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

Optimal. Leaf size=156 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a-b x) \sqrt{c+d x^2}}{x (a+b x)}+\frac{a \sqrt{d} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{a+b x}-\frac{b \sqrt{c} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.113191, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.229, Rules used = {1001, 813, 844, 217, 206, 266, 63, 208} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a-b x) \sqrt{c+d x^2}}{x (a+b x)}+\frac{a \sqrt{d} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{a+b x}-\frac{b \sqrt{c} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a+b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1001

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

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + 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 844

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/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{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+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+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+d x^2}}{x (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{-4 b^2 c-4 a b d x}{x \sqrt{c+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+d x^2}}{x (a+b x)}+\frac{\left (2 b^2 c \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{1}{x \sqrt{c+d x^2}} \, dx}{2 a b+2 b^2 x}+\frac{\left (2 a b d \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \frac{1}{\sqrt{c+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+d x^2}}{x (a+b x)}+\frac{\left (b^2 c \sqrt{a^2+2 a b x+b^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )}{2 a b+2 b^2 x}+\frac{\left (2 a b d \sqrt{a^2+2 a b x+b^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{1-d x^2} \, dx,x,\frac{x}{\sqrt{c+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+d x^2}}{x (a+b x)}+\frac{a \sqrt{d} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{a+b x}+\frac{\left (2 b^2 c \sqrt{a^2+2 a b x+b^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{d \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+d x^2}}{x (a+b x)}+\frac{a \sqrt{d} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{a+b x}-\frac{b \sqrt{c} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a+b x}\\ \end{align*}

Mathematica [A]  time = 0.283455, size = 118, normalized size = 0.76 \[ \frac{\sqrt{(a+b x)^2} \left (\frac{(b x-a) \sqrt{c+d x^2}}{x}+\frac{a \sqrt{c} \sqrt{d} \sqrt{\frac{d x^2}{c}+1} \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )}{\sqrt{c+d x^2}}-b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )\right )}{a+b x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.228, size = 120, normalized size = 0.8 \begin{align*} -{\frac{{\it csgn} \left ( bx+a \right ) }{cx} \left ({c}^{{\frac{3}{2}}}\ln \left ( 2\,{\frac{\sqrt{c}\sqrt{d{x}^{2}+c}+c}{x}} \right ) \sqrt{d}xb-{d}^{{\frac{3}{2}}}\sqrt{d{x}^{2}+c}{x}^{2}a+a \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}\sqrt{d}-\sqrt{d}\sqrt{d{x}^{2}+c}xbc-\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ) xacd \right ){\frac{1}{\sqrt{d}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d x^{2} + 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+c)^(1/2)/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.71983, size = 821, normalized size = 5.26 \begin{align*} \left [\frac{a \sqrt{d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + b \sqrt{c} x \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \, \sqrt{d x^{2} + c}{\left (b x - a\right )}}{2 \, x}, -\frac{2 \, a \sqrt{-d} x \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) - b \sqrt{c} x \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) - 2 \, \sqrt{d x^{2} + c}{\left (b x - a\right )}}{2 \, x}, \frac{2 \, b \sqrt{-c} x \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) + a \sqrt{d} x \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + 2 \, \sqrt{d x^{2} + c}{\left (b x - a\right )}}{2 \, x}, -\frac{a \sqrt{-d} x \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) - b \sqrt{-c} x \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) - \sqrt{d x^{2} + c}{\left (b x - a\right )}}{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+c)^(1/2)/x^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d x^{2}} \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+c)**(1/2)/x**2,x)

[Out]

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

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Giac [A]  time = 1.18116, size = 170, normalized size = 1.09 \begin{align*} \frac{2 \, b c \arctan \left (-\frac{\sqrt{d} x - \sqrt{d x^{2} + c}}{\sqrt{-c}}\right ) \mathrm{sgn}\left (b x + a\right )}{\sqrt{-c}} - a \sqrt{d} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right ) \mathrm{sgn}\left (b x + a\right ) + \sqrt{d x^{2} + c} b \mathrm{sgn}\left (b x + a\right ) + \frac{2 \, a c \sqrt{d} \mathrm{sgn}\left (b x + a\right )}{{\left (\sqrt{d} x - \sqrt{d x^{2} + 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+c)^(1/2)/x^2,x, algorithm="giac")

[Out]

2*b*c*arctan(-(sqrt(d)*x - sqrt(d*x^2 + c))/sqrt(-c))*sgn(b*x + a)/sqrt(-c) - a*sqrt(d)*log(abs(-sqrt(d)*x + s
qrt(d*x^2 + c)))*sgn(b*x + a) + sqrt(d*x^2 + c)*b*sgn(b*x + a) + 2*a*c*sqrt(d)*sgn(b*x + a)/((sqrt(d)*x - sqrt
(d*x^2 + c))^2 - c)