3.17.55 \(\int \frac {(d+e x)^{9/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\) [1655]

Optimal. Leaf size=172 \[ \frac {105 e^3 (b d-a e) \sqrt {d+e x}}{8 b^5}+\frac {35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}-\frac {105 e^3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{11/2}} \]

[Out]

35/8*e^3*(e*x+d)^(3/2)/b^4-21/8*e^2*(e*x+d)^(5/2)/b^3/(b*x+a)-3/4*e*(e*x+d)^(7/2)/b^2/(b*x+a)^2-1/3*(e*x+d)^(9
/2)/b/(b*x+a)^3-105/8*e^3*(-a*e+b*d)^(3/2)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(11/2)+105/8*e^3*
(-a*e+b*d)*(e*x+d)^(1/2)/b^5

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {27, 43, 52, 65, 214} \begin {gather*} -\frac {105 e^3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{11/2}}+\frac {105 e^3 \sqrt {d+e x} (b d-a e)}{8 b^5}-\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac {35 e^3 (d+e x)^{3/2}}{8 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(9/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(105*e^3*(b*d - a*e)*Sqrt[d + e*x])/(8*b^5) + (35*e^3*(d + e*x)^(3/2))/(8*b^4) - (21*e^2*(d + e*x)^(5/2))/(8*b
^3*(a + b*x)) - (3*e*(d + e*x)^(7/2))/(4*b^2*(a + b*x)^2) - (d + e*x)^(9/2)/(3*b*(a + b*x)^3) - (105*e^3*(b*d
- a*e)^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(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 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 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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*} \int \frac {(d+e x)^{9/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {(d+e x)^{9/2}}{(a+b x)^4} \, dx\\ &=-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac {(3 e) \int \frac {(d+e x)^{7/2}}{(a+b x)^3} \, dx}{2 b}\\ &=-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac {\left (21 e^2\right ) \int \frac {(d+e x)^{5/2}}{(a+b x)^2} \, dx}{8 b^2}\\ &=-\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac {\left (105 e^3\right ) \int \frac {(d+e x)^{3/2}}{a+b x} \, dx}{16 b^3}\\ &=\frac {35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac {\left (105 e^3 (b d-a e)\right ) \int \frac {\sqrt {d+e x}}{a+b x} \, dx}{16 b^4}\\ &=\frac {105 e^3 (b d-a e) \sqrt {d+e x}}{8 b^5}+\frac {35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac {\left (105 e^3 (b d-a e)^2\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 b^5}\\ &=\frac {105 e^3 (b d-a e) \sqrt {d+e x}}{8 b^5}+\frac {35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac {\left (105 e^2 (b d-a 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^5}\\ &=\frac {105 e^3 (b d-a e) \sqrt {d+e x}}{8 b^5}+\frac {35 e^3 (d+e x)^{3/2}}{8 b^4}-\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}-\frac {105 e^3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{11/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.86, size = 208, normalized size = 1.21 \begin {gather*} -\frac {\sqrt {d+e x} \left (315 a^4 e^4-420 a^3 b e^3 (d-2 e x)+63 a^2 b^2 e^2 \left (d^2-18 d e x+11 e^2 x^2\right )+18 a b^3 e \left (d^3+10 d^2 e x-53 d e^2 x^2+8 e^3 x^3\right )+b^4 \left (8 d^4+50 d^3 e x+165 d^2 e^2 x^2-208 d e^3 x^3-16 e^4 x^4\right )\right )}{24 b^5 (a+b x)^3}+\frac {105 e^3 (-b d+a e)^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{8 b^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(9/2)/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-1/24*(Sqrt[d + e*x]*(315*a^4*e^4 - 420*a^3*b*e^3*(d - 2*e*x) + 63*a^2*b^2*e^2*(d^2 - 18*d*e*x + 11*e^2*x^2) +
18*a*b^3*e*(d^3 + 10*d^2*e*x - 53*d*e^2*x^2 + 8*e^3*x^3) + b^4*(8*d^4 + 50*d^3*e*x + 165*d^2*e^2*x^2 - 208*d*
e^3*x^3 - 16*e^4*x^4)))/(b^5*(a + b*x)^3) + (105*e^3*(-(b*d) + a*e)^(3/2)*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[
-(b*d) + a*e]])/(8*b^(11/2))

________________________________________________________________________________________

Maple [A]
time = 0.80, size = 267, normalized size = 1.55

method	result	size
derivativedivides	\(2 e^{3} \left (-\frac {-\frac {\left (e x +d \right )^{\frac {3}{2}} b}{3}+4 a e \sqrt {e x +d}-4 d b \sqrt {e x +d}}{b^{5}}+\frac {\frac {\left (-\frac {55}{16} a^{2} b^{2} e^{2}+\frac {55}{8} a \,b^{3} d e -\frac {55}{16} b^{4} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}-\frac {35 b \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (-\frac {41}{16} e^{4} a^{4}+\frac {41}{4} a^{3} b d \,e^{3}-\frac {123}{8} a^{2} b^{2} d^{2} e^{2}+\frac {41}{4} a \,b^{3} d^{3} e -\frac {41}{16} b^{4} d^{4}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -b d \right )^{3}}+\frac {105 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 \sqrt {b \left (a e -b d \right )}}}{b^{5}}\right )\)	\(267\)
default	\(2 e^{3} \left (-\frac {-\frac {\left (e x +d \right )^{\frac {3}{2}} b}{3}+4 a e \sqrt {e x +d}-4 d b \sqrt {e x +d}}{b^{5}}+\frac {\frac {\left (-\frac {55}{16} a^{2} b^{2} e^{2}+\frac {55}{8} a \,b^{3} d e -\frac {55}{16} b^{4} d^{2}\right ) \left (e x +d \right )^{\frac {5}{2}}-\frac {35 b \left (e^{3} a^{3}-3 a^{2} b d \,e^{2}+3 a \,b^{2} d^{2} e -b^{3} d^{3}\right ) \left (e x +d \right )^{\frac {3}{2}}}{6}+\left (-\frac {41}{16} e^{4} a^{4}+\frac {41}{4} a^{3} b d \,e^{3}-\frac {123}{8} a^{2} b^{2} d^{2} e^{2}+\frac {41}{4} a \,b^{3} d^{3} e -\frac {41}{16} b^{4} d^{4}\right ) \sqrt {e x +d}}{\left (\left (e x +d \right ) b +a e -b d \right )^{3}}+\frac {105 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right ) \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right )}{16 \sqrt {b \left (a e -b d \right )}}}{b^{5}}\right )\)	\(267\)
risch	\(-\frac {2 e^{3} \left (-b e x +12 a e -13 b d \right ) \sqrt {e x +d}}{3 b^{5}}-\frac {55 e^{5} \left (e x +d \right )^{\frac {5}{2}} a^{2}}{8 b^{3} \left (b e x +a e \right )^{3}}+\frac {55 e^{4} \left (e x +d \right )^{\frac {5}{2}} a d}{4 b^{2} \left (b e x +a e \right )^{3}}-\frac {55 e^{3} \left (e x +d \right )^{\frac {5}{2}} d^{2}}{8 b \left (b e x +a e \right )^{3}}-\frac {35 e^{6} \left (e x +d \right )^{\frac {3}{2}} a^{3}}{3 b^{4} \left (b e x +a e \right )^{3}}+\frac {35 e^{5} \left (e x +d \right )^{\frac {3}{2}} a^{2} d}{b^{3} \left (b e x +a e \right )^{3}}-\frac {35 e^{4} \left (e x +d \right )^{\frac {3}{2}} a \,d^{2}}{b^{2} \left (b e x +a e \right )^{3}}+\frac {35 e^{3} \left (e x +d \right )^{\frac {3}{2}} d^{3}}{3 b \left (b e x +a e \right )^{3}}-\frac {41 e^{7} \sqrt {e x +d}\, a^{4}}{8 b^{5} \left (b e x +a e \right )^{3}}+\frac {41 e^{6} \sqrt {e x +d}\, a^{3} d}{2 b^{4} \left (b e x +a e \right )^{3}}-\frac {123 e^{5} \sqrt {e x +d}\, a^{2} d^{2}}{4 b^{3} \left (b e x +a e \right )^{3}}+\frac {41 e^{4} \sqrt {e x +d}\, a \,d^{3}}{2 b^{2} \left (b e x +a e \right )^{3}}-\frac {41 e^{3} \sqrt {e x +d}\, d^{4}}{8 b \left (b e x +a e \right )^{3}}+\frac {105 e^{5} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2}}{8 b^{5} \sqrt {b \left (a e -b d \right )}}-\frac {105 e^{4} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a d}{4 b^{4} \sqrt {b \left (a e -b d \right )}}+\frac {105 e^{3} \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) d^{2}}{8 b^{3} \sqrt {b \left (a e -b d \right )}}\)	\(507\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)

[Out]

2*e^3*(-1/b^5*(-1/3*(e*x+d)^(3/2)*b+4*a*e*(e*x+d)^(1/2)-4*d*b*(e*x+d)^(1/2))+1/b^5*(((-55/16*a^2*b^2*e^2+55/8*
a*b^3*d*e-55/16*b^4*d^2)*(e*x+d)^(5/2)-35/6*b*(a^3*e^3-3*a^2*b*d*e^2+3*a*b^2*d^2*e-b^3*d^3)*(e*x+d)^(3/2)+(-41
/16*e^4*a^4+41/4*a^3*b*d*e^3-123/8*a^2*b^2*d^2*e^2+41/4*a*b^3*d^3*e-41/16*b^4*d^4)*(e*x+d)^(1/2))/((e*x+d)*b+a
*e-b*d)^3+105/16*(a^2*e^2-2*a*b*d*e+b^2*d^2)/(b*(a*e-b*d))^(1/2)*arctan(b*(e*x+d)^(1/2)/(b*(a*e-b*d))^(1/2))))

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(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(b*d-%e*a>0)', see `assume?` fo
r more detai

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 333 vs. \(2 (146) = 292\).
time = 2.56, size = 677, normalized size = 3.94 \begin {gather*} \left [\frac {315 \, {\left ({\left (a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 3 \, a^{3} b x + a^{4}\right )} e^{4} - {\left (b^{4} d x^{3} + 3 \, a b^{3} d x^{2} + 3 \, a^{2} b^{2} d x + a^{3} b d\right )} e^{3}\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {2 \, b d + 2 \, \sqrt {x e + d} b \sqrt {\frac {b d - a e}{b}} + {\left (b x - a\right )} e}{b x + a}\right ) - 2 \, {\left (8 \, b^{4} d^{4} - {\left (16 \, b^{4} x^{4} - 144 \, a b^{3} x^{3} - 693 \, a^{2} b^{2} x^{2} - 840 \, a^{3} b x - 315 \, a^{4}\right )} e^{4} - 2 \, {\left (104 \, b^{4} d x^{3} + 477 \, a b^{3} d x^{2} + 567 \, a^{2} b^{2} d x + 210 \, a^{3} b d\right )} e^{3} + 3 \, {\left (55 \, b^{4} d^{2} x^{2} + 60 \, a b^{3} d^{2} x + 21 \, a^{2} b^{2} d^{2}\right )} e^{2} + 2 \, {\left (25 \, b^{4} d^{3} x + 9 \, a b^{3} d^{3}\right )} e\right )} \sqrt {x e + d}}{48 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}, \frac {315 \, {\left ({\left (a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + 3 \, a^{3} b x + a^{4}\right )} e^{4} - {\left (b^{4} d x^{3} + 3 \, a b^{3} d x^{2} + 3 \, a^{2} b^{2} d x + a^{3} b d\right )} e^{3}\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {x e + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (8 \, b^{4} d^{4} - {\left (16 \, b^{4} x^{4} - 144 \, a b^{3} x^{3} - 693 \, a^{2} b^{2} x^{2} - 840 \, a^{3} b x - 315 \, a^{4}\right )} e^{4} - 2 \, {\left (104 \, b^{4} d x^{3} + 477 \, a b^{3} d x^{2} + 567 \, a^{2} b^{2} d x + 210 \, a^{3} b d\right )} e^{3} + 3 \, {\left (55 \, b^{4} d^{2} x^{2} + 60 \, a b^{3} d^{2} x + 21 \, a^{2} b^{2} d^{2}\right )} e^{2} + 2 \, {\left (25 \, b^{4} d^{3} x + 9 \, a b^{3} d^{3}\right )} e\right )} \sqrt {x e + d}}{24 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/48*(315*((a*b^3*x^3 + 3*a^2*b^2*x^2 + 3*a^3*b*x + a^4)*e^4 - (b^4*d*x^3 + 3*a*b^3*d*x^2 + 3*a^2*b^2*d*x + a
^3*b*d)*e^3)*sqrt((b*d - a*e)/b)*log((2*b*d + 2*sqrt(x*e + d)*b*sqrt((b*d - a*e)/b) + (b*x - a)*e)/(b*x + a))
- 2*(8*b^4*d^4 - (16*b^4*x^4 - 144*a*b^3*x^3 - 693*a^2*b^2*x^2 - 840*a^3*b*x - 315*a^4)*e^4 - 2*(104*b^4*d*x^3
+ 477*a*b^3*d*x^2 + 567*a^2*b^2*d*x + 210*a^3*b*d)*e^3 + 3*(55*b^4*d^2*x^2 + 60*a*b^3*d^2*x + 21*a^2*b^2*d^2)
*e^2 + 2*(25*b^4*d^3*x + 9*a*b^3*d^3)*e)*sqrt(x*e + d))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5), 1/24*
(315*((a*b^3*x^3 + 3*a^2*b^2*x^2 + 3*a^3*b*x + a^4)*e^4 - (b^4*d*x^3 + 3*a*b^3*d*x^2 + 3*a^2*b^2*d*x + a^3*b*d
)*e^3)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(x*e + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (8*b^4*d^4 - (16*b^4*x
^4 - 144*a*b^3*x^3 - 693*a^2*b^2*x^2 - 840*a^3*b*x - 315*a^4)*e^4 - 2*(104*b^4*d*x^3 + 477*a*b^3*d*x^2 + 567*a
^2*b^2*d*x + 210*a^3*b*d)*e^3 + 3*(55*b^4*d^2*x^2 + 60*a*b^3*d^2*x + 21*a^2*b^2*d^2)*e^2 + 2*(25*b^4*d^3*x + 9
*a*b^3*d^3)*e)*sqrt(x*e + d))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5)]

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(9/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (146) = 292\).
time = 0.93, size = 360, normalized size = 2.09 \begin {gather*} \frac {105 \, {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, \sqrt {-b^{2} d + a b e} b^{5}} - \frac {165 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{4} d^{2} e^{3} - 280 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{3} + 123 \, \sqrt {x e + d} b^{4} d^{4} e^{3} - 330 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{3} d e^{4} + 840 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{4} - 492 \, \sqrt {x e + d} a b^{3} d^{3} e^{4} + 165 \, {\left (x e + d\right )}^{\frac {5}{2}} a^{2} b^{2} e^{5} - 840 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{5} + 738 \, \sqrt {x e + d} a^{2} b^{2} d^{2} e^{5} + 280 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{3} b e^{6} - 492 \, \sqrt {x e + d} a^{3} b d e^{6} + 123 \, \sqrt {x e + d} a^{4} e^{7}}{24 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{5}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{8} e^{3} + 12 \, \sqrt {x e + d} b^{8} d e^{3} - 12 \, \sqrt {x e + d} a b^{7} e^{4}\right )}}{3 \, b^{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(9/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

105/8*(b^2*d^2*e^3 - 2*a*b*d*e^4 + a^2*e^5)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)
*b^5) - 1/24*(165*(x*e + d)^(5/2)*b^4*d^2*e^3 - 280*(x*e + d)^(3/2)*b^4*d^3*e^3 + 123*sqrt(x*e + d)*b^4*d^4*e^
3 - 330*(x*e + d)^(5/2)*a*b^3*d*e^4 + 840*(x*e + d)^(3/2)*a*b^3*d^2*e^4 - 492*sqrt(x*e + d)*a*b^3*d^3*e^4 + 16
5*(x*e + d)^(5/2)*a^2*b^2*e^5 - 840*(x*e + d)^(3/2)*a^2*b^2*d*e^5 + 738*sqrt(x*e + d)*a^2*b^2*d^2*e^5 + 280*(x
*e + d)^(3/2)*a^3*b*e^6 - 492*sqrt(x*e + d)*a^3*b*d*e^6 + 123*sqrt(x*e + d)*a^4*e^7)/(((x*e + d)*b - b*d + a*e
)^3*b^5) + 2/3*((x*e + d)^(3/2)*b^8*e^3 + 12*sqrt(x*e + d)*b^8*d*e^3 - 12*sqrt(x*e + d)*a*b^7*e^4)/b^12

________________________________________________________________________________________

Mupad [B]
time = 0.68, size = 388, normalized size = 2.26 \begin {gather*} \frac {2\,e^3\,{\left (d+e\,x\right )}^{3/2}}{3\,b^4}-\frac {{\left (d+e\,x\right )}^{5/2}\,\left (\frac {55\,a^2\,b^2\,e^5}{8}-\frac {55\,a\,b^3\,d\,e^4}{4}+\frac {55\,b^4\,d^2\,e^3}{8}\right )+{\left (d+e\,x\right )}^{3/2}\,\left (\frac {35\,a^3\,b\,e^6}{3}-35\,a^2\,b^2\,d\,e^5+35\,a\,b^3\,d^2\,e^4-\frac {35\,b^4\,d^3\,e^3}{3}\right )+\sqrt {d+e\,x}\,\left (\frac {41\,a^4\,e^7}{8}-\frac {41\,a^3\,b\,d\,e^6}{2}+\frac {123\,a^2\,b^2\,d^2\,e^5}{4}-\frac {41\,a\,b^3\,d^3\,e^4}{2}+\frac {41\,b^4\,d^4\,e^3}{8}\right )}{b^8\,{\left (d+e\,x\right )}^3-\left (3\,b^8\,d-3\,a\,b^7\,e\right )\,{\left (d+e\,x\right )}^2+\left (d+e\,x\right )\,\left (3\,a^2\,b^6\,e^2-6\,a\,b^7\,d\,e+3\,b^8\,d^2\right )-b^8\,d^3+a^3\,b^5\,e^3-3\,a^2\,b^6\,d\,e^2+3\,a\,b^7\,d^2\,e}+\frac {2\,e^3\,\left (4\,b^4\,d-4\,a\,b^3\,e\right )\,\sqrt {d+e\,x}}{b^8}+\frac {105\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^3\,{\left (a\,e-b\,d\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^5-2\,a\,b\,d\,e^4+b^2\,d^2\,e^3}\right )\,{\left (a\,e-b\,d\right )}^{3/2}}{8\,b^{11/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(9/2)/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

(2*e^3*(d + e*x)^(3/2))/(3*b^4) - ((d + e*x)^(5/2)*((55*a^2*b^2*e^5)/8 + (55*b^4*d^2*e^3)/8 - (55*a*b^3*d*e^4)
/4) + (d + e*x)^(3/2)*((35*a^3*b*e^6)/3 - (35*b^4*d^3*e^3)/3 + 35*a*b^3*d^2*e^4 - 35*a^2*b^2*d*e^5) + (d + e*x
)^(1/2)*((41*a^4*e^7)/8 + (41*b^4*d^4*e^3)/8 - (41*a*b^3*d^3*e^4)/2 + (123*a^2*b^2*d^2*e^5)/4 - (41*a^3*b*d*e^
6)/2))/(b^8*(d + e*x)^3 - (3*b^8*d - 3*a*b^7*e)*(d + e*x)^2 + (d + e*x)*(3*b^8*d^2 + 3*a^2*b^6*e^2 - 6*a*b^7*d
*e) - b^8*d^3 + a^3*b^5*e^3 - 3*a^2*b^6*d*e^2 + 3*a*b^7*d^2*e) + (2*e^3*(4*b^4*d - 4*a*b^3*e)*(d + e*x)^(1/2))
/b^8 + (105*e^3*atan((b^(1/2)*e^3*(a*e - b*d)^(3/2)*(d + e*x)^(1/2))/(a^2*e^5 + b^2*d^2*e^3 - 2*a*b*d*e^4))*(a
*e - b*d)^(3/2))/(8*b^(11/2))

________________________________________________________________________________________