### 3.1715 $$\int \frac{\sqrt{d+e x}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx$$

Optimal. Leaf size=168 $\frac{e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{e \sqrt{d+e x}}{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{\sqrt{d+e x}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}$

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0875864, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 30, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {646, 47, 51, 63, 208} $\frac{e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{e \sqrt{d+e x}}{4 b \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac{\sqrt{d+e x}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [C]  time = 0.0253277, size = 67, normalized size = 0.4 $-\frac{2 e^2 (a+b x) (d+e x)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{b (d+e x)}{b d-a e}\right )}{3 \sqrt{(a+b x)^2} (b d-a e)^3}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-2*e^2*(a + b*x)*(d + e*x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, (b*(d + e*x))/(b*d - a*e)])/(3*(b*d - a*e)^3*
Sqrt[(a + b*x)^2])

________________________________________________________________________________________

Maple [A]  time = 0.271, size = 200, normalized size = 1.2 \begin{align*}{\frac{bx+a}{4\, \left ( ae-bd \right ) b} \left ( \arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){x}^{2}{b}^{2}{e}^{2}+2\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xab{e}^{2}+\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{{\frac{3}{2}}}b+\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){a}^{2}{e}^{2}-\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}ae+\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}bd \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x^2*b^2*e^2+2*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*x*a
*b*e^2+((a*e-b*d)*b)^(1/2)*(e*x+d)^(3/2)*b+arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a^2*e^2-((a*e-b*d)*b)^(
1/2)*(e*x+d)^(1/2)*a*e+((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*b*d)*(b*x+a)/((a*e-b*d)*b)^(1/2)/b/(a*e-b*d)/((b*x+a)
^2)^(3/2)

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.70585, size = 949, normalized size = 5.65 \begin{align*} \left [-\frac{{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt{b^{2} d - a b e} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{b^{2} d - a b e} \sqrt{e x + d}}{b x + a}\right ) + 2 \,{\left (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} +{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} +{\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \,{\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}, -\frac{{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt{-b^{2} d + a b e} \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) +{\left (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} +{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} +{\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \,{\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b
*e)*sqrt(e*x + d))/(b*x + a)) + 2*(2*b^3*d^2 - 3*a*b^2*d*e + a^2*b*e^2 + (b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d
))/(a^2*b^4*d^2 - 2*a^3*b^3*d*e + a^4*b^2*e^2 + (b^6*d^2 - 2*a*b^5*d*e + a^2*b^4*e^2)*x^2 + 2*(a*b^5*d^2 - 2*a
^2*b^4*d*e + a^3*b^3*e^2)*x), -1/4*((b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^
2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (2*b^3*d^2 - 3*a*b^2*d*e + a^2*b*e^2 + (b^3*d*e - a*b^2*e^2)*x)*sq
rt(e*x + d))/(a^2*b^4*d^2 - 2*a^3*b^3*d*e + a^4*b^2*e^2 + (b^6*d^2 - 2*a*b^5*d*e + a^2*b^4*e^2)*x^2 + 2*(a*b^5
*d^2 - 2*a^2*b^4*d*e + a^3*b^3*e^2)*x)]

________________________________________________________________________________________

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.21077, size = 297, normalized size = 1.77 \begin{align*} -\frac{\arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \,{\left (b^{2} d \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{-b^{2} d + a b e}} - \frac{{\left (x e + d\right )}^{\frac{3}{2}} b e^{2} + \sqrt{x e + d} b d e^{2} - \sqrt{x e + d} a e^{3}}{4 \,{\left (b^{2} d \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a b e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^2*d*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a*b*e*sgn((
x*e + d)*b*e - b*d*e + a*e^2))*sqrt(-b^2*d + a*b*e)) - 1/4*((x*e + d)^(3/2)*b*e^2 + sqrt(x*e + d)*b*d*e^2 - sq
rt(x*e + d)*a*e^3)/((b^2*d*sgn((x*e + d)*b*e - b*d*e + a*e^2) - a*b*e*sgn((x*e + d)*b*e - b*d*e + a*e^2))*((x*
e + d)*b - b*d + a*e)^2)