3.2134 \(\int \frac{a+b x}{\sqrt{d+e x} (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0885706, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {770, 21, 51, 63, 208} \[ \frac{e (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{b} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}}-\frac{\sqrt{d+e x}}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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{a+b x}{\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{a+b x}{\left (a b+b^2 x\right )^3 \sqrt{d+e x}} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{\sqrt{d+e x}}{(b d-a e) \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}{(a+b x) \sqrt{d+e x}} \, dx}{2 b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{\sqrt{d+e x}}{(b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (a b+b^2 x\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{\sqrt{d+e x}}{(b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{b} (b d-a e)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.093516, size = 85, normalized size = 0.75 \[ \frac{\frac{\sqrt{d+e x}}{a e-b d}+\frac{e (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )}{\sqrt{b} (a e-b d)^{3/2}}}{\sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 112, normalized size = 1. \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{ae-bd} \left ( \arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ) xbe+\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ) ae+\sqrt{ex+d}\sqrt{ \left ( ae-bd \right ) b} \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((b*x+a)/(b^2*x^2+2*a*b*x+a^2)^(3/2)/(e*x+d)^(1/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} \sqrt{e x + d}}\,{d x} \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)^(3/2)/(e*x+d)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.04387, size = 603, normalized size = 5.29 \begin{align*} \left [-\frac{\sqrt{b^{2} d - a b e}{\left (b e x + a e\right )} \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 (b^{2} d - a b e\right )} \sqrt{e x + d}}{2 \,{\left (a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} +{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x\right )}}, -\frac{\sqrt{-b^{2} d + a b e}{\left (b e x + a e\right )} \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) +{\left (b^{2} d - a b e\right )} \sqrt{e x + d}}{a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} +{\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x}\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)^(3/2)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Giac [B]  time = 1.1774, size = 255, normalized size = 2.24 \begin{align*} -\frac{\arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{{\left (b d e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a e^{2} \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{\sqrt{x e + d} e^{2}}{{\left (b d e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a e^{2} \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 )}} \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)^(3/2)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

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