\(\int \frac {1}{(d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1718]

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

Optimal result

Integrand size = 30, antiderivative size = 275 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 b^{3/2} e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {660, 44, 53, 65, 214} \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {35 b^{3/2} e^2 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{9/2}}+\frac {35 b e^2 (a+b x)}{4 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^4}+\frac {35 e^2 (a+b x)}{12 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^3}+\frac {7 e}{4 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)} \]

[In]

Int[1/((d + e*x)^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(7*e)/(4*(b*d - a*e)^2*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(2*(b*d - a*e)*(a + b*x)*(d + e*x)^(
3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (35*e^2*(a + b*x))/(12*(b*d - a*e)^3*(d + e*x)^(3/2)*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) + (35*b*e^2*(a + b*x))/(4*(b*d - a*e)^4*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (35*b^(3/2)
*e^2*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*(b*d - a*e)^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])

Rule 44

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] - Dist[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] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 53

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] - Dist[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] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[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] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^3 (d+e x)^{5/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (7 b e \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 (d+e x)^{5/2}} \, dx}{4 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{8 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 b e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{8 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 b^2 e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{8 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (35 b^2 e \left (a b+b^2 x\right )\right ) \text {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {7 e}{4 (b d-a e)^2 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{2 (b d-a e) (a+b x) (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 e^2 (a+b x)}{12 (b d-a e)^3 (d+e x)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {35 b e^2 (a+b x)}{4 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {35 b^{3/2} e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 (b d-a e)^{9/2} \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^2 (a+b x)^3 \left (\frac {-8 a^3 e^3+8 a^2 b e^2 (10 d+7 e x)+a b^2 e \left (39 d^2+238 d e x+175 e^2 x^2\right )+b^3 \left (-6 d^3+21 d^2 e x+140 d e^2 x^2+105 e^3 x^3\right )}{e^2 (b d-a e)^4 (a+b x)^2 (d+e x)^{3/2}}+\frac {105 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{9/2}}\right )}{12 \left ((a+b x)^2\right )^{3/2}} \]

[In]

Integrate[1/((d + e*x)^(5/2)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(e^2*(a + b*x)^3*((-8*a^3*e^3 + 8*a^2*b*e^2*(10*d + 7*e*x) + a*b^2*e*(39*d^2 + 238*d*e*x + 175*e^2*x^2) + b^3*
(-6*d^3 + 21*d^2*e*x + 140*d*e^2*x^2 + 105*e^3*x^3))/(e^2*(b*d - a*e)^4*(a + b*x)^2*(d + e*x)^(3/2)) + (105*b^
(3/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(9/2)))/(12*((a + b*x)^2)^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(387\) vs. \(2(192)=384\).

Time = 2.30 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.41

method result size
default \(-\frac {\left (-105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (e x +d \right )^{\frac {3}{2}} b^{4} e^{2} x^{2}-210 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (e x +d \right )^{\frac {3}{2}} a \,b^{3} e^{2} x -105 \sqrt {\left (a e -b d \right ) b}\, b^{3} e^{3} x^{3}-105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) \left (e x +d \right )^{\frac {3}{2}} a^{2} b^{2} e^{2}-175 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} e^{3} x^{2}-140 \sqrt {\left (a e -b d \right ) b}\, b^{3} d \,e^{2} x^{2}-56 \sqrt {\left (a e -b d \right ) b}\, a^{2} b \,e^{3} x -238 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d \,e^{2} x -21 \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{2} e x +8 \sqrt {\left (a e -b d \right ) b}\, a^{3} e^{3}-80 \sqrt {\left (a e -b d \right ) b}\, a^{2} b d \,e^{2}-39 \sqrt {\left (a e -b d \right ) b}\, a \,b^{2} d^{2} e +6 \sqrt {\left (a e -b d \right ) b}\, b^{3} d^{3}\right ) \left (b x +a \right )}{12 \sqrt {\left (a e -b d \right ) b}\, \left (e x +d \right )^{\frac {3}{2}} \left (a e -b d \right )^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\) \(388\)

[In]

int(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(-105*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*(e*x+d)^(3/2)*b^4*e^2*x^2-210*arctan(b*(e*x+d)^(1/2)/(
(a*e-b*d)*b)^(1/2))*(e*x+d)^(3/2)*a*b^3*e^2*x-105*((a*e-b*d)*b)^(1/2)*b^3*e^3*x^3-105*arctan(b*(e*x+d)^(1/2)/(
(a*e-b*d)*b)^(1/2))*(e*x+d)^(3/2)*a^2*b^2*e^2-175*((a*e-b*d)*b)^(1/2)*a*b^2*e^3*x^2-140*((a*e-b*d)*b)^(1/2)*b^
3*d*e^2*x^2-56*((a*e-b*d)*b)^(1/2)*a^2*b*e^3*x-238*((a*e-b*d)*b)^(1/2)*a*b^2*d*e^2*x-21*((a*e-b*d)*b)^(1/2)*b^
3*d^2*e*x+8*((a*e-b*d)*b)^(1/2)*a^3*e^3-80*((a*e-b*d)*b)^(1/2)*a^2*b*d*e^2-39*((a*e-b*d)*b)^(1/2)*a*b^2*d^2*e+
6*((a*e-b*d)*b)^(1/2)*b^3*d^3)*(b*x+a)/((a*e-b*d)*b)^(1/2)/(e*x+d)^(3/2)/(a*e-b*d)^4/((b*x+a)^2)^(3/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 608 vs. \(2 (192) = 384\).

Time = 0.47 (sec) , antiderivative size = 1226, normalized size of antiderivative = 4.46 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

[1/24*(105*(b^3*e^4*x^4 + a^2*b*d^2*e^2 + 2*(b^3*d*e^3 + a*b^2*e^4)*x^3 + (b^3*d^2*e^2 + 4*a*b^2*d*e^3 + a^2*b
*e^4)*x^2 + 2*(a*b^2*d^2*e^2 + a^2*b*d*e^3)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 2*(b*d - a*e)*sq
rt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) + 2*(105*b^3*e^3*x^3 - 6*b^3*d^3 + 39*a*b^2*d^2*e + 80*a^2*b*d*e^2
 - 8*a^3*e^3 + 35*(4*b^3*d*e^2 + 5*a*b^2*e^3)*x^2 + 7*(3*b^3*d^2*e + 34*a*b^2*d*e^2 + 8*a^2*b*e^3)*x)*sqrt(e*x
 + d))/(a^2*b^4*d^6 - 4*a^3*b^3*d^5*e + 6*a^4*b^2*d^4*e^2 - 4*a^5*b*d^3*e^3 + a^6*d^2*e^4 + (b^6*d^4*e^2 - 4*a
*b^5*d^3*e^3 + 6*a^2*b^4*d^2*e^4 - 4*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^4 + 2*(b^6*d^5*e - 3*a*b^5*d^4*e^2 + 2*a^2
*b^4*d^3*e^3 + 2*a^3*b^3*d^2*e^4 - 3*a^4*b^2*d*e^5 + a^5*b*e^6)*x^3 + (b^6*d^6 - 9*a^2*b^4*d^4*e^2 + 16*a^3*b^
3*d^3*e^3 - 9*a^4*b^2*d^2*e^4 + a^6*e^6)*x^2 + 2*(a*b^5*d^6 - 3*a^2*b^4*d^5*e + 2*a^3*b^3*d^4*e^2 + 2*a^4*b^2*
d^3*e^3 - 3*a^5*b*d^2*e^4 + a^6*d*e^5)*x), -1/12*(105*(b^3*e^4*x^4 + a^2*b*d^2*e^2 + 2*(b^3*d*e^3 + a*b^2*e^4)
*x^3 + (b^3*d^2*e^2 + 4*a*b^2*d*e^3 + a^2*b*e^4)*x^2 + 2*(a*b^2*d^2*e^2 + a^2*b*d*e^3)*x)*sqrt(-b/(b*d - a*e))
*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (105*b^3*e^3*x^3 - 6*b^3*d^3 + 39*a*b
^2*d^2*e + 80*a^2*b*d*e^2 - 8*a^3*e^3 + 35*(4*b^3*d*e^2 + 5*a*b^2*e^3)*x^2 + 7*(3*b^3*d^2*e + 34*a*b^2*d*e^2 +
 8*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^2*b^4*d^6 - 4*a^3*b^3*d^5*e + 6*a^4*b^2*d^4*e^2 - 4*a^5*b*d^3*e^3 + a^6*d^2
*e^4 + (b^6*d^4*e^2 - 4*a*b^5*d^3*e^3 + 6*a^2*b^4*d^2*e^4 - 4*a^3*b^3*d*e^5 + a^4*b^2*e^6)*x^4 + 2*(b^6*d^5*e
- 3*a*b^5*d^4*e^2 + 2*a^2*b^4*d^3*e^3 + 2*a^3*b^3*d^2*e^4 - 3*a^4*b^2*d*e^5 + a^5*b*e^6)*x^3 + (b^6*d^6 - 9*a^
2*b^4*d^4*e^2 + 16*a^3*b^3*d^3*e^3 - 9*a^4*b^2*d^2*e^4 + a^6*e^6)*x^2 + 2*(a*b^5*d^6 - 3*a^2*b^4*d^5*e + 2*a^3
*b^3*d^4*e^2 + 2*a^4*b^2*d^3*e^3 - 3*a^5*b*d^2*e^4 + a^6*d*e^5)*x)]

Sympy [F]

\[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {5}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(1/(e*x+d)**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Integral(1/((d + e*x)**(5/2)*((a + b*x)**2)**(3/2)), x)

Maxima [F]

\[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*(e*x + d)^(5/2)), x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 388 vs. \(2 (192) = 384\).

Time = 0.31 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.41 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {35 \, b^{2} e^{2} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{4 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, {\left (9 \, {\left (e x + d\right )} b e^{2} + b d e^{2} - a e^{3}\right )}}{3 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} {\left (e x + d\right )}^{\frac {3}{2}}} + \frac {11 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{3} e^{2} - 13 \, \sqrt {e x + d} b^{3} d e^{2} + 13 \, \sqrt {e x + d} a b^{2} e^{3}}{4 \, {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{2}} \]

[In]

integrate(1/(e*x+d)^(5/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

35/4*b^2*e^2*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)) +
2/3*(9*(e*x + d)*b*e^2 + b*d*e^2 - a*e^3)/((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)^(3/2)) + 1/4*(11*(e*x + d)^(3/
2)*b^3*e^2 - 13*sqrt(e*x + d)*b^3*d*e^2 + 13*sqrt(e*x + d)*a*b^2*e^3)/((b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*s
gn(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)^2)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (d+e\,x\right )}^{5/2}\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

[In]

int(1/((d + e*x)^(5/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)),x)

[Out]

int(1/((d + e*x)^(5/2)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2)), x)