3.93 \(\int \frac{(A+B x^2) \sqrt{b x^2+c x^4}}{x} \, dx\)

Optimal. Leaf size=100 \[ -\frac{b (b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{3/2}}-\frac{\sqrt{b x^2+c x^4} (b B-4 A c)}{8 c}+\frac{B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2} \]

[Out]

-((b*B - 4*A*c)*Sqrt[b*x^2 + c*x^4])/(8*c) + (B*(b*x^2 + c*x^4)^(3/2))/(4*c*x^2) - (b*(b*B - 4*A*c)*ArcTanh[(S
qrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(8*c^(3/2))

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Rubi [A]  time = 0.19702, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2034, 794, 664, 620, 206} \[ -\frac{b (b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{3/2}}-\frac{\sqrt{b x^2+c x^4} (b B-4 A c)}{8 c}+\frac{B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*B - 4*A*c)*Sqrt[b*x^2 + c*x^4])/(8*c) + (B*(b*x^2 + c*x^4)^(3/2))/(4*c*x^2) - (b*(b*B - 4*A*c)*ArcTanh[(S
qrt[c]*x^2)/Sqrt[b*x^2 + c*x^4]])/(8*c^(3/2))

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 794

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)
*(2*c*f - b*g))/(c*e*(m + 2*p + 2)), 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] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[m + 2*p + 2, 0] && (NeQ[m, 2] || Eq
Q[d, 0])

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*p]

Rule 620

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

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])

Rubi steps

\begin{align*} \int \frac{\left (A+B x^2\right ) \sqrt{b x^2+c x^4}}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(A+B x) \sqrt{b x+c x^2}}{x} \, dx,x,x^2\right )\\ &=\frac{B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2}+\frac{\left (b B-A c+\frac{3}{2} (-b B+2 A c)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{b x+c x^2}}{x} \, dx,x,x^2\right )}{4 c}\\ &=-\frac{(b B-4 A c) \sqrt{b x^2+c x^4}}{8 c}+\frac{B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2}-\frac{(b (b B-4 A c)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b x+c x^2}} \, dx,x,x^2\right )}{16 c}\\ &=-\frac{(b B-4 A c) \sqrt{b x^2+c x^4}}{8 c}+\frac{B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2}-\frac{(b (b B-4 A c)) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c}\\ &=-\frac{(b B-4 A c) \sqrt{b x^2+c x^4}}{8 c}+\frac{B \left (b x^2+c x^4\right )^{3/2}}{4 c x^2}-\frac{b (b B-4 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x^2}{\sqrt{b x^2+c x^4}}\right )}{8 c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.143771, size = 91, normalized size = 0.91 \[ \frac{\sqrt{x^2 \left (b+c x^2\right )} \left (\sqrt{c} \left (4 A c+b B+2 B c x^2\right )-\frac{\sqrt{b} (b B-4 A c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{x \sqrt{\frac{c x^2}{b}+1}}\right )}{8 c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[x^2*(b + c*x^2)]*(Sqrt[c]*(b*B + 4*A*c + 2*B*c*x^2) - (Sqrt[b]*(b*B - 4*A*c)*ArcSinh[(Sqrt[c]*x)/Sqrt[b]
])/(x*Sqrt[1 + (c*x^2)/b])))/(8*c^(3/2))

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Maple [A]  time = 0.006, size = 124, normalized size = 1.2 \begin{align*}{\frac{1}{8\,x}\sqrt{c{x}^{4}+b{x}^{2}} \left ( 2\,B\sqrt{c} \left ( c{x}^{2}+b \right ) ^{3/2}x+4\,A{c}^{3/2}\sqrt{c{x}^{2}+b}x-B\sqrt{c}\sqrt{c{x}^{2}+b}xb+4\,A\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ) bc-B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+b} \right ){b}^{2} \right ){c}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{c{x}^{2}+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x)

[Out]

1/8*(c*x^4+b*x^2)^(1/2)*(2*B*c^(1/2)*(c*x^2+b)^(3/2)*x+4*A*c^(3/2)*(c*x^2+b)^(1/2)*x-B*c^(1/2)*(c*x^2+b)^(1/2)
*x*b+4*A*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b*c-B*ln(x*c^(1/2)+(c*x^2+b)^(1/2))*b^2)/c^(3/2)/(c*x^2+b)^(1/2)/x

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.17224, size = 392, normalized size = 3.92 \begin{align*} \left [-\frac{{\left (B b^{2} - 4 \, A b c\right )} \sqrt{c} \log \left (-2 \, c x^{2} - b - 2 \, \sqrt{c x^{4} + b x^{2}} \sqrt{c}\right ) - 2 \,{\left (2 \, B c^{2} x^{2} + B b c + 4 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{16 \, c^{2}}, \frac{{\left (B b^{2} - 4 \, A b c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{4} + b x^{2}} \sqrt{-c}}{c x^{2} + b}\right ) +{\left (2 \, B c^{2} x^{2} + B b c + 4 \, A c^{2}\right )} \sqrt{c x^{4} + b x^{2}}}{8 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x, algorithm="fricas")

[Out]

[-1/16*((B*b^2 - 4*A*b*c)*sqrt(c)*log(-2*c*x^2 - b - 2*sqrt(c*x^4 + b*x^2)*sqrt(c)) - 2*(2*B*c^2*x^2 + B*b*c +
4*A*c^2)*sqrt(c*x^4 + b*x^2))/c^2, 1/8*((B*b^2 - 4*A*b*c)*sqrt(-c)*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-c)/(c*x^2
+ b)) + (2*B*c^2*x^2 + B*b*c + 4*A*c^2)*sqrt(c*x^4 + b*x^2))/c^2]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**(1/2)/x,x)

[Out]

Integral(sqrt(x**2*(b + c*x**2))*(A + B*x**2)/x, x)

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Giac [A]  time = 1.34825, size = 139, normalized size = 1.39 \begin{align*} \frac{1}{8} \,{\left (2 \, B x^{2} \mathrm{sgn}\left (x\right ) + \frac{B b c \mathrm{sgn}\left (x\right ) + 4 \, A c^{2} \mathrm{sgn}\left (x\right )}{c^{2}}\right )} \sqrt{c x^{2} + b} x + \frac{{\left (B b^{2} \mathrm{sgn}\left (x\right ) - 4 \, A b c \mathrm{sgn}\left (x\right )\right )} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + b} \right |}\right )}{8 \, c^{\frac{3}{2}}} - \frac{{\left (B b^{2} \log \left ({\left | b \right |}\right ) - 4 \, A b c \log \left ({\left | b \right |}\right )\right )} \mathrm{sgn}\left (x\right )}{16 \, c^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)^(1/2)/x,x, algorithm="giac")

[Out]

1/8*(2*B*x^2*sgn(x) + (B*b*c*sgn(x) + 4*A*c^2*sgn(x))/c^2)*sqrt(c*x^2 + b)*x + 1/8*(B*b^2*sgn(x) - 4*A*b*c*sgn
(x))*log(abs(-sqrt(c)*x + sqrt(c*x^2 + b)))/c^(3/2) - 1/16*(B*b^2*log(abs(b)) - 4*A*b*c*log(abs(b)))*sgn(x)/c^
(3/2)