3.6.24 \(\int \frac {(2+x^3) (1+x^3+x^6)}{x \sqrt {1+x^3}} \, dx\)

Optimal. Leaf size=40 \[ \frac {2}{45} \sqrt {x^3+1} \left (3 x^6+11 x^3+23\right )-\frac {4}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 54, normalized size of antiderivative = 1.35, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1821, 1612, 63, 207} \begin {gather*} \frac {2}{15} \left (x^3+1\right )^{5/2}+\frac {2}{9} \left (x^3+1\right )^{3/2}+\frac {2 \sqrt {x^3+1}}{3}-\frac {4}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 + x^3)*(1 + x^3 + x^6))/(x*Sqrt[1 + x^3]),x]

[Out]

(2*Sqrt[1 + x^3])/3 + (2*(1 + x^3)^(3/2))/9 + (2*(1 + x^3)^(5/2))/15 - (4*ArcTanh[Sqrt[1 + x^3]])/3

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 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

Rule 1821

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] -
1)*SubstFor[x^n, Pq, x]*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && PolyQ[Pq, x^n] && Intege
rQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (2+x^3\right ) \left (1+x^3+x^6\right )}{x \sqrt {1+x^3}} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {(2+x) \left (1+x+x^2\right )}{x \sqrt {1+x}} \, dx,x,x^3\right )\\ &=\frac {1}{3} \operatorname {Subst}\left (\int \left (\frac {1}{\sqrt {1+x}}+\frac {2}{x \sqrt {1+x}}+\sqrt {1+x}+(1+x)^{3/2}\right ) \, dx,x,x^3\right )\\ &=\frac {2 \sqrt {1+x^3}}{3}+\frac {2}{9} \left (1+x^3\right )^{3/2}+\frac {2}{15} \left (1+x^3\right )^{5/2}+\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,x^3\right )\\ &=\frac {2 \sqrt {1+x^3}}{3}+\frac {2}{9} \left (1+x^3\right )^{3/2}+\frac {2}{15} \left (1+x^3\right )^{5/2}+\frac {4}{3} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+x^3}\right )\\ &=\frac {2 \sqrt {1+x^3}}{3}+\frac {2}{9} \left (1+x^3\right )^{3/2}+\frac {2}{15} \left (1+x^3\right )^{5/2}-\frac {4}{3} \tanh ^{-1}\left (\sqrt {1+x^3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 40, normalized size = 1.00 \begin {gather*} \frac {2}{45} \sqrt {x^3+1} \left (3 x^6+11 x^3+23\right )-\frac {4}{3} \tanh ^{-1}\left (\sqrt {x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + x^3)*(1 + x^3 + x^6))/(x*Sqrt[1 + x^3]),x]

[Out]

(2*Sqrt[1 + x^3]*(23 + 11*x^3 + 3*x^6))/45 - (4*ArcTanh[Sqrt[1 + x^3]])/3

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IntegrateAlgebraic [A]  time = 0.04, size = 40, normalized size = 1.00 \begin {gather*} \frac {2}{45} \sqrt {1+x^3} \left (23+11 x^3+3 x^6\right )-\frac {4}{3} \tanh ^{-1}\left (\sqrt {1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((2 + x^3)*(1 + x^3 + x^6))/(x*Sqrt[1 + x^3]),x]

[Out]

(2*Sqrt[1 + x^3]*(23 + 11*x^3 + 3*x^6))/45 - (4*ArcTanh[Sqrt[1 + x^3]])/3

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fricas [A]  time = 0.45, size = 46, normalized size = 1.15 \begin {gather*} \frac {2}{45} \, {\left (3 \, x^{6} + 11 \, x^{3} + 23\right )} \sqrt {x^{3} + 1} - \frac {2}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {2}{3} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+2)*(x^6+x^3+1)/x/(x^3+1)^(1/2),x, algorithm="fricas")

[Out]

2/45*(3*x^6 + 11*x^3 + 23)*sqrt(x^3 + 1) - 2/3*log(sqrt(x^3 + 1) + 1) + 2/3*log(sqrt(x^3 + 1) - 1)

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giac [A]  time = 0.34, size = 53, normalized size = 1.32 \begin {gather*} \frac {2}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{2}} + \frac {2}{9} \, {\left (x^{3} + 1\right )}^{\frac {3}{2}} + \frac {2}{3} \, \sqrt {x^{3} + 1} - \frac {2}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {2}{3} \, \log \left ({\left | \sqrt {x^{3} + 1} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+2)*(x^6+x^3+1)/x/(x^3+1)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.25, size = 45, normalized size = 1.12

method	result	size
default	\(\frac {2 \sqrt {x^{3}+1}\, x^{6}}{15}+\frac {22 \sqrt {x^{3}+1}\, x^{3}}{45}+\frac {46 \sqrt {x^{3}+1}}{45}-\frac {4 \arctanh \left (\sqrt {x^{3}+1}\right )}{3}\)	\(45\)
elliptic	\(\frac {2 \sqrt {x^{3}+1}\, x^{6}}{15}+\frac {22 \sqrt {x^{3}+1}\, x^{3}}{45}+\frac {46 \sqrt {x^{3}+1}}{45}-\frac {4 \arctanh \left (\sqrt {x^{3}+1}\right )}{3}\)	\(45\)
trager	\(\left (\frac {2}{15} x^{6}+\frac {22}{45} x^{3}+\frac {46}{45}\right ) \sqrt {x^{3}+1}+\frac {2 \ln \left (-\frac {-x^{3}+2 \sqrt {x^{3}+1}-2}{x^{3}}\right )}{3}\)	\(46\)
meijerg	\(\frac {-\frac {16 \sqrt {\pi }}{15}+\frac {\sqrt {\pi }\, \left (6 x^{6}-8 x^{3}+16\right ) \sqrt {x^{3}+1}}{15}}{3 \sqrt {\pi }}+\frac {\frac {4 \sqrt {\pi }}{3}-\frac {\sqrt {\pi }\, \left (-4 x^{3}+8\right ) \sqrt {x^{3}+1}}{6}}{\sqrt {\pi }}+\frac {-2 \sqrt {\pi }+2 \sqrt {\pi }\, \sqrt {x^{3}+1}}{\sqrt {\pi }}+\frac {\frac {2 \left (-2 \ln \relax (2)+3 \ln \relax (x )\right ) \sqrt {\pi }}{3}-\frac {4 \ln \left (\frac {1}{2}+\frac {\sqrt {x^{3}+1}}{2}\right ) \sqrt {\pi }}{3}}{\sqrt {\pi }}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+2)*(x^6+x^3+1)/x/(x^3+1)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/15*(x^3+1)^(1/2)*x^6+22/45*(x^3+1)^(1/2)*x^3+46/45*(x^3+1)^(1/2)-4/3*arctanh((x^3+1)^(1/2))

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maxima [A]  time = 0.46, size = 52, normalized size = 1.30 \begin {gather*} \frac {2}{15} \, {\left (x^{3} + 1\right )}^{\frac {5}{2}} + \frac {2}{9} \, {\left (x^{3} + 1\right )}^{\frac {3}{2}} + \frac {2}{3} \, \sqrt {x^{3} + 1} - \frac {2}{3} \, \log \left (\sqrt {x^{3} + 1} + 1\right ) + \frac {2}{3} \, \log \left (\sqrt {x^{3} + 1} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+2)*(x^6+x^3+1)/x/(x^3+1)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.20, size = 198, normalized size = 4.95 \begin {gather*} \frac {46\,\sqrt {x^3+1}}{45}+\frac {22\,x^3\,\sqrt {x^3+1}}{45}+\frac {2\,x^6\,\sqrt {x^3+1}}{15}-\frac {4\,\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 )}{\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 + 2)*(x^3 + x^6 + 1))/(x*(x^3 + 1)^(1/2)),x)

[Out]

(46*(x^3 + 1)^(1/2))/45 + (22*x^3*(x^3 + 1)^(1/2))/45 + (2*x^6*(x^3 + 1)^(1/2))/15 - (4*((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)*1
i)/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)))/(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)

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sympy [A]  time = 27.31, size = 68, normalized size = 1.70 \begin {gather*} \frac {2 \left (x^{3} + 1\right )^{\frac {5}{2}}}{15} + \frac {2 \left (x^{3} + 1\right )^{\frac {3}{2}}}{9} + \frac {2 \sqrt {x^{3} + 1}}{3} + \frac {2 \log {\left (-1 + \frac {1}{\sqrt {x^{3} + 1}} \right )}}{3} - \frac {2 \log {\left (1 + \frac {1}{\sqrt {x^{3} + 1}} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+2)*(x**6+x**3+1)/x/(x**3+1)**(1/2),x)

[Out]

2*(x**3 + 1)**(5/2)/15 + 2*(x**3 + 1)**(3/2)/9 + 2*sqrt(x**3 + 1)/3 + 2*log(-1 + 1/sqrt(x**3 + 1))/3 - 2*log(1
+ 1/sqrt(x**3 + 1))/3

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