3.1.25 \(\int \frac {(2+3 x^2) (5+x^4)^{3/2}}{x^5} \, dx\) [25]

Optimal. Leaf size=86 \[ -\frac {3 \left (15-2 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (2-3 x^2\right ) \left (5+x^4\right )^{3/2}}{4 x^4}+\frac {45}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {3}{2} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right ) \]

[Out]

-1/4*(-3*x^2+2)*(x^4+5)^(3/2)/x^4+45/4*arcsinh(1/5*x^2*5^(1/2))-3/2*arctanh(1/5*(x^4+5)^(1/2)*5^(1/2))*5^(1/2)
-3/4*(-2*x^2+15)*(x^4+5)^(1/2)/x^2

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Rubi [A]
time = 0.05, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1266, 827, 858, 221, 272, 65, 213} \begin {gather*} -\frac {3}{2} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {x^4+5}}{\sqrt {5}}\right )+\frac {45}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {\left (2-3 x^2\right ) \left (x^4+5\right )^{3/2}}{4 x^4}-\frac {3 \left (15-2 x^2\right ) \sqrt {x^4+5}}{4 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(15 - 2*x^2)*Sqrt[5 + x^4])/(4*x^2) - ((2 - 3*x^2)*(5 + x^4)^(3/2))/(4*x^4) + (45*ArcSinh[x^2/Sqrt[5]])/4
- (3*Sqrt[5]*ArcTanh[Sqrt[5 + x^4]/Sqrt[5]])/2

Rule 65

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 213

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

Rule 221

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 272

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 827

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 858

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 1266

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^5} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {(2+3 x) \left (5+x^2\right )^{3/2}}{x^3} \, dx,x,x^2\right )\\ &=-\frac {\left (2-3 x^2\right ) \left (5+x^4\right )^{3/2}}{4 x^4}-\frac {3}{16} \text {Subst}\left (\int \frac {(-60-8 x) \sqrt {5+x^2}}{x^2} \, dx,x,x^2\right )\\ &=-\frac {3 \left (15-2 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (2-3 x^2\right ) \left (5+x^4\right )^{3/2}}{4 x^4}+\frac {3}{32} \text {Subst}\left (\int \frac {80+120 x}{x \sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {3 \left (15-2 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (2-3 x^2\right ) \left (5+x^4\right )^{3/2}}{4 x^4}+\frac {15}{2} \text {Subst}\left (\int \frac {1}{x \sqrt {5+x^2}} \, dx,x,x^2\right )+\frac {45}{4} \text {Subst}\left (\int \frac {1}{\sqrt {5+x^2}} \, dx,x,x^2\right )\\ &=-\frac {3 \left (15-2 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (2-3 x^2\right ) \left (5+x^4\right )^{3/2}}{4 x^4}+\frac {45}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+\frac {15}{4} \text {Subst}\left (\int \frac {1}{x \sqrt {5+x}} \, dx,x,x^4\right )\\ &=-\frac {3 \left (15-2 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (2-3 x^2\right ) \left (5+x^4\right )^{3/2}}{4 x^4}+\frac {45}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )+\frac {15}{2} \text {Subst}\left (\int \frac {1}{-5+x^2} \, dx,x,\sqrt {5+x^4}\right )\\ &=-\frac {3 \left (15-2 x^2\right ) \sqrt {5+x^4}}{4 x^2}-\frac {\left (2-3 x^2\right ) \left (5+x^4\right )^{3/2}}{4 x^4}+\frac {45}{4} \sinh ^{-1}\left (\frac {x^2}{\sqrt {5}}\right )-\frac {3}{2} \sqrt {5} \tanh ^{-1}\left (\frac {\sqrt {5+x^4}}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 81, normalized size = 0.94 \begin {gather*} \frac {\sqrt {5+x^4} \left (-10-30 x^2+4 x^4+3 x^6\right )}{4 x^4}+\frac {45}{4} \tanh ^{-1}\left (\frac {x^2}{\sqrt {5+x^4}}\right )+3 \sqrt {5} \tanh ^{-1}\left (\frac {x^2-\sqrt {5+x^4}}{\sqrt {5}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[5 + x^4]*(-10 - 30*x^2 + 4*x^4 + 3*x^6))/(4*x^4) + (45*ArcTanh[x^2/Sqrt[5 + x^4]])/4 + 3*Sqrt[5]*ArcTanh
[(x^2 - Sqrt[5 + x^4])/Sqrt[5]]

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Maple [A]
time = 0.27, size = 73, normalized size = 0.85

method	result	size
default	\(\sqrt {x^{4}+5}-\frac {5 \sqrt {x^{4}+5}}{2 x^{4}}-\frac {3 \sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}+\frac {3 x^{2} \sqrt {x^{4}+5}}{4}+\frac {45 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}-\frac {15 \sqrt {x^{4}+5}}{2 x^{2}}\)	\(73\)
elliptic	\(\sqrt {x^{4}+5}-\frac {5 \sqrt {x^{4}+5}}{2 x^{4}}-\frac {3 \sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}+\frac {3 x^{2} \sqrt {x^{4}+5}}{4}+\frac {45 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}-\frac {15 \sqrt {x^{4}+5}}{2 x^{2}}\)	\(73\)
risch	\(-\frac {5 \left (3 x^{6}+x^{4}+15 x^{2}+5\right )}{2 x^{4} \sqrt {x^{4}+5}}+\frac {3 x^{2} \sqrt {x^{4}+5}}{4}+\frac {45 \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}+\sqrt {x^{4}+5}-\frac {3 \sqrt {5}\, \arctanh \left (\frac {\sqrt {5}}{\sqrt {x^{4}+5}}\right )}{2}\)	\(76\)
trager	\(\frac {\left (3 x^{6}+4 x^{4}-30 x^{2}-10\right ) \sqrt {x^{4}+5}}{4 x^{4}}+\frac {3 \RootOf \left (\textit {\_Z}^{2}-5\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-5\right )-\sqrt {x^{4}+5}}{x^{2}}\right )}{2}+\frac {45 \ln \left (-x^{2}-\sqrt {x^{4}+5}\right )}{4}\)	\(79\)
meijerg	\(\frac {3 \sqrt {5}\, \left (\frac {5 \sqrt {\pi }\, \left (-\frac {12 x^{4}}{5}+8\right )}{6 x^{4}}-\frac {5 \sqrt {\pi }\, \left (8-\frac {16 x^{4}}{5}\right ) \sqrt {1+\frac {x^{4}}{5}}}{6 x^{4}}-4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1+\frac {x^{4}}{5}}}{2}\right )+2 \left (1-2 \ln \left (2\right )+4 \ln \left (x \right )-\ln \left (5\right )\right ) \sqrt {\pi }-\frac {20 \sqrt {\pi }}{3 x^{4}}\right )}{8 \sqrt {\pi }}+\frac {-\frac {15 \sqrt {\pi }\, \sqrt {5}\, \left (-\frac {x^{4}}{10}+1\right ) \sqrt {1+\frac {x^{4}}{5}}}{2 x^{2}}+\frac {45 \sqrt {\pi }\, \arcsinh \left (\frac {x^{2} \sqrt {5}}{5}\right )}{4}}{\sqrt {\pi }}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+2)*(x^4+5)^(3/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

(x^4+5)^(1/2)-5/2*(x^4+5)^(1/2)/x^4-3/2*5^(1/2)*arctanh(5^(1/2)/(x^4+5)^(1/2))+3/4*x^2*(x^4+5)^(1/2)+45/4*arcs
inh(1/5*x^2*5^(1/2))-15/2*(x^4+5)^(1/2)/x^2

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Maxima [A]
time = 0.49, size = 123, normalized size = 1.43 \begin {gather*} \frac {3}{4} \, \sqrt {5} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{\sqrt {5} + \sqrt {x^{4} + 5}}\right ) + \sqrt {x^{4} + 5} - \frac {15 \, \sqrt {x^{4} + 5}}{2 \, x^{2}} + \frac {15 \, \sqrt {x^{4} + 5}}{4 \, x^{2} {\left (\frac {x^{4} + 5}{x^{4}} - 1\right )}} - \frac {5 \, \sqrt {x^{4} + 5}}{2 \, x^{4}} + \frac {45}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} + 1\right ) - \frac {45}{8} \, \log \left (\frac {\sqrt {x^{4} + 5}}{x^{2}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*sqrt(5)*log(-(sqrt(5) - sqrt(x^4 + 5))/(sqrt(5) + sqrt(x^4 + 5))) + sqrt(x^4 + 5) - 15/2*sqrt(x^4 + 5)/x^2
+ 15/4*sqrt(x^4 + 5)/(x^2*((x^4 + 5)/x^4 - 1)) - 5/2*sqrt(x^4 + 5)/x^4 + 45/8*log(sqrt(x^4 + 5)/x^2 + 1) - 45
/8*log(sqrt(x^4 + 5)/x^2 - 1)

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Fricas [A]
time = 0.40, size = 82, normalized size = 0.95 \begin {gather*} \frac {6 \, \sqrt {5} x^{4} \log \left (-\frac {\sqrt {5} - \sqrt {x^{4} + 5}}{x^{2}}\right ) - 45 \, x^{4} \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) - 30 \, x^{4} + {\left (3 \, x^{6} + 4 \, x^{4} - 30 \, x^{2} - 10\right )} \sqrt {x^{4} + 5}}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(6*sqrt(5)*x^4*log(-(sqrt(5) - sqrt(x^4 + 5))/x^2) - 45*x^4*log(-x^2 + sqrt(x^4 + 5)) - 30*x^4 + (3*x^6 +
4*x^4 - 30*x^2 - 10)*sqrt(x^4 + 5))/x^4

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Sympy [A]
time = 6.47, size = 133, normalized size = 1.55 \begin {gather*} \frac {3 x^{6}}{4 \sqrt {x^{4} + 5}} - \frac {15 x^{2}}{4 \sqrt {x^{4} + 5}} + \sqrt {x^{4} + 5} + \frac {\sqrt {5} \log {\left (x^{4} \right )}}{2} - \sqrt {5} \log {\left (\sqrt {\frac {x^{4}}{5} + 1} + 1 \right )} - \frac {\sqrt {5} \operatorname {asinh}{\left (\frac {\sqrt {5}}{x^{2}} \right )}}{2} + \frac {45 \operatorname {asinh}{\left (\frac {\sqrt {5} x^{2}}{5} \right )}}{4} - \frac {5 \sqrt {1 + \frac {5}{x^{4}}}}{2 x^{2}} - \frac {75}{2 x^{2} \sqrt {x^{4} + 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x**6/(4*sqrt(x**4 + 5)) - 15*x**2/(4*sqrt(x**4 + 5)) + sqrt(x**4 + 5) + sqrt(5)*log(x**4)/2 - sqrt(5)*log(sq
rt(x**4/5 + 1) + 1) - sqrt(5)*asinh(sqrt(5)/x**2)/2 + 45*asinh(sqrt(5)*x**2/5)/4 - 5*sqrt(1 + 5/x**4)/(2*x**2)
- 75/(2*x**2*sqrt(x**4 + 5))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 146 vs. \(2 (68) = 136\).
time = 2.73, size = 146, normalized size = 1.70 \begin {gather*} \frac {1}{4} \, \sqrt {x^{4} + 5} {\left (3 \, x^{2} + 4\right )} + \frac {3}{2} \, \sqrt {5} \log \left (-\frac {x^{2} + \sqrt {5} - \sqrt {x^{4} + 5}}{x^{2} - \sqrt {5} - \sqrt {x^{4} + 5}}\right ) + \frac {5 \, {\left ({\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{3} + 15 \, {\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} + 5 \, x^{2} - 5 \, \sqrt {x^{4} + 5} - 75\right )}}{{\left ({\left (x^{2} - \sqrt {x^{4} + 5}\right )}^{2} - 5\right )}^{2}} - \frac {45}{4} \, \log \left (-x^{2} + \sqrt {x^{4} + 5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(x^4 + 5)*(3*x^2 + 4) + 3/2*sqrt(5)*log(-(x^2 + sqrt(5) - sqrt(x^4 + 5))/(x^2 - sqrt(5) - sqrt(x^4 + 5
))) + 5*((x^2 - sqrt(x^4 + 5))^3 + 15*(x^2 - sqrt(x^4 + 5))^2 + 5*x^2 - 5*sqrt(x^4 + 5) - 75)/((x^2 - sqrt(x^4
+ 5))^2 - 5)^2 - 45/4*log(-x^2 + sqrt(x^4 + 5))

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Mupad [B]
time = 0.55, size = 71, normalized size = 0.83 \begin {gather*} \frac {45\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x^2}{5}\right )}{4}+\sqrt {x^4+5}\,\left (\frac {3\,x^2}{4}+1\right )-\frac {15\,\sqrt {x^4+5}}{2\,x^2}-\frac {5\,\sqrt {x^4+5}}{2\,x^4}+\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {\sqrt {5}\,\sqrt {x^4+5}\,1{}\mathrm {i}}{5}\right )\,3{}\mathrm {i}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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