3.2098 \(\int \frac{x^3}{(a+\frac{b}{x^4})^{5/2}} \, dx\)

Optimal. Leaf size=88 \[ \frac{5 b}{4 a^3 \sqrt{a+\frac{b}{x^4}}}+\frac{5 b}{12 a^2 \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )}{4 a^{7/2}}+\frac{x^4}{4 a \left (a+\frac{b}{x^4}\right )^{3/2}} \]

[Out]

(5*b)/(12*a^2*(a + b/x^4)^(3/2)) + (5*b)/(4*a^3*Sqrt[a + b/x^4]) + x^4/(4*a*(a + b/x^4)^(3/2)) - (5*b*ArcTanh[
Sqrt[a + b/x^4]/Sqrt[a]])/(4*a^(7/2))

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Rubi [A]  time = 0.0465678, antiderivative size = 92, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ \frac{5 x^4 \sqrt{a+\frac{b}{x^4}}}{4 a^3}-\frac{5 x^4}{6 a^2 \sqrt{a+\frac{b}{x^4}}}-\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{x^4}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b/x^4)^(5/2),x]

[Out]

-x^4/(6*a*(a + b/x^4)^(3/2)) - (5*x^4)/(6*a^2*Sqrt[a + b/x^4]) + (5*Sqrt[a + b/x^4]*x^4)/(4*a^3) - (5*b*ArcTan
h[Sqrt[a + b/x^4]/Sqrt[a]])/(4*a^(7/2))

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{x^3}{\left (a+\frac{b}{x^4}\right )^{5/2}} \, dx &=-\left (\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{5/2}} \, dx,x,\frac{1}{x^4}\right )\right )\\ &=-\frac{x^4}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{3/2}} \, dx,x,\frac{1}{x^4}\right )}{12 a}\\ &=-\frac{x^4}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{5 x^4}{6 a^2 \sqrt{a+\frac{b}{x^4}}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x^4}\right )}{4 a^2}\\ &=-\frac{x^4}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{5 x^4}{6 a^2 \sqrt{a+\frac{b}{x^4}}}+\frac{5 \sqrt{a+\frac{b}{x^4}} x^4}{4 a^3}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^4}\right )}{8 a^3}\\ &=-\frac{x^4}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{5 x^4}{6 a^2 \sqrt{a+\frac{b}{x^4}}}+\frac{5 \sqrt{a+\frac{b}{x^4}} x^4}{4 a^3}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x^4}}\right )}{4 a^3}\\ &=-\frac{x^4}{6 a \left (a+\frac{b}{x^4}\right )^{3/2}}-\frac{5 x^4}{6 a^2 \sqrt{a+\frac{b}{x^4}}}+\frac{5 \sqrt{a+\frac{b}{x^4}} x^4}{4 a^3}-\frac{5 b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^4}}}{\sqrt{a}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.173985, size = 103, normalized size = 1.17 \[ \frac{\sqrt{a} \left (3 a^2 x^8+20 a b x^4+15 b^2\right )-\frac{15 b^{3/2} \left (a x^4+b\right ) \sqrt{\frac{a x^4}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{a} x^2}{\sqrt{b}}\right )}{x^2}}{12 a^{7/2} \sqrt{a+\frac{b}{x^4}} \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b/x^4)^(5/2),x]

[Out]

(Sqrt[a]*(15*b^2 + 20*a*b*x^4 + 3*a^2*x^8) - (15*b^(3/2)*(b + a*x^4)*Sqrt[1 + (a*x^4)/b]*ArcSinh[(Sqrt[a]*x^2)
/Sqrt[b]])/x^2)/(12*a^(7/2)*Sqrt[a + b/x^4]*(b + a*x^4))

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Maple [B]  time = 0.053, size = 282, normalized size = 3.2 \begin{align*}{\frac{1}{12\,{x}^{10}} \left ( a{x}^{4}+b \right ) ^{{\frac{5}{2}}} \left ( 3\,{a}^{15/2}\sqrt{a{x}^{4}+b}{x}^{10}+14\,{a}^{13/2}\sqrt{-{\frac{ \left ( -a{x}^{2}+\sqrt{-ab} \right ) \left ( a{x}^{2}+\sqrt{-ab} \right ) }{a}}}{x}^{6}b+6\,{a}^{13/2}b\sqrt{a{x}^{4}+b}{x}^{6}-15\,\ln \left ({x}^{2}\sqrt{a}+\sqrt{a{x}^{4}+b} \right ){x}^{8}{a}^{7}b+12\,{a}^{11/2}\sqrt{-{\frac{ \left ( -a{x}^{2}+\sqrt{-ab} \right ) \left ( a{x}^{2}+\sqrt{-ab} \right ) }{a}}}{b}^{2}{x}^{2}+3\,{a}^{11/2}{b}^{2}\sqrt{a{x}^{4}+b}{x}^{2}-30\,\ln \left ({x}^{2}\sqrt{a}+\sqrt{a{x}^{4}+b} \right ){a}^{6}{b}^{2}{x}^{4}-15\,\ln \left ({x}^{2}\sqrt{a}+\sqrt{a{x}^{4}+b} \right ){a}^{5}{b}^{3} \right ){a}^{-{\frac{13}{2}}} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{-{\frac{5}{2}}} \left ( -a{x}^{2}+\sqrt{-ab} \right ) ^{-2} \left ( a{x}^{2}+\sqrt{-ab} \right ) ^{-2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(a*x^4+b)^(5/2)*(3*a^(15/2)*(a*x^4+b)^(1/2)*x^10+14*a^(13/2)*(-1/a*(-a*x^2+(-a*b)^(1/2))*(a*x^2+(-a*b)^(1
/2)))^(1/2)*x^6*b+6*a^(13/2)*b*(a*x^4+b)^(1/2)*x^6-15*ln(x^2*a^(1/2)+(a*x^4+b)^(1/2))*x^8*a^7*b+12*a^(11/2)*(-
1/a*(-a*x^2+(-a*b)^(1/2))*(a*x^2+(-a*b)^(1/2)))^(1/2)*b^2*x^2+3*a^(11/2)*b^2*(a*x^4+b)^(1/2)*x^2-30*ln(x^2*a^(
1/2)+(a*x^4+b)^(1/2))*a^6*b^2*x^4-15*ln(x^2*a^(1/2)+(a*x^4+b)^(1/2))*a^5*b^3)/a^(13/2)/((a*x^4+b)/x^4)^(5/2)/x
^10/(-a*x^2+(-a*b)^(1/2))^2/(a*x^2+(-a*b)^(1/2))^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(x^3/(a+b/x^4)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57845, size = 570, normalized size = 6.48 \begin{align*} \left [\frac{15 \,{\left (a^{2} b x^{8} + 2 \, a b^{2} x^{4} + b^{3}\right )} \sqrt{a} \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^{3} x^{12} + 20 \, a^{2} b x^{8} + 15 \, a b^{2} x^{4}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{24 \,{\left (a^{6} x^{8} + 2 \, a^{5} b x^{4} + a^{4} b^{2}\right )}}, \frac{15 \,{\left (a^{2} b x^{8} + 2 \, a b^{2} x^{4} + b^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} x^{4} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) +{\left (3 \, a^{3} x^{12} + 20 \, a^{2} b x^{8} + 15 \, a b^{2} x^{4}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{12 \,{\left (a^{6} x^{8} + 2 \, a^{5} b x^{4} + a^{4} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(15*(a^2*b*x^8 + 2*a*b^2*x^4 + b^3)*sqrt(a)*log(-2*a*x^4 + 2*sqrt(a)*x^4*sqrt((a*x^4 + b)/x^4) - b) + 2*
(3*a^3*x^12 + 20*a^2*b*x^8 + 15*a*b^2*x^4)*sqrt((a*x^4 + b)/x^4))/(a^6*x^8 + 2*a^5*b*x^4 + a^4*b^2), 1/12*(15*
(a^2*b*x^8 + 2*a*b^2*x^4 + b^3)*sqrt(-a)*arctan(sqrt(-a)*x^4*sqrt((a*x^4 + b)/x^4)/(a*x^4 + b)) + (3*a^3*x^12
+ 20*a^2*b*x^8 + 15*a*b^2*x^4)*sqrt((a*x^4 + b)/x^4))/(a^6*x^8 + 2*a^5*b*x^4 + a^4*b^2)]

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Sympy [B]  time = 6.30963, size = 819, normalized size = 9.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*a**17*x**16*sqrt(1 + b/(a*x**4))/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*b**2*x**4 + 24*a**
(33/2)*b**3) + 46*a**16*b*x**12*sqrt(1 + b/(a*x**4))/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*
b**2*x**4 + 24*a**(33/2)*b**3) + 15*a**16*b*x**12*log(b/(a*x**4))/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 +
72*a**(35/2)*b**2*x**4 + 24*a**(33/2)*b**3) - 30*a**16*b*x**12*log(sqrt(1 + b/(a*x**4)) + 1)/(24*a**(39/2)*x**
12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*b**2*x**4 + 24*a**(33/2)*b**3) + 70*a**15*b**2*x**8*sqrt(1 + b/(a*x**4
))/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*b**2*x**4 + 24*a**(33/2)*b**3) + 45*a**15*b**2*x**
8*log(b/(a*x**4))/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*b**2*x**4 + 24*a**(33/2)*b**3) - 90
*a**15*b**2*x**8*log(sqrt(1 + b/(a*x**4)) + 1)/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*b**2*x
**4 + 24*a**(33/2)*b**3) + 30*a**14*b**3*x**4*sqrt(1 + b/(a*x**4))/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 +
72*a**(35/2)*b**2*x**4 + 24*a**(33/2)*b**3) + 45*a**14*b**3*x**4*log(b/(a*x**4))/(24*a**(39/2)*x**12 + 72*a**
(37/2)*b*x**8 + 72*a**(35/2)*b**2*x**4 + 24*a**(33/2)*b**3) - 90*a**14*b**3*x**4*log(sqrt(1 + b/(a*x**4)) + 1)
/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*b**2*x**4 + 24*a**(33/2)*b**3) + 15*a**13*b**4*log(b
/(a*x**4))/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*b**2*x**4 + 24*a**(33/2)*b**3) - 30*a**13*
b**4*log(sqrt(1 + b/(a*x**4)) + 1)/(24*a**(39/2)*x**12 + 72*a**(37/2)*b*x**8 + 72*a**(35/2)*b**2*x**4 + 24*a**
(33/2)*b**3)

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Giac [A]  time = 1.21167, size = 151, normalized size = 1.72 \begin{align*} \frac{1}{12} \, b{\left (\frac{2 \,{\left (a + \frac{6 \,{\left (a x^{4} + b\right )}}{x^{4}}\right )} x^{4}}{{\left (a x^{4} + b\right )} a^{3} \sqrt{\frac{a x^{4} + b}{x^{4}}}} + \frac{15 \, \arctan \left (\frac{\sqrt{\frac{a x^{4} + b}{x^{4}}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} - \frac{3 \, \sqrt{\frac{a x^{4} + b}{x^{4}}}}{{\left (a - \frac{a x^{4} + b}{x^{4}}\right )} a^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*b*(2*(a + 6*(a*x^4 + b)/x^4)*x^4/((a*x^4 + b)*a^3*sqrt((a*x^4 + b)/x^4)) + 15*arctan(sqrt((a*x^4 + b)/x^4
)/sqrt(-a))/(sqrt(-a)*a^3) - 3*sqrt((a*x^4 + b)/x^4)/((a - (a*x^4 + b)/x^4)*a^3))