3.2074 \(\int (a+\frac{b}{x^4})^{5/2} x^3 \, dx\)

Optimal. Leaf size=80 \[ \frac{5}{4} a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )+\frac{1}{4} x^4 \left (a+\frac{b}{x^4}\right )^{5/2}-\frac{5}{12} b \left (a+\frac{b}{x^4}\right )^{3/2}-\frac{5}{4} a b \sqrt{a+\frac{b}{x^4}} \]

[Out]

(-5*a*b*Sqrt[a + b/x^4])/4 - (5*b*(a + b/x^4)^(3/2))/12 + ((a + b/x^4)^(5/2)*x^4)/4 + (5*a^(3/2)*b*ArcTanh[Sqr
t[a + b/x^4]/Sqrt[a]])/4

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Rubi [A]  time = 0.0437962, antiderivative size = 80, 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} \[ \frac{5}{4} a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )+\frac{1}{4} x^4 \left (a+\frac{b}{x^4}\right )^{5/2}-\frac{5}{12} b \left (a+\frac{b}{x^4}\right )^{3/2}-\frac{5}{4} a b \sqrt{a+\frac{b}{x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(5/2)*x^3,x]

[Out]

(-5*a*b*Sqrt[a + b/x^4])/4 - (5*b*(a + b/x^4)^(3/2))/12 + ((a + b/x^4)^(5/2)*x^4)/4 + (5*a^(3/2)*b*ArcTanh[Sqr
t[a + b/x^4]/Sqrt[a]])/4

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^4}\right )^{5/2} x^3 \, dx &=-\left (\frac{1}{4} \operatorname{Subst}\left (\int \frac{(a+b x)^{5/2}}{x^2} \, dx,x,\frac{1}{x^4}\right )\right )\\ &=\frac{1}{4} \left (a+\frac{b}{x^4}\right )^{5/2} x^4-\frac{1}{8} (5 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x} \, dx,x,\frac{1}{x^4}\right )\\ &=-\frac{5}{12} b \left (a+\frac{b}{x^4}\right )^{3/2}+\frac{1}{4} \left (a+\frac{b}{x^4}\right )^{5/2} x^4-\frac{1}{8} (5 a b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\frac{1}{x^4}\right )\\ &=-\frac{5}{4} a b \sqrt{a+\frac{b}{x^4}}-\frac{5}{12} b \left (a+\frac{b}{x^4}\right )^{3/2}+\frac{1}{4} \left (a+\frac{b}{x^4}\right )^{5/2} x^4-\frac{1}{8} \left (5 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^4}\right )\\ &=-\frac{5}{4} a b \sqrt{a+\frac{b}{x^4}}-\frac{5}{12} b \left (a+\frac{b}{x^4}\right )^{3/2}+\frac{1}{4} \left (a+\frac{b}{x^4}\right )^{5/2} x^4-\frac{1}{4} \left (5 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x^4}}\right )\\ &=-\frac{5}{4} a b \sqrt{a+\frac{b}{x^4}}-\frac{5}{12} b \left (a+\frac{b}{x^4}\right )^{3/2}+\frac{1}{4} \left (a+\frac{b}{x^4}\right )^{5/2} x^4+\frac{5}{4} a^{3/2} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0153393, size = 54, normalized size = 0.68 \[ -\frac{b^2 \sqrt{a+\frac{b}{x^4}} \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};-\frac{a x^4}{b}\right )}{6 x^4 \sqrt{\frac{a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(5/2)*x^3,x]

[Out]

-(b^2*Sqrt[a + b/x^4]*Hypergeometric2F1[-5/2, -3/2, -1/2, -((a*x^4)/b)])/(6*x^4*Sqrt[1 + (a*x^4)/b])

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Maple [A]  time = 0.016, size = 103, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}}{12} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{{\frac{5}{2}}} \left ( 15\,b{a}^{3/2}\ln \left ({x}^{2}\sqrt{a}+\sqrt{a{x}^{4}+b} \right ){x}^{6}+3\,{a}^{2}{x}^{8}\sqrt{a{x}^{4}+b}-14\,ba\sqrt{a{x}^{4}+b}{x}^{4}-2\,{b}^{2}\sqrt{a{x}^{4}+b} \right ) \left ( a{x}^{4}+b \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^4)^(5/2)*x^3,x)

[Out]

1/12*((a*x^4+b)/x^4)^(5/2)*x^4*(15*b*a^(3/2)*ln(x^2*a^(1/2)+(a*x^4+b)^(1/2))*x^6+3*a^2*x^8*(a*x^4+b)^(1/2)-14*
b*a*(a*x^4+b)^(1/2)*x^4-2*b^2*(a*x^4+b)^(1/2))/(a*x^4+b)^(5/2)

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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^4)^(5/2)*x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60741, size = 396, normalized size = 4.95 \begin{align*} \left [\frac{15 \, a^{\frac{3}{2}} b x^{4} \log \left (-2 \, a x^{4} - 2 \, \sqrt{a} x^{4} \sqrt{\frac{a x^{4} + b}{x^{4}}} - b\right ) + 2 \,{\left (3 \, a^{2} x^{8} - 14 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{24 \, x^{4}}, -\frac{15 \, \sqrt{-a} a b x^{4} \arctan \left (\frac{\sqrt{-a} x^{4} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) -{\left (3 \, a^{2} x^{8} - 14 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{12 \, x^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^4)^(5/2)*x^3,x, algorithm="fricas")

[Out]

[1/24*(15*a^(3/2)*b*x^4*log(-2*a*x^4 - 2*sqrt(a)*x^4*sqrt((a*x^4 + b)/x^4) - b) + 2*(3*a^2*x^8 - 14*a*b*x^4 -
2*b^2)*sqrt((a*x^4 + b)/x^4))/x^4, -1/12*(15*sqrt(-a)*a*b*x^4*arctan(sqrt(-a)*x^4*sqrt((a*x^4 + b)/x^4)/(a*x^4
+ b)) - (3*a^2*x^8 - 14*a*b*x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^4]

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Sympy [A]  time = 5.38471, size = 112, normalized size = 1.4 \begin{align*} \frac{a^{\frac{5}{2}} x^{4} \sqrt{1 + \frac{b}{a x^{4}}}}{4} - \frac{7 a^{\frac{3}{2}} b \sqrt{1 + \frac{b}{a x^{4}}}}{6} - \frac{5 a^{\frac{3}{2}} b \log{\left (\frac{b}{a x^{4}} \right )}}{8} + \frac{5 a^{\frac{3}{2}} b \log{\left (\sqrt{1 + \frac{b}{a x^{4}}} + 1 \right )}}{4} - \frac{\sqrt{a} b^{2} \sqrt{1 + \frac{b}{a x^{4}}}}{6 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**4)**(5/2)*x**3,x)

[Out]

a**(5/2)*x**4*sqrt(1 + b/(a*x**4))/4 - 7*a**(3/2)*b*sqrt(1 + b/(a*x**4))/6 - 5*a**(3/2)*b*log(b/(a*x**4))/8 +
5*a**(3/2)*b*log(sqrt(1 + b/(a*x**4)) + 1)/4 - sqrt(a)*b**2*sqrt(1 + b/(a*x**4))/(6*x**4)

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Giac [B]  time = 1.14153, size = 192, normalized size = 2.4 \begin{align*} \frac{1}{4} \, \sqrt{a x^{4} + b} a^{2} x^{2} - \frac{5}{8} \, a^{\frac{3}{2}} b \log \left ({\left (\sqrt{a} x^{2} - \sqrt{a x^{4} + b}\right )}^{2}\right ) + \frac{9 \,{\left (\sqrt{a} x^{2} - \sqrt{a x^{4} + b}\right )}^{4} a^{\frac{3}{2}} b^{2} - 12 \,{\left (\sqrt{a} x^{2} - \sqrt{a x^{4} + b}\right )}^{2} a^{\frac{3}{2}} b^{3} + 7 \, a^{\frac{3}{2}} b^{4}}{3 \,{\left ({\left (\sqrt{a} x^{2} - \sqrt{a x^{4} + b}\right )}^{2} - b\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^4)^(5/2)*x^3,x, algorithm="giac")

[Out]

1/4*sqrt(a*x^4 + b)*a^2*x^2 - 5/8*a^(3/2)*b*log((sqrt(a)*x^2 - sqrt(a*x^4 + b))^2) + 1/3*(9*(sqrt(a)*x^2 - sqr
t(a*x^4 + b))^4*a^(3/2)*b^2 - 12*(sqrt(a)*x^2 - sqrt(a*x^4 + b))^2*a^(3/2)*b^3 + 7*a^(3/2)*b^4)/((sqrt(a)*x^2
- sqrt(a*x^4 + b))^2 - b)^3