Integrand size = 36, antiderivative size = 143 \[ \int \frac {-b+a x^3}{x^6 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\frac {4 \left (3 b-10 a x^3\right ) \left (-b x+a x^4\right )^{3/4}}{63 b^2 x^6}-\frac {2\ 2^{3/4} a^{7/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 b^2}-\frac {2\ 2^{3/4} a^{7/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )}{3 b^2} \] Output:
-4/63*(-10*a*x^3+3*b)*(a*x^4-b*x)^(3/4)/b^2/x^6-2/3*2^(3/4)*a^(7/4)*arctan (2^(1/4)*a^(1/4)*(a*x^4-b*x)^(3/4)/(a*x^3-b))/b^2-2/3*2^(3/4)*a^(7/4)*arct anh(2^(1/4)*a^(1/4)*(a*x^4-b*x)^(3/4)/(a*x^3-b))/b^2
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 5.82 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.92 \[ \int \frac {-b+a x^3}{x^6 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\frac {4 \left (-b x+a x^4\right )^{3/4} \left (3 b^2-7 a b x^3+4 a^2 x^6-6 a x^3 \left (-3 b+4 a x^3\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},1,\frac {5}{4},-\frac {2 a x^3}{b-a x^3}\right )-24 a x^3 \left (b+a x^3\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},2,\frac {5}{4},-\frac {2 a x^3}{b-a x^3}\right )\right )}{63 b^2 x^6 \left (b-a x^3\right )} \] Input:
Integrate[(-b + a*x^3)/(x^6*(b + a*x^3)*(-(b*x) + a*x^4)^(1/4)),x]
Output:
(-4*(-(b*x) + a*x^4)^(3/4)*(3*b^2 - 7*a*b*x^3 + 4*a^2*x^6 - 6*a*x^3*(-3*b + 4*a*x^3)*Hypergeometric2F1[1/4, 1, 5/4, (-2*a*x^3)/(b - a*x^3)] - 24*a*x ^3*(b + a*x^3)*Hypergeometric2F1[1/4, 2, 5/4, (-2*a*x^3)/(b - a*x^3)]))/(6 3*b^2*x^6*(b - a*x^3))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.65 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {2467, 966, 965, 1013, 1012}
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.
\(\displaystyle \int \frac {a x^3-b}{x^6 \left (a x^3+b\right ) \sqrt [4]{a x^4-b x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^3-b} \int \frac {\left (a x^3-b\right )^{3/4}}{x^{25/4} \left (a x^3+b\right )}dx}{\sqrt [4]{a x^4-b x}}\) |
\(\Big \downarrow \) 966 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3-b} \int \frac {\left (a x^3-b\right )^{3/4}}{x^{11/2} \left (a x^3+b\right )}d\sqrt [4]{x}}{\sqrt [4]{a x^4-b x}}\) |
\(\Big \downarrow \) 965 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^3-b} \int \frac {(a x-b)^{3/4}}{x^2 (b+a x)}dx^{3/4}}{3 \sqrt [4]{a x^4-b x}}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {4 \sqrt [4]{x} (a x-b)^{3/4} \sqrt [4]{a x^3-b} \int \frac {\left (1-\frac {a x}{b}\right )^{3/4}}{x^2 (b+a x)}dx^{3/4}}{3 \left (1-\frac {a x}{b}\right )^{3/4} \sqrt [4]{a x^4-b x}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {\operatorname {Gamma}\left (-\frac {3}{4}\right ) (a x-b)^{3/4} \left (1-\frac {a x}{b}\right ) \sqrt [4]{a x^3-b} \left (\left (4 a^2 x^2+7 a b x+3 b^2\right ) \operatorname {Hypergeometric2F1}\left (1,1,\frac {1}{4},\frac {2 a x}{b+a x}\right )-32 a x (b-a x) \operatorname {Hypergeometric2F1}\left (2,2,\frac {5}{4},\frac {2 a x}{b+a x}\right )\right )}{21 b x^{3/2} \operatorname {Gamma}\left (\frac {1}{4}\right ) (a x+b)^2 \sqrt [4]{a x^4-b x}}\) |
Input:
Int[(-b + a*x^3)/(x^6*(b + a*x^3)*(-(b*x) + a*x^4)^(1/4)),x]
Output:
((-b + a*x)^(3/4)*(1 - (a*x)/b)*(-b + a*x^3)^(1/4)*Gamma[-3/4]*((3*b^2 + 7 *a*b*x + 4*a^2*x^2)*Hypergeometric2F1[1, 1, 1/4, (2*a*x)/(b + a*x)] - 32*a *x*(b - a*x)*Hypergeometric2F1[2, 2, 5/4, (2*a*x)/(b + a*x)]))/(21*b*x^(3/ 2)*(b + a*x)^2*(-(b*x) + a*x^4)^(1/4)*Gamma[1/4])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) )^(q_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*( m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*x)^( 1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[ n, 0] && FractionQ[m] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
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]
Time = 0.62 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.09
method | result | size |
pseudoelliptic | \(\frac {\left (42 \arctan \left (\frac {{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}} 2^{\frac {3}{4}}}{2 x \,a^{\frac {1}{4}}}\right ) a^{2} x^{6}-21 \ln \left (\frac {-2^{\frac {1}{4}} a^{\frac {1}{4}} x -{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{2^{\frac {1}{4}} a^{\frac {1}{4}} x -{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}\right ) a^{2} x^{6}+20 {\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {3}{4}} a^{\frac {5}{4}} x^{3} 2^{\frac {1}{4}}-6 b 2^{\frac {1}{4}} a^{\frac {1}{4}} {\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {3}{4}}\right ) 2^{\frac {3}{4}}}{63 x^{6} a^{\frac {1}{4}} b^{2}}\) | \(156\) |
Input:
int((a*x^3-b)/x^6/(a*x^3+b)/(a*x^4-b*x)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/63/x^6*(42*arctan(1/2*(x*(a*x^3-b))^(1/4)/x*2^(3/4)/a^(1/4))*a^2*x^6-21* ln((-2^(1/4)*a^(1/4)*x-(x*(a*x^3-b))^(1/4))/(2^(1/4)*a^(1/4)*x-(x*(a*x^3-b ))^(1/4)))*a^2*x^6+20*(x*(a*x^3-b))^(3/4)*a^(5/4)*x^3*2^(1/4)-6*b*2^(1/4)* a^(1/4)*(x*(a*x^3-b))^(3/4))*2^(3/4)/a^(1/4)/b^2
Result contains complex when optimal does not.
Time = 79.87 (sec) , antiderivative size = 623, normalized size of antiderivative = 4.36 \[ \int \frac {-b+a x^3}{x^6 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx =\text {Too large to display} \] Input:
integrate((a*x^3-b)/x^6/(a*x^3+b)/(a*x^4-b*x)^(1/4),x, algorithm="fricas")
Output:
-1/126*(21*8^(1/4)*b^2*x^6*(a^7/b^8)^(1/4)*log((4*sqrt(2)*(a*x^4 - b*x)^(1 /4)*a^2*b^4*x^2*sqrt(a^7/b^8) + 8^(3/4)*sqrt(a*x^4 - b*x)*b^6*x*(a^7/b^8)^ (3/4) + 4*(a*x^4 - b*x)^(3/4)*a^5 + 8^(1/4)*(3*a^4*b^2*x^3 - a^3*b^3)*(a^7 /b^8)^(1/4))/(a*x^3 + b)) + 21*I*8^(1/4)*b^2*x^6*(a^7/b^8)^(1/4)*log(-(4*s qrt(2)*(a*x^4 - b*x)^(1/4)*a^2*b^4*x^2*sqrt(a^7/b^8) + I*8^(3/4)*sqrt(a*x^ 4 - b*x)*b^6*x*(a^7/b^8)^(3/4) - 4*(a*x^4 - b*x)^(3/4)*a^5 + 8^(1/4)*(-3*I *a^4*b^2*x^3 + I*a^3*b^3)*(a^7/b^8)^(1/4))/(a*x^3 + b)) - 21*I*8^(1/4)*b^2 *x^6*(a^7/b^8)^(1/4)*log(-(4*sqrt(2)*(a*x^4 - b*x)^(1/4)*a^2*b^4*x^2*sqrt( a^7/b^8) - I*8^(3/4)*sqrt(a*x^4 - b*x)*b^6*x*(a^7/b^8)^(3/4) - 4*(a*x^4 - b*x)^(3/4)*a^5 + 8^(1/4)*(3*I*a^4*b^2*x^3 - I*a^3*b^3)*(a^7/b^8)^(1/4))/(a *x^3 + b)) - 21*8^(1/4)*b^2*x^6*(a^7/b^8)^(1/4)*log((4*sqrt(2)*(a*x^4 - b* x)^(1/4)*a^2*b^4*x^2*sqrt(a^7/b^8) - 8^(3/4)*sqrt(a*x^4 - b*x)*b^6*x*(a^7/ b^8)^(3/4) + 4*(a*x^4 - b*x)^(3/4)*a^5 - 8^(1/4)*(3*a^4*b^2*x^3 - a^3*b^3) *(a^7/b^8)^(1/4))/(a*x^3 + b)) - 8*(a*x^4 - b*x)^(3/4)*(10*a*x^3 - 3*b))/( b^2*x^6)
\[ \int \frac {-b+a x^3}{x^6 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\int \frac {a x^{3} - b}{x^{6} \sqrt [4]{x \left (a x^{3} - b\right )} \left (a x^{3} + b\right )}\, dx \] Input:
integrate((a*x**3-b)/x**6/(a*x**3+b)/(a*x**4-b*x)**(1/4),x)
Output:
Integral((a*x**3 - b)/(x**6*(x*(a*x**3 - b))**(1/4)*(a*x**3 + b)), x)
\[ \int \frac {-b+a x^3}{x^6 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\int { \frac {a x^{3} - b}{{\left (a x^{4} - b x\right )}^{\frac {1}{4}} {\left (a x^{3} + b\right )} x^{6}} \,d x } \] Input:
integrate((a*x^3-b)/x^6/(a*x^3+b)/(a*x^4-b*x)^(1/4),x, algorithm="maxima")
Output:
integrate((a*x^3 - b)/((a*x^4 - b*x)^(1/4)*(a*x^3 + b)*x^6), x)
Leaf count of result is larger than twice the leaf count of optimal. 241 vs. \(2 (115) = 230\).
Time = 0.46 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.69 \[ \int \frac {-b+a x^3}{x^6 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\frac {2 \cdot 2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} a \arctan \left (\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, b^{2}} - \frac {2 \cdot 2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} a \arctan \left (-\frac {2^{\frac {1}{4}} {\left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{3 \, b^{2}} + \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} a \log \left (2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right )}{3 \, b^{2}} - \frac {2^{\frac {1}{4}} \left (-a\right )^{\frac {3}{4}} a \log \left (-2^{\frac {3}{4}} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {2} \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right )}{3 \, b^{2}} + \frac {4 \, {\left (3 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {7}{4}} b^{12} + 7 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {3}{4}} a b^{12}\right )}}{63 \, b^{14}} \] Input:
integrate((a*x^3-b)/x^6/(a*x^3+b)/(a*x^4-b*x)^(1/4),x, algorithm="giac")
Output:
-2/3*2^(1/4)*(-a)^(3/4)*a*arctan(1/2*2^(1/4)*(2^(3/4)*(-a)^(1/4) + 2*(a - b/x^3)^(1/4))/(-a)^(1/4))/b^2 - 2/3*2^(1/4)*(-a)^(3/4)*a*arctan(-1/2*2^(1/ 4)*(2^(3/4)*(-a)^(1/4) - 2*(a - b/x^3)^(1/4))/(-a)^(1/4))/b^2 + 1/3*2^(1/4 )*(-a)^(3/4)*a*log(2^(3/4)*(-a)^(1/4)*(a - b/x^3)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a - b/x^3))/b^2 - 1/3*2^(1/4)*(-a)^(3/4)*a*log(-2^(3/4)*(-a)^(1/4) *(a - b/x^3)^(1/4) + sqrt(2)*sqrt(-a) + sqrt(a - b/x^3))/b^2 + 4/63*(3*(a - b/x^3)^(7/4)*b^12 + 7*(a - b/x^3)^(3/4)*a*b^12)/b^14
Timed out. \[ \int \frac {-b+a x^3}{x^6 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=\int -\frac {b-a\,x^3}{x^6\,{\left (a\,x^4-b\,x\right )}^{1/4}\,\left (a\,x^3+b\right )} \,d x \] Input:
int(-(b - a*x^3)/(x^6*(a*x^4 - b*x)^(1/4)*(b + a*x^3)),x)
Output:
int(-(b - a*x^3)/(x^6*(a*x^4 - b*x)^(1/4)*(b + a*x^3)), x)
\[ \int \frac {-b+a x^3}{x^6 \left (b+a x^3\right ) \sqrt [4]{-b x+a x^4}} \, dx=-\left (\int \frac {1}{x^{\frac {37}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} a +x^{\frac {25}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} b}d x \right ) b +\left (\int \frac {1}{x^{\frac {25}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} a +x^{\frac {13}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} b}d x \right ) a \] Input:
int((a*x^3-b)/x^6/(a*x^3+b)/(a*x^4-b*x)^(1/4),x)
Output:
- int(1/(x**(1/4)*(a*x**3 - b)**(1/4)*a*x**9 + x**(1/4)*(a*x**3 - b)**(1/ 4)*b*x**6),x)*b + 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)*a