Integrand size = 35, antiderivative size = 208 \[ \int \frac {b+a x^6}{x^3 \left (-b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\frac {4 \left (b x+a x^4\right )^{3/4}}{9 b x^3}+\frac {2 \arctan \left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}}-\frac {2^{3/4} (a+b) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a} b}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a}}-\frac {2^{3/4} (a+b) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (b x+a x^4\right )^{3/4}}{b+a x^3}\right )}{3 \sqrt [4]{a} b} \] Output:
4/9*(a*x^4+b*x)^(3/4)/b/x^3+2/3*arctan(a^(1/4)*(a*x^4+b*x)^(3/4)/(a*x^3+b) )/a^(1/4)-1/3*2^(3/4)*(a+b)*arctan(2^(1/4)*a^(1/4)*(a*x^4+b*x)^(3/4)/(a*x^ 3+b))/a^(1/4)/b+2/3*arctanh(a^(1/4)*(a*x^4+b*x)^(3/4)/(a*x^3+b))/a^(1/4)-1 /3*2^(3/4)*(a+b)*arctanh(2^(1/4)*a^(1/4)*(a*x^4+b*x)^(3/4)/(a*x^3+b))/a^(1 /4)/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 12.30 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.90 \[ \int \frac {b+a x^6}{x^3 \left (-b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\frac {4 \left (-\frac {5 \left (b+a x^3\right )}{a}+15 x^3 \sqrt [4]{1+\frac {a x^3}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{4},\frac {5}{4},-\frac {a x^3}{b}\right )+\frac {4 (a+b) \left (b+a x^3\right ) \operatorname {Gamma}\left (\frac {5}{4}\right ) \left (5 \left (b^2+3 a b x^3-4 a^2 x^6\right ) \operatorname {Hypergeometric2F1}\left (1,1,\frac {5}{4},-\frac {2 a x^3}{b-a x^3}\right )-32 a x^3 \left (b+a x^3\right ) \operatorname {Hypergeometric2F1}\left (2,2,\frac {9}{4},-\frac {2 a x^3}{b-a x^3}\right )\right )}{a b \left (b-a x^3\right )^2 \operatorname {Gamma}\left (\frac {1}{4}\right )}\right )}{45 x^2 \sqrt [4]{x \left (b+a x^3\right )}} \] Input:
Integrate[(b + a*x^6)/(x^3*(-b + a*x^3)*(b*x + a*x^4)^(1/4)),x]
Output:
(4*((-5*(b + a*x^3))/a + 15*x^3*(1 + (a*x^3)/b)^(1/4)*Hypergeometric2F1[1/ 4, 1/4, 5/4, -((a*x^3)/b)] + (4*(a + b)*(b + a*x^3)*Gamma[5/4]*(5*(b^2 + 3 *a*b*x^3 - 4*a^2*x^6)*Hypergeometric2F1[1, 1, 5/4, (-2*a*x^3)/(b - a*x^3)] - 32*a*x^3*(b + a*x^3)*Hypergeometric2F1[2, 2, 9/4, (-2*a*x^3)/(b - a*x^3 )]))/(a*b*(b - a*x^3)^2*Gamma[1/4])))/(45*x^2*(x*(b + a*x^3))^(1/4))
Time = 1.31 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2467, 25, 2035, 7276, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {a x^6+b}{x^3 \left (a x^3-b\right ) \sqrt [4]{a x^4+b x}} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^3+b} \int -\frac {a x^6+b}{x^{13/4} \left (b-a x^3\right ) \sqrt [4]{a x^3+b}}dx}{\sqrt [4]{a x^4+b x}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{a x^3+b} \int \frac {a x^6+b}{x^{13/4} \left (b-a x^3\right ) \sqrt [4]{a x^3+b}}dx}{\sqrt [4]{a x^4+b x}}\) |
| \(\Big \downarrow \) 2035 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \int \frac {a x^6+b}{x^{5/2} \left (b-a x^3\right ) \sqrt [4]{a x^3+b}}d\sqrt [4]{x}}{\sqrt [4]{a x^4+b x}}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \int \left (-\frac {\sqrt {x} (a+b)}{\left (a x^3-b\right ) \sqrt [4]{a x^3+b}}-\frac {\sqrt {x}}{\sqrt [4]{a x^3+b}}+\frac {1}{x^{5/2} \sqrt [4]{a x^3+b}}\right )d\sqrt [4]{x}}{\sqrt [4]{a x^4+b x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^3+b} \left (-\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{6 \sqrt [4]{a}}+\frac {(a+b) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{6 \sqrt [4]{2} \sqrt [4]{a} b}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{6 \sqrt [4]{a}}+\frac {(a+b) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3+b}}\right )}{6 \sqrt [4]{2} \sqrt [4]{a} b}-\frac {\left (a x^3+b\right )^{3/4}}{9 b x^{9/4}}\right )}{\sqrt [4]{a x^4+b x}}\) |
Input:
Int[(b + a*x^6)/(x^3*(-b + a*x^3)*(b*x + a*x^4)^(1/4)),x]
Output:
(-4*x^(1/4)*(b + a*x^3)^(1/4)*(-1/9*(b + a*x^3)^(3/4)/(b*x^(9/4)) - ArcTan [(a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)]/(6*a^(1/4)) + ((a + b)*ArcTan[(2^(1/ 4)*a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(6*2^(1/4)*a^(1/4)*b) - ArcTanh[(a ^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)]/(6*a^(1/4)) + ((a + b)*ArcTanh[(2^(1/4) *a^(1/4)*x^(3/4))/(b + a*x^3)^(1/4)])/(6*2^(1/4)*a^(1/4)*b)))/(b*x + a*x^4 )^(1/4)
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 1.72 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.90
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {\ln \left (\frac {x 2^{\frac {1}{4}} a^{\frac {1}{4}}+{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}{-x 2^{\frac {1}{4}} a^{\frac {1}{4}}+{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}\right ) \left (a +b \right ) 2^{\frac {3}{4}} x^{3}}{2}+\ln \left (\frac {x \,a^{\frac {1}{4}}+{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}{-x \,a^{\frac {1}{4}}+{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}\right ) b \,x^{3}+\arctan \left (\frac {{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x \,a^{\frac {1}{4}}}\right ) \left (a +b \right ) 2^{\frac {3}{4}} x^{3}-2 \arctan \left (\frac {{\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}}\right ) b \,x^{3}+\frac {4 {\left (x \left (a \,x^{3}+b \right )\right )}^{\frac {3}{4}} a^{\frac {1}{4}}}{3}}{3 a^{\frac {1}{4}} x^{3} b}\) | \(187\) |
Input:
int((a*x^6+b)/x^3/(a*x^3-b)/(a*x^4+b*x)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/3/a^(1/4)*(-1/2*ln((x*2^(1/4)*a^(1/4)+(x*(a*x^3+b))^(1/4))/(-x*2^(1/4)*a ^(1/4)+(x*(a*x^3+b))^(1/4)))*(a+b)*2^(3/4)*x^3+ln((x*a^(1/4)+(x*(a*x^3+b)) ^(1/4))/(-x*a^(1/4)+(x*(a*x^3+b))^(1/4)))*b*x^3+arctan(1/2*(x*(a*x^3+b))^( 1/4)/x*2^(3/4)/a^(1/4))*(a+b)*2^(3/4)*x^3-2*arctan((x*(a*x^3+b))^(1/4)/x/a ^(1/4))*b*x^3+4/3*(x*(a*x^3+b))^(3/4)*a^(1/4))/x^3/b
Timed out. \[ \int \frac {b+a x^6}{x^3 \left (-b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\text {Timed out} \] Input:
integrate((a*x^6+b)/x^3/(a*x^3-b)/(a*x^4+b*x)^(1/4),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {b+a x^6}{x^3 \left (-b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\int \frac {a x^{6} + b}{x^{3} \sqrt [4]{x \left (a x^{3} + b\right )} \left (a x^{3} - b\right )}\, dx \] Input:
integrate((a*x**6+b)/x**3/(a*x**3-b)/(a*x**4+b*x)**(1/4),x)
Output:
Integral((a*x**6 + b)/(x**3*(x*(a*x**3 + b))**(1/4)*(a*x**3 - b)), x)
\[ \int \frac {b+a x^6}{x^3 \left (-b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\int { \frac {a x^{6} + b}{{\left (a x^{4} + b x\right )}^{\frac {1}{4}} {\left (a x^{3} - b\right )} x^{3}} \,d x } \] Input:
integrate((a*x^6+b)/x^3/(a*x^3-b)/(a*x^4+b*x)^(1/4),x, algorithm="maxima")
Output:
integrate((a*x^6 + b)/((a*x^4 + b*x)^(1/4)*(a*x^3 - b)*x^3), x)
Leaf count of result is larger than twice the leaf count of optimal. 437 vs. \(2 (164) = 328\).
Time = 0.91 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.10 \[ \int \frac {b+a x^6}{x^3 \left (-b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx =\text {Too large to display} \] Input:
integrate((a*x^6+b)/x^3/(a*x^3-b)/(a*x^4+b*x)^(1/4),x, algorithm="giac")
Output:
1/3*sqrt(2)*(-a)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a + b/x ^3)^(1/4))/(-a)^(1/4))/a + 1/3*sqrt(2)*(-a)^(3/4)*arctan(-1/2*sqrt(2)*(sqr t(2)*(-a)^(1/4) - 2*(a + b/x^3)^(1/4))/(-a)^(1/4))/a - 1/6*sqrt(2)*(-a)^(3 /4)*log(sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(-a) + sqrt(a + b/x^3)) /a + 1/6*sqrt(2)*(-a)^(3/4)*log(-sqrt(2)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sq rt(-a) + sqrt(a + b/x^3))/a - 1/6*2^(1/4)*(a + b)*log(2^(3/4)*(-a)^(1/4)*( a + b/x^3)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a + b/x^3))/((-a)^(1/4)*b) + 1/ 6*2^(1/4)*(a + b)*log(-2^(3/4)*(-a)^(1/4)*(a + b/x^3)^(1/4) + sqrt(2)*sqrt (-a) + sqrt(a + b/x^3))/((-a)^(1/4)*b) - 1/6*sqrt(2)*(2^(3/4)*(-a)^(3/4)*a + 2^(3/4)*(-a)^(3/4)*b)*arctan(1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) + 2*(a + b /x^3)^(1/4))/(-a)^(1/4))/(a*b) - 1/6*sqrt(2)*(2^(3/4)*(-a)^(3/4)*a + 2^(3/ 4)*(-a)^(3/4)*b)*arctan(-1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) - 2*(a + b/x^3)^( 1/4))/(-a)^(1/4))/(a*b) + 4/9*(a + b/x^3)^(3/4)/b
Timed out. \[ \int \frac {b+a x^6}{x^3 \left (-b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=-\int \frac {a\,x^6+b}{x^3\,{\left (a\,x^4+b\,x\right )}^{1/4}\,\left (b-a\,x^3\right )} \,d x \] Input:
int(-(b + a*x^6)/(x^3*(b*x + a*x^4)^(1/4)*(b - a*x^3)),x)
Output:
-int((b + a*x^6)/(x^3*(b*x + a*x^4)^(1/4)*(b - a*x^3)), x)
\[ \int \frac {b+a x^6}{x^3 \left (-b+a x^3\right ) \sqrt [4]{b x+a x^4}} \, dx=\left (\int \frac {x^{3}}{x^{\frac {13}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}} a -x^{\frac {1}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}} b}d x \right ) a +\left (\int \frac {1}{x^{\frac {25}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}} a -x^{\frac {13}{4}} \left (a \,x^{3}+b \right )^{\frac {1}{4}} b}d x \right ) b \] Input:
int((a*x^6+b)/x^3/(a*x^3-b)/(a*x^4+b*x)^(1/4),x)
Output:
int(x**3/(x**(1/4)*(a*x**3 + b)**(1/4)*a*x**3 - x**(1/4)*(a*x**3 + b)**(1/ 4)*b),x)*a + int(1/(x**(1/4)*(a*x**3 + b)**(1/4)*a*x**6 - x**(1/4)*(a*x**3 + b)**(1/4)*b*x**3),x)*b