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

Optimal. Leaf size=202 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a-b x) \sqrt{c+d x^2+e x}}{x (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (2 a d+b e) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{2 \sqrt{d} (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a e+2 b c) \tanh ^{-1}\left (\frac{2 c+e x}{2 \sqrt{c} \sqrt{c+d x^2+e x}}\right )}{2 \sqrt{c} (a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.194411, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {1000, 812, 843, 621, 206, 724} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a-b x) \sqrt{c+d x^2+e x}}{x (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (2 a d+b e) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+d x^2+e x}}\right )}{2 \sqrt{d} (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a e+2 b c) \tanh ^{-1}\left (\frac{2 c+e x}{2 \sqrt{c} \sqrt{c+d x^2+e x}}\right )}{2 \sqrt{c} (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1000

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

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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 621

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

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

Mathematica [A]  time = 0.195053, size = 155, normalized size = 0.77 \[ \frac{\sqrt{(a+b x)^2} \left (\sqrt{c} x (2 a d+b e) \tanh ^{-1}\left (\frac{2 d x+e}{2 \sqrt{d} \sqrt{c+x (d x+e)}}\right )+\sqrt{d} \left (2 \sqrt{c} (b x-a) \sqrt{c+x (d x+e)}-x (a e+2 b c) \tanh ^{-1}\left (\frac{2 c+e x}{2 \sqrt{c} \sqrt{c+x (d x+e)}}\right )\right )\right )}{2 \sqrt{c} \sqrt{d} x (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.216, size = 249, normalized size = 1.2 \begin{align*}{\frac{{\it csgn} \left ( bx+a \right ) }{2\,cx} \left ( 2\,{d}^{5/2}\sqrt{d{x}^{2}+ex+c}{x}^{2}a-2\,{d}^{3/2}{c}^{3/2}\ln \left ({\frac{2\,c+ex+2\,\sqrt{c}\sqrt{d{x}^{2}+ex+c}}{x}} \right ) xb-{d}^{{\frac{3}{2}}}\sqrt{c}\ln \left ({\frac{1}{x} \left ( 2\,c+ex+2\,\sqrt{c}\sqrt{d{x}^{2}+ex+c} \right ) } \right ) xae-2\,{d}^{3/2} \left ( d{x}^{2}+ex+c \right ) ^{3/2}a+2\,{d}^{3/2}\sqrt{d{x}^{2}+ex+c}xae+2\,{d}^{3/2}\sqrt{d{x}^{2}+ex+c}xbc+2\,\ln \left ( 1/2\,{\frac{2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e}{\sqrt{d}}} \right ) xac{d}^{2}+\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{d{x}^{2}+ex+c}\sqrt{d}+2\,dx+e \right ){\frac{1}{\sqrt{d}}}} \right ) dxbce \right ){d}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*csgn(b*x+a)*(2*d^(5/2)*(d*x^2+e*x+c)^(1/2)*x^2*a-2*d^(3/2)*c^(3/2)*ln((2*c+e*x+2*c^(1/2)*(d*x^2+e*x+c)^(1/
2))/x)*x*b-d^(3/2)*c^(1/2)*ln((2*c+e*x+2*c^(1/2)*(d*x^2+e*x+c)^(1/2))/x)*x*a*e-2*d^(3/2)*(d*x^2+e*x+c)^(3/2)*a
+2*d^(3/2)*(d*x^2+e*x+c)^(1/2)*x*a*e+2*d^(3/2)*(d*x^2+e*x+c)^(1/2)*x*b*c+2*ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/
2)+2*d*x+e)/d^(1/2))*x*a*c*d^2+ln(1/2*(2*(d*x^2+e*x+c)^(1/2)*d^(1/2)+2*d*x+e)/d^(1/2))*d*x*b*c*e)/x/c/d^(3/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 3.45487, size = 1562, normalized size = 7.73 \begin{align*} \left [\frac{{\left (2 \, a c d + b c e\right )} \sqrt{d} x \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt{d x^{2} + e x + c}{\left (2 \, d x + e\right )} \sqrt{d} + 4 \, c d + e^{2}\right ) +{\left (2 \, b c d + a d e\right )} \sqrt{c} x \log \left (\frac{8 \, c e x +{\left (4 \, c d + e^{2}\right )} x^{2} - 4 \, \sqrt{d x^{2} + e x + c}{\left (e x + 2 \, c\right )} \sqrt{c} + 8 \, c^{2}}{x^{2}}\right ) + 4 \,{\left (b c d x - a c d\right )} \sqrt{d x^{2} + e x + c}}{4 \, c d x}, -\frac{2 \,{\left (2 \, a c d + b c e\right )} \sqrt{-d} x \arctan \left (\frac{\sqrt{d x^{2} + e x + c}{\left (2 \, d x + e\right )} \sqrt{-d}}{2 \,{\left (d^{2} x^{2} + d e x + c d\right )}}\right ) -{\left (2 \, b c d + a d e\right )} \sqrt{c} x \log \left (\frac{8 \, c e x +{\left (4 \, c d + e^{2}\right )} x^{2} - 4 \, \sqrt{d x^{2} + e x + c}{\left (e x + 2 \, c\right )} \sqrt{c} + 8 \, c^{2}}{x^{2}}\right ) - 4 \,{\left (b c d x - a c d\right )} \sqrt{d x^{2} + e x + c}}{4 \, c d x}, \frac{2 \,{\left (2 \, b c d + a d e\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{d x^{2} + e x + c}{\left (e x + 2 \, c\right )} \sqrt{-c}}{2 \,{\left (c d x^{2} + c e x + c^{2}\right )}}\right ) +{\left (2 \, a c d + b c e\right )} \sqrt{d} x \log \left (8 \, d^{2} x^{2} + 8 \, d e x + 4 \, \sqrt{d x^{2} + e x + c}{\left (2 \, d x + e\right )} \sqrt{d} + 4 \, c d + e^{2}\right ) + 4 \,{\left (b c d x - a c d\right )} \sqrt{d x^{2} + e x + c}}{4 \, c d x}, \frac{{\left (2 \, b c d + a d e\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{d x^{2} + e x + c}{\left (e x + 2 \, c\right )} \sqrt{-c}}{2 \,{\left (c d x^{2} + c e x + c^{2}\right )}}\right ) -{\left (2 \, a c d + b c e\right )} \sqrt{-d} x \arctan \left (\frac{\sqrt{d x^{2} + e x + c}{\left (2 \, d x + e\right )} \sqrt{-d}}{2 \,{\left (d^{2} x^{2} + d e x + c d\right )}}\right ) + 2 \,{\left (b c d x - a c d\right )} \sqrt{d x^{2} + e x + c}}{2 \, c d x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*((2*a*c*d + b*c*e)*sqrt(d)*x*log(8*d^2*x^2 + 8*d*e*x + 4*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*
d + e^2) + (2*b*c*d + a*d*e)*sqrt(c)*x*log((8*c*e*x + (4*c*d + e^2)*x^2 - 4*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*
sqrt(c) + 8*c^2)/x^2) + 4*(b*c*d*x - a*c*d)*sqrt(d*x^2 + e*x + c))/(c*d*x), -1/4*(2*(2*a*c*d + b*c*e)*sqrt(-d)
*x*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) - (2*b*c*d + a*d*e)*sqrt(c)*
x*log((8*c*e*x + (4*c*d + e^2)*x^2 - 4*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*sqrt(c) + 8*c^2)/x^2) - 4*(b*c*d*x -
a*c*d)*sqrt(d*x^2 + e*x + c))/(c*d*x), 1/4*(2*(2*b*c*d + a*d*e)*sqrt(-c)*x*arctan(1/2*sqrt(d*x^2 + e*x + c)*(e
*x + 2*c)*sqrt(-c)/(c*d*x^2 + c*e*x + c^2)) + (2*a*c*d + b*c*e)*sqrt(d)*x*log(8*d^2*x^2 + 8*d*e*x + 4*sqrt(d*x
^2 + e*x + c)*(2*d*x + e)*sqrt(d) + 4*c*d + e^2) + 4*(b*c*d*x - a*c*d)*sqrt(d*x^2 + e*x + c))/(c*d*x), 1/2*((2
*b*c*d + a*d*e)*sqrt(-c)*x*arctan(1/2*sqrt(d*x^2 + e*x + c)*(e*x + 2*c)*sqrt(-c)/(c*d*x^2 + c*e*x + c^2)) - (2
*a*c*d + b*c*e)*sqrt(-d)*x*arctan(1/2*sqrt(d*x^2 + e*x + c)*(2*d*x + e)*sqrt(-d)/(d^2*x^2 + d*e*x + c*d)) + 2*
(b*c*d*x - a*c*d)*sqrt(d*x^2 + e*x + c))/(c*d*x)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.21186, size = 284, normalized size = 1.41 \begin{align*} \sqrt{d x^{2} + x e + c} b \mathrm{sgn}\left (b x + a\right ) + \frac{{\left (2 \, b c \mathrm{sgn}\left (b x + a\right ) + a e \mathrm{sgn}\left (b x + a\right )\right )} \arctan \left (-\frac{\sqrt{d} x - \sqrt{d x^{2} + x e + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} - \frac{{\left (2 \, a d \mathrm{sgn}\left (b x + a\right ) + b e \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + x e + c}\right )} \sqrt{d} + e \right |}\right )}{2 \, \sqrt{d}} + \frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + x e + c}\right )} a e \mathrm{sgn}\left (b x + a\right ) + 2 \, a c \sqrt{d} \mathrm{sgn}\left (b x + a\right )}{{\left (\sqrt{d} x - \sqrt{d x^{2} + x e + c}\right )}^{2} - c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(d*x^2 + x*e + c)*b*sgn(b*x + a) + (2*b*c*sgn(b*x + a) + a*e*sgn(b*x + a))*arctan(-(sqrt(d)*x - sqrt(d*x^2
+ x*e + c))/sqrt(-c))/sqrt(-c) - 1/2*(2*a*d*sgn(b*x + a) + b*e*sgn(b*x + a))*log(abs(2*(sqrt(d)*x - sqrt(d*x^
2 + x*e + c))*sqrt(d) + e))/sqrt(d) + ((sqrt(d)*x - sqrt(d*x^2 + x*e + c))*a*e*sgn(b*x + a) + 2*a*c*sqrt(d)*sg
n(b*x + a))/((sqrt(d)*x - sqrt(d*x^2 + x*e + c))^2 - c)