3.1232 \(\int \frac{(A+B x) \sqrt{d+e x}}{b x+c x^2} \, dx\)

Optimal. Leaf size=101 \[ -\frac{2 (b B-A c) \sqrt{c d-b e} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{3/2}}-\frac{2 A \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 B \sqrt{d+e x}}{c} \]

[Out]

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

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Rubi [A]  time = 0.175131, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {824, 826, 1166, 208} \[ -\frac{2 (b B-A c) \sqrt{c d-b e} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{3/2}}-\frac{2 A \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 B \sqrt{d+e x}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 824

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(g
*(d + e*x)^m)/(c*m), x] + Dist[1/c, Int[((d + e*x)^(m - 1)*Simp[c*d*f - a*e*g + (g*c*d - b*e*g + c*e*f)*x, x])
/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*
e^2, 0] && FractionQ[m] && GtQ[m, 0]

Rule 826

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2,
Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /
; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

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

Mathematica [A]  time = 0.118492, size = 101, normalized size = 1. \[ \frac{2 (A c-b B) \sqrt{c d-b e} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d+e x}}{\sqrt{c d-b e}}\right )}{b c^{3/2}}-\frac{2 A \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{b}+\frac{2 B \sqrt{d+e x}}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.011, size = 196, normalized size = 1.9 \begin{align*} 2\,{\frac{B\sqrt{ex+d}}{c}}-2\,{\frac{A\sqrt{d}}{b}{\it Artanh} \left ({\frac{\sqrt{ex+d}}{\sqrt{d}}} \right ) }+2\,{\frac{Ae}{\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{Acd}{b\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }-2\,{\frac{bBe}{c\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) }+2\,{\frac{Bd}{\sqrt{ \left ( be-cd \right ) c}}\arctan \left ({\frac{\sqrt{ex+d}c}{\sqrt{ \left ( be-cd \right ) c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(1/2)/(c*x^2+b*x),x)

[Out]

2*B*(e*x+d)^(1/2)/c-2*A*arctanh((e*x+d)^(1/2)/d^(1/2))*d^(1/2)/b+2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/
((b*e-c*d)*c)^(1/2))*A*e-2*c/b/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*A*d-2/c*b/((b*e
-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((b*e-c*d)*c)^(1/2))*B*e+2/((b*e-c*d)*c)^(1/2)*arctan((e*x+d)^(1/2)*c/((
b*e-c*d)*c)^(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)*(e*x+d)^(1/2)/(c*x^2+b*x),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.10021, size = 1013, normalized size = 10.03 \begin{align*} \left [\frac{A c \sqrt{d} \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) + 2 \, \sqrt{e x + d} B b -{\left (B b - A c\right )} \sqrt{\frac{c d - b e}{c}} \log \left (\frac{c e x + 2 \, c d - b e + 2 \, \sqrt{e x + d} c \sqrt{\frac{c d - b e}{c}}}{c x + b}\right )}{b c}, \frac{A c \sqrt{d} \log \left (\frac{e x - 2 \, \sqrt{e x + d} \sqrt{d} + 2 \, d}{x}\right ) + 2 \, \sqrt{e x + d} B b - 2 \,{\left (B b - A c\right )} \sqrt{-\frac{c d - b e}{c}} \arctan \left (-\frac{\sqrt{e x + d} c \sqrt{-\frac{c d - b e}{c}}}{c d - b e}\right )}{b c}, \frac{2 \, A c \sqrt{-d} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) + 2 \, \sqrt{e x + d} B b -{\left (B b - A c\right )} \sqrt{\frac{c d - b e}{c}} \log \left (\frac{c e x + 2 \, c d - b e + 2 \, \sqrt{e x + d} c \sqrt{\frac{c d - b e}{c}}}{c x + b}\right )}{b c}, \frac{2 \,{\left (A c \sqrt{-d} \arctan \left (\frac{\sqrt{e x + d} \sqrt{-d}}{d}\right ) + \sqrt{e x + d} B b -{\left (B b - A c\right )} \sqrt{-\frac{c d - b e}{c}} \arctan \left (-\frac{\sqrt{e x + d} c \sqrt{-\frac{c d - b e}{c}}}{c d - b e}\right )\right )}}{b c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(A*c*sqrt(d)*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*sqrt(e*x + d)*B*b - (B*b - A*c)*sqrt((c*d - b*e
)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/c))/(c*x + b)))/(b*c), (A*c*sqrt(d)*log((e*
x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + 2*sqrt(e*x + d)*B*b - 2*(B*b - A*c)*sqrt(-(c*d - b*e)/c)*arctan(-sqrt(
e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)))/(b*c), (2*A*c*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + 2*sqrt
(e*x + d)*B*b - (B*b - A*c)*sqrt((c*d - b*e)/c)*log((c*e*x + 2*c*d - b*e + 2*sqrt(e*x + d)*c*sqrt((c*d - b*e)/
c))/(c*x + b)))/(b*c), 2*(A*c*sqrt(-d)*arctan(sqrt(e*x + d)*sqrt(-d)/d) + sqrt(e*x + d)*B*b - (B*b - A*c)*sqrt
(-(c*d - b*e)/c)*arctan(-sqrt(e*x + d)*c*sqrt(-(c*d - b*e)/c)/(c*d - b*e)))/(b*c)]

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Sympy [A]  time = 12.2639, size = 104, normalized size = 1.03 \begin{align*} \frac{2 \left (\frac{A d e \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{- d}} \right )}}{b \sqrt{- d}} + \frac{B e \sqrt{d + e x}}{c} - \frac{e \left (- A c + B b\right ) \left (b e - c d\right ) \operatorname{atan}{\left (\frac{\sqrt{d + e x}}{\sqrt{\frac{b e - c d}{c}}} \right )}}{b c^{2} \sqrt{\frac{b e - c d}{c}}}\right )}{e} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(1/2)/(c*x**2+b*x),x)

[Out]

2*(A*d*e*atan(sqrt(d + e*x)/sqrt(-d))/(b*sqrt(-d)) + B*e*sqrt(d + e*x)/c - e*(-A*c + B*b)*(b*e - c*d)*atan(sqr
t(d + e*x)/sqrt((b*e - c*d)/c))/(b*c**2*sqrt((b*e - c*d)/c)))/e

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Giac [A]  time = 1.2155, size = 157, normalized size = 1.55 \begin{align*} \frac{2 \, A d \arctan \left (\frac{\sqrt{x e + d}}{\sqrt{-d}}\right )}{b \sqrt{-d}} + \frac{2 \, \sqrt{x e + d} B}{c} + \frac{2 \,{\left (B b c d - A c^{2} d - B b^{2} e + A b c e\right )} \arctan \left (\frac{\sqrt{x e + d} c}{\sqrt{-c^{2} d + b c e}}\right )}{\sqrt{-c^{2} d + b c e} b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(1/2)/(c*x^2+b*x),x, algorithm="giac")

[Out]

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