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

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

[Out]

5/12*b/a^2/(a+b/x^4)^(3/2)+1/4*x^4/a/(a+b/x^4)^(3/2)-5/4*b*arctanh((a+b/x^4)^(1/2)/a^(1/2))/a^(7/2)+5/4*b/a^3/
(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.05, 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 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]

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]]

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*}

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Mathematica [A]  time = 0.23, 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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fricas [A]  time = 0.85, size = 260, normalized size = 2.95 \[ \left [\frac {15 \, {\left (a^{2} b x^{8} + 2 \, a b^{2} x^{4} + b^{3}\right )} \sqrt {a} \log \left (-2 \, a x^{4} + 2 \, \sqrt {a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}} - b\right ) + 2 \, {\left (3 \, a^{3} x^{12} + 20 \, a^{2} b x^{8} + 15 \, a b^{2} x^{4}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{24 \, {\left (a^{6} x^{8} + 2 \, a^{5} b x^{4} + a^{4} b^{2}\right )}}, \frac {15 \, {\left (a^{2} b x^{8} + 2 \, a b^{2} x^{4} + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{4} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a x^{4} + b}\right ) + {\left (3 \, a^{3} x^{12} + 20 \, a^{2} b x^{8} + 15 \, a b^{2} x^{4}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{12 \, {\left (a^{6} x^{8} + 2 \, a^{5} b x^{4} + a^{4} b^{2}\right )}}\right ] \]

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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giac [A]  time = 0.23, size = 78, normalized size = 0.89 \[ \frac {{\left ({\left (\frac {3 \, x^{4}}{a} + \frac {20 \, b}{a^{2}}\right )} x^{4} + \frac {15 \, b^{2}}{a^{3}}\right )} x^{2}}{12 \, {\left (a x^{4} + b\right )}^{\frac {3}{2}}} + \frac {5 \, b \log \left ({\left | -\sqrt {a} x^{2} + \sqrt {a x^{4} + b} \right |}\right )}{4 \, a^{\frac {7}{2}}} - \frac {5 \, b \log \left ({\left | b \right |}\right )}{8 \, a^{\frac {7}{2}}} \]

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*((3*x^4/a + 20*b/a^2)*x^4 + 15*b^2/a^3)*x^2/(a*x^4 + b)^(3/2) + 5/4*b*log(abs(-sqrt(a)*x^2 + sqrt(a*x^4 +
b)))/a^(7/2) - 5/8*b*log(abs(b))/a^(7/2)

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maple [B]  time = 0.07, size = 282, normalized size = 3.20 \[ \frac {\left (a \,x^{4}+b \right )^{\frac {5}{2}} \left (3 \sqrt {a \,x^{4}+b}\, a^{\frac {15}{2}} x^{10}-15 a^{7} b \,x^{8} \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right )+14 \sqrt {-\frac {\left (-a \,x^{2}+\sqrt {-a b}\right ) \left (a \,x^{2}+\sqrt {-a b}\right )}{a}}\, a^{\frac {13}{2}} b \,x^{6}+6 \sqrt {a \,x^{4}+b}\, a^{\frac {13}{2}} b \,x^{6}-30 a^{6} b^{2} x^{4} \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right )+12 \sqrt {-\frac {\left (-a \,x^{2}+\sqrt {-a b}\right ) \left (a \,x^{2}+\sqrt {-a b}\right )}{a}}\, a^{\frac {11}{2}} b^{2} x^{2}+3 \sqrt {a \,x^{4}+b}\, a^{\frac {11}{2}} b^{2} x^{2}-15 a^{5} b^{3} \ln \left (\sqrt {a}\, x^{2}+\sqrt {a \,x^{4}+b}\right )\right )}{12 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (-a \,x^{2}+\sqrt {-a b}\right )^{2} \left (a \,x^{2}+\sqrt {-a b}\right )^{2} a^{\frac {13}{2}} x^{10}} \]

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)*(-(-a*x^2+(-a*b)^(1/2))/a*(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(a^(1/2)*x^2+(a*x^4+b)^(1/2))*x^8*a^7*b+12*a^(11/2)*(-(-
a*x^2+(-a*b)^(1/2))/a*(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(a^(1/2)*x^2
+(a*x^4+b)^(1/2))*a^6*b^2*x^4-15*ln(a^(1/2)*x^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 [A]  time = 1.98, size = 101, normalized size = 1.15 \[ \frac {15 \, {\left (a + \frac {b}{x^{4}}\right )}^{2} b - 10 \, {\left (a + \frac {b}{x^{4}}\right )} a b - 2 \, a^{2} b}{12 \, {\left ({\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}} a^{3} - {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{4}\right )}} + \frac {5 \, b \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{4}}} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(15*(a + b/x^4)^2*b - 10*(a + b/x^4)*a*b - 2*a^2*b)/((a + b/x^4)^(5/2)*a^3 - (a + b/x^4)^(3/2)*a^4) + 5/8
*b*log((sqrt(a + b/x^4) - sqrt(a))/(sqrt(a + b/x^4) + sqrt(a)))/a^(7/2)

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mupad [B]  time = 1.89, size = 73, normalized size = 0.83 \[ \frac {5\,b}{3\,a^2\,{\left (a+\frac {b}{x^4}\right )}^{3/2}}+\frac {x^4}{4\,a\,{\left (a+\frac {b}{x^4}\right )}^{3/2}}-\frac {5\,b\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^4}}}{\sqrt {a}}\right )}{4\,a^{7/2}}+\frac {5\,b^2}{4\,a^3\,x^4\,{\left (a+\frac {b}{x^4}\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(a + b/x^4)^(5/2),x)

[Out]

(5*b)/(3*a^2*(a + b/x^4)^(3/2)) + x^4/(4*a*(a + b/x^4)^(3/2)) - (5*b*atanh((a + b/x^4)^(1/2)/a^(1/2)))/(4*a^(7
/2)) + (5*b^2)/(4*a^3*x^4*(a + b/x^4)^(3/2))

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sympy [B]  time = 6.20, size = 819, normalized size = 9.31 \[ \frac {6 a^{17} x^{16} \sqrt {1 + \frac {b}{a x^{4}}}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} + \frac {46 a^{16} b x^{12} \sqrt {1 + \frac {b}{a x^{4}}}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} + \frac {15 a^{16} b x^{12} \log {\left (\frac {b}{a x^{4}} \right )}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} - \frac {30 a^{16} b x^{12} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} + \frac {70 a^{15} b^{2} x^{8} \sqrt {1 + \frac {b}{a x^{4}}}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} + \frac {45 a^{15} b^{2} x^{8} \log {\left (\frac {b}{a x^{4}} \right )}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} - \frac {90 a^{15} b^{2} x^{8} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} + \frac {30 a^{14} b^{3} x^{4} \sqrt {1 + \frac {b}{a x^{4}}}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} + \frac {45 a^{14} b^{3} x^{4} \log {\left (\frac {b}{a x^{4}} \right )}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} - \frac {90 a^{14} b^{3} x^{4} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} + \frac {15 a^{13} b^{4} \log {\left (\frac {b}{a x^{4}} \right )}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} - \frac {30 a^{13} b^{4} \log {\left (\sqrt {1 + \frac {b}{a x^{4}}} + 1 \right )}}{24 a^{\frac {39}{2}} x^{12} + 72 a^{\frac {37}{2}} b x^{8} + 72 a^{\frac {35}{2}} b^{2} x^{4} + 24 a^{\frac {33}{2}} b^{3}} \]

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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