\(\int \frac {1}{(d+e x)^{5/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\) [1662]

Optimal result

Integrand size = 28, antiderivative size = 200 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {105 b e^3}{8 (b d-a e)^5 \sqrt {d+e x}}+\frac {105 b^{3/2} e^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{11/2}} \]

[Out]

-35/8*e^3/(-a*e+b*d)^4/(e*x+d)^(3/2)-1/3/(-a*e+b*d)/(b*x+a)^3/(e*x+d)^(3/2)+3/4*e/(-a*e+b*d)^2/(b*x+a)^2/(e*x+
d)^(3/2)-21/8*e^2/(-a*e+b*d)^3/(b*x+a)/(e*x+d)^(3/2)+105/8*b^(3/2)*e^3*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d
)^(1/2))/(-a*e+b*d)^(11/2)-105/8*b*e^3/(-a*e+b*d)^5/(e*x+d)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 44, 53, 65, 214} \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {105 b^{3/2} e^3 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{11/2}}-\frac {105 b e^3}{8 \sqrt {d+e x} (b d-a e)^5}-\frac {35 e^3}{8 (d+e x)^{3/2} (b d-a e)^4}-\frac {21 e^2}{8 (a+b x) (d+e x)^{3/2} (b d-a e)^3}+\frac {3 e}{4 (a+b x)^2 (d+e x)^{3/2} (b d-a e)^2}-\frac {1}{3 (a+b x)^3 (d+e x)^{3/2} (b d-a e)} \]

[In]

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

[Out]

(-35*e^3)/(8*(b*d - a*e)^4*(d + e*x)^(3/2)) - 1/(3*(b*d - a*e)*(a + b*x)^3*(d + e*x)^(3/2)) + (3*e)/(4*(b*d -
a*e)^2*(a + b*x)^2*(d + e*x)^(3/2)) - (21*e^2)/(8*(b*d - a*e)^3*(a + b*x)*(d + e*x)^(3/2)) - (105*b*e^3)/(8*(b
*d - a*e)^5*Sqrt[d + e*x]) + (105*b^(3/2)*e^3*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*(b*d - a*e)
^(11/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(a+b x)^4 (d+e x)^{5/2}} \, dx \\ & = -\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}-\frac {(3 e) \int \frac {1}{(a+b x)^3 (d+e x)^{5/2}} \, dx}{2 (b d-a e)} \\ & = -\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}+\frac {\left (21 e^2\right ) \int \frac {1}{(a+b x)^2 (d+e x)^{5/2}} \, dx}{8 (b d-a e)^2} \\ & = -\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {\left (105 e^3\right ) \int \frac {1}{(a+b x) (d+e x)^{5/2}} \, dx}{16 (b d-a e)^3} \\ & = -\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {\left (105 b e^3\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{16 (b d-a e)^4} \\ & = -\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {105 b e^3}{8 (b d-a e)^5 \sqrt {d+e x}}-\frac {\left (105 b^2 e^3\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 (b d-a e)^5} \\ & = -\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {105 b e^3}{8 (b d-a e)^5 \sqrt {d+e x}}-\frac {\left (105 b^2 e^2\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{8 (b d-a e)^5} \\ & = -\frac {35 e^3}{8 (b d-a e)^4 (d+e x)^{3/2}}-\frac {1}{3 (b d-a e) (a+b x)^3 (d+e x)^{3/2}}+\frac {3 e}{4 (b d-a e)^2 (a+b x)^2 (d+e x)^{3/2}}-\frac {21 e^2}{8 (b d-a e)^3 (a+b x) (d+e x)^{3/2}}-\frac {105 b e^3}{8 (b d-a e)^5 \sqrt {d+e x}}+\frac {105 b^{3/2} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{11/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {1}{24} \left (\frac {-16 a^4 e^4+16 a^3 b e^3 (13 d+9 e x)+3 a^2 b^2 e^2 \left (55 d^2+318 d e x+231 e^2 x^2\right )+2 a b^3 e \left (-25 d^3+90 d^2 e x+567 d e^2 x^2+420 e^3 x^3\right )+b^4 \left (8 d^4-18 d^3 e x+63 d^2 e^2 x^2+420 d e^3 x^3+315 e^4 x^4\right )}{(-b d+a e)^5 (a+b x)^3 (d+e x)^{3/2}}+\frac {315 b^{3/2} e^3 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{11/2}}\right ) \]

[In]

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

[Out]

((-16*a^4*e^4 + 16*a^3*b*e^3*(13*d + 9*e*x) + 3*a^2*b^2*e^2*(55*d^2 + 318*d*e*x + 231*e^2*x^2) + 2*a*b^3*e*(-2
5*d^3 + 90*d^2*e*x + 567*d*e^2*x^2 + 420*e^3*x^3) + b^4*(8*d^4 - 18*d^3*e*x + 63*d^2*e^2*x^2 + 420*d*e^3*x^3 +
315*e^4*x^4))/((-(b*d) + a*e)^5*(a + b*x)^3*(d + e*x)^(3/2)) + (315*b^(3/2)*e^3*ArcTan[(Sqrt[b]*Sqrt[d + e*x]
)/Sqrt[-(b*d) + a*e]])/(-(b*d) + a*e)^(11/2))/24

Maple [A] (verified)

Time = 2.80 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(2 e^{3} \left (-\frac {1}{3 \left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}+\frac {4 b}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {b^{2} \left (\frac {\frac {41 b^{2} \left (e x +d \right )^{\frac {5}{2}}}{16}+\frac {35 \left (a e -b d \right ) b \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (\frac {55}{16} a^{2} e^{2}-\frac {55}{8} a b d e +\frac {55}{16} b^{2} d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{5}}\right )\)	\(177\)
default	\(2 e^{3} \left (-\frac {1}{3 \left (a e -b d \right )^{4} \left (e x +d \right )^{\frac {3}{2}}}+\frac {4 b}{\left (a e -b d \right )^{5} \sqrt {e x +d}}+\frac {b^{2} \left (\frac {\frac {41 b^{2} \left (e x +d \right )^{\frac {5}{2}}}{16}+\frac {35 \left (a e -b d \right ) b \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (\frac {55}{16} a^{2} e^{2}-\frac {55}{8} a b d e +\frac {55}{16} b^{2} d^{2}\right ) \sqrt {e x +d}}{\left (b \left (e x +d \right )+a e -b d \right )^{3}}+\frac {105 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16 \sqrt {\left (a e -b d \right ) b}}\right )}{\left (a e -b d \right )^{5}}\right )\)	\(177\)
pseudoelliptic	\(-\frac {2 \left (-\frac {315 b^{2} e^{3} \left (e x +d \right )^{\frac {3}{2}} \left (b x +a \right )^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{16}+\sqrt {\left (a e -b d \right ) b}\, \left (\left (-\frac {315}{16} e^{4} x^{4}-\frac {105}{4} d \,e^{3} x^{3}-\frac {63}{16} d^{2} e^{2} x^{2}+\frac {9}{8} d^{3} e x -\frac {1}{2} d^{4}\right ) b^{4}+\frac {25 e a \left (-\frac {84}{5} e^{3} x^{3}-\frac {567}{25} d \,e^{2} x^{2}-\frac {18}{5} d^{2} e x +d^{3}\right ) b^{3}}{8}-\frac {165 \left (\frac {21}{5} x^{2} e^{2}+\frac {318}{55} d e x +d^{2}\right ) e^{2} a^{2} b^{2}}{16}-13 e^{3} \left (\frac {9 e x}{13}+d \right ) a^{3} b +e^{4} a^{4}\right )\right )}{3 \left (e x +d \right )^{\frac {3}{2}} \sqrt {\left (a e -b d \right ) b}\, \left (b x +a \right )^{3} \left (a e -b d \right )^{5}}\)	\(228\)

[In]

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

[Out]

2*e^3*(-1/3/(a*e-b*d)^4/(e*x+d)^(3/2)+4/(a*e-b*d)^5*b/(e*x+d)^(1/2)+1/(a*e-b*d)^5*b^2*((41/16*b^2*(e*x+d)^(5/2
)+35/6*(a*e-b*d)*b*(e*x+d)^(3/2)+(55/16*a^2*e^2-55/8*a*b*d*e+55/16*b^2*d^2)*(e*x+d)^(1/2))/(b*(e*x+d)+a*e-b*d)
^3+105/16/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 915 vs. \(2 (168) = 336\).

Time = 0.65 (sec) , antiderivative size = 1840, normalized size of antiderivative = 9.20 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/48*(315*(b^4*e^5*x^5 + a^3*b*d^2*e^3 + (2*b^4*d*e^4 + 3*a*b^3*e^5)*x^4 + (b^4*d^2*e^3 + 6*a*b^3*d*e^4 + 3*
a^2*b^2*e^5)*x^3 + (3*a*b^3*d^2*e^3 + 6*a^2*b^2*d*e^4 + a^3*b*e^5)*x^2 + (3*a^2*b^2*d^2*e^3 + 2*a^3*b*d*e^4)*x
)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e - 2*(b*d - a*e)*sqrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) +
2*(315*b^4*e^4*x^4 + 8*b^4*d^4 - 50*a*b^3*d^3*e + 165*a^2*b^2*d^2*e^2 + 208*a^3*b*d*e^3 - 16*a^4*e^4 + 420*(b
^4*d*e^3 + 2*a*b^3*e^4)*x^3 + 63*(b^4*d^2*e^2 + 18*a*b^3*d*e^3 + 11*a^2*b^2*e^4)*x^2 - 18*(b^4*d^3*e - 10*a*b^
3*d^2*e^2 - 53*a^2*b^2*d*e^3 - 8*a^3*b*e^4)*x)*sqrt(e*x + d))/(a^3*b^5*d^7 - 5*a^4*b^4*d^6*e + 10*a^5*b^3*d^5*
e^2 - 10*a^6*b^2*d^4*e^3 + 5*a^7*b*d^3*e^4 - a^8*d^2*e^5 + (b^8*d^5*e^2 - 5*a*b^7*d^4*e^3 + 10*a^2*b^6*d^3*e^4
- 10*a^3*b^5*d^2*e^5 + 5*a^4*b^4*d*e^6 - a^5*b^3*e^7)*x^5 + (2*b^8*d^6*e - 7*a*b^7*d^5*e^2 + 5*a^2*b^6*d^4*e^
3 + 10*a^3*b^5*d^3*e^4 - 20*a^4*b^4*d^2*e^5 + 13*a^5*b^3*d*e^6 - 3*a^6*b^2*e^7)*x^4 + (b^8*d^7 + a*b^7*d^6*e -
17*a^2*b^6*d^5*e^2 + 35*a^3*b^5*d^4*e^3 - 25*a^4*b^4*d^3*e^4 - a^5*b^3*d^2*e^5 + 9*a^6*b^2*d*e^6 - 3*a^7*b*e^
7)*x^3 + (3*a*b^7*d^7 - 9*a^2*b^6*d^6*e + a^3*b^5*d^5*e^2 + 25*a^4*b^4*d^4*e^3 - 35*a^5*b^3*d^3*e^4 + 17*a^6*b
^2*d^2*e^5 - a^7*b*d*e^6 - a^8*e^7)*x^2 + (3*a^2*b^6*d^7 - 13*a^3*b^5*d^6*e + 20*a^4*b^4*d^5*e^2 - 10*a^5*b^3*
d^4*e^3 - 5*a^6*b^2*d^3*e^4 + 7*a^7*b*d^2*e^5 - 2*a^8*d*e^6)*x), 1/24*(315*(b^4*e^5*x^5 + a^3*b*d^2*e^3 + (2*b
^4*d*e^4 + 3*a*b^3*e^5)*x^4 + (b^4*d^2*e^3 + 6*a*b^3*d*e^4 + 3*a^2*b^2*e^5)*x^3 + (3*a*b^3*d^2*e^3 + 6*a^2*b^2
*d*e^4 + a^3*b*e^5)*x^2 + (3*a^2*b^2*d^2*e^3 + 2*a^3*b*d*e^4)*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)) - (315*b^4*e^4*x^4 + 8*b^4*d^4 - 50*a*b^3*d^3*e + 165*a^2*b^2*d^
2*e^2 + 208*a^3*b*d*e^3 - 16*a^4*e^4 + 420*(b^4*d*e^3 + 2*a*b^3*e^4)*x^3 + 63*(b^4*d^2*e^2 + 18*a*b^3*d*e^3 +
11*a^2*b^2*e^4)*x^2 - 18*(b^4*d^3*e - 10*a*b^3*d^2*e^2 - 53*a^2*b^2*d*e^3 - 8*a^3*b*e^4)*x)*sqrt(e*x + d))/(a^
3*b^5*d^7 - 5*a^4*b^4*d^6*e + 10*a^5*b^3*d^5*e^2 - 10*a^6*b^2*d^4*e^3 + 5*a^7*b*d^3*e^4 - a^8*d^2*e^5 + (b^8*d
^5*e^2 - 5*a*b^7*d^4*e^3 + 10*a^2*b^6*d^3*e^4 - 10*a^3*b^5*d^2*e^5 + 5*a^4*b^4*d*e^6 - a^5*b^3*e^7)*x^5 + (2*b
^8*d^6*e - 7*a*b^7*d^5*e^2 + 5*a^2*b^6*d^4*e^3 + 10*a^3*b^5*d^3*e^4 - 20*a^4*b^4*d^2*e^5 + 13*a^5*b^3*d*e^6 -
3*a^6*b^2*e^7)*x^4 + (b^8*d^7 + a*b^7*d^6*e - 17*a^2*b^6*d^5*e^2 + 35*a^3*b^5*d^4*e^3 - 25*a^4*b^4*d^3*e^4 - a
^5*b^3*d^2*e^5 + 9*a^6*b^2*d*e^6 - 3*a^7*b*e^7)*x^3 + (3*a*b^7*d^7 - 9*a^2*b^6*d^6*e + a^3*b^5*d^5*e^2 + 25*a^
4*b^4*d^4*e^3 - 35*a^5*b^3*d^3*e^4 + 17*a^6*b^2*d^2*e^5 - a^7*b*d*e^6 - a^8*e^7)*x^2 + (3*a^2*b^6*d^7 - 13*a^3
*b^5*d^6*e + 20*a^4*b^4*d^5*e^2 - 10*a^5*b^3*d^4*e^3 - 5*a^6*b^2*d^3*e^4 + 7*a^7*b*d^2*e^5 - 2*a^8*d*e^6)*x)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more detail

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 432 vs. \(2 (168) = 336\).

Time = 0.28 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.16 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {105 \, b^{2} e^{3} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} \sqrt {-b^{2} d + a b e}} - \frac {315 \, {\left (e x + d\right )}^{4} b^{4} e^{3} - 840 \, {\left (e x + d\right )}^{3} b^{4} d e^{3} + 693 \, {\left (e x + d\right )}^{2} b^{4} d^{2} e^{3} - 144 \, {\left (e x + d\right )} b^{4} d^{3} e^{3} - 16 \, b^{4} d^{4} e^{3} + 840 \, {\left (e x + d\right )}^{3} a b^{3} e^{4} - 1386 \, {\left (e x + d\right )}^{2} a b^{3} d e^{4} + 432 \, {\left (e x + d\right )} a b^{3} d^{2} e^{4} + 64 \, a b^{3} d^{3} e^{4} + 693 \, {\left (e x + d\right )}^{2} a^{2} b^{2} e^{5} - 432 \, {\left (e x + d\right )} a^{2} b^{2} d e^{5} - 96 \, a^{2} b^{2} d^{2} e^{5} + 144 \, {\left (e x + d\right )} a^{3} b e^{6} + 64 \, a^{3} b d e^{6} - 16 \, a^{4} e^{7}}{24 \, {\left (b^{5} d^{5} - 5 \, a b^{4} d^{4} e + 10 \, a^{2} b^{3} d^{3} e^{2} - 10 \, a^{3} b^{2} d^{2} e^{3} + 5 \, a^{4} b d e^{4} - a^{5} e^{5}\right )} {\left ({\left (e x + d\right )}^{\frac {3}{2}} b - \sqrt {e x + d} b d + \sqrt {e x + d} a e\right )}^{3}} \]

[In]

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

[Out]

-105/8*b^2*e^3*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 1
0*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*sqrt(-b^2*d + a*b*e)) - 1/24*(315*(e*x + d)^4*b^4*e^3 - 840*(e*x
+ d)^3*b^4*d*e^3 + 693*(e*x + d)^2*b^4*d^2*e^3 - 144*(e*x + d)*b^4*d^3*e^3 - 16*b^4*d^4*e^3 + 840*(e*x + d)^3*
a*b^3*e^4 - 1386*(e*x + d)^2*a*b^3*d*e^4 + 432*(e*x + d)*a*b^3*d^2*e^4 + 64*a*b^3*d^3*e^4 + 693*(e*x + d)^2*a^
2*b^2*e^5 - 432*(e*x + d)*a^2*b^2*d*e^5 - 96*a^2*b^2*d^2*e^5 + 144*(e*x + d)*a^3*b*e^6 + 64*a^3*b*d*e^6 - 16*a
^4*e^7)/((b^5*d^5 - 5*a*b^4*d^4*e + 10*a^2*b^3*d^3*e^2 - 10*a^3*b^2*d^2*e^3 + 5*a^4*b*d*e^4 - a^5*e^5)*((e*x +
d)^(3/2)*b - sqrt(e*x + d)*b*d + sqrt(e*x + d)*a*e)^3)

Mupad [B] (verification not implemented)

Time = 9.71 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.67 \[ \int \frac {1}{(d+e x)^{5/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\frac {231\,b^2\,e^3\,{\left (d+e\,x\right )}^2}{8\,{\left (a\,e-b\,d\right )}^3}-\frac {2\,e^3}{3\,\left (a\,e-b\,d\right )}+\frac {35\,b^3\,e^3\,{\left (d+e\,x\right )}^3}{{\left (a\,e-b\,d\right )}^4}+\frac {105\,b^4\,e^3\,{\left (d+e\,x\right )}^4}{8\,{\left (a\,e-b\,d\right )}^5}+\frac {6\,b\,e^3\,\left (d+e\,x\right )}{{\left (a\,e-b\,d\right )}^2}}{{\left (d+e\,x\right )}^{3/2}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )+b^3\,{\left (d+e\,x\right )}^{9/2}-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{7/2}+{\left (d+e\,x\right )}^{5/2}\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )}+\frac {105\,b^{3/2}\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^5\,e^5-5\,a^4\,b\,d\,e^4+10\,a^3\,b^2\,d^2\,e^3-10\,a^2\,b^3\,d^3\,e^2+5\,a\,b^4\,d^4\,e-b^5\,d^5\right )}{{\left (a\,e-b\,d\right )}^{11/2}}\right )}{8\,{\left (a\,e-b\,d\right )}^{11/2}} \]

[In]

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

[Out]

((231*b^2*e^3*(d + e*x)^2)/(8*(a*e - b*d)^3) - (2*e^3)/(3*(a*e - b*d)) + (35*b^3*e^3*(d + e*x)^3)/(a*e - b*d)^
4 + (105*b^4*e^3*(d + e*x)^4)/(8*(a*e - b*d)^5) + (6*b*e^3*(d + e*x))/(a*e - b*d)^2)/((d + e*x)^(3/2)*(a^3*e^3
- b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2*b*d*e^2) + b^3*(d + e*x)^(9/2) - (3*b^3*d - 3*a*b^2*e)*(d + e*x)^(7/2) + (d
+ e*x)^(5/2)*(3*b^3*d^2 + 3*a^2*b*e^2 - 6*a*b^2*d*e)) + (105*b^(3/2)*e^3*atan((b^(1/2)*(d + e*x)^(1/2)*(a^5*e
^5 - b^5*d^5 - 10*a^2*b^3*d^3*e^2 + 10*a^3*b^2*d^2*e^3 + 5*a*b^4*d^4*e - 5*a^4*b*d*e^4))/(a*e - b*d)^(11/2)))/
(8*(a*e - b*d)^(11/2))