3.1655 \(\int \frac {(d+e x)^{9/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

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

[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

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Rubi [A]  time = 0.10, 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, 47, 50, 63, 208} \[ -\frac {21 e^2 (d+e x)^{5/2}}{8 b^3 (a+b x)}+\frac {105 e^3 \sqrt {d+e x} (b d-a e)}{8 b^5}-\frac {105 e^3 (b d-a e)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{11/2}}-\frac {3 e (d+e x)^{7/2}}{4 b^2 (a+b x)^2}-\frac {(d+e x)^{9/2}}{3 b (a+b x)^3}+\frac {35 e^3 (d+e x)^{3/2}}{8 b^4} \]

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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

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 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, 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 ) \operatorname {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*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.30 \[ \frac {2 e^3 (d+e x)^{11/2} \, _2F_1\left (4,\frac {11}{2};\frac {13}{2};-\frac {b (d+e x)}{a e-b d}\right )}{11 (a e-b d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^3*(d + e*x)^(11/2)*Hypergeometric2F1[4, 11/2, 13/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(11*(-(b*d) + a*e)^
4)

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fricas [B]  time = 1.36, size = 730, normalized size = 4.24 \[ \left [-\frac {315 \, {\left (a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \sqrt {\frac {b d - a e}{b}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, \sqrt {e x + d} b \sqrt {\frac {b d - a e}{b}}}{b x + a}\right ) - 2 \, {\left (16 \, b^{4} e^{4} x^{4} - 8 \, b^{4} d^{4} - 18 \, a b^{3} d^{3} e - 63 \, a^{2} b^{2} d^{2} e^{2} + 420 \, a^{3} b d e^{3} - 315 \, a^{4} e^{4} + 16 \, {\left (13 \, b^{4} d e^{3} - 9 \, a b^{3} e^{4}\right )} x^{3} - 3 \, {\left (55 \, b^{4} d^{2} e^{2} - 318 \, a b^{3} d e^{3} + 231 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (25 \, b^{4} d^{3} e + 90 \, a b^{3} d^{2} e^{2} - 567 \, a^{2} b^{2} d e^{3} + 420 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}, -\frac {315 \, {\left (a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \sqrt {-\frac {b d - a e}{b}} \arctan \left (-\frac {\sqrt {e x + d} b \sqrt {-\frac {b d - a e}{b}}}{b d - a e}\right ) - {\left (16 \, b^{4} e^{4} x^{4} - 8 \, b^{4} d^{4} - 18 \, a b^{3} d^{3} e - 63 \, a^{2} b^{2} d^{2} e^{2} + 420 \, a^{3} b d e^{3} - 315 \, a^{4} e^{4} + 16 \, {\left (13 \, b^{4} d e^{3} - 9 \, a b^{3} e^{4}\right )} x^{3} - 3 \, {\left (55 \, b^{4} d^{2} e^{2} - 318 \, a b^{3} d e^{3} + 231 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (25 \, b^{4} d^{3} e + 90 \, a b^{3} d^{2} e^{2} - 567 \, a^{2} b^{2} d e^{3} + 420 \, a^{3} b e^{4}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}\right ] \]

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^3*b*d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 - a^2*b^2*e^4)*x^2 + 3*(a^2*
b^2*d*e^3 - a^3*b*e^4)*x)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e + 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b)
)/(b*x + a)) - 2*(16*b^4*e^4*x^4 - 8*b^4*d^4 - 18*a*b^3*d^3*e - 63*a^2*b^2*d^2*e^2 + 420*a^3*b*d*e^3 - 315*a^4
*e^4 + 16*(13*b^4*d*e^3 - 9*a*b^3*e^4)*x^3 - 3*(55*b^4*d^2*e^2 - 318*a*b^3*d*e^3 + 231*a^2*b^2*e^4)*x^2 - 2*(2
5*b^4*d^3*e + 90*a*b^3*d^2*e^2 - 567*a^2*b^2*d*e^3 + 420*a^3*b*e^4)*x)*sqrt(e*x + 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^3*b*d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 - a^
2*b^2*e^4)*x^2 + 3*(a^2*b^2*d*e^3 - a^3*b*e^4)*x)*sqrt(-(b*d - a*e)/b)*arctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*
e)/b)/(b*d - a*e)) - (16*b^4*e^4*x^4 - 8*b^4*d^4 - 18*a*b^3*d^3*e - 63*a^2*b^2*d^2*e^2 + 420*a^3*b*d*e^3 - 315
*a^4*e^4 + 16*(13*b^4*d*e^3 - 9*a*b^3*e^4)*x^3 - 3*(55*b^4*d^2*e^2 - 318*a*b^3*d*e^3 + 231*a^2*b^2*e^4)*x^2 -
2*(25*b^4*d^3*e + 90*a*b^3*d^2*e^2 - 567*a^2*b^2*d*e^3 + 420*a^3*b*e^4)*x)*sqrt(e*x + d))/(b^8*x^3 + 3*a*b^7*x
^2 + 3*a^2*b^6*x + a^3*b^5)]

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giac [B]  time = 0.22, size = 360, normalized size = 2.09 \[ \frac {105 \, {\left (b^{2} d^{2} e^{3} - 2 \, a b d e^{4} + a^{2} e^{5}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, \sqrt {-b^{2} d + a b e} b^{5}} - \frac {165 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{4} d^{2} e^{3} - 280 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{4} d^{3} e^{3} + 123 \, \sqrt {x e + d} b^{4} d^{4} e^{3} - 330 \, {\left (x e + d\right )}^{\frac {5}{2}} a b^{3} d e^{4} + 840 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{3} d^{2} e^{4} - 492 \, \sqrt {x e + d} a b^{3} d^{3} e^{4} + 165 \, {\left (x e + d\right )}^{\frac {5}{2}} a^{2} b^{2} e^{5} - 840 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{2} b^{2} d e^{5} + 738 \, \sqrt {x e + d} a^{2} b^{2} d^{2} e^{5} + 280 \, {\left (x e + d\right )}^{\frac {3}{2}} a^{3} b e^{6} - 492 \, \sqrt {x e + d} a^{3} b d e^{6} + 123 \, \sqrt {x e + d} a^{4} e^{7}}{24 \, {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3} b^{5}} + \frac {2 \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{8} e^{3} + 12 \, \sqrt {x e + d} b^{8} d e^{3} - 12 \, \sqrt {x e + d} a b^{7} e^{4}\right )}}{3 \, b^{12}} \]

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

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maple [B]  time = 0.07, size = 525, normalized size = 3.05 \[ -\frac {41 \sqrt {e x +d}\, a^{4} e^{7}}{8 \left (b e x +a e \right )^{3} b^{5}}+\frac {41 \sqrt {e x +d}\, a^{3} d \,e^{6}}{2 \left (b e x +a e \right )^{3} b^{4}}-\frac {123 \sqrt {e x +d}\, a^{2} d^{2} e^{5}}{4 \left (b e x +a e \right )^{3} b^{3}}+\frac {41 \sqrt {e x +d}\, a \,d^{3} e^{4}}{2 \left (b e x +a e \right )^{3} b^{2}}-\frac {41 \sqrt {e x +d}\, d^{4} e^{3}}{8 \left (b e x +a e \right )^{3} b}-\frac {35 \left (e x +d \right )^{\frac {3}{2}} a^{3} e^{6}}{3 \left (b e x +a e \right )^{3} b^{4}}+\frac {35 \left (e x +d \right )^{\frac {3}{2}} a^{2} d \,e^{5}}{\left (b e x +a e \right )^{3} b^{3}}-\frac {35 \left (e x +d \right )^{\frac {3}{2}} a \,d^{2} e^{4}}{\left (b e x +a e \right )^{3} b^{2}}+\frac {35 \left (e x +d \right )^{\frac {3}{2}} d^{3} e^{3}}{3 \left (b e x +a e \right )^{3} b}-\frac {55 \left (e x +d \right )^{\frac {5}{2}} a^{2} e^{5}}{8 \left (b e x +a e \right )^{3} b^{3}}+\frac {55 \left (e x +d \right )^{\frac {5}{2}} a d \,e^{4}}{4 \left (b e x +a e \right )^{3} b^{2}}-\frac {55 \left (e x +d \right )^{\frac {5}{2}} d^{2} e^{3}}{8 \left (b e x +a e \right )^{3} b}+\frac {105 a^{2} e^{5} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}\, b^{5}}-\frac {105 a d \,e^{4} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \sqrt {\left (a e -b d \right ) b}\, b^{4}}+\frac {105 d^{2} e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \sqrt {\left (a e -b d \right ) b}\, b^{3}}-\frac {8 \sqrt {e x +d}\, a \,e^{4}}{b^{5}}+\frac {8 \sqrt {e x +d}\, d \,e^{3}}{b^{4}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} e^{3}}{3 b^{4}} \]

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)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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

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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

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