3.546 \(\int \frac{(a+b x^2)^{5/2} (A+B x^2)}{x^3} \, dx\)

Optimal. Leaf size=135 \[ -\frac{1}{2} a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{\left (a+b x^2\right )^{5/2} (2 a B+5 A b)}{10 a}+\frac{1}{6} \left (a+b x^2\right )^{3/2} (2 a B+5 A b)+\frac{1}{2} a \sqrt{a+b x^2} (2 a B+5 A b)-\frac{A \left (a+b x^2\right )^{7/2}}{2 a x^2} \]

[Out]

(a*(5*A*b + 2*a*B)*Sqrt[a + b*x^2])/2 + ((5*A*b + 2*a*B)*(a + b*x^2)^(3/2))/6 + ((5*A*b + 2*a*B)*(a + b*x^2)^(
5/2))/(10*a) - (A*(a + b*x^2)^(7/2))/(2*a*x^2) - (a^(3/2)*(5*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2

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Rubi [A]  time = 0.100338, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{1}{2} a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )+\frac{\left (a+b x^2\right )^{5/2} (2 a B+5 A b)}{10 a}+\frac{1}{6} \left (a+b x^2\right )^{3/2} (2 a B+5 A b)+\frac{1}{2} a \sqrt{a+b x^2} (2 a B+5 A b)-\frac{A \left (a+b x^2\right )^{7/2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^3,x]

[Out]

(a*(5*A*b + 2*a*B)*Sqrt[a + b*x^2])/2 + ((5*A*b + 2*a*B)*(a + b*x^2)^(3/2))/6 + ((5*A*b + 2*a*B)*(a + b*x^2)^(
5/2))/(10*a) - (A*(a + b*x^2)^(7/2))/(2*a*x^2) - (a^(3/2)*(5*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 )^{5/2} \left (A+B x^2\right )}{x^3} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2} (A+B x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac{A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac{\left (\frac{5 A b}{2}+a B\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x} \, dx,x,x^2\right )}{2 a}\\ &=\frac{(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac{A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac{1}{4} (5 A b+2 a B) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac{(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac{A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac{1}{4} (a (5 A b+2 a B)) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{2} a (5 A b+2 a B) \sqrt{a+b x^2}+\frac{1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac{(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac{A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac{1}{4} \left (a^2 (5 A b+2 a B)\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=\frac{1}{2} a (5 A b+2 a B) \sqrt{a+b x^2}+\frac{1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac{(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac{A \left (a+b x^2\right )^{7/2}}{2 a x^2}+\frac{\left (a^2 (5 A b+2 a B)\right ) \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} a (5 A b+2 a B) \sqrt{a+b x^2}+\frac{1}{6} (5 A b+2 a B) \left (a+b x^2\right )^{3/2}+\frac{(5 A b+2 a B) \left (a+b x^2\right )^{5/2}}{10 a}-\frac{A \left (a+b x^2\right )^{7/2}}{2 a x^2}-\frac{1}{2} a^{3/2} (5 A b+2 a B) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0704792, size = 105, normalized size = 0.78 \[ \frac{\sqrt{a+b x^2} \left (a^2 \left (46 B x^2-15 A\right )+a \left (70 A b x^2+22 b B x^4\right )+2 b^2 x^4 \left (5 A+3 B x^2\right )\right )}{30 x^2}-\frac{1}{2} a^{3/2} (2 a B+5 A b) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^3,x]

[Out]

(Sqrt[a + b*x^2]*(2*b^2*x^4*(5*A + 3*B*x^2) + a^2*(-15*A + 46*B*x^2) + a*(70*A*b*x^2 + 22*b*B*x^4)))/(30*x^2)
- (a^(3/2)*(5*A*b + 2*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/2

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Maple [A]  time = 0.008, size = 161, normalized size = 1.2 \begin{align*}{\frac{B}{5} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{Ba}{3} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-B{a}^{{\frac{5}{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}^{2}-{\frac{A}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{7}{2}}}}+{\frac{Ab}{2\,a} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}+{\frac{5\,Ab}{6} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,Ab}{2}{a}^{{\frac{3}{2}}}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ) }+{\frac{5\,abA}{2}\sqrt{b{x}^{2}+a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^3,x)

[Out]

1/5*B*(b*x^2+a)^(5/2)+1/3*B*a*(b*x^2+a)^(3/2)-B*a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+B*(b*x^2+a)^(1/2
)*a^2-1/2*A*(b*x^2+a)^(7/2)/a/x^2+1/2*A*b/a*(b*x^2+a)^(5/2)+5/6*A*b*(b*x^2+a)^(3/2)-5/2*A*b*a^(3/2)*ln((2*a+2*
a^(1/2)*(b*x^2+a)^(1/2))/x)+5/2*A*b*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)^(5/2)*(B*x^2+A)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65179, size = 527, normalized size = 3.9 \begin{align*} \left [\frac{15 \,{\left (2 \, B a^{2} + 5 \, A 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 (6 \, B b^{2} x^{6} + 2 \,{\left (11 \, B a b + 5 \, A b^{2}\right )} x^{4} - 15 \, A a^{2} + 2 \,{\left (23 \, B a^{2} + 35 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{60 \, x^{2}}, \frac{15 \,{\left (2 \, B a^{2} + 5 \, A a b\right )} \sqrt{-a} x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (6 \, B b^{2} x^{6} + 2 \,{\left (11 \, B a b + 5 \, A b^{2}\right )} x^{4} - 15 \, A a^{2} + 2 \,{\left (23 \, B a^{2} + 35 \, A a b\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{30 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^3,x, algorithm="fricas")

[Out]

[1/60*(15*(2*B*a^2 + 5*A*a*b)*sqrt(a)*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(6*B*b^2*x^6
+ 2*(11*B*a*b + 5*A*b^2)*x^4 - 15*A*a^2 + 2*(23*B*a^2 + 35*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^2, 1/30*(15*(2*B*a^
2 + 5*A*a*b)*sqrt(-a)*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (6*B*b^2*x^6 + 2*(11*B*a*b + 5*A*b^2)*x^4 - 15*A*
a^2 + 2*(23*B*a^2 + 35*A*a*b)*x^2)*sqrt(b*x^2 + a))/x^2]

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Sympy [A]  time = 30.7282, size = 296, normalized size = 2.19 \begin{align*} - \frac{5 A a^{\frac{3}{2}} b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2} - \frac{A a^{2} \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{2 x} + \frac{2 A a^{2} \sqrt{b}}{x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{2 A a b^{\frac{3}{2}} x}{\sqrt{\frac{a}{b x^{2}} + 1}} + A b^{2} \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 ) - B a^{\frac{5}{2}} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )} + \frac{B a^{3}}{\sqrt{b} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{B a^{2} \sqrt{b} x}{\sqrt{\frac{a}{b x^{2}} + 1}} + 2 B a 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 ) + B b^{2} \left (\begin{cases} - \frac{2 a^{2} \sqrt{a + b x^{2}}}{15 b^{2}} + \frac{a x^{2} \sqrt{a + b x^{2}}}{15 b} + \frac{x^{4} \sqrt{a + b x^{2}}}{5} & \text{for}\: b \neq 0 \\\frac{\sqrt{a} x^{4}}{4} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**3,x)

[Out]

-5*A*a**(3/2)*b*asinh(sqrt(a)/(sqrt(b)*x))/2 - A*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(2*x) + 2*A*a**2*sqrt(b)/(x
*sqrt(a/(b*x**2) + 1)) + 2*A*a*b**(3/2)*x/sqrt(a/(b*x**2) + 1) + A*b**2*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)),
((a + b*x**2)**(3/2)/(3*b), True)) - B*a**(5/2)*asinh(sqrt(a)/(sqrt(b)*x)) + B*a**3/(sqrt(b)*x*sqrt(a/(b*x**2)
+ 1)) + B*a**2*sqrt(b)*x/sqrt(a/(b*x**2) + 1) + 2*B*a*b*Piecewise((sqrt(a)*x**2/2, Eq(b, 0)), ((a + b*x**2)**
(3/2)/(3*b), True)) + B*b**2*Piecewise((-2*a**2*sqrt(a + b*x**2)/(15*b**2) + a*x**2*sqrt(a + b*x**2)/(15*b) +
x**4*sqrt(a + b*x**2)/5, Ne(b, 0)), (sqrt(a)*x**4/4, True))

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Giac [A]  time = 1.14378, size = 188, normalized size = 1.39 \begin{align*} \frac{6 \,{\left (b x^{2} + a\right )}^{\frac{5}{2}} B b + 10 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} B a b + 30 \, \sqrt{b x^{2} + a} B a^{2} b + 10 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} A b^{2} + 60 \, \sqrt{b x^{2} + a} A a b^{2} - \frac{15 \, \sqrt{b x^{2} + a} A a^{2} b}{x^{2}} + \frac{15 \,{\left (2 \, B a^{3} b + 5 \, A a^{2} b^{2}\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}}}{30 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^3,x, algorithm="giac")

[Out]

1/30*(6*(b*x^2 + a)^(5/2)*B*b + 10*(b*x^2 + a)^(3/2)*B*a*b + 30*sqrt(b*x^2 + a)*B*a^2*b + 10*(b*x^2 + a)^(3/2)
*A*b^2 + 60*sqrt(b*x^2 + a)*A*a*b^2 - 15*sqrt(b*x^2 + a)*A*a^2*b/x^2 + 15*(2*B*a^3*b + 5*A*a^2*b^2)*arctan(sqr
t(b*x^2 + a)/sqrt(-a))/sqrt(-a))/b