3.527 \(\int \frac{(a+b x^2)^{3/2} (A+B x^2)}{x} \, dx\)

Optimal. Leaf size=76 \[ -a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{1}{3} A \left (a+b x^2\right )^{3/2}+a A \sqrt{a+b x^2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b} \]

[Out]

a*A*Sqrt[a + b*x^2] + (A*(a + b*x^2)^(3/2))/3 + (B*(a + b*x^2)^(5/2))/(5*b) - a^(3/2)*A*ArcTanh[Sqrt[a + b*x^2
]/Sqrt[a]]

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Rubi [A]  time = 0.0533293, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {446, 80, 50, 63, 208} \[ -a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{1}{3} A \left (a+b x^2\right )^{3/2}+a A \sqrt{a+b x^2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x,x]

[Out]

a*A*Sqrt[a + b*x^2] + (A*(a + b*x^2)^(3/2))/3 + (B*(a + b*x^2)^(5/2))/(5*b) - a^(3/2)*A*ArcTanh[Sqrt[a + b*x^2
]/Sqrt[a]]

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{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2} (A+B x)}{x} \, dx,x,x^2\right )\\ &=\frac{B \left (a+b x^2\right )^{5/2}}{5 b}+\frac{1}{2} A \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{3} A \left (a+b x^2\right )^{3/2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b}+\frac{1}{2} (a A) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=a A \sqrt{a+b x^2}+\frac{1}{3} A \left (a+b x^2\right )^{3/2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b}+\frac{1}{2} \left (a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=a A \sqrt{a+b x^2}+\frac{1}{3} A \left (a+b x^2\right )^{3/2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b}+\frac{\left (a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{b}\\ &=a A \sqrt{a+b x^2}+\frac{1}{3} A \left (a+b x^2\right )^{3/2}+\frac{B \left (a+b x^2\right )^{5/2}}{5 b}-a^{3/2} A \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.064648, size = 76, normalized size = 1. \[ \frac{1}{3} A \left (a+b x^2\right )^{3/2}+a A \left (\sqrt{a+b x^2}-\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\right )+\frac{B \left (a+b x^2\right )^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x,x]

[Out]

(A*(a + b*x^2)^(3/2))/3 + (B*(a + b*x^2)^(5/2))/(5*b) + a*A*(Sqrt[a + b*x^2] - Sqrt[a]*ArcTanh[Sqrt[a + b*x^2]
/Sqrt[a]])

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Maple [A]  time = 0.007, size = 70, normalized size = 0.9 \begin{align*}{\frac{B}{5\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{A}{3} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-A{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) +aA\sqrt{b{x}^{2}+a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/x,x)

[Out]

1/5*B*(b*x^2+a)^(5/2)/b+1/3*A*(b*x^2+a)^(3/2)-A*a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+a*A*(b*x^2+a)^(1
/2)

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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)^(3/2)*(B*x^2+A)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59311, size = 408, normalized size = 5.37 \begin{align*} \left [\frac{15 \, A a^{\frac{3}{2}} b \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (3 \, B b^{2} x^{4} + 3 \, B a^{2} + 20 \, A a b +{\left (6 \, B a b + 5 \, A b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{30 \, b}, \frac{15 \, A \sqrt{-a} a b \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (3 \, B b^{2} x^{4} + 3 \, B a^{2} + 20 \, A a b +{\left (6 \, B a b + 5 \, A b^{2}\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{15 \, b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x,x, algorithm="fricas")

[Out]

[1/30*(15*A*a^(3/2)*b*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*B*b^2*x^4 + 3*B*a^2 + 20*A*a*
b + (6*B*a*b + 5*A*b^2)*x^2)*sqrt(b*x^2 + a))/b, 1/15*(15*A*sqrt(-a)*a*b*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (3
*B*b^2*x^4 + 3*B*a^2 + 20*A*a*b + (6*B*a*b + 5*A*b^2)*x^2)*sqrt(b*x^2 + a))/b]

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Sympy [A]  time = 35.7311, size = 71, normalized size = 0.93 \begin{align*} \frac{A a^{2} \operatorname{atan}{\left (\frac{\sqrt{a + b x^{2}}}{\sqrt{- a}} \right )}}{\sqrt{- a}} + A a \sqrt{a + b x^{2}} + \frac{A \left (a + b x^{2}\right )^{\frac{3}{2}}}{3} + \frac{B \left (a + b x^{2}\right )^{\frac{5}{2}}}{5 b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x,x)

[Out]

A*a**2*atan(sqrt(a + b*x**2)/sqrt(-a))/sqrt(-a) + A*a*sqrt(a + b*x**2) + A*(a + b*x**2)**(3/2)/3 + B*(a + b*x*
*2)**(5/2)/(5*b)

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Giac [A]  time = 1.34425, size = 107, normalized size = 1.41 \begin{align*} \frac{A a^{2} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} + \frac{3 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B b^{4} + 5 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A b^{5} + 15 \, \sqrt{b x^{2} + a} A a b^{5}}{15 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x,x, algorithm="giac")

[Out]

A*a^2*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/15*(3*(b*x^2 + a)^(5/2)*B*b^4 + 5*(b*x^2 + a)^(3/2)*A*b^5
+ 15*sqrt(b*x^2 + a)*A*a*b^5)/b^5