3.1195 \(\int \frac{A+B x}{(d+e x) \sqrt{b x+c x^2}} \, dx\)

Optimal. Leaf size=113 \[ \frac{2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{\sqrt{c} e}-\frac{(B d-A e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{\sqrt{d} e \sqrt{c d-b e}} \]

[Out]

(2*B*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(Sqrt[c]*e) - ((B*d - A*e)*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqr
t[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(Sqrt[d]*e*Sqrt[c*d - b*e])

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Rubi [A]  time = 0.0816045, antiderivative size = 113, 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 = {843, 620, 206, 724} \[ \frac{2 B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{\sqrt{c} e}-\frac{(B d-A e) \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{\sqrt{d} e \sqrt{c d-b e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*B*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(Sqrt[c]*e) - ((B*d - A*e)*ArcTanh[(b*d + (2*c*d - b*e)*x)/(2*Sqr
t[d]*Sqrt[c*d - b*e]*Sqrt[b*x + c*x^2])])/(Sqrt[d]*e*Sqrt[c*d - b*e])

Rule 843

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

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

Rubi steps

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

Mathematica [A]  time = 0.181815, size = 131, normalized size = 1.16 \[ \frac{2 \sqrt{x} \left (\frac{\sqrt{b+c x} (A e-B d) \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{\sqrt{d} \sqrt{b e-c d}}+\frac{\sqrt{b} B \sqrt{\frac{c x}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{c}}\right )}{e \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*((Sqrt[b]*B*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/Sqrt[c] + ((-(B*d) + A*e)*Sqrt[b
+ c*x]*ArcTan[(Sqrt[-(c*d) + b*e]*Sqrt[x])/(Sqrt[d]*Sqrt[b + c*x])])/(Sqrt[d]*Sqrt[-(c*d) + b*e])))/(e*Sqrt[x*
(b + c*x)])

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Maple [B]  time = 0.007, size = 298, normalized size = 2.6 \begin{align*}{\frac{B}{e}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){\frac{1}{\sqrt{c}}}}-{\frac{A}{e}\ln \left ({ \left ( -2\,{\frac{d \left ( be-cd \right ) }{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ( x+{\frac{d}{e}} \right ) }+2\,\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}\sqrt{ \left ( x+{\frac{d}{e}} \right ) ^{2}c+{\frac{be-2\,cd}{e} \left ( x+{\frac{d}{e}} \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \right ) \left ( x+{\frac{d}{e}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}}}}+{\frac{Bd}{{e}^{2}}\ln \left ({ \left ( -2\,{\frac{d \left ( be-cd \right ) }{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ( x+{\frac{d}{e}} \right ) }+2\,\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}\sqrt{ \left ( x+{\frac{d}{e}} \right ) ^{2}c+{\frac{be-2\,cd}{e} \left ( x+{\frac{d}{e}} \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \right ) \left ( x+{\frac{d}{e}} \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8784, size = 1154, normalized size = 10.21 \begin{align*} \left [\frac{{\left (B c d^{2} - B b d e\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) -{\left (B c d - A c e\right )} \sqrt{c d^{2} - b d e} \log \left (\frac{b d +{\left (2 \, c d - b e\right )} x + 2 \, \sqrt{c d^{2} - b d e} \sqrt{c x^{2} + b x}}{e x + d}\right )}{c^{2} d^{2} e - b c d e^{2}}, -\frac{2 \,{\left (B c d - A c e\right )} \sqrt{-c d^{2} + b d e} \arctan \left (-\frac{\sqrt{-c d^{2} + b d e} \sqrt{c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) -{\left (B c d^{2} - B b d e\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right )}{c^{2} d^{2} e - b c d e^{2}}, -\frac{2 \,{\left (B c d^{2} - B b d e\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (B c d - A c e\right )} \sqrt{c d^{2} - b d e} \log \left (\frac{b d +{\left (2 \, c d - b e\right )} x + 2 \, \sqrt{c d^{2} - b d e} \sqrt{c x^{2} + b x}}{e x + d}\right )}{c^{2} d^{2} e - b c d e^{2}}, -\frac{2 \,{\left ({\left (B c d - A c e\right )} \sqrt{-c d^{2} + b d e} \arctan \left (-\frac{\sqrt{-c d^{2} + b d e} \sqrt{c x^{2} + b x}}{{\left (c d - b e\right )} x}\right ) +{\left (B c d^{2} - B b d e\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right )\right )}}{c^{2} d^{2} e - b c d e^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError