3.1907 \(\int (a+\frac{b}{x^2})^{5/2} x^3 \, dx\)

Optimal. Leaf size=86 \[ -\frac{15}{8} b^2 \sqrt{a+\frac{b}{x^2}}+\frac{15}{8} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )+\frac{1}{4} x^4 \left (a+\frac{b}{x^2}\right )^{5/2}+\frac{5}{8} b x^2 \left (a+\frac{b}{x^2}\right )^{3/2} \]

[Out]

(-15*b^2*Sqrt[a + b/x^2])/8 + (5*b*(a + b/x^2)^(3/2)*x^2)/8 + ((a + b/x^2)^(5/2)*x^4)/4 + (15*Sqrt[a]*b^2*ArcT
anh[Sqrt[a + b/x^2]/Sqrt[a]])/8

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Rubi [A]  time = 0.0421809, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 50, 63, 208} \[ -\frac{15}{8} b^2 \sqrt{a+\frac{b}{x^2}}+\frac{15}{8} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )+\frac{1}{4} x^4 \left (a+\frac{b}{x^2}\right )^{5/2}+\frac{5}{8} b x^2 \left (a+\frac{b}{x^2}\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(5/2)*x^3,x]

[Out]

(-15*b^2*Sqrt[a + b/x^2])/8 + (5*b*(a + b/x^2)^(3/2)*x^2)/8 + ((a + b/x^2)^(5/2)*x^4)/4 + (15*Sqrt[a]*b^2*ArcT
anh[Sqrt[a + b/x^2]/Sqrt[a]])/8

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 \left (a+\frac{b}{x^2}\right )^{5/2} x^3 \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^3} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{1}{4} \left (a+\frac{b}{x^2}\right )^{5/2} x^4-\frac{1}{8} (5 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^2} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{5}{8} b \left (a+\frac{b}{x^2}\right )^{3/2} x^2+\frac{1}{4} \left (a+\frac{b}{x^2}\right )^{5/2} x^4-\frac{1}{16} \left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{15}{8} b^2 \sqrt{a+\frac{b}{x^2}}+\frac{5}{8} b \left (a+\frac{b}{x^2}\right )^{3/2} x^2+\frac{1}{4} \left (a+\frac{b}{x^2}\right )^{5/2} x^4-\frac{1}{16} \left (15 a b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{15}{8} b^2 \sqrt{a+\frac{b}{x^2}}+\frac{5}{8} b \left (a+\frac{b}{x^2}\right )^{3/2} x^2+\frac{1}{4} \left (a+\frac{b}{x^2}\right )^{5/2} x^4-\frac{1}{8} (15 a b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x^2}}\right )\\ &=-\frac{15}{8} b^2 \sqrt{a+\frac{b}{x^2}}+\frac{5}{8} b \left (a+\frac{b}{x^2}\right )^{3/2} x^2+\frac{1}{4} \left (a+\frac{b}{x^2}\right )^{5/2} x^4+\frac{15}{8} \sqrt{a} b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0140904, size = 49, normalized size = 0.57 \[ -\frac{b^2 \sqrt{a+\frac{b}{x^2}} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};-\frac{a x^2}{b}\right )}{\sqrt{\frac{a x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(5/2)*x^3,x]

[Out]

-((b^2*Sqrt[a + b/x^2]*Hypergeometric2F1[-5/2, -1/2, 1/2, -((a*x^2)/b)])/Sqrt[1 + (a*x^2)/b])

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Maple [A]  time = 0.007, size = 127, normalized size = 1.5 \begin{align*}{\frac{{x}^{4}}{8\,b} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{{\frac{5}{2}}} \left ( 8\, \left ( a{x}^{2}+b \right ) ^{5/2}{a}^{3/2}{x}^{2}+10\,{a}^{3/2} \left ( a{x}^{2}+b \right ) ^{3/2}{x}^{2}b+15\,{a}^{3/2}\sqrt{a{x}^{2}+b}{x}^{2}{b}^{2}-8\, \left ( a{x}^{2}+b \right ) ^{7/2}\sqrt{a}+15\,\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ) xa{b}^{3} \right ) \left ( a{x}^{2}+b \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+1/x^2*b)^(5/2)*x^3,x)

[Out]

1/8*((a*x^2+b)/x^2)^(5/2)*x^4*(8*(a*x^2+b)^(5/2)*a^(3/2)*x^2+10*a^(3/2)*(a*x^2+b)^(3/2)*x^2*b+15*a^(3/2)*(a*x^
2+b)^(1/2)*x^2*b^2-8*(a*x^2+b)^(7/2)*a^(1/2)+15*ln(x*a^(1/2)+(a*x^2+b)^(1/2))*x*a*b^3)/(a*x^2+b)^(5/2)/b/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((a+b/x^2)^(5/2)*x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59693, size = 370, normalized size = 4.3 \begin{align*} \left [\frac{15}{16} \, \sqrt{a} b^{2} \log \left (-2 \, a x^{2} - 2 \, \sqrt{a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}} - b\right ) + \frac{1}{8} \,{\left (2 \, a^{2} x^{4} + 9 \, a b x^{2} - 8 \, b^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}, -\frac{15}{8} \, \sqrt{-a} b^{2} \arctan \left (\frac{\sqrt{-a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + \frac{1}{8} \,{\left (2 \, a^{2} x^{4} + 9 \, a b x^{2} - 8 \, b^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[15/16*sqrt(a)*b^2*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) + 1/8*(2*a^2*x^4 + 9*a*b*x^2 - 8*b^
2)*sqrt((a*x^2 + b)/x^2), -15/8*sqrt(-a)*b^2*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + 1/8*(2*a
^2*x^4 + 9*a*b*x^2 - 8*b^2)*sqrt((a*x^2 + b)/x^2)]

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Sympy [A]  time = 4.98454, size = 117, normalized size = 1.36 \begin{align*} \frac{15 \sqrt{a} b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} x}{\sqrt{b}} \right )}}{8} + \frac{a^{3} x^{5}}{4 \sqrt{b} \sqrt{\frac{a x^{2}}{b} + 1}} + \frac{11 a^{2} \sqrt{b} x^{3}}{8 \sqrt{\frac{a x^{2}}{b} + 1}} + \frac{a b^{\frac{3}{2}} x}{8 \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{b^{\frac{5}{2}}}{x \sqrt{\frac{a x^{2}}{b} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*sqrt(a)*b**2*asinh(sqrt(a)*x/sqrt(b))/8 + a**3*x**5/(4*sqrt(b)*sqrt(a*x**2/b + 1)) + 11*a**2*sqrt(b)*x**3/(
8*sqrt(a*x**2/b + 1)) + a*b**(3/2)*x/(8*sqrt(a*x**2/b + 1)) - b**(5/2)/(x*sqrt(a*x**2/b + 1))

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Giac [A]  time = 1.27763, size = 128, normalized size = 1.49 \begin{align*} -\frac{15}{16} \, \sqrt{a} b^{2} \log \left ({\left (\sqrt{a} x - \sqrt{a x^{2} + b}\right )}^{2}\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \, \sqrt{a} b^{3} \mathrm{sgn}\left (x\right )}{{\left (\sqrt{a} x - \sqrt{a x^{2} + b}\right )}^{2} - b} + \frac{1}{8} \,{\left (2 \, a^{2} x^{2} \mathrm{sgn}\left (x\right ) + 9 \, a b \mathrm{sgn}\left (x\right )\right )} \sqrt{a x^{2} + b} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/16*sqrt(a)*b^2*log((sqrt(a)*x - sqrt(a*x^2 + b))^2)*sgn(x) + 2*sqrt(a)*b^3*sgn(x)/((sqrt(a)*x - sqrt(a*x^2
+ b))^2 - b) + 1/8*(2*a^2*x^2*sgn(x) + 9*a*b*sgn(x))*sqrt(a*x^2 + b)*x