3.45 \(\int \frac{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{x^3} \, dx\)

Optimal. Leaf size=161 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+2 b x) \sqrt{c+d x^2}}{2 x^2 (a+b x)}+\frac{b \sqrt{d} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{a+b x}-\frac{a d \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 \sqrt{c} (a+b x)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.116269, antiderivative size = 161, 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, 811, 844, 217, 206, 266, 63, 208} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+2 b x) \sqrt{c+d x^2}}{2 x^2 (a+b x)}+\frac{b \sqrt{d} \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c+d x^2}}\right )}{a+b x}-\frac{a d \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 \sqrt{c} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/x^3,x]

[Out]

-((a + 2*b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/(2*x^2*(a + b*x)) + (b*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) - (a*d*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[Sqrt[
c + d*x^2]/Sqrt[c]])/(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 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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^3} \, 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^3} \, dx}{2 a b+2 b^2 x}\\ &=-\frac{(a+2 b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 x^2 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{-4 a b c d-8 b^2 c d x}{x \sqrt{c+d x^2}} \, dx}{4 c \left (2 a b+2 b^2 x\right )}\\ &=-\frac{(a+2 b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 x^2 (a+b x)}+\frac{\left (a b d \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 b^2 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+2 b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 x^2 (a+b x)}+\frac{\left (a b d \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 \left (2 a b+2 b^2 x\right )}+\frac{\left (2 b^2 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+2 b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 x^2 (a+b x)}+\frac{b \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 (a b \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 )}{2 a b+2 b^2 x}\\ &=-\frac{(a+2 b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{c+d x^2}}{2 x^2 (a+b x)}+\frac{b \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{a d \sqrt{a^2+2 a b x+b^2 x^2} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 \sqrt{c} (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.116849, size = 126, normalized size = 0.78 \[ -\frac{\sqrt{(a+b x)^2} \sqrt{c+d x^2} \left (c (a+2 b x) \sqrt{\frac{d x^2}{c}+1}+a d x^2 \tanh ^{-1}\left (\sqrt{\frac{d x^2}{c}+1}\right )-2 b \sqrt{c} \sqrt{d} x^2 \sinh ^{-1}\left (\frac{\sqrt{d} x}{\sqrt{c}}\right )\right )}{2 c x^2 (a+b x) \sqrt{\frac{d x^2}{c}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2])/x^3,x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.24, size = 141, normalized size = 0.9 \begin{align*} -{\frac{{\it csgn} \left ( bx+a \right ) }{2\,c{x}^{2}} \left ( \sqrt{c}\ln \left ( 2\,{\frac{\sqrt{c}\sqrt{d{x}^{2}+c}+c}{x}} \right ){d}^{{\frac{3}{2}}}{x}^{2}a-2\,{d}^{3/2}\sqrt{d{x}^{2}+c}{x}^{3}b+2\,\sqrt{d} \left ( d{x}^{2}+c \right ) ^{3/2}xb-{d}^{{\frac{3}{2}}}\sqrt{d{x}^{2}+c}{x}^{2}a-2\,\ln \left ( x\sqrt{d}+\sqrt{d{x}^{2}+c} \right ){x}^{2}bcd+a \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}\sqrt{d} \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^3,x)

[Out]

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

________________________________________________________________________________________

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^{3}}\,{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^3,x, algorithm="maxima")

[Out]

integrate(sqrt(d*x^2 + c)*sqrt((b*x + a)^2)/x^3, x)

________________________________________________________________________________________

Fricas [A]  time = 1.72258, size = 934, normalized size = 5.8 \begin{align*} \left [\frac{2 \, b c \sqrt{d} x^{2} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) + a \sqrt{c} d x^{2} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) - 2 \,{\left (2 \, b c x + a c\right )} \sqrt{d x^{2} + c}}{4 \, c x^{2}}, -\frac{4 \, b c \sqrt{-d} x^{2} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) - a \sqrt{c} d x^{2} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (2 \, b c x + a c\right )} \sqrt{d x^{2} + c}}{4 \, c x^{2}}, \frac{a \sqrt{-c} d x^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) + b c \sqrt{d} x^{2} \log \left (-2 \, d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{d} x - c\right ) -{\left (2 \, b c x + a c\right )} \sqrt{d x^{2} + c}}{2 \, c x^{2}}, -\frac{2 \, b c \sqrt{-d} x^{2} \arctan \left (\frac{\sqrt{-d} x}{\sqrt{d x^{2} + c}}\right ) - a \sqrt{-c} d x^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (2 \, b c x + a c\right )} \sqrt{d x^{2} + c}}{2 \, c x^{2}}\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^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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^{3}}\, 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**3,x)

[Out]

Integral(sqrt(c + d*x**2)*sqrt((a + b*x)**2)/x**3, x)

________________________________________________________________________________________

Giac [A]  time = 1.14724, size = 269, normalized size = 1.67 \begin{align*} \frac{a d \arctan \left (-\frac{\sqrt{d} x - \sqrt{d x^{2} + c}}{\sqrt{-c}}\right ) \mathrm{sgn}\left (b x + a\right )}{\sqrt{-c}} - b \sqrt{d} \log \left ({\left | -\sqrt{d} x + \sqrt{d x^{2} + c} \right |}\right ) \mathrm{sgn}\left (b x + a\right ) + \frac{{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{3} a d \mathrm{sgn}\left (b x + a\right ) + 2 \,{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} b c \sqrt{d} \mathrm{sgn}\left (b x + a\right ) +{\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )} a c d \mathrm{sgn}\left (b x + a\right ) - 2 \, b c^{2} \sqrt{d} \mathrm{sgn}\left (b x + a\right )}{{\left ({\left (\sqrt{d} x - \sqrt{d x^{2} + c}\right )}^{2} - c\right )}^{2}} \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^3,x, algorithm="giac")

[Out]

a*d*arctan(-(sqrt(d)*x - sqrt(d*x^2 + c))/sqrt(-c))*sgn(b*x + a)/sqrt(-c) - b*sqrt(d)*log(abs(-sqrt(d)*x + sqr
t(d*x^2 + c)))*sgn(b*x + a) + ((sqrt(d)*x - sqrt(d*x^2 + c))^3*a*d*sgn(b*x + a) + 2*(sqrt(d)*x - sqrt(d*x^2 +
c))^2*b*c*sqrt(d)*sgn(b*x + a) + (sqrt(d)*x - sqrt(d*x^2 + c))*a*c*d*sgn(b*x + a) - 2*b*c^2*sqrt(d)*sgn(b*x +
a))/((sqrt(d)*x - sqrt(d*x^2 + c))^2 - c)^2