3.1661 \(\int \frac {1}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=173 \[ -\frac {35 e^3}{8 \sqrt {d+e x} (b d-a e)^4}+\frac {35 \sqrt {b} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{9/2}}-\frac {35 e^2}{24 (a+b x) \sqrt {d+e x} (b d-a e)^3}+\frac {7 e}{12 (a+b x)^2 \sqrt {d+e x} (b d-a e)^2}-\frac {1}{3 (a+b x)^3 \sqrt {d+e x} (b d-a e)} \]

[Out]

35/8*e^3*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))*b^(1/2)/(-a*e+b*d)^(9/2)-35/8*e^3/(-a*e+b*d)^4/(e*x+d
)^(1/2)-1/3/(-a*e+b*d)/(b*x+a)^3/(e*x+d)^(1/2)+7/12*e/(-a*e+b*d)^2/(b*x+a)^2/(e*x+d)^(1/2)-35/24*e^2/(-a*e+b*d
)^3/(b*x+a)/(e*x+d)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 51, 63, 208} \[ -\frac {35 e^3}{8 \sqrt {d+e x} (b d-a e)^4}-\frac {35 e^2}{24 (a+b x) \sqrt {d+e x} (b d-a e)^3}+\frac {35 \sqrt {b} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{9/2}}+\frac {7 e}{12 (a+b x)^2 \sqrt {d+e x} (b d-a e)^2}-\frac {1}{3 (a+b x)^3 \sqrt {d+e x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35*e^3)/(8*(b*d - a*e)^4*Sqrt[d + e*x]) - 1/(3*(b*d - a*e)*(a + b*x)^3*Sqrt[d + e*x]) + (7*e)/(12*(b*d - a*e
)^2*(a + b*x)^2*Sqrt[d + e*x]) - (35*e^2)/(24*(b*d - a*e)^3*(a + b*x)*Sqrt[d + e*x]) + (35*Sqrt[b]*e^3*ArcTanh
[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*(b*d - a*e)^(9/2))

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 51

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 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 {1}{(d+e x)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {1}{(a+b x)^4 (d+e x)^{3/2}} \, dx\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 \sqrt {d+e x}}-\frac {(7 e) \int \frac {1}{(a+b x)^3 (d+e x)^{3/2}} \, dx}{6 (b d-a e)}\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 \sqrt {d+e x}}+\frac {7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x}}+\frac {\left (35 e^2\right ) \int \frac {1}{(a+b x)^2 (d+e x)^{3/2}} \, dx}{24 (b d-a e)^2}\\ &=-\frac {1}{3 (b d-a e) (a+b x)^3 \sqrt {d+e x}}+\frac {7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x}}-\frac {35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt {d+e x}}-\frac {\left (35 e^3\right ) \int \frac {1}{(a+b x) (d+e x)^{3/2}} \, dx}{16 (b d-a e)^3}\\ &=-\frac {35 e^3}{8 (b d-a e)^4 \sqrt {d+e x}}-\frac {1}{3 (b d-a e) (a+b x)^3 \sqrt {d+e x}}+\frac {7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x}}-\frac {35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt {d+e x}}-\frac {\left (35 b e^3\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 (b d-a e)^4}\\ &=-\frac {35 e^3}{8 (b d-a e)^4 \sqrt {d+e x}}-\frac {1}{3 (b d-a e) (a+b x)^3 \sqrt {d+e x}}+\frac {7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x}}-\frac {35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt {d+e x}}-\frac {\left (35 b 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 d-a e)^4}\\ &=-\frac {35 e^3}{8 (b d-a e)^4 \sqrt {d+e x}}-\frac {1}{3 (b d-a e) (a+b x)^3 \sqrt {d+e x}}+\frac {7 e}{12 (b d-a e)^2 (a+b x)^2 \sqrt {d+e x}}-\frac {35 e^2}{24 (b d-a e)^3 (a+b x) \sqrt {d+e x}}+\frac {35 \sqrt {b} e^3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 (b d-a e)^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 50, normalized size = 0.29 \[ -\frac {2 e^3 \, _2F_1\left (-\frac {1}{2},4;\frac {1}{2};-\frac {b (d+e x)}{a e-b d}\right )}{\sqrt {d+e x} (a e-b d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*e^3*Hypergeometric2F1[-1/2, 4, 1/2, -((b*(d + e*x))/(-(b*d) + a*e))])/((-(b*d) + a*e)^4*Sqrt[d + e*x])

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fricas [B]  time = 1.30, size = 1204, normalized size = 6.96 \[ \left [\frac {105 \, {\left (b^{3} e^{4} x^{4} + a^{3} d e^{3} + {\left (b^{3} d e^{3} + 3 \, a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (a b^{2} d e^{3} + a^{2} b e^{4}\right )} x^{2} + {\left (3 \, a^{2} b d e^{3} + a^{3} e^{4}\right )} x\right )} \sqrt {\frac {b}{b d - a e}} \log \left (\frac {b e x + 2 \, b d - a e + 2 \, {\left (b d - a e\right )} \sqrt {e x + d} \sqrt {\frac {b}{b d - a e}}}{b x + a}\right ) - 2 \, {\left (105 \, b^{3} e^{3} x^{3} + 8 \, b^{3} d^{3} - 38 \, a b^{2} d^{2} e + 87 \, a^{2} b d e^{2} + 48 \, a^{3} e^{3} + 35 \, {\left (b^{3} d e^{2} + 8 \, a b^{2} e^{3}\right )} x^{2} - 7 \, {\left (2 \, b^{3} d^{2} e - 14 \, a b^{2} d e^{2} - 33 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{4} d^{5} - 4 \, a^{4} b^{3} d^{4} e + 6 \, a^{5} b^{2} d^{3} e^{2} - 4 \, a^{6} b d^{2} e^{3} + a^{7} d e^{4} + {\left (b^{7} d^{4} e - 4 \, a b^{6} d^{3} e^{2} + 6 \, a^{2} b^{5} d^{2} e^{3} - 4 \, a^{3} b^{4} d e^{4} + a^{4} b^{3} e^{5}\right )} x^{4} + {\left (b^{7} d^{5} - a b^{6} d^{4} e - 6 \, a^{2} b^{5} d^{3} e^{2} + 14 \, a^{3} b^{4} d^{2} e^{3} - 11 \, a^{4} b^{3} d e^{4} + 3 \, a^{5} b^{2} e^{5}\right )} x^{3} + 3 \, {\left (a b^{6} d^{5} - 3 \, a^{2} b^{5} d^{4} e + 2 \, a^{3} b^{4} d^{3} e^{2} + 2 \, a^{4} b^{3} d^{2} e^{3} - 3 \, a^{5} b^{2} d e^{4} + a^{6} b e^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} d^{5} - 11 \, a^{3} b^{4} d^{4} e + 14 \, a^{4} b^{3} d^{3} e^{2} - 6 \, a^{5} b^{2} d^{2} e^{3} - a^{6} b d e^{4} + a^{7} e^{5}\right )} x\right )}}, \frac {105 \, {\left (b^{3} e^{4} x^{4} + a^{3} d e^{3} + {\left (b^{3} d e^{3} + 3 \, a b^{2} e^{4}\right )} x^{3} + 3 \, {\left (a b^{2} d e^{3} + a^{2} b e^{4}\right )} x^{2} + {\left (3 \, a^{2} b d e^{3} + a^{3} e^{4}\right )} x\right )} \sqrt {-\frac {b}{b d - a e}} \arctan \left (-\frac {{\left (b d - a e\right )} \sqrt {e x + d} \sqrt {-\frac {b}{b d - a e}}}{b e x + b d}\right ) - {\left (105 \, b^{3} e^{3} x^{3} + 8 \, b^{3} d^{3} - 38 \, a b^{2} d^{2} e + 87 \, a^{2} b d e^{2} + 48 \, a^{3} e^{3} + 35 \, {\left (b^{3} d e^{2} + 8 \, a b^{2} e^{3}\right )} x^{2} - 7 \, {\left (2 \, b^{3} d^{2} e - 14 \, a b^{2} d e^{2} - 33 \, a^{2} b e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{4} d^{5} - 4 \, a^{4} b^{3} d^{4} e + 6 \, a^{5} b^{2} d^{3} e^{2} - 4 \, a^{6} b d^{2} e^{3} + a^{7} d e^{4} + {\left (b^{7} d^{4} e - 4 \, a b^{6} d^{3} e^{2} + 6 \, a^{2} b^{5} d^{2} e^{3} - 4 \, a^{3} b^{4} d e^{4} + a^{4} b^{3} e^{5}\right )} x^{4} + {\left (b^{7} d^{5} - a b^{6} d^{4} e - 6 \, a^{2} b^{5} d^{3} e^{2} + 14 \, a^{3} b^{4} d^{2} e^{3} - 11 \, a^{4} b^{3} d e^{4} + 3 \, a^{5} b^{2} e^{5}\right )} x^{3} + 3 \, {\left (a b^{6} d^{5} - 3 \, a^{2} b^{5} d^{4} e + 2 \, a^{3} b^{4} d^{3} e^{2} + 2 \, a^{4} b^{3} d^{2} e^{3} - 3 \, a^{5} b^{2} d e^{4} + a^{6} b e^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{5} d^{5} - 11 \, a^{3} b^{4} d^{4} e + 14 \, a^{4} b^{3} d^{3} e^{2} - 6 \, a^{5} b^{2} d^{2} e^{3} - a^{6} b d e^{4} + a^{7} e^{5}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(105*(b^3*e^4*x^4 + a^3*d*e^3 + (b^3*d*e^3 + 3*a*b^2*e^4)*x^3 + 3*(a*b^2*d*e^3 + a^2*b*e^4)*x^2 + (3*a^2
*b*d*e^3 + a^3*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*(105*b^3*e^3*x^3 + 8*b^3*d^3 - 38*a*b^2*d^2*e + 87*a^2*b*d*e^2 + 48*a^3*e^3 + 35*(b^3*
d*e^2 + 8*a*b^2*e^3)*x^2 - 7*(2*b^3*d^2*e - 14*a*b^2*d*e^2 - 33*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^3*b^4*d^5 - 4*
a^4*b^3*d^4*e + 6*a^5*b^2*d^3*e^2 - 4*a^6*b*d^2*e^3 + a^7*d*e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 + 6*a^2*b^5*d^2
*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^4 + (b^7*d^5 - a*b^6*d^4*e - 6*a^2*b^5*d^3*e^2 + 14*a^3*b^4*d^2*e^3 -
11*a^4*b^3*d*e^4 + 3*a^5*b^2*e^5)*x^3 + 3*(a*b^6*d^5 - 3*a^2*b^5*d^4*e + 2*a^3*b^4*d^3*e^2 + 2*a^4*b^3*d^2*e^3
- 3*a^5*b^2*d*e^4 + a^6*b*e^5)*x^2 + (3*a^2*b^5*d^5 - 11*a^3*b^4*d^4*e + 14*a^4*b^3*d^3*e^2 - 6*a^5*b^2*d^2*e
^3 - a^6*b*d*e^4 + a^7*e^5)*x), 1/24*(105*(b^3*e^4*x^4 + a^3*d*e^3 + (b^3*d*e^3 + 3*a*b^2*e^4)*x^3 + 3*(a*b^2*
d*e^3 + a^2*b*e^4)*x^2 + (3*a^2*b*d*e^3 + a^3*e^4)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*s
qrt(-b/(b*d - a*e))/(b*e*x + b*d)) - (105*b^3*e^3*x^3 + 8*b^3*d^3 - 38*a*b^2*d^2*e + 87*a^2*b*d*e^2 + 48*a^3*e
^3 + 35*(b^3*d*e^2 + 8*a*b^2*e^3)*x^2 - 7*(2*b^3*d^2*e - 14*a*b^2*d*e^2 - 33*a^2*b*e^3)*x)*sqrt(e*x + d))/(a^3
*b^4*d^5 - 4*a^4*b^3*d^4*e + 6*a^5*b^2*d^3*e^2 - 4*a^6*b*d^2*e^3 + a^7*d*e^4 + (b^7*d^4*e - 4*a*b^6*d^3*e^2 +
6*a^2*b^5*d^2*e^3 - 4*a^3*b^4*d*e^4 + a^4*b^3*e^5)*x^4 + (b^7*d^5 - a*b^6*d^4*e - 6*a^2*b^5*d^3*e^2 + 14*a^3*b
^4*d^2*e^3 - 11*a^4*b^3*d*e^4 + 3*a^5*b^2*e^5)*x^3 + 3*(a*b^6*d^5 - 3*a^2*b^5*d^4*e + 2*a^3*b^4*d^3*e^2 + 2*a^
4*b^3*d^2*e^3 - 3*a^5*b^2*d*e^4 + a^6*b*e^5)*x^2 + (3*a^2*b^5*d^5 - 11*a^3*b^4*d^4*e + 14*a^4*b^3*d^3*e^2 - 6*
a^5*b^2*d^2*e^3 - a^6*b*d*e^4 + a^7*e^5)*x)]

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giac [B]  time = 0.22, size = 324, normalized size = 1.87 \[ -\frac {35 \, b \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{3}}{8 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt {-b^{2} d + a b e}} - \frac {2 \, e^{3}}{{\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \sqrt {x e + d}} - \frac {57 \, {\left (x e + d\right )}^{\frac {5}{2}} b^{3} e^{3} - 136 \, {\left (x e + d\right )}^{\frac {3}{2}} b^{3} d e^{3} + 87 \, \sqrt {x e + d} b^{3} d^{2} e^{3} + 136 \, {\left (x e + d\right )}^{\frac {3}{2}} a b^{2} e^{4} - 174 \, \sqrt {x e + d} a b^{2} d e^{4} + 87 \, \sqrt {x e + d} a^{2} b e^{5}}{24 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-35/8*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^3/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3
*b*d*e^3 + a^4*e^4)*sqrt(-b^2*d + a*b*e)) - 2*e^3/((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^
3 + a^4*e^4)*sqrt(x*e + d)) - 1/24*(57*(x*e + d)^(5/2)*b^3*e^3 - 136*(x*e + d)^(3/2)*b^3*d*e^3 + 87*sqrt(x*e +
d)*b^3*d^2*e^3 + 136*(x*e + d)^(3/2)*a*b^2*e^4 - 174*sqrt(x*e + d)*a*b^2*d*e^4 + 87*sqrt(x*e + d)*a^2*b*e^5)/
((b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*((x*e + d)*b - b*d + a*e)^3)

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maple [B]  time = 0.06, size = 292, normalized size = 1.69 \[ -\frac {29 \sqrt {e x +d}\, a^{2} b \,e^{5}}{8 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{3}}+\frac {29 \sqrt {e x +d}\, a \,b^{2} d \,e^{4}}{4 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{3}}-\frac {29 \sqrt {e x +d}\, b^{3} d^{2} e^{3}}{8 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{3}}-\frac {17 \left (e x +d \right )^{\frac {3}{2}} a \,b^{2} e^{4}}{3 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{3}}+\frac {17 \left (e x +d \right )^{\frac {3}{2}} b^{3} d \,e^{3}}{3 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{3}}-\frac {19 \left (e x +d \right )^{\frac {5}{2}} b^{3} e^{3}}{8 \left (a e -b d \right )^{4} \left (b e x +a e \right )^{3}}-\frac {35 b \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a e -b d \right )^{4} \sqrt {\left (a e -b d \right ) b}}-\frac {2 e^{3}}{\left (a e -b d \right )^{4} \sqrt {e x +d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-19/8*e^3/(a*e-b*d)^4*b^3/(b*e*x+a*e)^3*(e*x+d)^(5/2)-17/3*e^4/(a*e-b*d)^4*b^2/(b*e*x+a*e)^3*(e*x+d)^(3/2)*a+1
7/3*e^3/(a*e-b*d)^4*b^3/(b*e*x+a*e)^3*(e*x+d)^(3/2)*d-29/8*e^5/(a*e-b*d)^4*b/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a^2+2
9/4*e^4/(a*e-b*d)^4*b^2/(b*e*x+a*e)^3*(e*x+d)^(1/2)*a*d-29/8*e^3/(a*e-b*d)^4*b^3/(b*e*x+a*e)^3*(e*x+d)^(1/2)*d
^2-35/8*e^3/(a*e-b*d)^4*b/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)-2*e^3/(a*e-b*d)^4/(e
*x+d)^(1/2)

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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(1/(e*x+d)^(3/2)/(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 = 0.78, size = 294, normalized size = 1.70 \[ -\frac {\frac {2\,e^3}{a\,e-b\,d}+\frac {35\,b^2\,e^3\,{\left (d+e\,x\right )}^2}{3\,{\left (a\,e-b\,d\right )}^3}+\frac {35\,b^3\,e^3\,{\left (d+e\,x\right )}^3}{8\,{\left (a\,e-b\,d\right )}^4}+\frac {77\,b\,e^3\,\left (d+e\,x\right )}{8\,{\left (a\,e-b\,d\right )}^2}}{\sqrt {d+e\,x}\,\left (a^3\,e^3-3\,a^2\,b\,d\,e^2+3\,a\,b^2\,d^2\,e-b^3\,d^3\right )+b^3\,{\left (d+e\,x\right )}^{7/2}-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^{5/2}+{\left (d+e\,x\right )}^{3/2}\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )}-\frac {35\,\sqrt {b}\,e^3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}\,\left (a^4\,e^4-4\,a^3\,b\,d\,e^3+6\,a^2\,b^2\,d^2\,e^2-4\,a\,b^3\,d^3\,e+b^4\,d^4\right )}{{\left (a\,e-b\,d\right )}^{9/2}}\right )}{8\,{\left (a\,e-b\,d\right )}^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*e^3)/(a*e - b*d) + (35*b^2*e^3*(d + e*x)^2)/(3*(a*e - b*d)^3) + (35*b^3*e^3*(d + e*x)^3)/(8*(a*e - b*d)^
4) + (77*b*e^3*(d + e*x))/(8*(a*e - b*d)^2))/((d + e*x)^(1/2)*(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)^(7/2) - (3*b^3*d - 3*a*b^2*e)*(d + e*x)^(5/2) + (d + e*x)^(3/2)*(3*b^3*d^2 + 3*a^2*b*e^2 -
6*a*b^2*d*e)) - (35*b^(1/2)*e^3*atan((b^(1/2)*(d + e*x)^(1/2)*(a^4*e^4 + b^4*d^4 + 6*a^2*b^2*d^2*e^2 - 4*a*b^
3*d^3*e - 4*a^3*b*d*e^3))/(a*e - b*d)^(9/2)))/(8*(a*e - b*d)^(9/2))

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

[Out]

Timed out

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