\(\int \frac {1}{(d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}})^{5/2}} \, dx\) [18]

Optimal result

Integrand size = 30, antiderivative size = 335 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=-\frac {4 \left (d^2 e-b d f^2+a e f^2\right )}{3 \left (2 d e-b f^2\right )^2 \left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}}-\frac {4 f^2 \left (4 a e^2-b^2 f^2\right )}{\left (2 d e-b f^2\right )^3 \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}-\frac {2 e f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\left (2 d e-b f^2\right )^3 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac {5 \sqrt {2} \sqrt {e} f^2 \left (4 a e^2-b^2 f^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\sqrt {2 d e-b f^2}}\right )}{\left (2 d e-b f^2\right )^{7/2}} \] Output:

1/3*(-4*a*e*f^2+4*b*d*f^2-4*d^2*e)/(-b*f^2+2*d*e)^2/(d+e*x+f*(a+b*x+e^2*x^ 
2/f^2)^(1/2))^(3/2)-4*f^2*(-b^2*f^2+4*a*e^2)/(-b*f^2+2*d*e)^3/(d+e*x+f*(a+ 
b*x+e^2*x^2/f^2)^(1/2))^(1/2)-2*e*f^2*(-b^2*f^2+4*a*e^2)*(d+e*x+f*(a+b*x+e 
^2*x^2/f^2)^(1/2))^(1/2)/(-b*f^2+2*d*e)^3/(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^ 
2*x)/f^2)^(1/2)))+5*2^(1/2)*e^(1/2)*f^2*(-b^2*f^2+4*a*e^2)*arctanh(2^(1/2) 
*e^(1/2)*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(1/2)/(-b*f^2+2*d*e)^(1/2))/( 
-b*f^2+2*d*e)^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 4.78 (sec) , antiderivative size = 557, normalized size of antiderivative = 1.66 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\frac {2 b^3 f^6 \left (4 d+21 e x+6 f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+2 b^2 e f^4 \left (9 d^2+17 a f^2+14 d \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+30 e x \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )-8 b e^2 f^2 \left (d^3+7 a d f^2-3 d^2 \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+5 a f^2 \left (4 e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )-8 e^3 \left (15 a^2 f^4+2 d^3 \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+a f^2 \left (3 d^2+20 d \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+30 e x \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )\right )}{3 \left (2 d e-b f^2\right )^3 \left (d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )^{3/2} \left (b f^2+2 e \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}+\frac {20 \sqrt {2} a e^{5/2} f^2 \arctan \left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{\left (-2 d e+b f^2\right )^{7/2}}-\frac {5 \sqrt {2} b^2 \sqrt {e} f^4 \arctan \left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{\left (-2 d e+b f^2\right )^{7/2}} \] Input:

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

Output:

(2*b^3*f^6*(4*d + 21*e*x + 6*f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 2*b^2*e*f^ 
4*(9*d^2 + 17*a*f^2 + 14*d*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 30*e* 
x*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) - 8*b*e^2*f^2*(d^3 + 7*a*d*f^2 
- 3*d^2*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + 5*a*f^2*(4*e*x + f*Sqrt[ 
a + x*(b + (e^2*x)/f^2)])) - 8*e^3*(15*a^2*f^4 + 2*d^3*(e*x + f*Sqrt[a + x 
*(b + (e^2*x)/f^2)]) + a*f^2*(3*d^2 + 20*d*(e*x + f*Sqrt[a + x*(b + (e^2*x 
)/f^2)]) + 30*e*x*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))))/(3*(2*d*e - b 
*f^2)^3*(d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])^(3/2)*(b*f^2 + 2*e*(e* 
x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))) + (20*Sqrt[2]*a*e^(5/2)*f^2*ArcTan[ 
(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[-2* 
d*e + b*f^2]])/(-2*d*e + b*f^2)^(7/2) - (5*Sqrt[2]*b^2*Sqrt[e]*f^4*ArcTan[ 
(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[-2* 
d*e + b*f^2]])/(-2*d*e + b*f^2)^(7/2)
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.06, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2541, 1192, 1582, 27, 1584, 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.

\(\displaystyle \int \frac {1}{\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2541

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

\(\Big \downarrow \) 1192

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

\(\Big \downarrow \) 1582

\(\displaystyle 4 \left (\frac {\int \frac {4 \left (f^2 \left (4 a e^2-b^2 f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2 e^3+2 \left (2 d e-b f^2\right )^2 \left (e d^2-b f^2 d+a e f^2\right ) e^2-2 \left (2 d e-b f^2\right ) \left (b^2 f^4-2 a e^2 f^2-2 b d e f^2+2 d^2 e^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right ) e^2\right )}{\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2 \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}}{8 e^2 \left (2 d e-b f^2\right )^3}+\frac {e f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{2 \left (2 d e-b f^2\right )^3 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}\right )\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1584

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

4*((e*f^2*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^ 
2]])/(2*(2*d*e - b*f^2)^3*(2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + 
 (e^2*x^2)/f^2]))) + ((-2*e^2*(2*d*e - b*f^2)*(d^2*e - b*d*f^2 + a*e*f^2)) 
/(3*(d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2])^(3/2)) - (2*e^2*f^2*(4*a*e 
^2 - b^2*f^2))/Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]] + (5*e^(5/2 
)*f^2*(4*a*e^2 - b^2*f^2)*ArcTanh[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a 
 + b*x + (e^2*x^2)/f^2]])/Sqrt[2*d*e - b*f^2]])/(Sqrt[2]*Sqrt[2*d*e - b*f^ 
2]))/(2*e^2*(2*d*e - b*f^2)^3))
 

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 1192

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1)   Subst[Int[x^( 
2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + 
 c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && 
IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
 

rule 1582

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 
4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d 
+ e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ 
(2*p)*(q + 1))   Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e 
*x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - 
 b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre 
eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] 
&& ILtQ[m/2, 0]
 

rule 1584

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( 
c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* 
(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ 
b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
 

rule 2009

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

rule 2541

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Simp[2   Subst[Int[(g + h*x^n)^p*((d 
^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2*e*x 
)^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e 
, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
 

Maple [F]

\[\int \frac {1}{{\left (d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}^{\frac {5}{2}}}d x\]

Input:

int(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2),x)
 

Output:

int(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1105 vs. \(2 (298) = 596\).

Time = 1.71 (sec) , antiderivative size = 2396, normalized size of antiderivative = 7.15 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2),x, algorithm="fricas 
")
 

Output:

[-1/6*(15*sqrt(2)*(a^2*b^2*f^8 - 4*a*d^4*e^2*f^2 - 2*(a*b^2*d^2 + 2*a^3*e^ 
2)*f^6 + (b^2*d^4 + 8*a^2*d^2*e^2)*f^4 + (b^4*f^8 - 16*a*d^2*e^4*f^2 - 4*( 
b^3*d*e + a*b^2*e^2)*f^6 + 4*(b^2*d^2*e^2 + 4*a*b*d*e^3)*f^4)*x^2 + 2*(a*b 
^3*f^8 - 8*a*d^3*e^3*f^2 - (b^3*d^2 + 2*a*b^2*d*e + 4*a^2*b*e^2)*f^6 + 2*( 
b^2*d^3*e + 2*a*b*d^2*e^2 + 4*a^2*d*e^3)*f^4)*x)*sqrt(-e/(b*f^2 - 2*d*e))* 
log(-b^2*f^4 + 4*(b*d*e - a*e^2)*f^2 - 4*(b*e^2*f^2 - 2*d*e^3)*x - 2*(2*sq 
rt(2)*(b*e*f^3 - 2*d*e^2*f)*sqrt(-e/(b*f^2 - 2*d*e))*sqrt((b*f^2*x + e^2*x 
^2 + a*f^2)/f^2) - sqrt(2)*(b^2*f^4 - 2*b*d*e*f^2 + 2*(b*e^2*f^2 - 2*d*e^3 
)*x)*sqrt(-e/(b*f^2 - 2*d*e)))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^ 
2)/f^2) + d) + 4*(b*e*f^3 - 2*d*e^2*f)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^ 
2)) - 4*(4*d^5*e^2 + (8*a*b^2*d - 5*a^2*b*e)*f^6 - 2*(2*b^2*d^3 + a*b*d^2* 
e + 10*a^2*d*e^2)*f^4 - 6*(b^2*e^3*f^4 - 4*b*d*e^4*f^2 + 4*d^2*e^5)*x^3 - 
(9*b*d^4*e - 32*a*d^3*e^2)*f^2 + (3*b^3*e*f^6 - 16*d^3*e^4 + 4*(b^2*d*e^2 
- 10*a*b*e^3)*f^4 - 4*(3*b*d^2*e^3 - 20*a*d*e^4)*f^2)*x^2 + 2*(d^4*e^3 + ( 
4*b^3*d - a*b^2*e)*f^6 - (7*b^2*d^2*e + 6*a*b*d*e^2 + 15*a^2*e^3)*f^4 - 2* 
(5*b*d^3*e^2 - 23*a*d^2*e^3)*f^2)*x - 2*(3*a*b^2*f^7 + d^4*e^2*f - (b^2*d^ 
2 + 2*a*b*d*e + 15*a^2*e^2)*f^5 - 2*(3*b*d^3*e - 11*a*d^2*e^2)*f^3 - 3*(b^ 
2*e^2*f^5 - 4*b*d*e^3*f^3 + 4*d^2*e^4*f)*x^2 + (3*b^3*f^7 + 40*a*d*e^3*f^3 
 - 8*d^3*e^3*f - 4*(b^2*d*e + 5*a*b*e^2)*f^5)*x)*sqrt((b*f^2*x + e^2*x^2 + 
 a*f^2)/f^2))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/...
 

Sympy [F]

\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int \frac {1}{\left (d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/(d+e*x+f*(a+b*x+e**2*x**2/f**2)**(1/2))**(5/2),x)
 

Output:

Integral((d + e*x + f*sqrt(a + b*x + e**2*x**2/f**2))**(-5/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2),x, algorithm="maxima 
")
 

Output:

integrate((e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d)^(-5/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int { \frac {1}{{\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2),x, algorithm="giac")
 

Output:

integrate((e*x + sqrt(b*x + e^2*x^2/f^2 + a)*f + d)^(-5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^{5/2}} \,d x \] Input:

int(1/(d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^(5/2),x)
 

Output:

int(1/(d + e*x + f*(a + b*x + (e^2*x^2)/f^2)^(1/2))^(5/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{5/2}} \, dx=\int \frac {1}{{\left (d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}^{\frac {5}{2}}}d x \] Input:

int(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2),x)
 

Output:

int(1/(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^(5/2),x)