3.5.78 \(\int \frac {\sqrt {1+x^3}}{x^7} \, dx\)

Optimal. Leaf size=38 \[ \frac {1}{12} \tanh ^{-1}\left (\sqrt {x^3+1}\right )+\frac {\sqrt {x^3+1} \left (-x^3-2\right )}{12 x^6} \]

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.24, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {266, 47, 51, 63, 207} \begin {gather*} -\frac {\sqrt {x^3+1}}{12 x^3}+\frac {1}{12} \tanh ^{-1}\left (\sqrt {x^3+1}\right )-\frac {\sqrt {x^3+1}}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 + x^3]/x^7,x]

[Out]

-1/6*Sqrt[1 + x^3]/x^6 - Sqrt[1 + x^3]/(12*x^3) + ArcTanh[Sqrt[1 + x^3]]/12

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 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 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 {\sqrt {1+x^3}}{x^7} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{x^3} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {1+x^3}}{6 x^6}+\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+x}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {1+x^3}}{6 x^6}-\frac {\sqrt {1+x^3}}{12 x^3}-\frac {1}{24} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt {1+x^3}}{6 x^6}-\frac {\sqrt {1+x^3}}{12 x^3}-\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^3}\right )\\ &=-\frac {\sqrt {1+x^3}}{6 x^6}-\frac {\sqrt {1+x^3}}{12 x^3}+\frac {1}{12} \tanh ^{-1}\left (\sqrt {1+x^3}\right )\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 26, normalized size = 0.68 \begin {gather*} -\frac {2}{9} \left (x^3+1\right )^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};x^3+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 + x^3]/x^7,x]

[Out]

(-2*(1 + x^3)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + x^3])/9

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IntegrateAlgebraic [A]  time = 0.06, size = 38, normalized size = 1.00 \begin {gather*} \frac {\left (-2-x^3\right ) \sqrt {1+x^3}}{12 x^6}+\frac {1}{12} \tanh ^{-1}\left (\sqrt {1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[1 + x^3]/x^7,x]

[Out]

((-2 - x^3)*Sqrt[1 + x^3])/(12*x^6) + ArcTanh[Sqrt[1 + x^3]]/12

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fricas [A]  time = 0.45, size = 49, normalized size = 1.29 \begin {gather*} \frac {x^{6} \log \left (\sqrt {x^{3} + 1} + 1\right ) - x^{6} \log \left (\sqrt {x^{3} + 1} - 1\right ) - 2 \, {\left (x^{3} + 2\right )} \sqrt {x^{3} + 1}}{24 \, x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(1/2)/x^7,x, algorithm="fricas")

[Out]

1/24*(x^6*log(sqrt(x^3 + 1) + 1) - x^6*log(sqrt(x^3 + 1) - 1) - 2*(x^3 + 2)*sqrt(x^3 + 1))/x^6

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giac [B]  time = 0.33, size = 78, normalized size = 2.05 \begin {gather*} -\frac {\sqrt {x^{3} + 1} + \frac {1}{\sqrt {x^{3} + 1}}}{12 \, {\left ({\left (\sqrt {x^{3} + 1} + \frac {1}{\sqrt {x^{3} + 1}}\right )}^{2} - 4\right )}} + \frac {1}{48} \, \log \left (\sqrt {x^{3} + 1} + \frac {1}{\sqrt {x^{3} + 1}} + 2\right ) - \frac {1}{48} \, \log \left ({\left | \sqrt {x^{3} + 1} + \frac {1}{\sqrt {x^{3} + 1}} - 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(1/2)/x^7,x, algorithm="giac")

[Out]

-1/12*(sqrt(x^3 + 1) + 1/sqrt(x^3 + 1))/((sqrt(x^3 + 1) + 1/sqrt(x^3 + 1))^2 - 4) + 1/48*log(sqrt(x^3 + 1) + 1
/sqrt(x^3 + 1) + 2) - 1/48*log(abs(sqrt(x^3 + 1) + 1/sqrt(x^3 + 1) - 2))

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maple [A]  time = 0.21, size = 34, normalized size = 0.89

method	result	size
risch	\(-\frac {x^{6}+3 x^{3}+2}{12 x^{6} \sqrt {x^{3}+1}}+\frac {\arctanh \left (\sqrt {x^{3}+1}\right )}{12}\)	\(34\)
default	\(-\frac {\sqrt {x^{3}+1}}{6 x^{6}}-\frac {\sqrt {x^{3}+1}}{12 x^{3}}+\frac {\arctanh \left (\sqrt {x^{3}+1}\right )}{12}\)	\(36\)
elliptic	\(-\frac {\sqrt {x^{3}+1}}{6 x^{6}}-\frac {\sqrt {x^{3}+1}}{12 x^{3}}+\frac {\arctanh \left (\sqrt {x^{3}+1}\right )}{12}\)	\(36\)
trager	\(-\frac {\left (x^{3}+2\right ) \sqrt {x^{3}+1}}{12 x^{6}}+\frac {\ln \left (-\frac {x^{3}+2 \sqrt {x^{3}+1}+2}{x^{3}}\right )}{24}\)	\(41\)
meijerg	\(-\frac {\frac {\sqrt {\pi }}{x^{6}}+\frac {\sqrt {\pi }}{x^{3}}+\frac {\left (\frac {1}{2}-2 \ln \relax (2)+3 \ln \relax (x )\right ) \sqrt {\pi }}{4}-\frac {\sqrt {\pi }\, \left (x^{6}+8 x^{3}+8\right )}{8 x^{6}}+\frac {\sqrt {\pi }\, \left (4 x^{3}+8\right ) \sqrt {x^{3}+1}}{8 x^{6}}-\frac {\ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right ) \sqrt {\pi }}{2}}{6 \sqrt {\pi }}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)^(1/2)/x^7,x,method=_RETURNVERBOSE)

[Out]

-1/12*(x^6+3*x^3+2)/x^6/(x^3+1)^(1/2)+1/12*arctanh((x^3+1)^(1/2))

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maxima [B]  time = 0.32, size = 60, normalized size = 1.58 \begin {gather*} \frac {{\left (x^{3} + 1\right )}^{\frac {3}{2}} + \sqrt {x^{3} + 1}}{12 \, {\left (2 \, x^{3} - {\left (x^{3} + 1\right )}^{2} + 1\right )}} + \frac {1}{24} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) - \frac {1}{24} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)^(1/2)/x^7,x, algorithm="maxima")

[Out]

1/12*((x^3 + 1)^(3/2) + sqrt(x^3 + 1))/(2*x^3 - (x^3 + 1)^2 + 1) + 1/24*log(sqrt(x^3 + 1) + 1) - 1/24*log(sqrt
(x^3 + 1) - 1)

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mupad [B]  time = 0.06, size = 189, normalized size = 4.97 \begin {gather*} -\frac {\sqrt {x^3+1}}{12\,x^3}-\frac {\sqrt {x^3+1}}{6\,x^6}+\frac {\left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\sqrt {\frac {x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\sqrt {\frac {\frac {1}{2}-x+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\,\Pi \left (\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2};\mathrm {asin}\left (\sqrt {\frac {x+1}{\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}\right )\middle |-\frac {\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}{-\frac {3}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\right )}{4\,\sqrt {x^3+\left (-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )-1\right )\,x-\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 + 1)^(1/2)/x^7,x)

[Out]

(((3^(1/2)*1i)/2 + 3/2)*((x + (3^(1/2)*1i)/2 - 1/2)/((3^(1/2)*1i)/2 - 3/2))^(1/2)*((x + 1)/((3^(1/2)*1i)/2 + 3
/2))^(1/2)*(((3^(1/2)*1i)/2 - x + 1/2)/((3^(1/2)*1i)/2 + 3/2))^(1/2)*ellipticPi((3^(1/2)*1i)/2 + 3/2, asin(((x
+ 1)/((3^(1/2)*1i)/2 + 3/2))^(1/2)), -((3^(1/2)*1i)/2 + 3/2)/((3^(1/2)*1i)/2 - 3/2)))/(4*(x^3 - x*(((3^(1/2)*
1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2) + 1) - ((3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/2 + 1/2))^(1/2)) - (x^3 + 1)^(
1/2)/(6*x^6) - (x^3 + 1)^(1/2)/(12*x^3)

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sympy [B]  time = 2.26, size = 65, normalized size = 1.71 \begin {gather*} \frac {\operatorname {asinh}{\left (\frac {1}{x^{\frac {3}{2}}} \right )}}{12} - \frac {1}{12 x^{\frac {3}{2}} \sqrt {1 + \frac {1}{x^{3}}}} - \frac {1}{4 x^{\frac {9}{2}} \sqrt {1 + \frac {1}{x^{3}}}} - \frac {1}{6 x^{\frac {15}{2}} \sqrt {1 + \frac {1}{x^{3}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+1)**(1/2)/x**7,x)

[Out]

asinh(x**(-3/2))/12 - 1/(12*x**(3/2)*sqrt(1 + x**(-3))) - 1/(4*x**(9/2)*sqrt(1 + x**(-3))) - 1/(6*x**(15/2)*sq
rt(1 + x**(-3)))

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