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

Optimal. Leaf size=129 \[ -\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{\sqrt{c} e^2}+\frac{\sqrt{d} \sqrt{c d-b e} \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{e^2}+\frac{\sqrt{b x+c x^2}}{e} \]

[Out]

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

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Rubi [A]  time = 0.141262, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {734, 843, 620, 206, 724} \[ -\frac{(2 c d-b e) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{\sqrt{c} e^2}+\frac{\sqrt{d} \sqrt{c d-b e} \tanh ^{-1}\left (\frac{x (2 c d-b e)+b d}{2 \sqrt{d} \sqrt{b x+c x^2} \sqrt{c d-b e}}\right )}{e^2}+\frac{\sqrt{b x+c x^2}}{e} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[b*x + c*x^2]/(d + e*x),x]

[Out]

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

Rule 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

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

Mathematica [A]  time = 0.523773, size = 137, normalized size = 1.06 \[ \frac{\sqrt{x (b+c x)} \left (-\frac{2 \sqrt{d} \sqrt{b e-c d} \tan ^{-1}\left (\frac{\sqrt{x} \sqrt{b e-c d}}{\sqrt{d} \sqrt{b+c x}}\right )}{\sqrt{b+c x}}+\frac{(b e-2 c d) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{b} \sqrt{c} \sqrt{\frac{c x}{b}+1}}+e \sqrt{x}\right )}{e^2 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.249, size = 490, normalized size = 3.8 \begin{align*}{\frac{1}{e}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}}+{\frac{b}{2\,e}\ln \left ({ \left ({\frac{be-2\,cd}{2\,e}}+ \left ({\frac{d}{e}}+x \right ) c \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \right ){\frac{1}{\sqrt{c}}}}-{\frac{d}{{e}^{2}}\ln \left ({ \left ({\frac{be-2\,cd}{2\,e}}+ \left ({\frac{d}{e}}+x \right ) c \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \right ) \sqrt{c}}+{\frac{bd}{{e}^{2}}\ln \left ({ \left ( -2\,{\frac{d \left ( be-cd \right ) }{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \right ) ^{-1}} \right ){\frac{1}{\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}}}}-{\frac{c{d}^{2}}{{e}^{3}}\ln \left ({ \left ( -2\,{\frac{d \left ( be-cd \right ) }{{e}^{2}}}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }+2\,\sqrt{-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}}\sqrt{c \left ({\frac{d}{e}}+x \right ) ^{2}+{\frac{be-2\,cd}{e} \left ({\frac{d}{e}}+x \right ) }-{\frac{d \left ( be-cd \right ) }{{e}^{2}}}} \right ) \left ({\frac{d}{e}}+x \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((c*x^2+b*x)^(1/2)/(e*x+d),x)

[Out]

1/e*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2)+1/2/e*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(1/2)+(
c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/c^(1/2)*b-1/e^2*ln((1/2*(b*e-2*c*d)/e+(d/e+x)*c)/c^(
1/2)+(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))*c^(1/2)*d+1/e^2*d/(-d*(b*e-c*d)/e^2)^(1/2)*ln(
(-2*d*(b*e-c*d)/e^2+(b*e-2*c*d)/e*(d/e+x)+2*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e
-c*d)/e^2)^(1/2))/(d/e+x))*b-1/e^3*d^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*(d/e+x)+2
*(-d*(b*e-c*d)/e^2)^(1/2)*(c*(d/e+x)^2+(b*e-2*c*d)/e*(d/e+x)-d*(b*e-c*d)/e^2)^(1/2))/(d/e+x))*c

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.59809, size = 1102, normalized size = 8.54 \begin{align*} \left [\frac{2 \, \sqrt{c x^{2} + b x} c e -{\left (2 \, c d - b e\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \, \sqrt{c d^{2} - b d e} c \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 )}{2 \, c e^{2}}, \frac{2 \, \sqrt{c x^{2} + b x} c e + 4 \, \sqrt{-c d^{2} + b d e} c \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 (2 \, c d - b e\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right )}{2 \, c e^{2}}, \frac{\sqrt{c x^{2} + b x} c e +{\left (2 \, c d - b e\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) + \sqrt{c d^{2} - b d e} c \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 e^{2}}, \frac{\sqrt{c x^{2} + b x} c e + 2 \, \sqrt{-c d^{2} + b d e} c \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 (2 \, c d - b e\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right )}{c e^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: TypeError