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

Optimal. Leaf size=152 \[ \frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{3 d \left (a+b x^2\right )}+\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2}}{a+b x^2}-\frac{a \sqrt{c} \sqrt{a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a+b x^2} \]

[Out]

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

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Rubi [A]  time = 0.0948647, antiderivative size = 152, 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, 80, 50, 63, 208} \[ \frac{b \sqrt{a^2+2 a b x^2+b^2 x^4} \left (c+d x^2\right )^{3/2}}{3 d \left (a+b x^2\right )}+\frac{a \sqrt{a^2+2 a b x^2+b^2 x^4} \sqrt{c+d x^2}}{a+b x^2}-\frac{a \sqrt{c} \sqrt{a^2+2 a b x^2+b^2 x^4} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )}{a+b x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sqrt[c + d*x^2]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(a + b*x^2) + (b*(c + d*x^2)^(3/2)*Sqrt[a^2 + 2*a*b*x^2 +
b^2*x^4])/(3*d*(a + b*x^2)) - (a*Sqrt[c]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*ArcTanh[Sqrt[c + d*x^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 80

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

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

Mathematica [A]  time = 0.0460664, size = 83, normalized size = 0.55 \[ \frac{\sqrt{\left (a+b x^2\right )^2} \left (\sqrt{c+d x^2} \left (3 a d+b \left (c+d x^2\right )\right )-3 a \sqrt{c} d \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )\right )}{3 d \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 80, normalized size = 0.5 \begin{align*} -{\frac{1}{ \left ( 3\,b{x}^{2}+3\,a \right ) d}\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}} \left ( 3\,\sqrt{c}\ln \left ( 2\,{\frac{\sqrt{c}\sqrt{d{x}^{2}+c}+c}{x}} \right ) ad-b \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}-3\,\sqrt{d{x}^{2}+c}ad \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,x)

[Out]

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

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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}\,{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,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.83764, size = 300, normalized size = 1.97 \begin{align*} \left [\frac{3 \, a \sqrt{c} d \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (b d x^{2} + b c + 3 \, a d\right )} \sqrt{d x^{2} + c}}{6 \, d}, \frac{3 \, a \sqrt{-c} d \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (b d x^{2} + b c + 3 \, a d\right )} \sqrt{d x^{2} + c}}{3 \, d}\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,x, algorithm="fricas")

[Out]

[1/6*(3*a*sqrt(c)*d*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(b*d*x^2 + b*c + 3*a*d)*sqrt(d*x^2
+ c))/d, 1/3*(3*a*sqrt(-c)*d*arctan(sqrt(-c)/sqrt(d*x^2 + c)) + (b*d*x^2 + b*c + 3*a*d)*sqrt(d*x^2 + c))/d]

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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^{2}\right )^{2}}}{x}\, dx \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,x)

[Out]

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

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Giac [A]  time = 1.1103, size = 113, normalized size = 0.74 \begin{align*} \frac{a c \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right ) \mathrm{sgn}\left (b x^{2} + a\right )}{\sqrt{-c}} + \frac{{\left (d x^{2} + c\right )}^{\frac{3}{2}} b d^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 3 \, \sqrt{d x^{2} + c} a d^{3} \mathrm{sgn}\left (b x^{2} + a\right )}{3 \, d^{3}} \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,x, algorithm="giac")

[Out]

a*c*arctan(sqrt(d*x^2 + c)/sqrt(-c))*sgn(b*x^2 + a)/sqrt(-c) + 1/3*((d*x^2 + c)^(3/2)*b*d^2*sgn(b*x^2 + a) + 3
*sqrt(d*x^2 + c)*a*d^3*sgn(b*x^2 + a))/d^3