\(\int \frac {\sqrt {x+x^4} (-b+a x^6)}{x^6} \, dx\) [658]

Optimal result

Integrand size = 22, antiderivative size = 52 \[ \int \frac {\sqrt {x+x^4} \left (-b+a x^6\right )}{x^6} \, dx=\frac {\sqrt {x+x^4} \left (2 b+2 b x^3+3 a x^6\right )}{9 x^5}+\frac {1}{3} a \text {arctanh}\left (\frac {x^2}{\sqrt {x+x^4}}\right ) \]

[Out]

1/9*(x^4+x)^(1/2)*(3*a*x^6+2*b*x^3+2*b)/x^5+1/3*a*arctanh(x^2/(x^4+x)^(1/2))

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.

Time = 0.20 (sec) , antiderivative size = 163, normalized size of antiderivative = 3.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2073, 2077, 2039, 2045, 2036, 335, 231} \[ \int \frac {\sqrt {x+x^4} \left (-b+a x^6\right )}{x^6} \, dx=\frac {3^{3/4} a x (x+1) \sqrt {\frac {x^2-x+1}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\left (1-\sqrt {3}\right ) x+1}{\left (1+\sqrt {3}\right ) x+1}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{5 \sqrt {\frac {x (x+1)}{\left (\left (1+\sqrt {3}\right ) x+1\right )^2}} \sqrt {x^4+x}}+\frac {a \left (x^4+x\right )^{3/2}}{3 x^3}-\frac {2 a \sqrt {x^4+x}}{5 x^3}+\frac {2 b \left (x^4+x\right )^{3/2}}{9 x^6} \]

[In]

Int[(Sqrt[x + x^4]*(-b + a*x^6))/x^6,x]

[Out]

(-2*a*Sqrt[x + x^4])/(5*x^3) + (2*b*(x + x^4)^(3/2))/(9*x^6) + (a*(x + x^4)^(3/2))/(3*x^3) + (3^(3/4)*a*x*(1 +
x)*Sqrt[(1 - x + x^2)/(1 + (1 + Sqrt[3])*x)^2]*EllipticF[ArcCos[(1 + (1 - Sqrt[3])*x)/(1 + (1 + Sqrt[3])*x)],
(2 + Sqrt[3])/4])/(5*Sqrt[(x*(1 + x))/(1 + (1 + Sqrt[3])*x)^2]*Sqrt[x + x^4])

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 2039

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

Rule 2045

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

Rule 2073

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]},
With[{Pqq = Coeff[Pq, x, q]}, Int[(c*x)^m*ExpandToSum[Pq - Pqq*x^q - a*Pqq*(m + q - n + 1)*(x^(q - n)/(b*(m +
q + n*p + 1))), x]*(a*x^j + b*x^n)^p, x] + Simp[Pqq*(c*x)^(m + q - n + 1)*((a*x^j + b*x^n)^(p + 1)/(b*c^(q - n
+ 1)*(m + q + n*p + 1))), x]] /; GtQ[q, n - 1] && NeQ[m + q + n*p + 1, 0] && (IntegerQ[2*p] || IntegerQ[p + (
q + 1)/(2*n)])] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] &&  !IntegerQ[p] && IGtQ[j, 0] && IGtQ[n, 0] && L
tQ[j, n]

Rule 2077

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

Rubi steps \begin{align*} \text {integral}& = \frac {a \left (x+x^4\right )^{3/2}}{3 x^3}+\int \frac {\left (-b+a x^2\right ) \sqrt {x+x^4}}{x^6} \, dx \\ & = \frac {a \left (x+x^4\right )^{3/2}}{3 x^3}+\int \left (-\frac {b \sqrt {x+x^4}}{x^6}+\frac {a \sqrt {x+x^4}}{x^4}\right ) \, dx \\ & = \frac {a \left (x+x^4\right )^{3/2}}{3 x^3}+a \int \frac {\sqrt {x+x^4}}{x^4} \, dx-b \int \frac {\sqrt {x+x^4}}{x^6} \, dx \\ & = -\frac {2 a \sqrt {x+x^4}}{5 x^3}+\frac {2 b \left (x+x^4\right )^{3/2}}{9 x^6}+\frac {a \left (x+x^4\right )^{3/2}}{3 x^3}+\frac {1}{5} (3 a) \int \frac {1}{\sqrt {x+x^4}} \, dx \\ & = -\frac {2 a \sqrt {x+x^4}}{5 x^3}+\frac {2 b \left (x+x^4\right )^{3/2}}{9 x^6}+\frac {a \left (x+x^4\right )^{3/2}}{3 x^3}+\frac {\left (3 a \sqrt {x} \sqrt {1+x^3}\right ) \int \frac {1}{\sqrt {x} \sqrt {1+x^3}} \, dx}{5 \sqrt {x+x^4}} \\ & = -\frac {2 a \sqrt {x+x^4}}{5 x^3}+\frac {2 b \left (x+x^4\right )^{3/2}}{9 x^6}+\frac {a \left (x+x^4\right )^{3/2}}{3 x^3}+\frac {\left (6 a \sqrt {x} \sqrt {1+x^3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^6}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {x+x^4}} \\ & = -\frac {2 a \sqrt {x+x^4}}{5 x^3}+\frac {2 b \left (x+x^4\right )^{3/2}}{9 x^6}+\frac {a \left (x+x^4\right )^{3/2}}{3 x^3}+\frac {3^{3/4} a 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 )}{5 \sqrt {\frac {x (1+x)}{\left (1+\left (1+\sqrt {3}\right ) x\right )^2}} \sqrt {x+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.48 \[ \int \frac {\sqrt {x+x^4} \left (-b+a x^6\right )}{x^6} \, dx=\frac {\sqrt {x+x^4} \left (2 b+2 b x^3+3 a x^6\right )}{9 x^5}+\frac {a \sqrt {x+x^4} \log \left (x^{3/2}+\sqrt {1+x^3}\right )}{3 \sqrt {x} \sqrt {1+x^3}} \]

[In]

Integrate[(Sqrt[x + x^4]*(-b + a*x^6))/x^6,x]

[Out]

(Sqrt[x + x^4]*(2*b + 2*b*x^3 + 3*a*x^6))/(9*x^5) + (a*Sqrt[x + x^4]*Log[x^(3/2) + Sqrt[1 + x^3]])/(3*Sqrt[x]*
Sqrt[1 + x^3])

Maple [A] (verified)

Time = 3.31 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.88

method	result	size
meijerg	\(-\frac {a \left (-2 \sqrt {\pi }\, x^{\frac {3}{2}} \sqrt {x^{3}+1}-2 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {3}{2}}\right )\right )}{6 \sqrt {\pi }}+\frac {2 b \left (x^{3}+1\right )^{\frac {3}{2}}}{9 x^{\frac {9}{2}}}\)	\(46\)
trager	\(\frac {\sqrt {x^{4}+x}\, \left (3 a \,x^{6}+2 b \,x^{3}+2 b \right )}{9 x^{5}}+\frac {a \ln \left (-2 x^{3}-2 x \sqrt {x^{4}+x}-1\right )}{6}\)	\(51\)
risch	\(\frac {\left (x^{3}+1\right ) \left (3 a \,x^{6}+2 b \,x^{3}+2 b \right )}{9 x^{4} \sqrt {x \left (x^{3}+1\right )}}-\frac {a \ln \left (2 x^{3}-2 x \sqrt {x^{4}+x}+1\right )}{6}\)	\(58\)
pseudoelliptic	\(\frac {\left (6 a \,x^{6}+4 b \,x^{3}+4 b \right ) \sqrt {x^{4}+x}-3 x^{5} \left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )\right ) a}{18 x^{5}}\)	\(74\)
default	\(a \left (\frac {x \sqrt {x^{4}+x}}{3}+\frac {\ln \left (\frac {x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{6}-\frac {\ln \left (\frac {-x^{2}+\sqrt {x^{4}+x}}{x^{2}}\right )}{6}\right )-b \left (-\frac {2 \sqrt {x^{4}+x}}{9 x^{5}}-\frac {2 \sqrt {x^{4}+x}}{9 x^{2}}\right )\)	\(81\)
elliptic	\(\frac {2 b \sqrt {x^{4}+x}}{9 x^{5}}+\frac {2 b \sqrt {x^{4}+x}}{9 x^{2}}+\frac {a x \sqrt {x^{4}+x}}{3}-\frac {a \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}\, \left (1+x \right )^{2} \sqrt {-\frac {x -\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\left (\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}\, \sqrt {-\frac {x -\frac {1}{2}-\frac {i \sqrt {3}}{2}}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}\, \left (-\operatorname {EllipticF}\left (\sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}, \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )+\operatorname {EllipticPi}\left (\sqrt {\frac {\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) x}{\left (\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (1+x \right )}}, \frac {\frac {1}{2}+\frac {i \sqrt {3}}{2}}{\frac {3}{2}+\frac {i \sqrt {3}}{2}}, \sqrt {\frac {\left (-\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}{\left (-\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (-\frac {3}{2}-\frac {i \sqrt {3}}{2}\right )}}\right )\right )}{\left (\frac {3}{2}+\frac {i \sqrt {3}}{2}\right ) \sqrt {x \left (1+x \right ) \left (x -\frac {1}{2}+\frac {i \sqrt {3}}{2}\right ) \left (x -\frac {1}{2}-\frac {i \sqrt {3}}{2}\right )}}\)	\(329\)

[In]

int((x^4+x)^(1/2)*(a*x^6-b)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/6*a/Pi^(1/2)*(-2*Pi^(1/2)*x^(3/2)*(x^3+1)^(1/2)-2*Pi^(1/2)*arcsinh(x^(3/2)))+2/9*b/x^(9/2)*(x^3+1)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.06 \[ \int \frac {\sqrt {x+x^4} \left (-b+a x^6\right )}{x^6} \, dx=\frac {3 \, a x^{5} \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} + x} x - 1\right ) + 2 \, {\left (3 \, a x^{6} + 2 \, b x^{3} + 2 \, b\right )} \sqrt {x^{4} + x}}{18 \, x^{5}} \]

[In]

integrate((x^4+x)^(1/2)*(a*x^6-b)/x^6,x, algorithm="fricas")

[Out]

1/18*(3*a*x^5*log(-2*x^3 - 2*sqrt(x^4 + x)*x - 1) + 2*(3*a*x^6 + 2*b*x^3 + 2*b)*sqrt(x^4 + x))/x^5

Sympy [F]

\[ \int \frac {\sqrt {x+x^4} \left (-b+a x^6\right )}{x^6} \, dx=\int \frac {\sqrt {x \left (x + 1\right ) \left (x^{2} - x + 1\right )} \left (a x^{6} - b\right )}{x^{6}}\, dx \]

[In]

integrate((x**4+x)**(1/2)*(a*x**6-b)/x**6,x)

[Out]

Integral(sqrt(x*(x + 1)*(x**2 - x + 1))*(a*x**6 - b)/x**6, x)

Maxima [F]

\[ \int \frac {\sqrt {x+x^4} \left (-b+a x^6\right )}{x^6} \, dx=\int { \frac {{\left (a x^{6} - b\right )} \sqrt {x^{4} + x}}{x^{6}} \,d x } \]

[In]

integrate((x^4+x)^(1/2)*(a*x^6-b)/x^6,x, algorithm="maxima")

[Out]

integrate((a*x^6 - b)*sqrt(x^4 + x)/x^6, x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {x+x^4} \left (-b+a x^6\right )}{x^6} \, dx=\frac {1}{3} \, \sqrt {x^{4} + x} a x + \frac {2}{9} \, b {\left (\frac {1}{x^{3}} + 1\right )}^{\frac {3}{2}} + \frac {1}{6} \, a \log \left (\sqrt {\frac {1}{x^{3}} + 1} + 1\right ) - \frac {1}{6} \, a \log \left ({\left | \sqrt {\frac {1}{x^{3}} + 1} - 1 \right |}\right ) \]

[In]

integrate((x^4+x)^(1/2)*(a*x^6-b)/x^6,x, algorithm="giac")

[Out]

1/3*sqrt(x^4 + x)*a*x + 2/9*b*(1/x^3 + 1)^(3/2) + 1/6*a*log(sqrt(1/x^3 + 1) + 1) - 1/6*a*log(abs(sqrt(1/x^3 +
1) - 1))

Mupad [B] (verification not implemented)

Time = 6.33 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {x+x^4} \left (-b+a x^6\right )}{x^6} \, dx=\frac {2\,b\,\left (x^3+1\right )\,\sqrt {x^4+x}}{9\,x^5}+\frac {2\,a\,x\,\sqrt {x^4+x}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{2},\frac {1}{2};\ \frac {3}{2};\ -x^3\right )}{3\,\sqrt {x^3+1}} \]

[In]

int(-((b - a*x^6)*(x + x^4)^(1/2))/x^6,x)

[Out]

(2*b*(x^3 + 1)*(x + x^4)^(1/2))/(9*x^5) + (2*a*x*(x + x^4)^(1/2)*hypergeom([-1/2, 1/2], 3/2, -x^3))/(3*(x^3 +
1)^(1/2))