3.26 \(\int \frac {(2+3 x^2) (5+x^4)^{3/2}}{x^7} \, dx\)

Optimal. Leaf size=82 \[ -\frac {9}{4} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {\left (4-9 x^2\right ) \sqrt {x^4+5}}{4 x^2}-\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{12 x^6} \]

[Out]

-1/12*(9*x^2+4)*(x^4+5)^(3/2)/x^6+arcsinh(1/5*x^2*5^(1/2))-9/4*arctanh(1/5*(x^4+5)^(1/2)*5^(1/2))*5^(1/2)-1/4*
(-9*x^2+4)*(x^4+5)^(1/2)/x^2

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Rubi [A]  time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1252, 811, 813, 844, 215, 266, 63, 207} \[ -\frac {\left (9 x^2+4\right ) \left (x^4+5\right )^{3/2}}{12 x^6}-\frac {\left (4-9 x^2\right ) \sqrt {x^4+5}}{4 x^2}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {9}{4} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[((2 + 3*x^2)*(5 + x^4)^(3/2))/x^7,x]

[Out]

-((4 - 9*x^2)*Sqrt[5 + x^4])/(4*x^2) - ((4 + 9*x^2)*(5 + x^4)^(3/2))/(12*x^6) + ArcSinh[x^2/Sqrt[5]] - (9*Sqrt
[5]*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]])/4

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 207

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

Rule 215

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

Rule 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rubi steps

\begin {align*} \int \frac {\left (2+3 x^2\right ) \left (5+x^4\right )^{3/2}}{x^7} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(2+3 x) \left (5+x^2\right )^{3/2}}{x^4} \, dx,x,x^2\right )\\ &=-\frac {\left (4+9 x^2\right ) \left (5+x^4\right )^{3/2}}{12 x^6}-\frac {1}{40} \operatorname {Subst}\left (\int \frac {(-40-90 x) \sqrt {5+x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (4-9 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (4+9 x^2\right ) \left (5+x^4\right )^{3/2}}{12 x^6}+\frac {1}{80} \operatorname {Subst}\left (\int \frac {900+80 x}{x \sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\left (4-9 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (4+9 x^2\right ) \left (5+x^4\right )^{3/2}}{12 x^6}+\frac {45}{4} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {5+x^2}} \, dx,x,x^2\right )+\operatorname {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {\left (4-9 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (4+9 x^2\right ) \left (5+x^4\right )^{3/2}}{12 x^6}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+\frac {45}{8} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {5+x}} \, dx,x,x^4\right )\\ &=-\frac {\left (4-9 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (4+9 x^2\right ) \left (5+x^4\right )^{3/2}}{12 x^6}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+\frac {45}{4} \operatorname {Subst}\left (\int \frac {1}{-5+x^2} \, dx,x,\sqrt {5+x^4}\right )\\ &=-\frac {\left (4-9 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (4+9 x^2\right ) \left (5+x^4\right )^{3/2}}{12 x^6}+\sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {9}{4} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 60, normalized size = 0.73 \[ \frac {3}{250} \left (x^4+5\right )^{5/2} \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};\frac {x^4}{5}+1\right )-\frac {5 \sqrt {5} \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};-\frac {x^4}{5}\right )}{3 x^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((2 + 3*x^2)*(5 + x^4)^(3/2))/x^7,x]

[Out]

(-5*Sqrt[5]*Hypergeometric2F1[-3/2, -3/2, -1/2, -1/5*x^4])/(3*x^6) + (3*(5 + x^4)^(5/2)*Hypergeometric2F1[2, 5
/2, 7/2, 1 + x^4/5])/250

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fricas [A]  time = 0.70, size = 82, normalized size = 1.00 \[ \frac {27 \, \sqrt {5} x^{6} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - 12 \, x^{6} \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) - 16 \, x^{6} + {\left (18 \, x^{6} - 16 \, x^{4} - 45 \, x^{2} - 20\right )} \sqrt {x^{4} + 5}}{12 \, x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(3/2)/x^7,x, algorithm="fricas")

[Out]

1/12*(27*sqrt(5)*x^6*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) - 12*x^6*log(-x^2 + sqrt(x^4 + 5)) - 16*x^6 + (18*x^6
- 16*x^4 - 45*x^2 - 20)*sqrt(x^4 + 5))/x^6

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giac [B]  time = 0.27, size = 158, normalized size = 1.93 \[ \frac {9}{4} \, \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) + \frac {3}{2} \, \sqrt {x^{4} + 5} + \frac {5 \, {\left (9 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{5} + 24 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{4} - 120 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 225 \, x^{2} + 225 \, \sqrt {x^{4} + 5} + 400\right )}}{6 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 5\right )}^{3}} - \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(3/2)/x^7,x, algorithm="giac")

[Out]

9/4*sqrt(5)*log(-(x^2 + sqrt(5) - sqrt(x^4 + 5))/(x^2 - sqrt(5) - sqrt(x^4 + 5))) + 3/2*sqrt(x^4 + 5) + 5/6*(9
*(x^2 - sqrt(x^4 + 5))^5 + 24*(x^2 - sqrt(x^4 + 5))^4 - 120*(x^2 - sqrt(x^4 + 5))^2 - 225*x^2 + 225*sqrt(x^4 +
5) + 400)/((x^2 - sqrt(x^4 + 5))^2 - 5)^3 - log(-x^2 + sqrt(x^4 + 5))

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maple [A]  time = 0.02, size = 73, normalized size = 0.89 \[ \arcsinh \left (\frac {\sqrt {5}\, x^{2}}{5}\right )-\frac {9 \sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{4}-\frac {4 \sqrt {x^{4}+5}}{3 x^{2}}-\frac {15 \sqrt {x^{4}+5}}{4 x^{4}}-\frac {5 \sqrt {x^{4}+5}}{3 x^{6}}+\frac {3 \sqrt {x^{4}+5}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+2)*(x^4+5)^(3/2)/x^7,x)

[Out]

3/2*(x^4+5)^(1/2)-9/4*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))-15/4*(x^4+5)^(1/2)/x^4+arcsinh(1/5*5^(1/2)*x^2)-4
/3*(x^4+5)^(1/2)/x^2-5/3*(x^4+5)^(1/2)/x^6

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maxima [A]  time = 1.35, size = 112, normalized size = 1.37 \[ \frac {9}{8} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \frac {3}{2} \, \sqrt {x^{4} + 5} - \frac {\sqrt {x^{4} + 5}}{x^{2}} - \frac {15 \, \sqrt {x^{4} + 5}}{4 \, x^{4}} - \frac {{\left (x^{4} + 5\right )}^{\frac {3}{2}}}{3 \, x^{6}} + \frac {1}{2} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {1}{2} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2)*(x^4+5)^(3/2)/x^7,x, algorithm="maxima")

[Out]

9/8*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + 3/2*sqrt(x^4 + 5) - sqrt(x^4 + 5)/x^2
- 15/4*sqrt(x^4 + 5)/x^4 - 1/3*(x^4 + 5)^(3/2)/x^6 + 1/2*log(sqrt(x^4 + 5)/x^2 + 1) - 1/2*log(sqrt(x^4 + 5)/x^
2 - 1)

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mupad [B]  time = 0.95, size = 82, normalized size = 1.00 \[ \mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )+\frac {3\,\sqrt {x^4+5}}{2}+\sqrt {x^4+5}\,\left (\frac {2}{3\,x^2}-\frac {5}{3\,x^6}\right )-\frac {2\,\sqrt {x^4+5}}{x^2}-\frac {15\,\sqrt {x^4+5}}{4\,x^4}+\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}\,1{}\mathrm {i}}{5}\right )\,9{}\mathrm {i}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^4 + 5)^(3/2)*(3*x^2 + 2))/x^7,x)

[Out]

asinh((5^(1/2)*x^2)/5) + (5^(1/2)*atan((5^(1/2)*(x^4 + 5)^(1/2)*1i)/5)*9i)/4 + (3*(x^4 + 5)^(1/2))/2 + (x^4 +
5)^(1/2)*(2/(3*x^2) - 5/(3*x^6)) - (2*(x^4 + 5)^(1/2))/x^2 - (15*(x^4 + 5)^(1/2))/(4*x^4)

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sympy [A]  time = 12.56, size = 148, normalized size = 1.80 \[ - \frac {x^{2}}{\sqrt {x^{4} + 5}} - \frac {\sqrt {1 + \frac {5}{x^{4}}}}{3} + \frac {3 \sqrt {x^{4} + 5}}{2} + \frac {3 \sqrt {5} \log {\left (x^{4} \right )}}{4} - \frac {3 \sqrt {5} \log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )}}{2} - \frac {3 \sqrt {5} \operatorname {asinh}{\left (\frac {\sqrt {5}}{x^{2}} \right )}}{4} + \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )} - \frac {15 \sqrt {1 + \frac {5}{x^{4}}}}{4 x^{2}} - \frac {5}{x^{2} \sqrt {x^{4} + 5}} - \frac {5 \sqrt {1 + \frac {5}{x^{4}}}}{3 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+2)*(x**4+5)**(3/2)/x**7,x)

[Out]

-x**2/sqrt(x**4 + 5) - sqrt(1 + 5/x**4)/3 + 3*sqrt(x**4 + 5)/2 + 3*sqrt(5)*log(x**4)/4 - 3*sqrt(5)*log(sqrt(x*
*4/5 + 1) + 1)/2 - 3*sqrt(5)*asinh(sqrt(5)/x**2)/4 + asinh(sqrt(5)*x**2/5) - 15*sqrt(1 + 5/x**4)/(4*x**2) - 5/
(x**2*sqrt(x**4 + 5)) - 5*sqrt(1 + 5/x**4)/(3*x**4)

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