Integrand size = 29, antiderivative size = 148 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {\sqrt [4]{-b x^3+a x^4} \left (160 a b^2 d-32 a^2 b d x-128 a^3 d x^2-45 b^3 c x^3+180 a b^2 c x^4\right )}{360 a b^2 x^3}+\frac {3 b^2 c \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^3+a x^4}}\right )}{16 a^{7/4}}-\frac {3 b^2 c \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b x^3+a x^4}}\right )}{16 a^{7/4}} \]
1/360*(a*x^4-b*x^3)^(1/4)*(180*a*b^2*c*x^4-45*b^3*c*x^3-128*a^3*d*x^2-32*a ^2*b*d*x+160*a*b^2*d)/a/b^2/x^3+3/16*b^2*c*arctan(a^(1/4)*x/(a*x^4-b*x^3)^ (1/4))/a^(7/4)-3/16*b^2*c*arctanh(a^(1/4)*x/(a*x^4-b*x^3)^(1/4))/a^(7/4)
Time = 0.66 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {(-b+a x)^{3/4} \left (2 a^{3/4} \sqrt [4]{-b+a x} \left (-32 a^2 b d x-128 a^3 d x^2-45 b^3 c x^3+20 a b^2 \left (8 d+9 c x^4\right )\right )+135 b^4 c x^{9/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )-135 b^4 c x^{9/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{-b+a x}}\right )\right )}{720 a^{7/4} b^2 \left (x^3 (-b+a x)\right )^{3/4}} \]
((-b + a*x)^(3/4)*(2*a^(3/4)*(-b + a*x)^(1/4)*(-32*a^2*b*d*x - 128*a^3*d*x ^2 - 45*b^3*c*x^3 + 20*a*b^2*(8*d + 9*c*x^4)) + 135*b^4*c*x^(9/4)*ArcTan[( a^(1/4)*x^(1/4))/(-b + a*x)^(1/4)] - 135*b^4*c*x^(9/4)*ArcTanh[(a^(1/4)*x^ (1/4))/(-b + a*x)^(1/4)]))/(720*a^(7/4)*b^2*(x^3*(-b + a*x))^(3/4))
Time = 0.48 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.61, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2449, 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.
\(\displaystyle \int \frac {\sqrt [4]{a x^4-b x^3} \left (c x^4-d\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2449 |
\(\displaystyle \int \left (c \sqrt [4]{a x^4-b x^3}-\frac {d \sqrt [4]{a x^4-b x^3}}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 b^2 c x^{9/4} (a x-b)^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x-b}}\right )}{16 a^{7/4} \left (a x^4-b x^3\right )^{3/4}}-\frac {3 b^2 c x^{9/4} (a x-b)^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{x}}{\sqrt [4]{a x-b}}\right )}{16 a^{7/4} \left (a x^4-b x^3\right )^{3/4}}-\frac {16 a d \left (a x^4-b x^3\right )^{5/4}}{45 b^2 x^5}+\frac {1}{2} c x \sqrt [4]{a x^4-b x^3}-\frac {b c \sqrt [4]{a x^4-b x^3}}{8 a}-\frac {4 d \left (a x^4-b x^3\right )^{5/4}}{9 b x^6}\) |
-1/8*(b*c*(-(b*x^3) + a*x^4)^(1/4))/a + (c*x*(-(b*x^3) + a*x^4)^(1/4))/2 - (4*d*(-(b*x^3) + a*x^4)^(5/4))/(9*b*x^6) - (16*a*d*(-(b*x^3) + a*x^4)^(5/ 4))/(45*b^2*x^5) + (3*b^2*c*x^(9/4)*(-b + a*x)^(3/4)*ArcTan[(a^(1/4)*x^(1/ 4))/(-b + a*x)^(1/4)])/(16*a^(7/4)*(-(b*x^3) + a*x^4)^(3/4)) - (3*b^2*c*x^ (9/4)*(-b + a*x)^(3/4)*ArcTanh[(a^(1/4)*x^(1/4))/(-b + a*x)^(1/4)])/(16*a^ (7/4)*(-(b*x^3) + a*x^4)^(3/4))
3.21.61.3.1 Defintions of rubi rules used
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_S ymbol] :> 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]) && !Integer Q[p] && NeQ[n, j]
Time = 0.44 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.07
method | result | size |
pseudoelliptic | \(\frac {-\frac {3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}\right ) b^{4} c \,x^{3}}{16}-\frac {3 \arctan \left (\frac {\left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b^{4} c \,x^{3}}{8}+\left (b^{2} \left (c \,x^{4}+\frac {8 d}{9}\right ) a^{\frac {7}{4}}-\frac {8 x \left (\frac {45 b^{3} c \,x^{2} a^{\frac {3}{4}}}{32}+a^{\frac {11}{4}} b d +4 a^{\frac {15}{4}} d x \right )}{45}\right ) \left (x^{3} \left (a x -b \right )\right )^{\frac {1}{4}}}{2 a^{\frac {7}{4}} x^{3} b^{2}}\) | \(159\) |
1/2*(-3/16*ln((-a^(1/4)*x-(x^3*(a*x-b))^(1/4))/(a^(1/4)*x-(x^3*(a*x-b))^(1 /4)))*b^4*c*x^3-3/8*arctan(1/a^(1/4)/x*(x^3*(a*x-b))^(1/4))*b^4*c*x^3+(b^2 *(c*x^4+8/9*d)*a^(7/4)-8/45*x*(45/32*b^3*c*x^2*a^(3/4)+a^(11/4)*b*d+4*a^(1 5/4)*d*x))*(x^3*(a*x-b))^(1/4))/a^(7/4)/x^3/b^2
Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 331, normalized size of antiderivative = 2.24 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=-\frac {135 \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \log \left (\frac {3 \, {\left ({\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} b^{2} c + \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a^{2} x\right )}}{x}\right ) + 135 i \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \log \left (\frac {3 \, {\left ({\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} b^{2} c + i \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a^{2} x\right )}}{x}\right ) - 135 i \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \log \left (\frac {3 \, {\left ({\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} b^{2} c - i \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a^{2} x\right )}}{x}\right ) - 135 \, \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a b^{2} x^{3} \log \left (\frac {3 \, {\left ({\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} b^{2} c - \left (\frac {b^{8} c^{4}}{a^{7}}\right )^{\frac {1}{4}} a^{2} x\right )}}{x}\right ) - 4 \, {\left (180 \, a b^{2} c x^{4} - 45 \, b^{3} c x^{3} - 128 \, a^{3} d x^{2} - 32 \, a^{2} b d x + 160 \, a b^{2} d\right )} {\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}}}{1440 \, a b^{2} x^{3}} \]
-1/1440*(135*(b^8*c^4/a^7)^(1/4)*a*b^2*x^3*log(3*((a*x^4 - b*x^3)^(1/4)*b^ 2*c + (b^8*c^4/a^7)^(1/4)*a^2*x)/x) + 135*I*(b^8*c^4/a^7)^(1/4)*a*b^2*x^3* log(3*((a*x^4 - b*x^3)^(1/4)*b^2*c + I*(b^8*c^4/a^7)^(1/4)*a^2*x)/x) - 135 *I*(b^8*c^4/a^7)^(1/4)*a*b^2*x^3*log(3*((a*x^4 - b*x^3)^(1/4)*b^2*c - I*(b ^8*c^4/a^7)^(1/4)*a^2*x)/x) - 135*(b^8*c^4/a^7)^(1/4)*a*b^2*x^3*log(3*((a* x^4 - b*x^3)^(1/4)*b^2*c - (b^8*c^4/a^7)^(1/4)*a^2*x)/x) - 4*(180*a*b^2*c* x^4 - 45*b^3*c*x^3 - 128*a^3*d*x^2 - 32*a^2*b*d*x + 160*a*b^2*d)*(a*x^4 - b*x^3)^(1/4))/(a*b^2*x^3)
\[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\int \frac {\sqrt [4]{x^{3} \left (a x - b\right )} \left (c x^{4} - d\right )}{x^{4}}\, dx \]
\[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\int { \frac {{\left (a x^{4} - b x^{3}\right )}^{\frac {1}{4}} {\left (c x^{4} - d\right )}}{x^{4}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (128) = 256\).
Time = 0.31 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.00 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {\frac {270 \, \sqrt {2} b^{3} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {270 \, \sqrt {2} b^{3} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {135 \, \sqrt {2} b^{3} c \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x}}\right )}{\left (-a\right )^{\frac {3}{4}} a} + \frac {135 \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} b^{3} c \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x}}\right )}{a^{2}} + \frac {360 \, {\left ({\left (a - \frac {b}{x}\right )}^{\frac {5}{4}} b^{3} c + 3 \, {\left (a - \frac {b}{x}\right )}^{\frac {1}{4}} a b^{3} c\right )} x^{2}}{a b^{2}} + \frac {256 \, {\left (5 \, {\left (a - \frac {b}{x}\right )}^{\frac {9}{4}} b^{8} d - 9 \, {\left (a - \frac {b}{x}\right )}^{\frac {5}{4}} a b^{8} d\right )}}{b^{9}}}{2880 \, b} \]
1/2880*(270*sqrt(2)*b^3*c*arctan(1/2*sqrt(2)*(sqrt(2)*(-a)^(1/4) + 2*(a - b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a) + 270*sqrt(2)*b^3*c*arctan(-1/2*sqr t(2)*(sqrt(2)*(-a)^(1/4) - 2*(a - b/x)^(1/4))/(-a)^(1/4))/((-a)^(3/4)*a) + 135*sqrt(2)*b^3*c*log(sqrt(2)*(-a)^(1/4)*(a - b/x)^(1/4) + sqrt(-a) + sqr t(a - b/x))/((-a)^(3/4)*a) + 135*sqrt(2)*(-a)^(1/4)*b^3*c*log(-sqrt(2)*(-a )^(1/4)*(a - b/x)^(1/4) + sqrt(-a) + sqrt(a - b/x))/a^2 + 360*((a - b/x)^( 5/4)*b^3*c + 3*(a - b/x)^(1/4)*a*b^3*c)*x^2/(a*b^2) + 256*(5*(a - b/x)^(9/ 4)*b^8*d - 9*(a - b/x)^(5/4)*a*b^8*d)/b^9)/b
Time = 6.70 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.75 \[ \int \frac {\sqrt [4]{-b x^3+a x^4} \left (-d+c x^4\right )}{x^4} \, dx=\frac {4\,d\,{\left (a\,x^4-b\,x^3\right )}^{1/4}}{9\,x^3}-\frac {16\,a^2\,d\,{\left (a\,x^4-b\,x^3\right )}^{1/4}}{45\,b^2\,x}+\frac {4\,c\,x\,{\left (a\,x^4-b\,x^3\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {7}{4};\ \frac {11}{4};\ \frac {a\,x}{b}\right )}{7\,{\left (1-\frac {a\,x}{b}\right )}^{1/4}}-\frac {4\,a\,d\,{\left (a\,x^4-b\,x^3\right )}^{1/4}}{45\,b\,x^2} \]