### 3.2300 $$\int \frac{1}{(d+e x)^{3/2} (a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=604 $-\frac{e \left (-2 c e (5 a e+b d)+3 b^2 e^2+2 c^2 d^2\right )}{\left (b^2-4 a c\right ) \sqrt{d+e x} \left (a e^2-b d e+c d^2\right )^2}+\frac{\sqrt{c} \left (-2 c^2 d e \left (-d \sqrt{b^2-4 a c}-16 a e+6 b d\right )-2 c e^2 \left (b d \sqrt{b^2-4 a c}+5 a e \sqrt{b^2-4 a c}+8 a b e+b^2 d\right )+3 b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )+8 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )^2}-\frac{\sqrt{c} \left (-2 c^2 d e \left (d \sqrt{b^2-4 a c}-16 a e+6 b d\right )-2 c e^2 \left (-b d \sqrt{b^2-4 a c}-5 a e \sqrt{b^2-4 a c}+8 a b e+b^2 d\right )+3 b^2 e^3 \left (b-\sqrt{b^2-4 a c}\right )+8 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )^2}-\frac{2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \sqrt{d+e x} \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}$

[Out]

-((e*(2*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(b*d + 5*a*e)))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)^2*Sqrt[d + e*x])) -
(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x]*(a + b*x +
c*x^2)) + (Sqrt[c]*(8*c^3*d^3 + 3*b^2*(b + Sqrt[b^2 - 4*a*c])*e^3 - 2*c^2*d*e*(6*b*d - Sqrt[b^2 - 4*a*c]*d -
16*a*e) - 2*c*e^2*(b^2*d + b*Sqrt[b^2 - 4*a*c]*d + 8*a*b*e + 5*a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c
]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*(b^2 - 4*a*c)^(3/2)*Sqrt[2*c*d - (b - Sqrt
[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)^2) - (Sqrt[c]*(8*c^3*d^3 + 3*b^2*(b - Sqrt[b^2 - 4*a*c])*e^3 - 2*c^2
*d*e*(6*b*d + Sqrt[b^2 - 4*a*c]*d - 16*a*e) - 2*c*e^2*(b^2*d - b*Sqrt[b^2 - 4*a*c]*d + 8*a*b*e - 5*a*Sqrt[b^2
- 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*(b^2 -
4*a*c)^(3/2)*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)^2)

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Rubi [A]  time = 4.94267, antiderivative size = 604, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.227, Rules used = {740, 828, 826, 1166, 208} $-\frac{e \left (-2 c e (5 a e+b d)+3 b^2 e^2+2 c^2 d^2\right )}{\left (b^2-4 a c\right ) \sqrt{d+e x} \left (a e^2-b d e+c d^2\right )^2}+\frac{\sqrt{c} \left (-2 c^2 d e \left (-d \sqrt{b^2-4 a c}-16 a e+6 b d\right )-2 c e^2 \left (b d \sqrt{b^2-4 a c}+5 a e \sqrt{b^2-4 a c}+8 a b e+b^2 d\right )+3 b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )+8 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )^2}-\frac{\sqrt{c} \left (-2 c^2 d e \left (d \sqrt{b^2-4 a c}-16 a e+6 b d\right )-2 c e^2 \left (-b d \sqrt{b^2-4 a c}-5 a e \sqrt{b^2-4 a c}+8 a b e+b^2 d\right )+3 b^2 e^3 \left (b-\sqrt{b^2-4 a c}\right )+8 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )^2}-\frac{2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d}{\left (b^2-4 a c\right ) \sqrt{d+e x} \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((e*(2*c^2*d^2 + 3*b^2*e^2 - 2*c*e*(b*d + 5*a*e)))/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)^2*Sqrt[d + e*x])) -
(b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*Sqrt[d + e*x]*(a + b*x +
c*x^2)) + (Sqrt[c]*(8*c^3*d^3 + 3*b^2*(b + Sqrt[b^2 - 4*a*c])*e^3 - 2*c^2*d*e*(6*b*d - Sqrt[b^2 - 4*a*c]*d -
16*a*e) - 2*c*e^2*(b^2*d + b*Sqrt[b^2 - 4*a*c]*d + 8*a*b*e + 5*a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c
]*Sqrt[d + e*x])/Sqrt[2*c*d - (b - Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*(b^2 - 4*a*c)^(3/2)*Sqrt[2*c*d - (b - Sqrt
[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)^2) - (Sqrt[c]*(8*c^3*d^3 + 3*b^2*(b - Sqrt[b^2 - 4*a*c])*e^3 - 2*c^2
*d*e*(6*b*d + Sqrt[b^2 - 4*a*c]*d - 16*a*e) - 2*c*e^2*(b^2*d - b*Sqrt[b^2 - 4*a*c]*d + 8*a*b*e - 5*a*Sqrt[b^2
- 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/(Sqrt[2]*(b^2 -
4*a*c)^(3/2)*Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]*(c*d^2 - b*d*e + a*e^2)^2)

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 828

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[((
e*f - d*g)*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/(c*d^2 - b*d*e + a*e^2), Int[((d
+ e*x)^(m + 1)*Simp[c*d*f - f*b*e + a*e*g - c*(e*f - d*g)*x, x])/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && FractionQ[m] && LtQ[m, -1]

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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+b x+c x^2\right )^2} \, dx &=-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x} \left (a+b x+c x^2\right )}-\frac{\int \frac{\frac{1}{2} \left (4 c^2 d^2-b c d e-3 b^2 e^2+10 a c e^2\right )+\frac{3}{2} c e (2 c d-b e) x}{(d+e x)^{3/2} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=-\frac{e \left (2 c^2 d^2+3 b^2 e^2-2 c e (b d+5 a e)\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x} \left (a+b x+c x^2\right )}-\frac{\int \frac{\frac{1}{2} \left (4 c^3 d^3+3 b^3 e^3-c^2 d e (5 b d-16 a e)-b c e^2 (2 b d+13 a e)\right )+\frac{1}{2} c e \left (2 c^2 d^2+3 b^2 e^2-2 c e (b d+5 a e)\right ) x}{\sqrt{d+e x} \left (a+b x+c x^2\right )} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{e \left (2 c^2 d^2+3 b^2 e^2-2 c e (b d+5 a e)\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x} \left (a+b x+c x^2\right )}-\frac{2 \operatorname{Subst}\left (\int \frac{-\frac{1}{2} c d e \left (2 c^2 d^2+3 b^2 e^2-2 c e (b d+5 a e)\right )+\frac{1}{2} e \left (4 c^3 d^3+3 b^3 e^3-c^2 d e (5 b d-16 a e)-b c e^2 (2 b d+13 a e)\right )+\frac{1}{2} c e \left (2 c^2 d^2+3 b^2 e^2-2 c e (b d+5 a e)\right ) x^2}{c d^2-b d e+a e^2+(-2 c d+b e) x^2+c x^4} \, dx,x,\sqrt{d+e x}\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{e \left (2 c^2 d^2+3 b^2 e^2-2 c e (b d+5 a e)\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x} \left (a+b x+c x^2\right )}+\frac{\left (c \left (8 c^3 d^3+3 b^2 \left (b-\sqrt{b^2-4 a c}\right ) e^3-2 c^2 d e \left (6 b d+\sqrt{b^2-4 a c} d-16 a e\right )-2 c e^2 \left (b^2 d-b \sqrt{b^2-4 a c} d+8 a b e-5 a \sqrt{b^2-4 a c} e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{2 \left (b^2-4 a c\right )^{3/2} \left (c d^2-b d e+a e^2\right )^2}-\frac{\left (c \left (8 c^3 d^3+3 b^2 \left (b+\sqrt{b^2-4 a c}\right ) e^3-2 c^2 d e \left (6 b d-\sqrt{b^2-4 a c} d-16 a e\right )-2 c e^2 \left (b^2 d+b \sqrt{b^2-4 a c} d+8 a b e+5 a \sqrt{b^2-4 a c} e\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2} \sqrt{b^2-4 a c} e+\frac{1}{2} (-2 c d+b e)+c x^2} \, dx,x,\sqrt{d+e x}\right )}{2 \left (b^2-4 a c\right )^{3/2} \left (c d^2-b d e+a e^2\right )^2}\\ &=-\frac{e \left (2 c^2 d^2+3 b^2 e^2-2 c e (b d+5 a e)\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 \sqrt{d+e x}}-\frac{b c d-b^2 e+2 a c e+c (2 c d-b e) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \sqrt{d+e x} \left (a+b x+c x^2\right )}+\frac{\sqrt{c} \left (8 c^3 d^3+3 b^2 \left (b+\sqrt{b^2-4 a c}\right ) e^3-2 c^2 d e \left (6 b d-\sqrt{b^2-4 a c} d-16 a e\right )-2 c e^2 \left (b^2 d+b \sqrt{b^2-4 a c} d+8 a b e+5 a \sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )^2}-\frac{\sqrt{c} \left (8 c^3 d^3+3 b^2 \left (b-\sqrt{b^2-4 a c}\right ) e^3-2 c^2 d e \left (6 b d+\sqrt{b^2-4 a c} d-16 a e\right )-2 c e^2 \left (b^2 d-b \sqrt{b^2-4 a c} d+8 a b e-5 a \sqrt{b^2-4 a c} e\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e} \left (c d^2-b d e+a e^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 2.93336, size = 552, normalized size = 0.91 $\frac{\frac{e \left (2 c e (5 a e+b d)-3 b^2 e^2-2 c^2 d^2\right )}{\sqrt{d+e x} \left (e (a e-b d)+c d^2\right )}-\frac{\sqrt{c} \left (\frac{\left (2 c^2 d e \left (d \sqrt{b^2-4 a c}+16 a e-6 b d\right )-2 c e^2 \left (b d \sqrt{b^2-4 a c}+5 a e \sqrt{b^2-4 a c}+8 a b e+b^2 d\right )+3 b^2 e^3 \left (\sqrt{b^2-4 a c}+b\right )+8 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\left (-2 c^2 d e \left (d \sqrt{b^2-4 a c}-16 a e+6 b d\right )+2 c e^2 \left (b d \sqrt{b^2-4 a c}+5 a e \sqrt{b^2-4 a c}-8 a b e+b^2 (-d)\right )+3 b^2 e^3 \left (b-\sqrt{b^2-4 a c}\right )+8 c^3 d^3\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \sqrt{b^2-4 a c} \left (e (b d-a e)-c d^2\right )}+\frac{-2 c (a e+c d x)+b^2 e+b c (e x-d)}{\sqrt{d+e x} (a+x (b+c x))}}{\left (b^2-4 a c\right ) \left (e (a e-b d)+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((e*(-2*c^2*d^2 - 3*b^2*e^2 + 2*c*e*(b*d + 5*a*e)))/((c*d^2 + e*(-(b*d) + a*e))*Sqrt[d + e*x]) + (b^2*e - 2*c*
(a*e + c*d*x) + b*c*(-d + e*x))/(Sqrt[d + e*x]*(a + x*(b + c*x))) - (Sqrt[c]*(((8*c^3*d^3 + 3*b^2*(b + Sqrt[b^
2 - 4*a*c])*e^3 + 2*c^2*d*e*(-6*b*d + Sqrt[b^2 - 4*a*c]*d + 16*a*e) - 2*c*e^2*(b^2*d + b*Sqrt[b^2 - 4*a*c]*d +
8*a*b*e + 5*a*Sqrt[b^2 - 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - b*e + Sqrt[b^2 - 4*a
*c]*e]])/Sqrt[2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e] - ((8*c^3*d^3 + 3*b^2*(b - Sqrt[b^2 - 4*a*c])*e^3 - 2*c^2*d*
e*(6*b*d + Sqrt[b^2 - 4*a*c]*d - 16*a*e) + 2*c*e^2*(-(b^2*d) + b*Sqrt[b^2 - 4*a*c]*d - 8*a*b*e + 5*a*Sqrt[b^2
- 4*a*c]*e))*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[d + e*x])/Sqrt[2*c*d - (b + Sqrt[b^2 - 4*a*c])*e]])/Sqrt[2*c*d - (b
+ Sqrt[b^2 - 4*a*c])*e]))/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*(-(c*d^2) + e*(b*d - a*e))))/((b^2 - 4*a*c)*(c*d^2 + e*(
-(b*d) + a*e)))

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Maple [B]  time = 0.3, size = 3212, normalized size = 5.3 \begin{align*} \text{output too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*e/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^4/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1
/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d^3-4*e/(a*e^2-
b*d*e+c*d^2)^2/(4*a*c-b^2)*c^4/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)
*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d^3-e^2/(a*e^2-b*d*e+c*d^2)^2/
(4*a*c-b^2)*c^2*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*
c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d-3/2*e^4/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c/(-e^2*(4*a*c-b^2))^(1/
2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*
a*c-b^2))^(1/2))*c)^(1/2))*b^3-3/2*e^4/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((
-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^
(1/2))*c)^(1/2))*b^3+e^2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^2*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*
c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d-2*e^3/(a*e^2-b*d
*e+c*d^2)^2/(e*x+d)^(1/2)+8*e^4/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2
*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c
)^(1/2))*a*b+e^3/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a
*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*d-
16*e^3/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^3/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^
(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*d+2*e/(a*e^
2-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)*c^3/(4*a*c-b^2)*(e*x+d)^(3/2)*d^2-2*e/(a*e^2-b*d*e+c*d^2)^2/(c*e^2*
x^2+b*e^2*x+a*e^2)/(4*a*c-b^2)*(e*x+d)^(1/2)*c^3*d^3-2*e^3/(a*e^2-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)*c^2
/(4*a*c-b^2)*(e*x+d)^(3/2)*a+e^3/(a*e^2-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)*c/(4*a*c-b^2)*(e*x+d)^(3/2)*b
^2-e/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^3*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*
x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*d^2+e/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*
c^3*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4
*a*c-b^2))^(1/2))*c)^(1/2))*d^2-3*e^4/(a*e^2-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)/(4*a*c-b^2)*(e*x+d)^(1/2
)*a*b*c+3*e^2/(a*e^2-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)/(4*a*c-b^2)*(e*x+d)^(1/2)*b*c^2*d^2-2*e^2/(a*e^2
-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)*c^2/(4*a*c-b^2)*(e*x+d)^(3/2)*b*d+e^3/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b
^2)*c^2/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)
*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2*d-16*e^3/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^3
/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2
)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*d+6*e^2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^3/(-e^2*(4*a*c
-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*
c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d^2+6*e^3/(a*e^2-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)/(4*a*c-b^2
)*(e*x+d)^(1/2)*c^2*a*d-3*e^3/(a*e^2-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)/(4*a*c-b^2)*(e*x+d)^(1/2)*b^2*c*
d-3/2*e^3/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh(
(e*x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2-5*e^3/(a*e^2-b*d*e+c*d^2)^2/(4*a*
c-b^2)*c^2*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(
-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a+3/2*e^3/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*
a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b^2+5
*e^3/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^2*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*
x+d)^(1/2)*c*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a+6*e^2/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2
)*c^3/(-e^2*(4*a*c-b^2))^(1/2)*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctan((e*x+d)^(1/2)*c*2
^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*b*d^2+8*e^4/(a*e^2-b*d*e+c*d^2)^2/(4*a*c-b^2)*c^2/(-e^2
*(4*a*c-b^2))^(1/2)*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arctanh((e*x+d)^(1/2)*c*2^(1/2)/((
-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2))*a*b+e^4/(a*e^2-b*d*e+c*d^2)^2/(c*e^2*x^2+b*e^2*x+a*e^2)/(4*a*c-
b^2)*(e*x+d)^(1/2)*b^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x + a\right )}^{2}{\left (e x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out