\(\int \frac {(d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^3} \, dx\) [1670]

Optimal result

Integrand size = 28, antiderivative size = 208 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {3 e \sqrt {d+e x}}{40 b^2 (a+b x)^4}-\frac {e^2 \sqrt {d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac {e^3 \sqrt {d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac {3 e^4 \sqrt {d+e x}}{128 b^2 (b d-a e)^3 (a+b x)}-\frac {(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac {3 e^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{5/2} (b d-a e)^{7/2}} \]

[Out]

-1/5*(e*x+d)^(3/2)/b/(b*x+a)^5+3/128*e^5*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(5/2)/(-a*e+b*d)^(7
/2)-3/40*e*(e*x+d)^(1/2)/b^2/(b*x+a)^4-1/80*e^2*(e*x+d)^(1/2)/b^2/(-a*e+b*d)/(b*x+a)^3+1/64*e^3*(e*x+d)^(1/2)/
b^2/(-a*e+b*d)^2/(b*x+a)^2-3/128*e^4*(e*x+d)^(1/2)/b^2/(-a*e+b*d)^3/(b*x+a)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 208, 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, 43, 44, 65, 214} \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {3 e^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{5/2} (b d-a e)^{7/2}}-\frac {3 e^4 \sqrt {d+e x}}{128 b^2 (a+b x) (b d-a e)^3}+\frac {e^3 \sqrt {d+e x}}{64 b^2 (a+b x)^2 (b d-a e)^2}-\frac {e^2 \sqrt {d+e x}}{80 b^2 (a+b x)^3 (b d-a e)}-\frac {3 e \sqrt {d+e x}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{3/2}}{5 b (a+b x)^5} \]

[In]

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

[Out]

(-3*e*Sqrt[d + e*x])/(40*b^2*(a + b*x)^4) - (e^2*Sqrt[d + e*x])/(80*b^2*(b*d - a*e)*(a + b*x)^3) + (e^3*Sqrt[d
+ e*x])/(64*b^2*(b*d - a*e)^2*(a + b*x)^2) - (3*e^4*Sqrt[d + e*x])/(128*b^2*(b*d - a*e)^3*(a + b*x)) - (d + e
*x)^(3/2)/(5*b*(a + b*x)^5) + (3*e^5*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(128*b^(5/2)*(b*d - a*e
)^(7/2))

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 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 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 {(d+e x)^{3/2}}{(a+b x)^6} \, dx \\ & = -\frac {(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac {(3 e) \int \frac {\sqrt {d+e x}}{(a+b x)^5} \, dx}{10 b} \\ & = -\frac {3 e \sqrt {d+e x}}{40 b^2 (a+b x)^4}-\frac {(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac {\left (3 e^2\right ) \int \frac {1}{(a+b x)^4 \sqrt {d+e x}} \, dx}{80 b^2} \\ & = -\frac {3 e \sqrt {d+e x}}{40 b^2 (a+b x)^4}-\frac {e^2 \sqrt {d+e x}}{80 b^2 (b d-a e) (a+b x)^3}-\frac {(d+e x)^{3/2}}{5 b (a+b x)^5}-\frac {e^3 \int \frac {1}{(a+b x)^3 \sqrt {d+e x}} \, dx}{32 b^2 (b d-a e)} \\ & = -\frac {3 e \sqrt {d+e x}}{40 b^2 (a+b x)^4}-\frac {e^2 \sqrt {d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac {e^3 \sqrt {d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac {(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac {\left (3 e^4\right ) \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx}{128 b^2 (b d-a e)^2} \\ & = -\frac {3 e \sqrt {d+e x}}{40 b^2 (a+b x)^4}-\frac {e^2 \sqrt {d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac {e^3 \sqrt {d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac {3 e^4 \sqrt {d+e x}}{128 b^2 (b d-a e)^3 (a+b x)}-\frac {(d+e x)^{3/2}}{5 b (a+b x)^5}-\frac {\left (3 e^5\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{256 b^2 (b d-a e)^3} \\ & = -\frac {3 e \sqrt {d+e x}}{40 b^2 (a+b x)^4}-\frac {e^2 \sqrt {d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac {e^3 \sqrt {d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac {3 e^4 \sqrt {d+e x}}{128 b^2 (b d-a e)^3 (a+b x)}-\frac {(d+e x)^{3/2}}{5 b (a+b x)^5}-\frac {\left (3 e^4\right ) \text {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{128 b^2 (b d-a e)^3} \\ & = -\frac {3 e \sqrt {d+e x}}{40 b^2 (a+b x)^4}-\frac {e^2 \sqrt {d+e x}}{80 b^2 (b d-a e) (a+b x)^3}+\frac {e^3 \sqrt {d+e x}}{64 b^2 (b d-a e)^2 (a+b x)^2}-\frac {3 e^4 \sqrt {d+e x}}{128 b^2 (b d-a e)^3 (a+b x)}-\frac {(d+e x)^{3/2}}{5 b (a+b x)^5}+\frac {3 e^5 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{128 b^{5/2} (b d-a e)^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.54 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {\sqrt {b} \sqrt {d+e x} \left (-15 a^4 e^4-10 a^3 b e^3 (d+7 e x)+2 a^2 b^2 e^2 \left (124 d^2+233 d e x+64 e^2 x^2\right )-2 a b^3 e \left (168 d^3+256 d^2 e x+23 d e^2 x^2-35 e^3 x^3\right )+b^4 \left (128 d^4+176 d^3 e x+8 d^2 e^2 x^2-10 d e^3 x^3+15 e^4 x^4\right )\right )}{(-b d+a e)^3 (a+b x)^5}+\frac {15 e^5 \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{(-b d+a e)^{7/2}}}{640 b^{5/2}} \]

[In]

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

[Out]

((Sqrt[b]*Sqrt[d + e*x]*(-15*a^4*e^4 - 10*a^3*b*e^3*(d + 7*e*x) + 2*a^2*b^2*e^2*(124*d^2 + 233*d*e*x + 64*e^2*
x^2) - 2*a*b^3*e*(168*d^3 + 256*d^2*e*x + 23*d*e^2*x^2 - 35*e^3*x^3) + b^4*(128*d^4 + 176*d^3*e*x + 8*d^2*e^2*
x^2 - 10*d*e^3*x^3 + 15*e^4*x^4)))/((-(b*d) + a*e)^3*(a + b*x)^5) + (15*e^5*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqr
t[-(b*d) + a*e]])/(-(b*d) + a*e)^(7/2))/(640*b^(5/2))

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.05

method	result	size
pseudoelliptic	\(-\frac {3 \left (\left (\left (-b^{4} x^{4}-\frac {14}{3} a \,b^{3} x^{3}-\frac {128}{15} a^{2} b^{2} x^{2}+\frac {14}{3} a^{3} b x +a^{4}\right ) e^{4}+\frac {2 b d \left (b^{3} x^{3}+\frac {23}{5} a \,b^{2} x^{2}-\frac {233}{5} a^{2} b x +a^{3}\right ) e^{3}}{3}-\frac {248 b^{2} \left (\frac {1}{31} b^{2} x^{2}-\frac {64}{31} a b x +a^{2}\right ) d^{2} e^{2}}{15}+\frac {112 b^{3} \left (-\frac {11 b x}{21}+a \right ) d^{3} e}{5}-\frac {128 b^{4} d^{4}}{15}\right ) \sqrt {\left (a e -b d \right ) b}\, \sqrt {e x +d}-e^{5} \left (b x +a \right )^{5} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )\right )}{128 \sqrt {\left (a e -b d \right ) b}\, \left (b x +a \right )^{5} b^{2} \left (a e -b d \right )^{3}}\)	\(219\)
derivativedivides	\(2 e^{5} \left (\frac {\frac {3 b^{2} \left (e x +d \right )^{\frac {9}{2}}}{256 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {7 b \left (e x +d \right )^{\frac {7}{2}}}{128 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {\left (e x +d \right )^{\frac {5}{2}}}{10 a e -10 b d}-\frac {7 \left (e x +d \right )^{\frac {3}{2}}}{128 b}-\frac {3 \left (a e -b d \right ) \sqrt {e x +d}}{256 b^{2}}}{\left (b \left (e x +d \right )+a e -b d \right )^{5}}+\frac {3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{256 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) b^{2} \sqrt {\left (a e -b d \right ) b}}\right )\)	\(237\)
default	\(2 e^{5} \left (\frac {\frac {3 b^{2} \left (e x +d \right )^{\frac {9}{2}}}{256 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right )}+\frac {7 b \left (e x +d \right )^{\frac {7}{2}}}{128 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {\left (e x +d \right )^{\frac {5}{2}}}{10 a e -10 b d}-\frac {7 \left (e x +d \right )^{\frac {3}{2}}}{128 b}-\frac {3 \left (a e -b d \right ) \sqrt {e x +d}}{256 b^{2}}}{\left (b \left (e x +d \right )+a e -b d \right )^{5}}+\frac {3 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{256 \left (a^{3} e^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) b^{2} \sqrt {\left (a e -b d \right ) b}}\right )\)	\(237\)

[In]

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

[Out]

-3/128*(((-b^4*x^4-14/3*a*b^3*x^3-128/15*a^2*b^2*x^2+14/3*a^3*b*x+a^4)*e^4+2/3*b*d*(b^3*x^3+23/5*a*b^2*x^2-233
/5*a^2*b*x+a^3)*e^3-248/15*b^2*(1/31*b^2*x^2-64/31*a*b*x+a^2)*d^2*e^2+112/5*b^3*(-11/21*b*x+a)*d^3*e-128/15*b^
4*d^4)*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)-e^5*(b*x+a)^5*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)))/((a*e-b*d)
*b)^(1/2)/(b*x+a)^5/b^2/(a*e-b*d)^3

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 739 vs. \(2 (176) = 352\).

Time = 0.51 (sec) , antiderivative size = 1492, normalized size of antiderivative = 7.17 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

[-1/1280*(15*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 10*a^3*b^2*e^5*x^2 + 5*a^4*b*e^5*x + a^5*e^
5)*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*(128*b^6
*d^5 - 464*a*b^5*d^4*e + 584*a^2*b^4*d^3*e^2 - 258*a^3*b^3*d^2*e^3 - 5*a^4*b^2*d*e^4 + 15*a^5*b*e^5 + 15*(b^6*
d*e^4 - a*b^5*e^5)*x^4 - 10*(b^6*d^2*e^3 - 8*a*b^5*d*e^4 + 7*a^2*b^4*e^5)*x^3 + 2*(4*b^6*d^3*e^2 - 27*a*b^5*d^
2*e^3 + 87*a^2*b^4*d*e^4 - 64*a^3*b^3*e^5)*x^2 + 2*(88*b^6*d^4*e - 344*a*b^5*d^3*e^2 + 489*a^2*b^4*d^2*e^3 - 2
68*a^3*b^3*d*e^4 + 35*a^4*b^2*e^5)*x)*sqrt(e*x + d))/(a^5*b^7*d^4 - 4*a^6*b^6*d^3*e + 6*a^7*b^5*d^2*e^2 - 4*a^
8*b^4*d*e^3 + a^9*b^3*e^4 + (b^12*d^4 - 4*a*b^11*d^3*e + 6*a^2*b^10*d^2*e^2 - 4*a^3*b^9*d*e^3 + a^4*b^8*e^4)*x
^5 + 5*(a*b^11*d^4 - 4*a^2*b^10*d^3*e + 6*a^3*b^9*d^2*e^2 - 4*a^4*b^8*d*e^3 + a^5*b^7*e^4)*x^4 + 10*(a^2*b^10*
d^4 - 4*a^3*b^9*d^3*e + 6*a^4*b^8*d^2*e^2 - 4*a^5*b^7*d*e^3 + a^6*b^6*e^4)*x^3 + 10*(a^3*b^9*d^4 - 4*a^4*b^8*d
^3*e + 6*a^5*b^7*d^2*e^2 - 4*a^6*b^6*d*e^3 + a^7*b^5*e^4)*x^2 + 5*(a^4*b^8*d^4 - 4*a^5*b^7*d^3*e + 6*a^6*b^6*d
^2*e^2 - 4*a^7*b^5*d*e^3 + a^8*b^4*e^4)*x), -1/640*(15*(b^5*e^5*x^5 + 5*a*b^4*e^5*x^4 + 10*a^2*b^3*e^5*x^3 + 1
0*a^3*b^2*e^5*x^2 + 5*a^4*b*e^5*x + a^5*e^5)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b
*e*x + b*d)) + (128*b^6*d^5 - 464*a*b^5*d^4*e + 584*a^2*b^4*d^3*e^2 - 258*a^3*b^3*d^2*e^3 - 5*a^4*b^2*d*e^4 +
15*a^5*b*e^5 + 15*(b^6*d*e^4 - a*b^5*e^5)*x^4 - 10*(b^6*d^2*e^3 - 8*a*b^5*d*e^4 + 7*a^2*b^4*e^5)*x^3 + 2*(4*b^
6*d^3*e^2 - 27*a*b^5*d^2*e^3 + 87*a^2*b^4*d*e^4 - 64*a^3*b^3*e^5)*x^2 + 2*(88*b^6*d^4*e - 344*a*b^5*d^3*e^2 +
489*a^2*b^4*d^2*e^3 - 268*a^3*b^3*d*e^4 + 35*a^4*b^2*e^5)*x)*sqrt(e*x + d))/(a^5*b^7*d^4 - 4*a^6*b^6*d^3*e + 6
*a^7*b^5*d^2*e^2 - 4*a^8*b^4*d*e^3 + a^9*b^3*e^4 + (b^12*d^4 - 4*a*b^11*d^3*e + 6*a^2*b^10*d^2*e^2 - 4*a^3*b^9
*d*e^3 + a^4*b^8*e^4)*x^5 + 5*(a*b^11*d^4 - 4*a^2*b^10*d^3*e + 6*a^3*b^9*d^2*e^2 - 4*a^4*b^8*d*e^3 + a^5*b^7*e
^4)*x^4 + 10*(a^2*b^10*d^4 - 4*a^3*b^9*d^3*e + 6*a^4*b^8*d^2*e^2 - 4*a^5*b^7*d*e^3 + a^6*b^6*e^4)*x^3 + 10*(a^
3*b^9*d^4 - 4*a^4*b^8*d^3*e + 6*a^5*b^7*d^2*e^2 - 4*a^6*b^6*d*e^3 + a^7*b^5*e^4)*x^2 + 5*(a^4*b^8*d^4 - 4*a^5*
b^7*d^3*e + 6*a^6*b^6*d^2*e^2 - 4*a^7*b^5*d*e^3 + a^8*b^4*e^4)*x)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^3,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. 410 vs. \(2 (176) = 352\).

Time = 0.29 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.97 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {3 \, e^{5} \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{128 \, {\left (b^{5} d^{3} - 3 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\right )} \sqrt {-b^{2} d + a b e}} - \frac {15 \, {\left (e x + d\right )}^{\frac {9}{2}} b^{4} e^{5} - 70 \, {\left (e x + d\right )}^{\frac {7}{2}} b^{4} d e^{5} + 128 \, {\left (e x + d\right )}^{\frac {5}{2}} b^{4} d^{2} e^{5} + 70 \, {\left (e x + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{5} - 15 \, \sqrt {e x + d} b^{4} d^{4} e^{5} + 70 \, {\left (e x + d\right )}^{\frac {7}{2}} a b^{3} e^{6} - 256 \, {\left (e x + d\right )}^{\frac {5}{2}} a b^{3} d e^{6} - 210 \, {\left (e x + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{6} + 60 \, \sqrt {e x + d} a b^{3} d^{3} e^{6} + 128 \, {\left (e x + d\right )}^{\frac {5}{2}} a^{2} b^{2} e^{7} + 210 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{7} - 90 \, \sqrt {e x + d} a^{2} b^{2} d^{2} e^{7} - 70 \, {\left (e x + d\right )}^{\frac {3}{2}} a^{3} b e^{8} + 60 \, \sqrt {e x + d} a^{3} b d e^{8} - 15 \, \sqrt {e x + d} a^{4} e^{9}}{640 \, {\left (b^{5} d^{3} - 3 \, a b^{4} d^{2} e + 3 \, a^{2} b^{3} d e^{2} - a^{3} b^{2} e^{3}\right )} {\left ({\left (e x + d\right )} b - b d + a e\right )}^{5}} \]

[In]

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

[Out]

-3/128*e^5*arctan(sqrt(e*x + d)*b/sqrt(-b^2*d + a*b*e))/((b^5*d^3 - 3*a*b^4*d^2*e + 3*a^2*b^3*d*e^2 - a^3*b^2*
e^3)*sqrt(-b^2*d + a*b*e)) - 1/640*(15*(e*x + d)^(9/2)*b^4*e^5 - 70*(e*x + d)^(7/2)*b^4*d*e^5 + 128*(e*x + d)^
(5/2)*b^4*d^2*e^5 + 70*(e*x + d)^(3/2)*b^4*d^3*e^5 - 15*sqrt(e*x + d)*b^4*d^4*e^5 + 70*(e*x + d)^(7/2)*a*b^3*e
^6 - 256*(e*x + d)^(5/2)*a*b^3*d*e^6 - 210*(e*x + d)^(3/2)*a*b^3*d^2*e^6 + 60*sqrt(e*x + d)*a*b^3*d^3*e^6 + 12
8*(e*x + d)^(5/2)*a^2*b^2*e^7 + 210*(e*x + d)^(3/2)*a^2*b^2*d*e^7 - 90*sqrt(e*x + d)*a^2*b^2*d^2*e^7 - 70*(e*x
+ d)^(3/2)*a^3*b*e^8 + 60*sqrt(e*x + d)*a^3*b*d*e^8 - 15*sqrt(e*x + d)*a^4*e^9)/((b^5*d^3 - 3*a*b^4*d^2*e + 3
*a^2*b^3*d*e^2 - a^3*b^2*e^3)*((e*x + d)*b - b*d + a*e)^5)

Mupad [B] (verification not implemented)

Time = 9.58 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.91 \[ \int \frac {(d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {\frac {e^5\,{\left (d+e\,x\right )}^{5/2}}{5\,\left (a\,e-b\,d\right )}-\frac {7\,e^5\,{\left (d+e\,x\right )}^{3/2}}{64\,b}+\frac {3\,b^2\,e^5\,{\left (d+e\,x\right )}^{9/2}}{128\,{\left (a\,e-b\,d\right )}^3}-\frac {3\,e^5\,\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}}{128\,b^2}+\frac {7\,b\,e^5\,{\left (d+e\,x\right )}^{7/2}}{64\,{\left (a\,e-b\,d\right )}^2}}{\left (d+e\,x\right )\,\left (5\,a^4\,b\,e^4-20\,a^3\,b^2\,d\,e^3+30\,a^2\,b^3\,d^2\,e^2-20\,a\,b^4\,d^3\,e+5\,b^5\,d^4\right )-{\left (d+e\,x\right )}^2\,\left (-10\,a^3\,b^2\,e^3+30\,a^2\,b^3\,d\,e^2-30\,a\,b^4\,d^2\,e+10\,b^5\,d^3\right )+b^5\,{\left (d+e\,x\right )}^5-\left (5\,b^5\,d-5\,a\,b^4\,e\right )\,{\left (d+e\,x\right )}^4+a^5\,e^5-b^5\,d^5+{\left (d+e\,x\right )}^3\,\left (10\,a^2\,b^3\,e^2-20\,a\,b^4\,d\,e+10\,b^5\,d^2\right )-10\,a^2\,b^3\,d^3\,e^2+10\,a^3\,b^2\,d^2\,e^3+5\,a\,b^4\,d^4\,e-5\,a^4\,b\,d\,e^4}+\frac {3\,e^5\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{128\,b^{5/2}\,{\left (a\,e-b\,d\right )}^{7/2}} \]

[In]

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

[Out]

((e^5*(d + e*x)^(5/2))/(5*(a*e - b*d)) - (7*e^5*(d + e*x)^(3/2))/(64*b) + (3*b^2*e^5*(d + e*x)^(9/2))/(128*(a*
e - b*d)^3) - (3*e^5*(a*e - b*d)*(d + e*x)^(1/2))/(128*b^2) + (7*b*e^5*(d + e*x)^(7/2))/(64*(a*e - b*d)^2))/((
d + e*x)*(5*b^5*d^4 + 5*a^4*b*e^4 - 20*a^3*b^2*d*e^3 + 30*a^2*b^3*d^2*e^2 - 20*a*b^4*d^3*e) - (d + e*x)^2*(10*
b^5*d^3 - 10*a^3*b^2*e^3 + 30*a^2*b^3*d*e^2 - 30*a*b^4*d^2*e) + b^5*(d + e*x)^5 - (5*b^5*d - 5*a*b^4*e)*(d + e
*x)^4 + a^5*e^5 - b^5*d^5 + (d + e*x)^3*(10*b^5*d^2 + 10*a^2*b^3*e^2 - 20*a*b^4*d*e) - 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) + (3*e^5*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/
(128*b^(5/2)*(a*e - b*d)^(7/2))