3.2079 \(\int \frac{a+b x}{\sqrt{d+e x} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=114 \[ -\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{b} (b d-a e)^{5/2}}+\frac{3 e \sqrt{d+e x}}{4 (a+b x) (b d-a e)^2}-\frac{\sqrt{d+e x}}{2 (a+b x)^2 (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0493291, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 51, 63, 208} \[ -\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{b} (b d-a e)^{5/2}}+\frac{3 e \sqrt{d+e x}}{4 (a+b x) (b d-a e)^2}-\frac{\sqrt{d+e x}}{2 (a+b x)^2 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sqrt[d + e*x]/(2*(b*d - a*e)*(a + b*x)^2) + (3*e*Sqrt[d + e*x])/(4*(b*d - a*e)^2*(a + b*x)) - (3*e^2*ArcTanh[
(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*Sqrt[b]*(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{a+b x}{\sqrt{d+e x} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{1}{(a+b x)^3 \sqrt{d+e x}} \, dx\\ &=-\frac{\sqrt{d+e x}}{2 (b d-a e) (a+b x)^2}-\frac{(3 e) \int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx}{4 (b d-a e)}\\ &=-\frac{\sqrt{d+e x}}{2 (b d-a e) (a+b x)^2}+\frac{3 e \sqrt{d+e x}}{4 (b d-a e)^2 (a+b x)}+\frac{\left (3 e^2\right ) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{8 (b d-a e)^2}\\ &=-\frac{\sqrt{d+e x}}{2 (b d-a e) (a+b x)^2}+\frac{3 e \sqrt{d+e x}}{4 (b d-a e)^2 (a+b x)}+\frac{(3 e) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 (b d-a e)^2}\\ &=-\frac{\sqrt{d+e x}}{2 (b d-a e) (a+b x)^2}+\frac{3 e \sqrt{d+e x}}{4 (b d-a e)^2 (a+b x)}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{b} (b d-a e)^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.0108233, size = 50, normalized size = 0.44 \[ \frac{2 e^2 \sqrt{d+e x} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};-\frac{b (d+e x)}{a e-b d}\right )}{(a e-b d)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e^2*Sqrt[d + e*x]*Hypergeometric2F1[1/2, 3, 3/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(-(b*d) + a*e)^3

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Maple [A]  time = 0.01, size = 115, normalized size = 1. \begin{align*}{\frac{{e}^{2}}{ \left ( 2\,ae-2\,bd \right ) \left ( bex+ae \right ) ^{2}}\sqrt{ex+d}}+{\frac{3\,{e}^{2}}{4\, \left ( ae-bd \right ) ^{2} \left ( bex+ae \right ) }\sqrt{ex+d}}+{\frac{3\,{e}^{2}}{4\, \left ( ae-bd \right ) ^{2}}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^2*(e*x+d)^(1/2)/(a*e-b*d)/(b*e*x+a*e)^2+3/4*e^2/(a*e-b*d)^2*(e*x+d)^(1/2)/(b*e*x+a*e)+3/4*e^2/(a*e-b*d)^
2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(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((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.09301, size = 1119, normalized size = 9.82 \begin{align*} \left [\frac{3 \,{\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} - 7 \, a b^{2} d e + 5 \, a^{2} b e^{2} - 3 \,{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} b^{4} d^{3} - 3 \, a^{3} b^{3} d^{2} e + 3 \, a^{4} b^{2} d e^{2} - a^{5} b e^{3} +{\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} x^{2} + 2 \,{\left (a b^{5} d^{3} - 3 \, a^{2} b^{4} d^{2} e + 3 \, a^{3} b^{3} d e^{2} - a^{4} b^{2} e^{3}\right )} x\right )}}, \frac{3 \,{\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} - 7 \, a b^{2} d e + 5 \, a^{2} b e^{2} - 3 \,{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} b^{4} d^{3} - 3 \, a^{3} b^{3} d^{2} e + 3 \, a^{4} b^{2} d e^{2} - a^{5} b e^{3} +{\left (b^{6} d^{3} - 3 \, a b^{5} d^{2} e + 3 \, a^{2} b^{4} d e^{2} - a^{3} b^{3} e^{3}\right )} x^{2} + 2 \,{\left (a b^{5} d^{3} - 3 \, a^{2} b^{4} d^{2} e + 3 \, a^{3} b^{3} d e^{2} - a^{4} b^{2} e^{3}\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(3*(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 - 7*a*b^2*d*e + 5*a^2*b*e^2 - 3*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*
x + d))/(a^2*b^4*d^3 - 3*a^3*b^3*d^2*e + 3*a^4*b^2*d*e^2 - a^5*b*e^3 + (b^6*d^3 - 3*a*b^5*d^2*e + 3*a^2*b^4*d*
e^2 - a^3*b^3*e^3)*x^2 + 2*(a*b^5*d^3 - 3*a^2*b^4*d^2*e + 3*a^3*b^3*d*e^2 - a^4*b^2*e^3)*x), 1/4*(3*(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 - 7*a*b^2*d*e + 5*a^2*b*e^2 - 3*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d))/(a^2*b^4*d^3 - 3*a^3*b^3*d^2*
e + 3*a^4*b^2*d*e^2 - a^5*b*e^3 + (b^6*d^3 - 3*a*b^5*d^2*e + 3*a^2*b^4*d*e^2 - a^3*b^3*e^3)*x^2 + 2*(a*b^5*d^3
- 3*a^2*b^4*d^2*e + 3*a^3*b^3*d*e^2 - a^4*b^2*e^3)*x)]

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

[Out]

Timed out

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Giac [A]  time = 1.1946, size = 205, normalized size = 1.8 \begin{align*} \frac{3 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \,{\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 )}^{\frac{3}{2}} b e^{2} - 5 \, \sqrt{x e + d} b d e^{2} + 5 \, \sqrt{x e + d} a e^{3}}{4 \,{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\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((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^2/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

3/4*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^2*d^2 - 2*a*b*d*e + a^2*e^2)*sqrt(-b^2*d + a*b*e)) +
1/4*(3*(x*e + d)^(3/2)*b*e^2 - 5*sqrt(x*e + d)*b*d*e^2 + 5*sqrt(x*e + d)*a*e^3)/((b^2*d^2 - 2*a*b*d*e + a^2*e^
2)*((x*e + d)*b - b*d + a*e)^2)