3.1812 \(\int \frac{A+B x}{\sqrt{d+e x} (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=103 \[ -\frac{(-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} (b d-a e)^{3/2}}-\frac{\sqrt{d+e x} (A b-a B)}{b (a+b x) (b d-a e)} \]

[Out]

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

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Rubi [A]  time = 0.0953529, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {27, 78, 63, 208} \[ -\frac{(-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} (b d-a e)^{3/2}}-\frac{\sqrt{d+e x} (A b-a B)}{b (a+b x) (b d-a e)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 78

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

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

Mathematica [A]  time = 0.0821406, size = 102, normalized size = 0.99 \[ \frac{\sqrt{d+e x} (a B-A b)}{b (a+b x) (b d-a e)}-\frac{(-a B e-A b e+2 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{3/2} (b d-a e)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.016, size = 195, normalized size = 1.9 \begin{align*}{\frac{ \left ( Ab-aB \right ) e}{ \left ( ae-bd \right ) b \left ( b \left ( ex+d \right ) +ae-bd \right ) }\sqrt{ex+d}}+{\frac{Ae}{ae-bd}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}+{\frac{aBe}{ \left ( ae-bd \right ) b}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}}-2\,{\frac{Bd}{ \left ( ae-bd \right ) \sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \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)/(e*x+d)^(1/2),x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.42362, size = 833, normalized size = 8.09 \begin{align*} \left [\frac{{\left (2 \, B a b d -{\left (B a^{2} + A a b\right )} e +{\left (2 \, B b^{2} d -{\left (B a b + A b^{2}\right )} e\right )} x\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 ({\left (B a b^{2} - A b^{3}\right )} d -{\left (B a^{2} b - A a b^{2}\right )} e\right )} \sqrt{e x + d}}{2 \,{\left (a b^{4} d^{2} - 2 \, a^{2} b^{3} d e + a^{3} b^{2} e^{2} +{\left (b^{5} d^{2} - 2 \, a b^{4} d e + a^{2} b^{3} e^{2}\right )} x\right )}}, \frac{{\left (2 \, B a b d -{\left (B a^{2} + A a b\right )} e +{\left (2 \, B b^{2} d -{\left (B a b + A b^{2}\right )} e\right )} x\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 ({\left (B a b^{2} - A b^{3}\right )} d -{\left (B a^{2} b - A a b^{2}\right )} e\right )} \sqrt{e x + d}}{a b^{4} d^{2} - 2 \, a^{2} b^{3} d e + a^{3} b^{2} e^{2} +{\left (b^{5} d^{2} - 2 \, a b^{4} d e + a^{2} b^{3} 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)/(e*x+d)^(1/2),x, algorithm="fricas")

[Out]

[1/2*((2*B*a*b*d - (B*a^2 + A*a*b)*e + (2*B*b^2*d - (B*a*b + A*b^2)*e)*x)*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*((B*a*b^2 - A*b^3)*d - (B*a^2*b - A*a*b^2)*e)*s
qrt(e*x + d))/(a*b^4*d^2 - 2*a^2*b^3*d*e + a^3*b^2*e^2 + (b^5*d^2 - 2*a*b^4*d*e + a^2*b^3*e^2)*x), ((2*B*a*b*d
- (B*a^2 + A*a*b)*e + (2*B*b^2*d - (B*a*b + A*b^2)*e)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqr
t(e*x + d)/(b*e*x + b*d)) + ((B*a*b^2 - A*b^3)*d - (B*a^2*b - A*a*b^2)*e)*sqrt(e*x + d))/(a*b^4*d^2 - 2*a^2*b^
3*d*e + a^3*b^2*e^2 + (b^5*d^2 - 2*a*b^4*d*e + a^2*b^3*e^2)*x)]

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

[Out]

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

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Giac [A]  time = 1.14629, size = 182, normalized size = 1.77 \begin{align*} \frac{{\left (2 \, B b d - B a e - A b e\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{{\left (b^{2} d - a b e\right )} \sqrt{-b^{2} d + a b e}} + \frac{\sqrt{x e + d} B a e - \sqrt{x e + d} A b e}{{\left (b^{2} d - a b e\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)/(e*x+d)^(1/2),x, algorithm="giac")

[Out]

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