 + 4*a**2*b**3*d*e*m*x**3 - 18*a* 
*2*b**3*d*e*x**3 + 4*a**2*b**3*e**2*m*x**4 - 12*a*b**4*d**2*x**3 + a*b**4* 
d*e*m*x**4 - 12*a*b**4*d*e*x**4 + a*b**4*e**2*m*x**5 - 3*b**5*d**2*x**4 - 
3*b**5*d*e*x**5),x)*a**5*e**3*m**2 - int(((d + e*x)**m*x)/(a**5*d*e*m + a* 
*5*e**2*m*x - 3*a**4*b*d**2 + 4*a**4*b*d*e*m*x - 3*a**4*b*d*e*x + 4*a**4*b 
*e**2*m*x**2 - 12*a**3*b**2*d**2*x + 6*a**3*b**2*d*e*m*x**2 - 12*a**3*b**2 
*d*e*x**2 + 6*a**3*b**2*e**2*m*x**3 - 18*a**2*b**3*d**2*x**2 + 4*a**2*b**3 
*d*e*m*x**3 - 18*a**2*b**3*d*e*x**3 + 4*a**2*b**3*e**2*m*x**4 - 12*a*b**4* 
d**2*x**3 + a*b**4*d*e*m*x**4 - 12*a*b**4*d*e*x**4 + a*b**4*e**2*m*x**5 - 
3*b**5*d**2*x**4 - 3*b**5*d*e*x**5),x)*a**4*b*d*e**2*m**2 - 3*int(((d + e* 
x)**m*x)/(a**5*d*e*m + a**5*e**2*m*x - 3*a**4*b*d**2 + 4*a**4*b*d*e*m*x - 
3*a**4*b*d*e*x + 4*a**4*b*e**2*m*x**2 - 12*a**3*b**2*d**2*x + 6*a**3*b**2* 
d*e*m*x**2 - 12*a**3*b**2*d*e*x**2 + 6*a**3*b**2*e**2*m*x**3 - 18*a**2*b** 
3*d**2*x**2 + 4*a**2*b**3*d*e*m*x**3 - 18*a**2*b**3*d*e*x**3 + 4*a**2*b**3 
*e**2*m*x**4 - 12*a*b**4*d**2*x**3 + a*b**4*d*e*m*x**4 - 12*a*b**4*d*e*x** 
4 + a*b**4*e**2*m*x**5 - 3*b**5*d**2*x**4 - 3*b**5*d*e*x**5),x)*a**4*b*d*e 
**2*m + 3*int(((d + e*x)**m*x)/(a**5*d*e*m + a**5*e**2*m*x - 3*a**4*b*d...