### 3.2299 $$\int \frac{1}{\sqrt{d+e x} (a+b x+c x^2)^2} \, dx$$

Optimal. Leaf size=428 $\frac{\sqrt{c} \left (-2 c e \left (-d \sqrt{b^2-4 a c}-6 a e+4 b d\right )-b e^2 \left (\sqrt{b^2-4 a c}+b\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac{\sqrt{c} \left (-2 c e \left (d \sqrt{b^2-4 a c}-6 a e+4 b d\right )-b e^2 \left (b-\sqrt{b^2-4 a c}\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}-\frac{\sqrt{d+e x} \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}$

[Out]

-((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)*(a + b*
x + c*x^2))) + (Sqrt[c]*(8*c^2*d^2 - b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*(4*b*d - Sqrt[b^2 - 4*a*c]*d - 6*a*
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)) - (Sqrt[c]*(8*c^2*d^2 - b*(b - Sqrt[b^2
- 4*a*c])*e^2 - 2*c*e*(4*b*d + Sqrt[b^2 - 4*a*c]*d - 6*a*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))

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Rubi [A]  time = 1.47212, antiderivative size = 428, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {740, 826, 1166, 208} $\frac{\sqrt{c} \left (-2 c e \left (-d \sqrt{b^2-4 a c}-6 a e+4 b d\right )-b e^2 \left (\sqrt{b^2-4 a c}+b\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )} \left (a e^2-b d e+c d^2\right )}-\frac{\sqrt{c} \left (-2 c e \left (d \sqrt{b^2-4 a c}-6 a e+4 b d\right )-b e^2 \left (b-\sqrt{b^2-4 a c}\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \left (b^2-4 a c\right )^{3/2} \sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )} \left (a e^2-b d e+c d^2\right )}-\frac{\sqrt{d+e x} \left (2 a c e+b^2 (-e)+c x (2 c d-b e)+b c d\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((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)*(a + b*
x + c*x^2))) + (Sqrt[c]*(8*c^2*d^2 - b*(b + Sqrt[b^2 - 4*a*c])*e^2 - 2*c*e*(4*b*d - Sqrt[b^2 - 4*a*c]*d - 6*a*
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)) - (Sqrt[c]*(8*c^2*d^2 - b*(b - Sqrt[b^2
- 4*a*c])*e^2 - 2*c*e*(4*b*d + Sqrt[b^2 - 4*a*c]*d - 6*a*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))

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

Mathematica [A]  time = 1.44925, size = 366, normalized size = 0.86 $\frac{\frac{\sqrt{c} \left (\frac{\left (2 c e \left (d \sqrt{b^2-4 a c}+6 a e-4 b d\right )-b e^2 \left (\sqrt{b^2-4 a c}+b\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{e \sqrt{b^2-4 a c}-b e+2 c d}}\right )}{\sqrt{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}}-\frac{\left (-2 c e \left (d \sqrt{b^2-4 a c}-6 a e+4 b d\right )+b e^2 \left (\sqrt{b^2-4 a c}-b\right )+8 c^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d+e x}}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}}\right )}{\sqrt{2} \sqrt{b^2-4 a c}}+\frac{\sqrt{d+e x} \left (-2 c (a e+c d x)+b^2 e+b c (e x-d)\right )}{a+x (b+c x)}}{\left (b^2-4 a c\right ) \left (e (a e-b d)+c d^2\right )}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [B]  time = 0.283, size = 1120, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*e*c/(-e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2)/(-b*e+2*c*d+(-4*a*c*e^2+b^2*e^2)^(1/2))*(-4*a*c*e^2+b^2*e^2)^(1/2)*
(e*x+d)^(1/2)/(e*x+1/2*b*e/c-1/2/c*(e^2*(-4*a*c+b^2))^(1/2))+8*e^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2)/(-
2*b*e+4*c*d+2*(-4*a*c*e^2+b^2*e^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-16*e*c^3/(-e^2*(4*a*c-b^2))^(1/2)/(4*a*
c-b^2)/(-2*b*e+4*c*d+2*(-4*a*c*e^2+b^2*e^2)^(1/2))*2^(1/2)/((-b*e+2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arc
tanh((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-12*e*c^2/(-e^2*(4*a*c-b^2))^(1
/2)/(4*a*c-b^2)/(-2*b*e+4*c*d+2*(-4*a*c*e^2+b^2*e^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))*(-4*a*c*e^2+b^2*e^2)^(1
/2)+2*e*c/(-e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2)/(-b*e+2*c*d-(-4*a*c*e^2+b^2*e^2)^(1/2))*(-4*a*c*e^2+b^2*e^2)^(1
/2)*(e*x+d)^(1/2)/(e*x+1/2*b*e/c+1/2/c*(e^2*(-4*a*c+b^2))^(1/2))-8*e^2*c^2/(-e^2*(4*a*c-b^2))^(1/2)/(4*a*c-b^2
)/(2*b*e-4*c*d+2*(-4*a*c*e^2+b^2*e^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+16*e*c^3/(-e^2*(4*a*c-b^2))^(1/2)/(4*a*
c-b^2)/(2*b*e-4*c*d+2*(-4*a*c*e^2+b^2*e^2)^(1/2))*2^(1/2)/((b*e-2*c*d+(-e^2*(4*a*c-b^2))^(1/2))*c)^(1/2)*arcta
n((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-12*e*c^2/(-e^2*(4*a*c-b^2))^(1/2)/
(4*a*c-b^2)/(2*b*e-4*c*d+2*(-4*a*c*e^2+b^2*e^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))*(-4*a*c*e^2+b^2*e^2)^(1/2)

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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} \sqrt{e x + d}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out