\(\int \frac {1}{\sqrt {d+e x} (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [395]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 30, antiderivative size = 278 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {35 e^3 \sqrt {d+e x}}{64 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{4 (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {7 e \sqrt {d+e x}}{24 (b d-a e)^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^2 \sqrt {d+e x}}{96 (b d-a e)^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 e^4 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 \sqrt {b} (b d-a e)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \] Output:

35/64*e^3*(e*x+d)^(1/2)/(-a*e+b*d)^4/((b*x+a)^2)^(1/2)-1/4*(e*x+d)^(1/2)/( 
-a*e+b*d)/(b*x+a)^3/((b*x+a)^2)^(1/2)+7/24*e*(e*x+d)^(1/2)/(-a*e+b*d)^2/(b 
*x+a)^2/((b*x+a)^2)^(1/2)-35/96*e^2*(e*x+d)^(1/2)/(-a*e+b*d)^3/(b*x+a)/((b 
*x+a)^2)^(1/2)-35/64*e^4*(b*x+a)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^ 
(1/2))/b^(1/2)/(-a*e+b*d)^(9/2)/((b*x+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 0.57 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.67 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {e^4 (a+b x)^5 \left (\frac {\sqrt {d+e x} \left (279 a^3 e^3+a^2 b e^2 (-326 d+511 e x)+a b^2 e \left (200 d^2-252 d e x+385 e^2 x^2\right )+b^3 \left (-48 d^3+56 d^2 e x-70 d e^2 x^2+105 e^3 x^3\right )\right )}{e^4 (b d-a e)^4 (a+b x)^4}+\frac {105 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {b} (-b d+a e)^{9/2}}\right )}{192 \left ((a+b x)^2\right )^{5/2}} \] Input:

Integrate[1/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
 

Output:

(e^4*(a + b*x)^5*((Sqrt[d + e*x]*(279*a^3*e^3 + a^2*b*e^2*(-326*d + 511*e* 
x) + a*b^2*e*(200*d^2 - 252*d*e*x + 385*e^2*x^2) + b^3*(-48*d^3 + 56*d^2*e 
*x - 70*d*e^2*x^2 + 105*e^3*x^3)))/(e^4*(b*d - a*e)^4*(a + b*x)^4) + (105* 
ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(Sqrt[b]*(-(b*d) + a*e 
)^(9/2))))/(192*((a + b*x)^2)^(5/2))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.86, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1102, 27, 52, 52, 52, 52, 73, 221}

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.

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

\(\Big \downarrow \) 1102

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 52

\(\displaystyle \frac {(a+b x) \left (-\frac {7 e \int \frac {1}{(a+b x)^4 \sqrt {d+e x}}dx}{8 (b d-a e)}-\frac {\sqrt {d+e x}}{4 (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (-\frac {5 e \int \frac {1}{(a+b x)^3 \sqrt {d+e x}}dx}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{8 (b d-a e)}-\frac {\sqrt {d+e x}}{4 (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (-\frac {5 e \left (-\frac {3 e \int \frac {1}{(a+b x)^2 \sqrt {d+e x}}dx}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{8 (b d-a e)}-\frac {\sqrt {d+e x}}{4 (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (-\frac {5 e \left (-\frac {3 e \left (-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}}dx}{2 (b d-a e)}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{8 (b d-a e)}-\frac {\sqrt {d+e x}}{4 (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (-\frac {5 e \left (-\frac {3 e \left (-\frac {\int \frac {1}{a+\frac {b (d+e x)}{e}-\frac {b d}{e}}d\sqrt {d+e x}}{b d-a e}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{8 (b d-a e)}-\frac {\sqrt {d+e x}}{4 (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(a+b x) \left (-\frac {7 e \left (-\frac {5 e \left (-\frac {3 e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)}\right )}{4 (b d-a e)}-\frac {\sqrt {d+e x}}{2 (a+b x)^2 (b d-a e)}\right )}{6 (b d-a e)}-\frac {\sqrt {d+e x}}{3 (a+b x)^3 (b d-a e)}\right )}{8 (b d-a e)}-\frac {\sqrt {d+e x}}{4 (a+b x)^4 (b d-a e)}\right )}{\sqrt {a^2+2 a b x+b^2 x^2}}\)

Input:

Int[1/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]
 

Output:

((a + b*x)*(-1/4*Sqrt[d + e*x]/((b*d - a*e)*(a + b*x)^4) - (7*e*(-1/3*Sqrt 
[d + e*x]/((b*d - a*e)*(a + b*x)^3) - (5*e*(-1/2*Sqrt[d + e*x]/((b*d - a*e 
)*(a + b*x)^2) - (3*e*(-(Sqrt[d + e*x]/((b*d - a*e)*(a + b*x))) + (e*ArcTa 
nh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(Sqrt[b]*(b*d - a*e)^(3/2)))) 
/(4*(b*d - a*e))))/(6*(b*d - a*e))))/(8*(b*d - a*e))))/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 52
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(496\) vs. \(2(195)=390\).

Time = 1.52 (sec) , antiderivative size = 497, normalized size of antiderivative = 1.79

method result size
default \(\frac {\left (105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{4} e^{4} x^{4}+420 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a \,b^{3} e^{4} x^{3}+105 b^{3} e^{3} x^{3} \sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}+630 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} b^{2} e^{4} x^{2}+385 a \,b^{2} e^{3} x^{2} \sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}-70 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b^{3} d \,e^{2} x^{2}+420 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{3} b \,e^{4} x +511 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a^{2} b \,e^{3} x -252 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a \,b^{2} d \,e^{2} x +56 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b^{3} d^{2} e x +105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{4} e^{4}+279 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a^{3} e^{3}-326 \sqrt {e x +d}\, a^{2} d b \,e^{2} \sqrt {b \left (a e -b d \right )}+200 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, a \,b^{2} d^{2} e -48 \sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b^{3} d^{3}\right ) \left (b x +a \right )}{192 \sqrt {b \left (a e -b d \right )}\, \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) \(497\)

Input:

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

Output:

1/192*(105*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*b^4*e^4*x^4+420*arc 
tan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))*a*b^3*e^4*x^3+105*b^3*e^3*x^3*(b* 
(a*e-b*d))^(1/2)*(e*x+d)^(1/2)+630*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1 
/2))*a^2*b^2*e^4*x^2+385*a*b^2*e^3*x^2*(b*(a*e-b*d))^(1/2)*(e*x+d)^(1/2)-7 
0*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*b^3*d*e^2*x^2+420*arctan(b*(e*x+d)^(1/ 
2)/(b*(a*e-b*d))^(1/2))*a^3*b*e^4*x+511*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)* 
a^2*b*e^3*x-252*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*a*b^2*d*e^2*x+56*(e*x+d) 
^(1/2)*(b*(a*e-b*d))^(1/2)*b^3*d^2*e*x+105*arctan(b*(e*x+d)^(1/2)/(b*(a*e- 
b*d))^(1/2))*a^4*e^4+279*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*a^3*e^3-326*(e* 
x+d)^(1/2)*a^2*d*b*e^2*(b*(a*e-b*d))^(1/2)+200*(e*x+d)^(1/2)*(b*(a*e-b*d)) 
^(1/2)*a*b^2*d^2*e-48*(e*x+d)^(1/2)*(b*(a*e-b*d))^(1/2)*b^3*d^3)*(b*x+a)/( 
b*(a*e-b*d))^(1/2)/(a*e-b*d)^4/((b*x+a)^2)^(5/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 656 vs. \(2 (195) = 390\).

Time = 0.14 (sec) , antiderivative size = 1325, normalized size of antiderivative = 4.77 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/384*(105*(b^4*e^4*x^4 + 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e 
^4*x + a^4*e^4)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2* 
d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(48*b^5*d^4 - 248*a*b^4*d^3*e + 5 
26*a^2*b^3*d^2*e^2 - 605*a^3*b^2*d*e^3 + 279*a^4*b*e^4 - 105*(b^5*d*e^3 - 
a*b^4*e^4)*x^3 + 35*(2*b^5*d^2*e^2 - 13*a*b^4*d*e^3 + 11*a^2*b^3*e^4)*x^2 
- 7*(8*b^5*d^3*e - 44*a*b^4*d^2*e^2 + 109*a^2*b^3*d*e^3 - 73*a^3*b^2*e^4)* 
x)*sqrt(e*x + d))/(a^4*b^6*d^5 - 5*a^5*b^5*d^4*e + 10*a^6*b^4*d^3*e^2 - 10 
*a^7*b^3*d^2*e^3 + 5*a^8*b^2*d*e^4 - a^9*b*e^5 + (b^10*d^5 - 5*a*b^9*d^4*e 
 + 10*a^2*b^8*d^3*e^2 - 10*a^3*b^7*d^2*e^3 + 5*a^4*b^6*d*e^4 - a^5*b^5*e^5 
)*x^4 + 4*(a*b^9*d^5 - 5*a^2*b^8*d^4*e + 10*a^3*b^7*d^3*e^2 - 10*a^4*b^6*d 
^2*e^3 + 5*a^5*b^5*d*e^4 - a^6*b^4*e^5)*x^3 + 6*(a^2*b^8*d^5 - 5*a^3*b^7*d 
^4*e + 10*a^4*b^6*d^3*e^2 - 10*a^5*b^5*d^2*e^3 + 5*a^6*b^4*d*e^4 - a^7*b^3 
*e^5)*x^2 + 4*(a^3*b^7*d^5 - 5*a^4*b^6*d^4*e + 10*a^5*b^5*d^3*e^2 - 10*a^6 
*b^4*d^2*e^3 + 5*a^7*b^3*d*e^4 - a^8*b^2*e^5)*x), 1/192*(105*(b^4*e^4*x^4 
+ 4*a*b^3*e^4*x^3 + 6*a^2*b^2*e^4*x^2 + 4*a^3*b*e^4*x + a^4*e^4)*sqrt(-b^2 
*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (48 
*b^5*d^4 - 248*a*b^4*d^3*e + 526*a^2*b^3*d^2*e^2 - 605*a^3*b^2*d*e^3 + 279 
*a^4*b*e^4 - 105*(b^5*d*e^3 - a*b^4*e^4)*x^3 + 35*(2*b^5*d^2*e^2 - 13*a*b^ 
4*d*e^3 + 11*a^2*b^3*e^4)*x^2 - 7*(8*b^5*d^3*e - 44*a*b^4*d^2*e^2 + 109*a^ 
2*b^3*d*e^3 - 73*a^3*b^2*e^4)*x)*sqrt(e*x + d))/(a^4*b^6*d^5 - 5*a^5*b^...
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt(d + e*x)*((a + b*x)**2)**(5/2)), x)
 

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (195) = 390\).

Time = 0.14 (sec) , antiderivative size = 391, normalized size of antiderivative = 1.41 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {35 \, e^{4} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{64 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} + \frac {105 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{3} e^{4} - 385 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{3} d e^{4} + 511 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} d^{2} e^{4} - 279 \, \sqrt {e x + d} b^{3} d^{3} e^{4} + 385 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{2} e^{5} - 1022 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{2} d e^{5} + 837 \, \sqrt {e x + d} a b^{2} d^{2} e^{5} + 511 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b e^{6} - 837 \, \sqrt {e x + d} a^{2} b d e^{6} + 279 \, \sqrt {e x + d} a^{3} e^{7}}{192 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{4}} \] Input:

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

Output:

35/64*e^4*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((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))*sqrt(-b^2*d + a*b*e)) + 1/192 
*(105*(e*x + d)^(7/2)*b^3*e^4 - 385*(e*x + d)^(5/2)*b^3*d*e^4 + 511*(e*x + 
 d)^(3/2)*b^3*d^2*e^4 - 279*sqrt(e*x + d)*b^3*d^3*e^4 + 385*(e*x + d)^(5/2 
)*a*b^2*e^5 - 1022*(e*x + d)^(3/2)*a*b^2*d*e^5 + 837*sqrt(e*x + d)*a*b^2*d 
^2*e^5 + 511*(e*x + d)^(3/2)*a^2*b*e^6 - 837*sqrt(e*x + d)*a^2*b*d*e^6 + 2 
79*sqrt(e*x + d)*a^3*e^7)/((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))*((e*x + d)*b - b*d + a*e)^4)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 903, normalized size of antiderivative = 3.25 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(105*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b* 
d)))*a**4*e**4 + 420*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt( 
b)*sqrt(a*e - b*d)))*a**3*b*e**4*x + 630*sqrt(b)*sqrt(a*e - b*d)*atan((sqr 
t(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a**2*b**2*e**4*x**2 + 420*sqrt(b) 
*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sqrt(a*e - b*d)))*a*b**3* 
e**4*x**3 + 105*sqrt(b)*sqrt(a*e - b*d)*atan((sqrt(d + e*x)*b)/(sqrt(b)*sq 
rt(a*e - b*d)))*b**4*e**4*x**4 + 279*sqrt(d + e*x)*a**4*b*e**4 - 605*sqrt( 
d + e*x)*a**3*b**2*d*e**3 + 511*sqrt(d + e*x)*a**3*b**2*e**4*x + 526*sqrt( 
d + e*x)*a**2*b**3*d**2*e**2 - 763*sqrt(d + e*x)*a**2*b**3*d*e**3*x + 385* 
sqrt(d + e*x)*a**2*b**3*e**4*x**2 - 248*sqrt(d + e*x)*a*b**4*d**3*e + 308* 
sqrt(d + e*x)*a*b**4*d**2*e**2*x - 455*sqrt(d + e*x)*a*b**4*d*e**3*x**2 + 
105*sqrt(d + e*x)*a*b**4*e**4*x**3 + 48*sqrt(d + e*x)*b**5*d**4 - 56*sqrt( 
d + e*x)*b**5*d**3*e*x + 70*sqrt(d + e*x)*b**5*d**2*e**2*x**2 - 105*sqrt(d 
 + e*x)*b**5*d*e**3*x**3)/(192*b*(a**9*e**5 - 5*a**8*b*d*e**4 + 4*a**8*b*e 
**5*x + 10*a**7*b**2*d**2*e**3 - 20*a**7*b**2*d*e**4*x + 6*a**7*b**2*e**5* 
x**2 - 10*a**6*b**3*d**3*e**2 + 40*a**6*b**3*d**2*e**3*x - 30*a**6*b**3*d* 
e**4*x**2 + 4*a**6*b**3*e**5*x**3 + 5*a**5*b**4*d**4*e - 40*a**5*b**4*d**3 
*e**2*x + 60*a**5*b**4*d**2*e**3*x**2 - 20*a**5*b**4*d*e**4*x**3 + a**5*b* 
*4*e**5*x**4 - a**4*b**5*d**5 + 20*a**4*b**5*d**4*e*x - 60*a**4*b**5*d**3* 
e**2*x**2 + 40*a**4*b**5*d**2*e**3*x**3 - 5*a**4*b**5*d*e**4*x**4 - 4*a...