### 3.1651 $$\int \frac{1}{(d+e x)^{3/2} (a^2+2 a b x+b^2 x^2)} \, dx$$

Optimal. Leaf size=99 $-\frac{3 e}{\sqrt{d+e x} (b d-a e)^2}-\frac{1}{(a+b x) \sqrt{d+e x} (b d-a e)}+\frac{3 \sqrt{b} e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}}$

[Out]

(-3*e)/((b*d - a*e)^2*Sqrt[d + e*x]) - 1/((b*d - a*e)*(a + b*x)*Sqrt[d + e*x]) + (3*Sqrt[b]*e*ArcTanh[(Sqrt[b]
*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(5/2)

________________________________________________________________________________________

Rubi [A]  time = 0.0539409, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.143, Rules used = {27, 51, 63, 208} $-\frac{3 e}{\sqrt{d+e x} (b d-a e)^2}-\frac{1}{(a+b x) \sqrt{d+e x} (b d-a e)}+\frac{3 \sqrt{b} e \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-3*e)/((b*d - a*e)^2*Sqrt[d + e*x]) - 1/((b*d - a*e)*(a + b*x)*Sqrt[d + e*x]) + (3*Sqrt[b]*e*ArcTanh[(Sqrt[b]
*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b*d - a*e)^(5/2)

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

Mathematica [C]  time = 0.0131378, size = 48, normalized size = 0.48 $-\frac{2 e \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};-\frac{b (d+e x)}{a e-b d}\right )}{\sqrt{d+e x} (a e-b d)^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.207, size = 101, normalized size = 1. \begin{align*} -2\,{\frac{e}{ \left ( ae-bd \right ) ^{2}\sqrt{ex+d}}}-{\frac{be}{ \left ( ae-bd \right ) ^{2} \left ( bxe+ae \right ) }\sqrt{ex+d}}-3\,{\frac{be}{ \left ( ae-bd \right ) ^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}

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),x)

[Out]

-2*e/(a*e-b*d)^2/(e*x+d)^(1/2)-e*b/(a*e-b*d)^2*(e*x+d)^(1/2)/(b*e*x+a*e)-3*e*b/(a*e-b*d)^2/((a*e-b*d)*b)^(1/2)
*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.01781, size = 883, normalized size = 8.92 \begin{align*} \left [\frac{3 \,{\left (b e^{2} x^{2} + a d e +{\left (b d e + a e^{2}\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 (3 \, b e x + b d + 2 \, a e\right )} \sqrt{e x + d}}{2 \,{\left (a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} +{\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} +{\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x\right )}}, \frac{3 \,{\left (b e^{2} x^{2} + a d e +{\left (b d e + a e^{2}\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 (3 \, b e x + b d + 2 \, a e\right )} \sqrt{e x + d}}{a b^{2} d^{3} - 2 \, a^{2} b d^{2} e + a^{3} d e^{2} +{\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x^{2} +{\left (b^{3} d^{3} - a b^{2} d^{2} e - a^{2} b d e^{2} + a^{3} e^{3}\right )} x}\right ] \end{align*}

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),x, algorithm="fricas")

[Out]

[1/2*(3*(b*e^2*x^2 + a*d*e + (b*d*e + a*e^2)*x)*sqrt(b/(b*d - a*e))*log((b*e*x + 2*b*d - a*e + 2*(b*d - a*e)*s
qrt(e*x + d)*sqrt(b/(b*d - a*e)))/(b*x + a)) - 2*(3*b*e*x + b*d + 2*a*e)*sqrt(e*x + d))/(a*b^2*d^3 - 2*a^2*b*d
^2*e + a^3*d*e^2 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x^2 + (b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^
3)*x), (3*(b*e^2*x^2 + a*d*e + (b*d*e + a*e^2)*x)*sqrt(-b/(b*d - a*e))*arctan(-(b*d - a*e)*sqrt(e*x + d)*sqrt(
-b/(b*d - a*e))/(b*e*x + b*d)) - (3*b*e*x + b*d + 2*a*e)*sqrt(e*x + d))/(a*b^2*d^3 - 2*a^2*b*d^2*e + a^3*d*e^2
+ (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2*b*e^3)*x^2 + (b^3*d^3 - a*b^2*d^2*e - a^2*b*d*e^2 + a^3*e^3)*x)]

________________________________________________________________________________________

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

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),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.18709, size = 207, normalized size = 2.09 \begin{align*} -\frac{3 \, b \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \sqrt{-b^{2} d + a b e}} - \frac{3 \,{\left (x e + d\right )} b e - 2 \, b d e + 2 \, a e^{2}}{{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )}{\left ({\left (x e + d\right )}^{\frac{3}{2}} b - \sqrt{x e + d} b d + \sqrt{x e + d} a e\right )}} \end{align*}

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),x, algorithm="giac")

[Out]

-3*b*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e/((b^2*d^2 - 2*a*b*d*e + a^2*e^2)*sqrt(-b^2*d + a*b*e)) - (
3*(x*e + d)*b*e - 2*b*d*e + 2*a*e^2)/((b^2*d^2 - 2*a*b*d*e + a^2*e^2)*((x*e + d)^(3/2)*b - sqrt(x*e + d)*b*d +
sqrt(x*e + d)*a*e))