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

Optimal. Leaf size=110 \[ \frac{\left (a+b x^2\right )^{3/2} (2 a B+3 A b)}{6 a}+\frac{1}{2} \sqrt{a+b x^2} (2 a B+3 A b)-\frac{1}{2} \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{A \left (a+b x^2\right )^{5/2}}{2 a x^2} \]

[Out]

((3*A*b + 2*a*B)*Sqrt[a + b*x^2])/2 + ((3*A*b + 2*a*B)*(a + b*x^2)^(3/2))/(6*a) - (A*(a + b*x^2)^(5/2))/(2*a*x
^2) - (Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2

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Rubi [A]  time = 0.0801891, antiderivative size = 110, 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, 78, 50, 63, 208} \[ \frac{\left (a+b x^2\right )^{3/2} (2 a B+3 A b)}{6 a}+\frac{1}{2} \sqrt{a+b x^2} (2 a B+3 A b)-\frac{1}{2} \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )-\frac{A \left (a+b x^2\right )^{5/2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A*b + 2*a*B)*Sqrt[a + b*x^2])/2 + ((3*A*b + 2*a*B)*(a + b*x^2)^(3/2))/(6*a) - (A*(a + b*x^2)^(5/2))/(2*a*x
^2) - (Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2

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{\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2} (A+B x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac{A \left (a+b x^2\right )^{5/2}}{2 a x^2}+\frac{\left (\frac{3 A b}{2}+a B\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )}{2 a}\\ &=\frac{(3 A b+2 a B) \left (a+b x^2\right )^{3/2}}{6 a}-\frac{A \left (a+b x^2\right )^{5/2}}{2 a x^2}+\frac{1}{4} (3 A b+2 a B) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{2} (3 A b+2 a B) \sqrt{a+b x^2}+\frac{(3 A b+2 a B) \left (a+b x^2\right )^{3/2}}{6 a}-\frac{A \left (a+b x^2\right )^{5/2}}{2 a x^2}+\frac{1}{4} (a (3 A b+2 a B)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{1}{2} (3 A b+2 a B) \sqrt{a+b x^2}+\frac{(3 A b+2 a B) \left (a+b x^2\right )^{3/2}}{6 a}-\frac{A \left (a+b x^2\right )^{5/2}}{2 a x^2}+\frac{(a (3 A b+2 a B)) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{2 b}\\ &=\frac{1}{2} (3 A b+2 a B) \sqrt{a+b x^2}+\frac{(3 A b+2 a B) \left (a+b x^2\right )^{3/2}}{6 a}-\frac{A \left (a+b x^2\right )^{5/2}}{2 a x^2}-\frac{1}{2} \sqrt{a} (3 A b+2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0497146, size = 80, normalized size = 0.73 \[ \frac{1}{6} \left (\frac{\sqrt{a+b x^2} \left (-3 a A+8 a B x^2+6 A b x^2+2 b B x^4\right )}{x^2}-3 \sqrt{a} (2 a B+3 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a + b*x^2]*(-3*a*A + 6*A*b*x^2 + 8*a*B*x^2 + 2*b*B*x^4))/x^2 - 3*Sqrt[a]*(3*A*b + 2*a*B)*ArcTanh[Sqrt[a
+ b*x^2]/Sqrt[a]])/6

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Maple [A]  time = 0.008, size = 132, normalized size = 1.2 \begin{align*}{\frac{B}{3} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-B{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) +B\sqrt{b{x}^{2}+a}a-{\frac{A}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Ab}{2\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{3\,Ab}{2}\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{3\,Ab}{2}\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^3,x)

[Out]

1/3*B*(b*x^2+a)^(3/2)-B*a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+B*(b*x^2+a)^(1/2)*a-1/2*A*(b*x^2+a)^(5/2
)/a/x^2+1/2*A*b/a*(b*x^2+a)^(3/2)-3/2*A*b*a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+3/2*A*b*(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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.68712, size = 404, normalized size = 3.67 \begin{align*} \left [\frac{3 \,{\left (2 \, B a + 3 \, A b\right )} \sqrt{a} x^{2} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (2 \, B b x^{4} + 2 \,{\left (4 \, B a + 3 \, A b\right )} x^{2} - 3 \, A a\right )} \sqrt{b x^{2} + a}}{12 \, x^{2}}, \frac{3 \,{\left (2 \, B a + 3 \, A b\right )} \sqrt{-a} x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (2 \, B b x^{4} + 2 \,{\left (4 \, B a + 3 \, A b\right )} x^{2} - 3 \, A a\right )} \sqrt{b x^{2} + a}}{6 \, x^{2}}\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^3,x, algorithm="fricas")

[Out]

[1/12*(3*(2*B*a + 3*A*b)*sqrt(a)*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(2*B*b*x^4 + 2*(4
*B*a + 3*A*b)*x^2 - 3*A*a)*sqrt(b*x^2 + a))/x^2, 1/6*(3*(2*B*a + 3*A*b)*sqrt(-a)*x^2*arctan(sqrt(-a)/sqrt(b*x^
2 + a)) + (2*B*b*x^4 + 2*(4*B*a + 3*A*b)*x^2 - 3*A*a)*sqrt(b*x^2 + a))/x^2]

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Sympy [A]  time = 25.6166, size = 184, normalized size = 1.67 \begin{align*} - \frac{3 A \sqrt{a} b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2} - \frac{A a \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{2 x} + \frac{A a \sqrt{b}}{x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{A b^{\frac{3}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} - B a^{\frac{3}{2}} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )} + \frac{B a^{2}}{\sqrt{b} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{B a \sqrt{b} x}{\sqrt{\frac{a}{b x^{2}} + 1}} + B b \left (\begin{cases} \frac{\sqrt{a} x^{2}}{2} & \text{for}\: b = 0 \\\frac{\left (a + b x^{2}\right )^{\frac{3}{2}}}{3 b} & \text{otherwise} \end{cases}\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**3,x)

[Out]

-3*A*sqrt(a)*b*asinh(sqrt(a)/(sqrt(b)*x))/2 - A*a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(2*x) + A*a*sqrt(b)/(x*sqrt(a/(
b*x**2) + 1)) + A*b**(3/2)*x/sqrt(a/(b*x**2) + 1) - B*a**(3/2)*asinh(sqrt(a)/(sqrt(b)*x)) + B*a**2/(sqrt(b)*x*
sqrt(a/(b*x**2) + 1)) + B*a*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + B*b*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b
*x**2)**(3/2)/(3*b), True))

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Giac [A]  time = 1.12249, size = 139, normalized size = 1.26 \begin{align*} \frac{2 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B b + 6 \, \sqrt{b x^{2} + a} B a b + 6 \, \sqrt{b x^{2} + a} A b^{2} - \frac{3 \, \sqrt{b x^{2} + a} A a b}{x^{2}} + \frac{3 \,{\left (2 \, B a^{2} b + 3 \, A a b^{2}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}}}{6 \, 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^3,x, algorithm="giac")

[Out]

1/6*(2*(b*x^2 + a)^(3/2)*B*b + 6*sqrt(b*x^2 + a)*B*a*b + 6*sqrt(b*x^2 + a)*A*b^2 - 3*sqrt(b*x^2 + a)*A*a*b/x^2
+ 3*(2*B*a^2*b + 3*A*a*b^2)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a))/b