3.2075 \(\int (a+\frac {b}{x^4})^{5/2} x \, dx\)

Optimal. Leaf size=91 \[ -\frac {15}{16} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^4}}}\right )+\frac {1}{2} x^2 \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {5 b \left (a+\frac {b}{x^4}\right )^{3/2}}{8 x^2}-\frac {15 a b \sqrt {a+\frac {b}{x^4}}}{16 x^2} \]

[Out]

-5/8*b*(a+b/x^4)^(3/2)/x^2+1/2*(a+b/x^4)^(5/2)*x^2-15/16*a^2*arctanh(b^(1/2)/x^2/(a+b/x^4)^(1/2))*b^(1/2)-15/1
6*a*b*(a+b/x^4)^(1/2)/x^2

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Rubi [A]  time = 0.07, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {335, 275, 277, 195, 217, 206} \[ -\frac {15}{16} a^2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b}}{x^2 \sqrt {a+\frac {b}{x^4}}}\right )+\frac {1}{2} x^2 \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {5 b \left (a+\frac {b}{x^4}\right )^{3/2}}{8 x^2}-\frac {15 a b \sqrt {a+\frac {b}{x^4}}}{16 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(5/2)*x,x]

[Out]

(-15*a*b*Sqrt[a + b/x^4])/(16*x^2) - (5*b*(a + b/x^4)^(3/2))/(8*x^2) + ((a + b/x^4)^(5/2)*x^2)/2 - (15*a^2*Sqr
t[b]*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^4]*x^2)])/16

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.54 \[ -\frac {a^2 x^{10} \left (a+\frac {b}{x^4}\right )^{5/2} \left (a x^4+b\right ) \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {a x^4}{b}+1\right )}{14 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(5/2)*x,x]

[Out]

-1/14*(a^2*(a + b/x^4)^(5/2)*x^10*(b + a*x^4)*Hypergeometric2F1[3, 7/2, 9/2, 1 + (a*x^4)/b])/b^3

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fricas [A]  time = 0.88, size = 169, normalized size = 1.86 \[ \left [\frac {15 \, a^{2} \sqrt {b} x^{6} \log \left (\frac {a x^{4} - 2 \, \sqrt {b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ) + 2 \, {\left (8 \, a^{2} x^{8} - 9 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{32 \, x^{6}}, \frac {15 \, a^{2} \sqrt {-b} x^{6} \arctan \left (\frac {\sqrt {-b} x^{2} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{b}\right ) + {\left (8 \, a^{2} x^{8} - 9 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{16 \, x^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(15*a^2*sqrt(b)*x^6*log((a*x^4 - 2*sqrt(b)*x^2*sqrt((a*x^4 + b)/x^4) + 2*b)/x^4) + 2*(8*a^2*x^8 - 9*a*b*
x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^6, 1/16*(15*a^2*sqrt(-b)*x^6*arctan(sqrt(-b)*x^2*sqrt((a*x^4 + b)/x^4)/b
) + (8*a^2*x^8 - 9*a*b*x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^6]

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giac [A]  time = 0.19, size = 88, normalized size = 0.97 \[ \frac {\frac {15 \, a^{3} b \arctan \left (\frac {\sqrt {a x^{4} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 8 \, \sqrt {a x^{4} + b} a^{3} - \frac {9 \, {\left (a x^{4} + b\right )}^{\frac {3}{2}} a^{3} b - 7 \, \sqrt {a x^{4} + b} a^{3} b^{2}}{a^{2} x^{8}}}{16 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(15*a^3*b*arctan(sqrt(a*x^4 + b)/sqrt(-b))/sqrt(-b) + 8*sqrt(a*x^4 + b)*a^3 - (9*(a*x^4 + b)^(3/2)*a^3*b
- 7*sqrt(a*x^4 + b)*a^3*b^2)/(a^2*x^8))/a

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maple [A]  time = 0.02, size = 108, normalized size = 1.19 \[ \frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (-15 a^{2} \sqrt {b}\, x^{8} \ln \left (\frac {2 b +2 \sqrt {a \,x^{4}+b}\, \sqrt {b}}{x^{2}}\right )+8 \sqrt {a \,x^{4}+b}\, a^{2} x^{8}-9 \sqrt {a \,x^{4}+b}\, a b \,x^{4}-2 \sqrt {a \,x^{4}+b}\, b^{2}\right ) x^{2}}{16 \left (a \,x^{4}+b \right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((a*x^4+b)/x^4)^(5/2)*x^2*(-15*a^2*b^(1/2)*ln(2*(b+(a*x^4+b)^(1/2)*b^(1/2))/x^2)*x^8+8*(a*x^4+b)^(1/2)*a^
2*x^8-9*(a*x^4+b)^(1/2)*a*b*x^4-2*(a*x^4+b)^(1/2)*b^2)/(a*x^4+b)^(5/2)

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maxima [A]  time = 1.96, size = 139, normalized size = 1.53 \[ \frac {1}{2} \, \sqrt {a + \frac {b}{x^{4}}} a^{2} x^{2} + \frac {15}{32} \, a^{2} \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{x^{4}}} x^{2} - \sqrt {b}}{\sqrt {a + \frac {b}{x^{4}}} x^{2} + \sqrt {b}}\right ) - \frac {9 \, {\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}} a^{2} b x^{6} - 7 \, \sqrt {a + \frac {b}{x^{4}}} a^{2} b^{2} x^{2}}{16 \, {\left ({\left (a + \frac {b}{x^{4}}\right )}^{2} x^{8} - 2 \, {\left (a + \frac {b}{x^{4}}\right )} b x^{4} + b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(a + b/x^4)*a^2*x^2 + 15/32*a^2*sqrt(b)*log((sqrt(a + b/x^4)*x^2 - sqrt(b))/(sqrt(a + b/x^4)*x^2 + sqr
t(b))) - 1/16*(9*(a + b/x^4)^(3/2)*a^2*b*x^6 - 7*sqrt(a + b/x^4)*a^2*b^2*x^2)/((a + b/x^4)^2*x^8 - 2*(a + b/x^
4)*b*x^4 + b^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x\,{\left (a+\frac {b}{x^4}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/x^4)^(5/2),x)

[Out]

int(x*(a + b/x^4)^(5/2), x)

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sympy [A]  time = 4.97, size = 124, normalized size = 1.36 \[ \frac {a^{\frac {5}{2}} x^{2}}{2 \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {a^{\frac {3}{2}} b}{16 x^{2} \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {11 \sqrt {a} b^{2}}{16 x^{6} \sqrt {1 + \frac {b}{a x^{4}}}} - \frac {15 a^{2} \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} x^{2}} \right )}}{16} - \frac {b^{3}}{8 \sqrt {a} x^{10} \sqrt {1 + \frac {b}{a x^{4}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/2)*x**2/(2*sqrt(1 + b/(a*x**4))) - a**(3/2)*b/(16*x**2*sqrt(1 + b/(a*x**4))) - 11*sqrt(a)*b**2/(16*x**6*
sqrt(1 + b/(a*x**4))) - 15*a**2*sqrt(b)*asinh(sqrt(b)/(sqrt(a)*x**2))/16 - b**3/(8*sqrt(a)*x**10*sqrt(1 + b/(a
*x**4)))

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