\(\int \frac {1}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [1728]

Optimal result

Integrand size = 30, antiderivative size = 329 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {105 e^3}{64 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e}{8 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {21 e^2}{32 (b d-a e)^3 (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {315 e^4 (a+b x)}{64 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {315 \sqrt {b} e^4 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 (b d-a e)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

-315/64*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)^(11/2)/((b*x+a)^2)^(1/2
)+105/64*e^3/(-a*e+b*d)^4/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)-1/4/(-a*e+b*d)/(b*x+a)^3/(e*x+d)^(1/2)/((b*x+a)^2)^(
1/2)+3/8*e/(-a*e+b*d)^2/(b*x+a)^2/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)-21/32*e^2/(-a*e+b*d)^3/(b*x+a)/(e*x+d)^(1/2)
/((b*x+a)^2)^(1/2)+315/64*e^4*(b*x+a)/(-a*e+b*d)^5/(e*x+d)^(1/2)/((b*x+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {660, 44, 53, 65, 214} \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {315 \sqrt {b} e^4 (a+b x) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{11/2}}+\frac {315 e^4 (a+b x)}{64 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^5}+\frac {105 e^3}{64 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^4}-\frac {21 e^2}{32 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^3}+\frac {3 e}{8 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)^2}-\frac {1}{4 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {d+e x} (b d-a e)} \]

[In]

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

[Out]

(105*e^3)/(64*(b*d - a*e)^4*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - 1/(4*(b*d - a*e)*(a + b*x)^3*Sqrt[d
+ e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*e)/(8*(b*d - a*e)^2*(a + b*x)^2*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) - (21*e^2)/(32*(b*d - a*e)^3*(a + b*x)*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (315*e^4*(a +
b*x))/(64*(b*d - a*e)^5*Sqrt[d + e*x]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (315*Sqrt[b]*e^4*(a + b*x)*ArcTanh[(Sq
rt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(64*(b*d - a*e)^(11/2)*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^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^5 (d+e x)^{3/2}} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (9 b^3 e \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^4 (d+e x)^{3/2}} \, dx}{8 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e}{8 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (21 b^2 e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^3 (d+e x)^{3/2}} \, dx}{16 (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e}{8 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {21 e^2}{32 (b d-a e)^3 (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (105 b e^3 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 (d+e x)^{3/2}} \, dx}{64 (b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {105 e^3}{64 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e}{8 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {21 e^2}{32 (b d-a e)^3 (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (315 e^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{128 (b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {105 e^3}{64 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e}{8 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {21 e^2}{32 (b d-a e)^3 (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {315 e^4 (a+b x)}{64 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (315 b e^4 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{128 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {105 e^3}{64 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e}{8 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {21 e^2}{32 (b d-a e)^3 (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {315 e^4 (a+b x)}{64 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (315 b e^3 \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 )}{64 (b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {105 e^3}{64 (b d-a e)^4 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {1}{4 (b d-a e) (a+b x)^3 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e}{8 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {21 e^2}{32 (b d-a e)^3 (a+b x) \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {315 e^4 (a+b x)}{64 (b d-a e)^5 \sqrt {d+e x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {315 \sqrt {b} e^4 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{64 (b d-a e)^{11/2} \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {-b d+a e} \left (128 a^4 e^4+a^3 b e^3 (325 d+837 e x)+3 a^2 b^2 e^2 \left (-70 d^2+185 d e x+511 e^2 x^2\right )+a b^3 e \left (88 d^3-156 d^2 e x+399 d e^2 x^2+1155 e^3 x^3\right )+b^4 \left (-16 d^4+24 d^3 e x-42 d^2 e^2 x^2+105 d e^3 x^3+315 e^4 x^4\right )\right )+315 \sqrt {b} e^4 (a+b x)^4 \sqrt {d+e x} \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{64 (-b d+a e)^{11/2} (a+b x)^3 \sqrt {(a+b x)^2} \sqrt {d+e x}} \]

[In]

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

[Out]

-1/64*(Sqrt[-(b*d) + a*e]*(128*a^4*e^4 + a^3*b*e^3*(325*d + 837*e*x) + 3*a^2*b^2*e^2*(-70*d^2 + 185*d*e*x + 51
1*e^2*x^2) + a*b^3*e*(88*d^3 - 156*d^2*e*x + 399*d*e^2*x^2 + 1155*e^3*x^3) + b^4*(-16*d^4 + 24*d^3*e*x - 42*d^
2*e^2*x^2 + 105*d*e^3*x^3 + 315*e^4*x^4)) + 315*Sqrt[b]*e^4*(a + b*x)^4*Sqrt[d + e*x]*ArcTan[(Sqrt[b]*Sqrt[d +
e*x])/Sqrt[-(b*d) + a*e]])/((-(b*d) + a*e)^(11/2)*(a + b*x)^3*Sqrt[(a + b*x)^2]*Sqrt[d + e*x])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(601\) vs. \(2(231)=462\).

Time = 2.29 (sec) , antiderivative size = 602, normalized size of antiderivative = 1.83

method	result	size
default	\(-\frac {\left (315 \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) b^{5} e^{4} x^{4}+1260 \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a \,b^{4} e^{4} x^{3}+1890 \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{2} b^{3} e^{4} x^{2}+315 \sqrt {\left (a e -b d \right ) b}\, b^{4} e^{4} x^{4}+1260 \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{3} b^{2} e^{4} x +1155 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} e^{4} x^{3}+105 \sqrt {\left (a e -b d \right ) b}\, b^{4} d \,e^{3} x^{3}+315 \sqrt {e x +d}\, \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right ) a^{4} b \,e^{4}+1533 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} e^{4} x^{2}+399 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d \,e^{3} x^{2}-42 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{2} e^{2} x^{2}+837 \sqrt {\left (a e -b d \right ) b}\, a^{3} b \,e^{4} x +555 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} d \,e^{3} x -156 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d^{2} e^{2} x +24 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{3} e x +128 \sqrt {\left (a e -b d \right ) b}\, a^{4} e^{4}+325 \sqrt {\left (a e -b d \right ) b}\, a^{3} b d \,e^{3}-210 \sqrt {\left (a e -b d \right ) b}\, a^{2} b^{2} d^{2} e^{2}+88 \sqrt {\left (a e -b d \right ) b}\, a \,b^{3} d^{3} e -16 \sqrt {\left (a e -b d \right ) b}\, b^{4} d^{4}\right ) \left (b x +a \right )}{64 \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}\, \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(602\)

[In]

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

[Out]

-1/64*(315*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*b^5*e^4*x^4+1260*(e*x+d)^(1/2)*arctan(b*(
e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*b^4*e^4*x^3+1890*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))
*a^2*b^3*e^4*x^2+315*((a*e-b*d)*b)^(1/2)*b^4*e^4*x^4+1260*(e*x+d)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(
1/2))*a^3*b^2*e^4*x+1155*((a*e-b*d)*b)^(1/2)*a*b^3*e^4*x^3+105*((a*e-b*d)*b)^(1/2)*b^4*d*e^3*x^3+315*(e*x+d)^(
1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^4*b*e^4+1533*((a*e-b*d)*b)^(1/2)*a^2*b^2*e^4*x^2+399*((a*e-
b*d)*b)^(1/2)*a*b^3*d*e^3*x^2-42*((a*e-b*d)*b)^(1/2)*b^4*d^2*e^2*x^2+837*((a*e-b*d)*b)^(1/2)*a^3*b*e^4*x+555*(
(a*e-b*d)*b)^(1/2)*a^2*b^2*d*e^3*x-156*((a*e-b*d)*b)^(1/2)*a*b^3*d^2*e^2*x+24*((a*e-b*d)*b)^(1/2)*b^4*d^3*e*x+
128*((a*e-b*d)*b)^(1/2)*a^4*e^4+325*((a*e-b*d)*b)^(1/2)*a^3*b*d*e^3-210*((a*e-b*d)*b)^(1/2)*a^2*b^2*d^2*e^2+88
*((a*e-b*d)*b)^(1/2)*a*b^3*d^3*e-16*((a*e-b*d)*b)^(1/2)*b^4*d^4)*(b*x+a)/((a*e-b*d)*b)^(1/2)/(e*x+d)^(1/2)/(a*
e-b*d)^5/((b*x+a)^2)^(5/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 862 vs. \(2 (231) = 462\).

Time = 0.54 (sec) , antiderivative size = 1734, normalized size of antiderivative = 5.27 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/128*(315*(b^4*e^5*x^5 + a^4*d*e^4 + (b^4*d*e^4 + 4*a*b^3*e^5)*x^4 + 2*(2*a*b^3*d*e^4 + 3*a^2*b^2*e^5)*x^3
+ 2*(3*a^2*b^2*d*e^4 + 2*a^3*b*e^5)*x^2 + (4*a^3*b*d*e^4 + a^4*e^5)*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 - 16*b^4*d^4 + 88*a*b
^3*d^3*e - 210*a^2*b^2*d^2*e^2 + 325*a^3*b*d*e^3 + 128*a^4*e^4 + 105*(b^4*d*e^3 + 11*a*b^3*e^4)*x^3 - 21*(2*b^
4*d^2*e^2 - 19*a*b^3*d*e^3 - 73*a^2*b^2*e^4)*x^2 + 3*(8*b^4*d^3*e - 52*a*b^3*d^2*e^2 + 185*a^2*b^2*d*e^3 + 279
*a^3*b*e^4)*x)*sqrt(e*x + d))/(a^4*b^5*d^6 - 5*a^5*b^4*d^5*e + 10*a^6*b^3*d^4*e^2 - 10*a^7*b^2*d^3*e^3 + 5*a^8
*b*d^2*e^4 - a^9*d*e^5 + (b^9*d^5*e - 5*a*b^8*d^4*e^2 + 10*a^2*b^7*d^3*e^3 - 10*a^3*b^6*d^2*e^4 + 5*a^4*b^5*d*
e^5 - a^5*b^4*e^6)*x^5 + (b^9*d^6 - a*b^8*d^5*e - 10*a^2*b^7*d^4*e^2 + 30*a^3*b^6*d^3*e^3 - 35*a^4*b^5*d^2*e^4
+ 19*a^5*b^4*d*e^5 - 4*a^6*b^3*e^6)*x^4 + 2*(2*a*b^8*d^6 - 7*a^2*b^7*d^5*e + 5*a^3*b^6*d^4*e^2 + 10*a^4*b^5*d
^3*e^3 - 20*a^5*b^4*d^2*e^4 + 13*a^6*b^3*d*e^5 - 3*a^7*b^2*e^6)*x^3 + 2*(3*a^2*b^7*d^6 - 13*a^3*b^6*d^5*e + 20
*a^4*b^5*d^4*e^2 - 10*a^5*b^4*d^3*e^3 - 5*a^6*b^3*d^2*e^4 + 7*a^7*b^2*d*e^5 - 2*a^8*b*e^6)*x^2 + (4*a^3*b^6*d^
6 - 19*a^4*b^5*d^5*e + 35*a^5*b^4*d^4*e^2 - 30*a^6*b^3*d^3*e^3 + 10*a^7*b^2*d^2*e^4 + a^8*b*d*e^5 - a^9*e^6)*x
), -1/64*(315*(b^4*e^5*x^5 + a^4*d*e^4 + (b^4*d*e^4 + 4*a*b^3*e^5)*x^4 + 2*(2*a*b^3*d*e^4 + 3*a^2*b^2*e^5)*x^3
+ 2*(3*a^2*b^2*d*e^4 + 2*a^3*b*e^5)*x^2 + (4*a^3*b*d*e^4 + a^4*e^5)*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 - 16*b^4*d^4 + 88*a*b^3*d^3*e - 210*a^
2*b^2*d^2*e^2 + 325*a^3*b*d*e^3 + 128*a^4*e^4 + 105*(b^4*d*e^3 + 11*a*b^3*e^4)*x^3 - 21*(2*b^4*d^2*e^2 - 19*a*
b^3*d*e^3 - 73*a^2*b^2*e^4)*x^2 + 3*(8*b^4*d^3*e - 52*a*b^3*d^2*e^2 + 185*a^2*b^2*d*e^3 + 279*a^3*b*e^4)*x)*sq
rt(e*x + d))/(a^4*b^5*d^6 - 5*a^5*b^4*d^5*e + 10*a^6*b^3*d^4*e^2 - 10*a^7*b^2*d^3*e^3 + 5*a^8*b*d^2*e^4 - a^9*
d*e^5 + (b^9*d^5*e - 5*a*b^8*d^4*e^2 + 10*a^2*b^7*d^3*e^3 - 10*a^3*b^6*d^2*e^4 + 5*a^4*b^5*d*e^5 - a^5*b^4*e^6
)*x^5 + (b^9*d^6 - a*b^8*d^5*e - 10*a^2*b^7*d^4*e^2 + 30*a^3*b^6*d^3*e^3 - 35*a^4*b^5*d^2*e^4 + 19*a^5*b^4*d*e
^5 - 4*a^6*b^3*e^6)*x^4 + 2*(2*a*b^8*d^6 - 7*a^2*b^7*d^5*e + 5*a^3*b^6*d^4*e^2 + 10*a^4*b^5*d^3*e^3 - 20*a^5*b
^4*d^2*e^4 + 13*a^6*b^3*d*e^5 - 3*a^7*b^2*e^6)*x^3 + 2*(3*a^2*b^7*d^6 - 13*a^3*b^6*d^5*e + 20*a^4*b^5*d^4*e^2
- 10*a^5*b^4*d^3*e^3 - 5*a^6*b^3*d^2*e^4 + 7*a^7*b^2*d*e^5 - 2*a^8*b*e^6)*x^2 + (4*a^3*b^6*d^6 - 19*a^4*b^5*d^
5*e + 35*a^5*b^4*d^4*e^2 - 30*a^6*b^3*d^3*e^3 + 10*a^7*b^2*d^2*e^4 + a^8*b*d*e^5 - a^9*e^6)*x)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 553 vs. \(2 (231) = 462\).

Time = 0.30 (sec) , antiderivative size = 553, normalized size of antiderivative = 1.68 \[ \int \frac {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\frac {315 \, b e^{4} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{64 \, {\left (b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} + \frac {2 \, e^{4}}{{\left (b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {e x + d}} + \frac {187 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{4} e^{4} - 643 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{4} d e^{4} + 765 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d^{2} e^{4} - 325 \, \sqrt {e x + d} b^{4} d^{3} e^{4} + 643 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{3} e^{5} - 1530 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} d e^{5} + 975 \, \sqrt {e x + d} a b^{3} d^{2} e^{5} + 765 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b^{2} e^{6} - 975 \, \sqrt {e x + d} a^{2} b^{2} d e^{6} + 325 \, \sqrt {e x + d} a^{3} b e^{7}}{64 \, {\left (b^{5} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{5} \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{4}} \]

[In]

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

[Out]

315/64*b*e^4*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e*sgn(b*x + a)
+ 10*a^2*b^3*d^3*e^2*sgn(b*x + a) - 10*a^3*b^2*d^2*e^3*sgn(b*x + a) + 5*a^4*b*d*e^4*sgn(b*x + a) - a^5*e^5*sgn
(b*x + a))*sqrt(-b^2*d + a*b*e)) + 2*e^4/((b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e*sgn(b*x + a) + 10*a^2*b^3*d^3*
e^2*sgn(b*x + a) - 10*a^3*b^2*d^2*e^3*sgn(b*x + a) + 5*a^4*b*d*e^4*sgn(b*x + a) - a^5*e^5*sgn(b*x + a))*sqrt(e
*x + d)) + 1/64*(187*(e*x + d)^(7/2)*b^4*e^4 - 643*(e*x + d)^(5/2)*b^4*d*e^4 + 765*(e*x + d)^(3/2)*b^4*d^2*e^4
- 325*sqrt(e*x + d)*b^4*d^3*e^4 + 643*(e*x + d)^(5/2)*a*b^3*e^5 - 1530*(e*x + d)^(3/2)*a*b^3*d*e^5 + 975*sqrt
(e*x + d)*a*b^3*d^2*e^5 + 765*(e*x + d)^(3/2)*a^2*b^2*e^6 - 975*sqrt(e*x + d)*a^2*b^2*d*e^6 + 325*sqrt(e*x + d
)*a^3*b*e^7)/((b^5*d^5*sgn(b*x + a) - 5*a*b^4*d^4*e*sgn(b*x + a) + 10*a^2*b^3*d^3*e^2*sgn(b*x + a) - 10*a^3*b^
2*d^2*e^3*sgn(b*x + a) + 5*a^4*b*d*e^4*sgn(b*x + a) - a^5*e^5*sgn(b*x + a))*((e*x + d)*b - b*d + a*e)^4)

Mupad [F(-1)]

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

[In]

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

[Out]

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