3.272 \(\int \frac{\sqrt{c+d x^2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{x^3} \, dx\)

Optimal. Leaf size=177 \[ -\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{2 c x^2 \left (a+b x^2\right )}+\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2} (a d+2 b c)}{2 c \left (a+b x^2\right )}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} (a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 \sqrt{c} \left (a+b x^2\right )} \]

[Out]

((2*b*c + a*d)*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*c*(a + b*x^2)) - (a*(c + d*x^2)^(3/2)*Sqrt[
a^2 + 2*a*b*x^2 + b^2*x^4])/(2*c*x^2*(a + b*x^2)) - ((2*b*c + a*d)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[Sqr
t[c + d*x^2]/Sqrt[c]])/(2*Sqrt[c]*(a + b*x^2))

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Rubi [A]  time = 0.118521, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {1250, 446, 78, 50, 63, 208} \[ -\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{2 c x^2 \left (a+b x^2\right )}+\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2} (a d+2 b c)}{2 c \left (a+b x^2\right )}-\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} (a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{2 \sqrt{c} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^3,x]

[Out]

((2*b*c + a*d)*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(2*c*(a + b*x^2)) - (a*(c + d*x^2)^(3/2)*Sqrt[
a^2 + 2*a*b*x^2 + b^2*x^4])/(2*c*x^2*(a + b*x^2)) - ((2*b*c + a*d)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[Sqr
t[c + d*x^2]/Sqrt[c]])/(2*Sqrt[c]*(a + b*x^2))

Rule 1250

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

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A]  time = 0.0449428, size = 90, normalized size = 0.51 \[ -\frac{\sqrt{\left (a+b x^2\right )^2} \left (\sqrt{c} \left (a-2 b x^2\right ) \sqrt{c+d x^2}+x^2 (a d+2 b c) \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )\right )}{2 \sqrt{c} x^2 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/x^3,x]

[Out]

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

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Maple [A]  time = 0.01, size = 133, normalized size = 0.8 \begin{align*} -{\frac{1}{ \left ( 2\,b{x}^{2}+2\,a \right ) c{x}^{2}}\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( \sqrt{c}\ln \left ( 2\,{\frac{\sqrt{c}\sqrt{d{x}^{2}+c}+c}{x}} \right ){x}^{2}ad+2\,{c}^{3/2}\ln \left ( 2\,{\frac{\sqrt{c}\sqrt{d{x}^{2}+c}+c}{x}} \right ){x}^{2}b-\sqrt{d{x}^{2}+c}{x}^{2}ad-2\,\sqrt{d{x}^{2}+c}{x}^{2}bc+ \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x^2+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^3,x)

[Out]

-1/2*((b*x^2+a)^2)^(1/2)*(c^(1/2)*ln(2*(c^(1/2)*(d*x^2+c)^(1/2)+c)/x)*x^2*a*d+2*c^(3/2)*ln(2*(c^(1/2)*(d*x^2+c
)^(1/2)+c)/x)*x^2*b-(d*x^2+c)^(1/2)*x^2*a*d-2*(d*x^2+c)^(1/2)*x^2*b*c+(d*x^2+c)^(3/2)*a)/(b*x^2+a)/c/x^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^{2} + a\right )}^{2}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.02218, size = 332, normalized size = 1.88 \begin{align*} \left [\frac{{\left (2 \, b c + a d\right )} \sqrt{c} 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^{2} - a c\right )} \sqrt{d x^{2} + c}}{4 \, c x^{2}}, \frac{{\left (2 \, b c + a d\right )} \sqrt{-c} x^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (2 \, b c x^{2} - 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((d*x^2+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x**2+c)**(1/2)*((b*x**2+a)**2)**(1/2)/x**3,x)

[Out]

Timed out

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Giac [A]  time = 1.1049, size = 135, normalized size = 0.76 \begin{align*} \frac{2 \, \sqrt{d x^{2} + c} b d \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{{\left (2 \, b c d \mathrm{sgn}\left (b x^{2} + a\right ) + a d^{2} \mathrm{sgn}\left (b x^{2} + a\right )\right )} \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} - \frac{\sqrt{d x^{2} + c} a d \mathrm{sgn}\left (b x^{2} + a\right )}{x^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x^2+c)^(1/2)*((b*x^2+a)^2)^(1/2)/x^3,x, algorithm="giac")

[Out]

1/2*(2*sqrt(d*x^2 + c)*b*d*sgn(b*x^2 + a) + (2*b*c*d*sgn(b*x^2 + a) + a*d^2*sgn(b*x^2 + a))*arctan(sqrt(d*x^2
+ c)/sqrt(-c))/sqrt(-c) - sqrt(d*x^2 + c)*a*d*sgn(b*x^2 + a)/x^2)/d