\(\int \frac {(1+4 x^3) (1+2 x+2 x^4)}{\sqrt {x+x^4}} \, dx\) [231]

Optimal result

Integrand size = 27, antiderivative size = 23 \[ \int \frac {\left (1+4 x^3\right ) \left (1+2 x+2 x^4\right )}{\sqrt {x+x^4}} \, dx=\frac {2}{3} \sqrt {x+x^4} \left (3+2 x+2 x^4\right ) \]

[Out]

2/3*(x^4+x)^(1/2)*(2*x^4+2*x+3)

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83, number of steps used = 18, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2078, 2036, 335, 231, 2054, 212, 2049} \[ \int \frac {\left (1+4 x^3\right ) \left (1+2 x+2 x^4\right )}{\sqrt {x+x^4}} \, dx=\frac {4}{3} \sqrt {x^4+x} x^4+\frac {4}{3} \sqrt {x^4+x} x+2 \sqrt {x^4+x} \]

[In]

Int[((1 + 4*x^3)*(1 + 2*x + 2*x^4))/Sqrt[x + x^4],x]

[Out]

2*Sqrt[x + x^4] + (4*x*Sqrt[x + x^4])/3 + (4*x^4*Sqrt[x + x^4])/3

Rule 212

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

Rule 231

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[x*(s +
r*x^2)*(Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[r*x^2*(
(s + r*x^2)/(s + (1 + Sqrt[3])*r*x^2)^2)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rule 335

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

Rule 2036

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2049

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

Rule 2054

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rule 2078

Int[(Pq_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[Pq*(a*x^j + b*x^n)^p, x]
, x] /; FreeQ[{a, b, j, n, p}, x] && (PolyQ[Pq, x] || PolyQ[Pq, x^n]) &&  !IntegerQ[p] && NeQ[n, j]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt {x+x^4}}+\frac {2 x}{\sqrt {x+x^4}}+\frac {4 x^3}{\sqrt {x+x^4}}+\frac {10 x^4}{\sqrt {x+x^4}}+\frac {8 x^7}{\sqrt {x+x^4}}\right ) \, dx \\ & = 2 \int \frac {x}{\sqrt {x+x^4}} \, dx+4 \int \frac {x^3}{\sqrt {x+x^4}} \, dx+8 \int \frac {x^7}{\sqrt {x+x^4}} \, dx+10 \int \frac {x^4}{\sqrt {x+x^4}} \, dx+\int \frac {1}{\sqrt {x+x^4}} \, dx \\ & = 2 \sqrt {x+x^4}+\frac {10}{3} x \sqrt {x+x^4}+\frac {4}{3} x^4 \sqrt {x+x^4}+\frac {4}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^2}{\sqrt {x+x^4}}\right )-5 \int \frac {x}{\sqrt {x+x^4}} \, dx-6 \int \frac {x^4}{\sqrt {x+x^4}} \, dx+\frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}}-\int \frac {1}{\sqrt {x+x^4}} \, dx \\ & = 2 \sqrt {x+x^4}+\frac {4}{3} x \sqrt {x+x^4}+\frac {4}{3} x^4 \sqrt {x+x^4}+\frac {4}{3} \text {arctanh}\left (\frac {x^2}{\sqrt {x+x^4}}\right )+3 \int \frac {x}{\sqrt {x+x^4}} \, dx-\frac {10}{3} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^2}{\sqrt {x+x^4}}\right )-\frac {\left (\sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+x^3}} \, dx}{\sqrt {x+x^4}}+\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}} \\ & = 2 \sqrt {x+x^4}+\frac {4}{3} x \sqrt {x+x^4}+\frac {4}{3} x^4 \sqrt {x+x^4}-2 \text {arctanh}\left (\frac {x^2}{\sqrt {x+x^4}}\right )+\frac {x (1+x) \sqrt {\frac {1-x+x^2}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x}{1+\left (1+\sqrt {3}\right ) x}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{\sqrt [4]{3} \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}}+2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x^2}{\sqrt {x+x^4}}\right )-\frac {\left (2 \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{\sqrt {x+x^4}} \\ & = 2 \sqrt {x+x^4}+\frac {4}{3} x \sqrt {x+x^4}+\frac {4}{3} x^4 \sqrt {x+x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 10.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+4 x^3\right ) \left (1+2 x+2 x^4\right )}{\sqrt {x+x^4}} \, dx=\frac {2}{3} \sqrt {x+x^4} \left (3+2 x+2 x^4\right ) \]

[In]

Integrate[((1 + 4*x^3)*(1 + 2*x + 2*x^4))/Sqrt[x + x^4],x]

[Out]

(2*Sqrt[x + x^4]*(3 + 2*x + 2*x^4))/3

Maple [A] (verified)

Time = 3.37 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83

method	result	size
trager	\(\left (\frac {4}{3} x^{4}+\frac {4}{3} x +2\right ) \sqrt {x^{4}+x}\)	\(19\)
pseudoelliptic	\(\frac {2 \sqrt {x^{4}+x}\, \left (2 x^{4}+2 x +3\right )}{3}\)	\(20\)
risch	\(\frac {2 \left (2 x^{4}+2 x +3\right ) x \left (x^{3}+1\right )}{3 \sqrt {x \left (x^{3}+1\right )}}\)	\(28\)
gosper	\(\frac {2 x \left (1+x \right ) \left (x^{2}-x +1\right ) \left (2 x^{4}+2 x +3\right )}{3 \sqrt {x^{4}+x}}\)	\(32\)
elliptic	\(\frac {4 x^{4} \sqrt {x^{4}+x}}{3}+\frac {4 x \sqrt {x^{4}+x}}{3}+2 \sqrt {x^{4}+x}\)	\(33\)
meijerg	\(\frac {-\frac {2 \sqrt {\pi }\, x^{\frac {3}{2}} \left (-10 x^{3}+15\right ) \sqrt {x^{3}+1}}{15}+2 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {3}{2}}\right )}{\sqrt {\pi }}+\frac {\frac {10 \sqrt {\pi }\, x^{\frac {3}{2}} \sqrt {x^{3}+1}}{3}-\frac {10 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {3}{2}}\right )}{3}}{\sqrt {\pi }}+\frac {8 x^{\frac {7}{2}} \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {7}{6}\right ], \left [\frac {13}{6}\right ], -x^{3}\right )}{7}+\frac {4 \,\operatorname {arcsinh}\left (x^{\frac {3}{2}}\right )}{3}+2 \sqrt {x}\, \operatorname {hypergeom}\left (\left [\frac {1}{6}, \frac {1}{2}\right ], \left [\frac {7}{6}\right ], -x^{3}\right )\)	\(106\)
default	\(-\frac {2 \ln \left (2 x^{3}-2 x \sqrt {x^{4}+x}+1\right )}{3}-\frac {x^{2} \left (\left (-4 x^{4}+6 x \right ) \sqrt {x^{4}+x}+3 \ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )-3 \ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )\right )}{3 \left (x^{2}+\sqrt {x^{4}+x}\right )^{2} \left (x^{2}-\sqrt {x^{4}+x}\right )^{2}}+2 \sqrt {x^{4}+x}+\frac {10 x \sqrt {x^{4}+x}}{3}-\frac {5 \ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{3}+\frac {5 \ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{3}\)	\(168\)

[In]

int((4*x^3+1)*(2*x^4+2*x+1)/(x^4+x)^(1/2),x,method=_RETURNVERBOSE)

[Out]

(4/3*x^4+4/3*x+2)*(x^4+x)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (1+4 x^3\right ) \left (1+2 x+2 x^4\right )}{\sqrt {x+x^4}} \, dx=\frac {2}{3} \, {\left (2 \, x^{4} + 2 \, x + 3\right )} \sqrt {x^{4} + x} \]

[In]

integrate((4*x^3+1)*(2*x^4+2*x+1)/(x^4+x)^(1/2),x, algorithm="fricas")

[Out]

2/3*(2*x^4 + 2*x + 3)*sqrt(x^4 + x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.61 \[ \int \frac {\left (1+4 x^3\right ) \left (1+2 x+2 x^4\right )}{\sqrt {x+x^4}} \, dx=\frac {4 x^{4} \sqrt {x^{4} + x}}{3} + \frac {4 x \sqrt {x^{4} + x}}{3} + 2 \sqrt {x^{4} + x} \]

[In]

integrate((4*x**3+1)*(2*x**4+2*x+1)/(x**4+x)**(1/2),x)

[Out]

4*x**4*sqrt(x**4 + x)/3 + 4*x*sqrt(x**4 + x)/3 + 2*sqrt(x**4 + x)

Maxima [F]

\[ \int \frac {\left (1+4 x^3\right ) \left (1+2 x+2 x^4\right )}{\sqrt {x+x^4}} \, dx=\int { \frac {{\left (2 \, x^{4} + 2 \, x + 1\right )} {\left (4 \, x^{3} + 1\right )}}{\sqrt {x^{4} + x}} \,d x } \]

[In]

integrate((4*x^3+1)*(2*x^4+2*x+1)/(x^4+x)^(1/2),x, algorithm="maxima")

[Out]

integrate((2*x^4 + 2*x + 1)*(4*x^3 + 1)/sqrt(x^4 + x), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (1+4 x^3\right ) \left (1+2 x+2 x^4\right )}{\sqrt {x+x^4}} \, dx=\frac {4}{3} \, {\left (x^{4} + x\right )}^{\frac {3}{2}} + 2 \, \sqrt {x^{4} + x} \]

[In]

integrate((4*x^3+1)*(2*x^4+2*x+1)/(x^4+x)^(1/2),x, algorithm="giac")

[Out]

4/3*(x^4 + x)^(3/2) + 2*sqrt(x^4 + x)

Mupad [B] (verification not implemented)

Time = 5.65 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {\left (1+4 x^3\right ) \left (1+2 x+2 x^4\right )}{\sqrt {x+x^4}} \, dx=\frac {2\,\sqrt {x^4+x}\,\left (2\,x^4+2\,x+3\right )}{3} \]

[In]

int(((4*x^3 + 1)*(2*x + 2*x^4 + 1))/(x + x^4)^(1/2),x)

[Out]

(2*(x + x^4)^(1/2)*(2*x + 2*x^4 + 3))/3