### 3.1957 $$\int \frac{(d+e x)^2}{(a d e+(c d^2+a e^2) x+c d e x^2)^{3/2}} \, dx$$

Optimal. Leaf size=125 $\frac{\sqrt{e} \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c^{3/2} d^{3/2}}-\frac{2 (d+e x)}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

[Out]

(-2*(d + e*x))/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (Sqrt[e]*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x
)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(c^(3/2)*d^(3/2))

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Rubi [A]  time = 0.0582064, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 37, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.081, Rules used = {652, 621, 206} $\frac{\sqrt{e} \tanh ^{-1}\left (\frac{a e^2+c d^2+2 c d e x}{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{c^{3/2} d^{3/2}}-\frac{2 (d+e x)}{c d \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(d + e*x))/(c*d*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) + (Sqrt[e]*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x
)/(2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(c^(3/2)*d^(3/2))

Rule 652

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

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.356946, size = 164, normalized size = 1.31 $\frac{2 \sqrt{c} \sqrt{d} \sqrt{e} \sqrt{c d^2-a e^2} \sqrt{a e+c d x} \sqrt{\frac{c d (d+e x)}{c d^2-a e^2}} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{e} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d^2-a e^2}}\right )-2 (c d)^{3/2} (d+e x)}{(c d)^{5/2} \sqrt{(d+e x) (a e+c d x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*(c*d)^(3/2)*(d + e*x) + 2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[c*d^2 - a*e^2]*Sqrt[a*e + c*d*x]*Sqrt[(c*d*(d + e*x
))/(c*d^2 - a*e^2)]*ArcSinh[(Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/((c*
d)^(5/2)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e*x/d/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+1/2*e^2/d^2/c^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a-3/2
/c/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+e^5/d/c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d
*e*x^2)^(1/2)*x*a^2-2*e^3*d/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x*a-3*e*d
^3*c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*x+1/2*e^6/d^2/c^2/(-a^2*e^4+2*a*
c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^3-1/2*e^4/c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d
*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)*a^2-5/2*e^2*d^2/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*
d*e*x^2)^(1/2)*a-3/2*d^4*c/(-a^2*e^4+2*a*c*d^2*e^2-c^2*d^4)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)+e/d/c*ln((
1/2*a*e^2+1/2*c*d^2+c*d*e*x)/(d*e*c)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(d*e*c)^(1/2)+2*d^2*(2*c*d
*e*x+a*e^2+c*d^2)/(4*a*c*d^2*e^2-(a*e^2+c*d^2)^2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.69143, size = 737, normalized size = 5.9 \begin{align*} \left [\frac{{\left (c d x + a e\right )} \sqrt{\frac{e}{c d}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 8 \,{\left (c^{2} d^{3} e + a c d e^{3}\right )} x + 4 \,{\left (2 \, c^{2} d^{2} e x + c^{2} d^{3} + a c d e^{2}\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{\frac{e}{c d}}\right ) - 4 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{2 \,{\left (c^{2} d^{2} x + a c d e\right )}}, -\frac{{\left (c d x + a e\right )} \sqrt{-\frac{e}{c d}} \arctan \left (\frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt{-\frac{e}{c d}}}{2 \,{\left (c d e^{2} x^{2} + a d e^{2} +{\left (c d^{2} e + a e^{3}\right )} x\right )}}\right ) + 2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}}{c^{2} d^{2} x + a c d e}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((c*d*x + a*e)*sqrt(e/(c*d))*log(8*c^2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 8*(c^2*d^3*e + a
*c*d*e^3)*x + 4*(2*c^2*d^2*e*x + c^2*d^3 + a*c*d*e^2)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(e/(c*d)
)) - 4*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*x + a*c*d*e), -((c*d*x + a*e)*sqrt(-e/(c*d))*arct
an(1/2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-e/(c*d))/(c*d*e^2*x^2 + a
*d*e^2 + (c*d^2*e + a*e^3)*x)) + 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c^2*d^2*x + a*c*d*e)]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**2/((d + e*x)*(a*e + c*d*x))**(3/2), x)

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Giac [B]  time = 1.29553, size = 309, normalized size = 2.47 \begin{align*} -\frac{2 \,{\left (\frac{{\left (c^{2} d^{4} e - 2 \, a c d^{2} e^{3} + a^{2} e^{5}\right )} x}{c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}} + \frac{c^{2} d^{5} - 2 \, a c d^{3} e^{2} + a^{2} d e^{4}}{c^{3} d^{5} - 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}}\right )}}{\sqrt{c d x^{2} e + a d e +{\left (c d^{2} + a e^{2}\right )} x}} - \frac{\sqrt{c d} e^{\frac{1}{2}} \log \left ({\left | -\sqrt{c d} c d^{2} e^{\frac{1}{2}} - 2 \,{\left (\sqrt{c d} x e^{\frac{1}{2}} - \sqrt{c d x^{2} e + a d e +{\left (c d^{2} + a e^{2}\right )} x}\right )} c d e - \sqrt{c d} a e^{\frac{5}{2}} \right |}\right )}{c^{2} d^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((c^2*d^4*e - 2*a*c*d^2*e^3 + a^2*e^5)*x/(c^3*d^5 - 2*a*c^2*d^3*e^2 + a^2*c*d*e^4) + (c^2*d^5 - 2*a*c*d^3*e
^2 + a^2*d*e^4)/(c^3*d^5 - 2*a*c^2*d^3*e^2 + a^2*c*d*e^4))/sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e^2)*x) - sqrt(
c*d)*e^(1/2)*log(abs(-sqrt(c*d)*c*d^2*e^(1/2) - 2*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x^2*e + a*d*e + (c*d^2 + a*e
^2)*x))*c*d*e - sqrt(c*d)*a*e^(5/2)))/(c^2*d^2)