Integrand size = 29, antiderivative size = 162 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{a-b x^4} \, dx=-\frac {d x}{b}-\frac {e x^2}{2 b}-\frac {f x^3}{3 b}+\frac {\sqrt [4]{a} \left (\sqrt {b} d-\sqrt {a} f\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 b^{7/4}}+\frac {\sqrt [4]{a} \left (\sqrt {b} d+\sqrt {a} f\right ) \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )}{2 b^{7/4}}+\frac {\sqrt {a} e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{3/2}}-\frac {c \log \left (a-b x^4\right )}{4 b} \]
-d*x/b-1/2*e*x^2/b-1/3*f*x^3/b-1/4*c*ln(-b*x^4+a)/b+1/2*e*arctanh(x^2*b^(1 /2)/a^(1/2))*a^(1/2)/b^(3/2)+1/2*a^(1/4)*arctan(b^(1/4)*x/a^(1/4))*(-f*a^( 1/2)+d*b^(1/2))/b^(7/4)+1/2*a^(1/4)*arctanh(b^(1/4)*x/a^(1/4))*(f*a^(1/2)+ d*b^(1/2))/b^(7/4)
Time = 0.08 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.36 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{a-b x^4} \, dx=\frac {-12 b^{3/4} d x-6 b^{3/4} e x^2-4 b^{3/4} f x^3+6 \left (\sqrt [4]{a} \sqrt {b} d-a^{3/4} f\right ) \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )-3 \left (\sqrt [4]{a} \sqrt {b} d+\sqrt {a} \sqrt [4]{b} e+a^{3/4} f\right ) \log \left (\sqrt [4]{a}-\sqrt [4]{b} x\right )+3 \left (\sqrt [4]{a} \sqrt {b} d-\sqrt {a} \sqrt [4]{b} e+a^{3/4} f\right ) \log \left (\sqrt [4]{a}+\sqrt [4]{b} x\right )+3 \sqrt {a} \sqrt [4]{b} e \log \left (\sqrt {a}+\sqrt {b} x^2\right )-3 b^{3/4} c \log \left (a-b x^4\right )}{12 b^{7/4}} \]
(-12*b^(3/4)*d*x - 6*b^(3/4)*e*x^2 - 4*b^(3/4)*f*x^3 + 6*(a^(1/4)*Sqrt[b]* d - a^(3/4)*f)*ArcTan[(b^(1/4)*x)/a^(1/4)] - 3*(a^(1/4)*Sqrt[b]*d + Sqrt[a ]*b^(1/4)*e + a^(3/4)*f)*Log[a^(1/4) - b^(1/4)*x] + 3*(a^(1/4)*Sqrt[b]*d - Sqrt[a]*b^(1/4)*e + a^(3/4)*f)*Log[a^(1/4) + b^(1/4)*x] + 3*Sqrt[a]*b^(1/ 4)*e*Log[Sqrt[a] + Sqrt[b]*x^2] - 3*b^(3/4)*c*Log[a - b*x^4])/(12*b^(7/4))
Time = 0.40 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2370, 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.
\(\displaystyle \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{a-b x^4} \, dx\) |
\(\Big \downarrow \) 2370 |
\(\displaystyle \int \left (\frac {x^3 \left (c+e x^2\right )}{a-b x^4}+\frac {x^4 \left (d+f x^2\right )}{a-b x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt {b} d-\sqrt {a} f\right )}{2 b^{7/4}}+\frac {\sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ) \left (\sqrt {a} f+\sqrt {b} d\right )}{2 b^{7/4}}+\frac {\sqrt {a} e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )}{2 b^{3/2}}-\frac {c \log \left (a-b x^4\right )}{4 b}-\frac {d x}{b}-\frac {e x^2}{2 b}-\frac {f x^3}{3 b}\) |
-((d*x)/b) - (e*x^2)/(2*b) - (f*x^3)/(3*b) + (a^(1/4)*(Sqrt[b]*d - Sqrt[a] *f)*ArcTan[(b^(1/4)*x)/a^(1/4)])/(2*b^(7/4)) + (a^(1/4)*(Sqrt[b]*d + Sqrt[ a]*f)*ArcTanh[(b^(1/4)*x)/a^(1/4)])/(2*b^(7/4)) + (Sqrt[a]*e*ArcTanh[(Sqrt [b]*x^2)/Sqrt[a]])/(2*b^(3/2)) - (c*Log[a - b*x^4])/(4*b)
3.5.86.3.1 Defintions of rubi rules used
Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[ {v = Sum[(c*x)^(m + ii)*((Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2) )/(c^ii*(a + b*x^n))), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{ a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.55 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.49
method | result | size |
risch | \(-\frac {f \,x^{3}}{3 b}-\frac {e \,x^{2}}{2 b}-\frac {d x}{b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4} b -a \right )}{\sum }\frac {\left (-\textit {\_R}^{3} b c -\textit {\_R}^{2} a f -\textit {\_R} a e -a d \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}}{4 b^{2}}\) | \(79\) |
default | \(-\frac {\frac {1}{3} f \,x^{3}+\frac {1}{2} e \,x^{2}+d x}{b}+\frac {\frac {d \left (\frac {a}{b}\right )^{\frac {1}{4}} \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4}+\frac {a e \ln \left (\frac {a +x^{2} \sqrt {a b}}{a -x^{2} \sqrt {a b}}\right )}{4 \sqrt {a b}}-\frac {a f \left (2 \arctan \left (\frac {x}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )\right )}{4 b \left (\frac {a}{b}\right )^{\frac {1}{4}}}-\frac {c \ln \left (-b \,x^{4}+a \right )}{4}}{b}\) | \(176\) |
-1/3*f*x^3/b-1/2*e*x^2/b-d*x/b+1/4/b^2*sum((-_R^3*b*c-_R^2*a*f-_R*a*e-a*d) /_R^3*ln(x-_R),_R=RootOf(_Z^4*b-a))
Result contains complex when optimal does not.
Time = 6.17 (sec) , antiderivative size = 220680, normalized size of antiderivative = 1362.22 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{a-b x^4} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{a-b x^4} \, dx=\text {Timed out} \]
Time = 0.29 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.28 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{a-b x^4} \, dx=-\frac {2 \, f x^{3} + 3 \, e x^{2} + 6 \, d x}{6 \, b} + \frac {\frac {2 \, {\left (a \sqrt {b} d - a^{\frac {3}{2}} f\right )} \arctan \left (\frac {\sqrt {b} x}{\sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {{\left (\sqrt {a} b c - a \sqrt {b} e\right )} \log \left (\sqrt {b} x^{2} + \sqrt {a}\right )}{\sqrt {a} b} - \frac {{\left (\sqrt {a} b c + a \sqrt {b} e\right )} \log \left (\sqrt {b} x^{2} - \sqrt {a}\right )}{\sqrt {a} b} - \frac {{\left (a \sqrt {b} d + a^{\frac {3}{2}} f\right )} \log \left (\frac {\sqrt {b} x - \sqrt {\sqrt {a} \sqrt {b}}}{\sqrt {b} x + \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}}}{4 \, b} \]
-1/6*(2*f*x^3 + 3*e*x^2 + 6*d*x)/b + 1/4*(2*(a*sqrt(b)*d - a^(3/2)*f)*arct an(sqrt(b)*x/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*sqrt(b) ) - (sqrt(a)*b*c - a*sqrt(b)*e)*log(sqrt(b)*x^2 + sqrt(a))/(sqrt(a)*b) - ( sqrt(a)*b*c + a*sqrt(b)*e)*log(sqrt(b)*x^2 - sqrt(a))/(sqrt(a)*b) - (a*sqr t(b)*d + a^(3/2)*f)*log((sqrt(b)*x - sqrt(sqrt(a)*sqrt(b)))/(sqrt(b)*x + s qrt(sqrt(a)*sqrt(b))))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))*sqrt(b)))/b
Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (118) = 236\).
Time = 0.28 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.01 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{a-b x^4} \, dx=-\frac {c \log \left ({\left | b x^{4} - a \right |}\right )}{4 \, b} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-a b} b^{2} e - \left (-a b^{3}\right )^{\frac {1}{4}} b^{2} d - \left (-a b^{3}\right )^{\frac {3}{4}} f\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x + \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, b^{4}} - \frac {\sqrt {2} {\left (\sqrt {2} \sqrt {-a b} b^{2} e - \left (-a b^{3}\right )^{\frac {1}{4}} b^{2} d - \left (-a b^{3}\right )^{\frac {3}{4}} f\right )} \arctan \left (\frac {\sqrt {2} {\left (2 \, x - \sqrt {2} \left (-\frac {a}{b}\right )^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{4 \, b^{4}} + \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} d - \left (-a b^{3}\right )^{\frac {3}{4}} f\right )} \log \left (x^{2} + \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{8 \, b^{4}} - \frac {\sqrt {2} {\left (\left (-a b^{3}\right )^{\frac {1}{4}} b^{2} d - \left (-a b^{3}\right )^{\frac {3}{4}} f\right )} \log \left (x^{2} - \sqrt {2} x \left (-\frac {a}{b}\right )^{\frac {1}{4}} + \sqrt {-\frac {a}{b}}\right )}{8 \, b^{4}} - \frac {2 \, b^{2} f x^{3} + 3 \, b^{2} e x^{2} + 6 \, b^{2} d x}{6 \, b^{3}} \]
-1/4*c*log(abs(b*x^4 - a))/b - 1/4*sqrt(2)*(sqrt(2)*sqrt(-a*b)*b^2*e - (-a *b^3)^(1/4)*b^2*d - (-a*b^3)^(3/4)*f)*arctan(1/2*sqrt(2)*(2*x + sqrt(2)*(- a/b)^(1/4))/(-a/b)^(1/4))/b^4 - 1/4*sqrt(2)*(sqrt(2)*sqrt(-a*b)*b^2*e - (- a*b^3)^(1/4)*b^2*d - (-a*b^3)^(3/4)*f)*arctan(1/2*sqrt(2)*(2*x - sqrt(2)*( -a/b)^(1/4))/(-a/b)^(1/4))/b^4 + 1/8*sqrt(2)*((-a*b^3)^(1/4)*b^2*d - (-a*b ^3)^(3/4)*f)*log(x^2 + sqrt(2)*x*(-a/b)^(1/4) + sqrt(-a/b))/b^4 - 1/8*sqrt (2)*((-a*b^3)^(1/4)*b^2*d - (-a*b^3)^(3/4)*f)*log(x^2 - sqrt(2)*x*(-a/b)^( 1/4) + sqrt(-a/b))/b^4 - 1/6*(2*b^2*f*x^3 + 3*b^2*e*x^2 + 6*b^2*d*x)/b^3
Time = 9.02 (sec) , antiderivative size = 846, normalized size of antiderivative = 5.22 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{a-b x^4} \, dx=\left (\sum _{k=1}^4\ln \left (-\frac {a^4\,f^3-2\,a^3\,b\,c\,e\,f-a^3\,b\,d^2\,f+a^3\,b\,d\,e^2+a^2\,b^2\,c^2\,d}{b^2}-\mathrm {root}\left (256\,b^7\,z^4+256\,b^6\,c\,z^3-64\,a\,b^4\,d\,f\,z^2-32\,a\,b^4\,e^2\,z^2+96\,b^5\,c^2\,z^2-32\,a\,b^3\,c\,d\,f\,z+16\,a^2\,b^2\,e\,f^2\,z+16\,a\,b^3\,d^2\,e\,z-16\,a\,b^3\,c\,e^2\,z+16\,b^4\,c^3\,z-4\,a^2\,b\,d\,e^2\,f+4\,a^2\,b\,c\,e\,f^2-4\,a\,b^2\,c^2\,d\,f+4\,a\,b^2\,c\,d^2\,e+2\,a^2\,b\,d^2\,f^2-2\,a\,b^2\,c^2\,e^2+a^2\,b\,e^4+b^3\,c^4-a\,b^2\,d^4-a^3\,f^4,z,k\right )\,\left (\mathrm {root}\left (256\,b^7\,z^4+256\,b^6\,c\,z^3-64\,a\,b^4\,d\,f\,z^2-32\,a\,b^4\,e^2\,z^2+96\,b^5\,c^2\,z^2-32\,a\,b^3\,c\,d\,f\,z+16\,a^2\,b^2\,e\,f^2\,z+16\,a\,b^3\,d^2\,e\,z-16\,a\,b^3\,c\,e^2\,z+16\,b^4\,c^3\,z-4\,a^2\,b\,d\,e^2\,f+4\,a^2\,b\,c\,e\,f^2-4\,a\,b^2\,c^2\,d\,f+4\,a\,b^2\,c\,d^2\,e+2\,a^2\,b\,d^2\,f^2-2\,a\,b^2\,c^2\,e^2+a^2\,b\,e^4+b^3\,c^4-a\,b^2\,d^4-a^3\,f^4,z,k\right )\,\left (16\,a^2\,b^2\,d-16\,a^2\,b^2\,e\,x\right )+\frac {8\,a^2\,b^3\,c\,d-8\,a^3\,b^2\,e\,f}{b^2}+\frac {x\,\left (4\,a^3\,b\,f^2+4\,a^2\,b^2\,d^2-8\,c\,e\,a^2\,b^2\right )}{b}\right )-\frac {x\,\left (a^3\,c\,f^2-2\,a^3\,d\,e\,f+a^3\,e^3-b\,a^2\,c^2\,e+b\,a^2\,c\,d^2\right )}{b}\right )\,\mathrm {root}\left (256\,b^7\,z^4+256\,b^6\,c\,z^3-64\,a\,b^4\,d\,f\,z^2-32\,a\,b^4\,e^2\,z^2+96\,b^5\,c^2\,z^2-32\,a\,b^3\,c\,d\,f\,z+16\,a^2\,b^2\,e\,f^2\,z+16\,a\,b^3\,d^2\,e\,z-16\,a\,b^3\,c\,e^2\,z+16\,b^4\,c^3\,z-4\,a^2\,b\,d\,e^2\,f+4\,a^2\,b\,c\,e\,f^2-4\,a\,b^2\,c^2\,d\,f+4\,a\,b^2\,c\,d^2\,e+2\,a^2\,b\,d^2\,f^2-2\,a\,b^2\,c^2\,e^2+a^2\,b\,e^4+b^3\,c^4-a\,b^2\,d^4-a^3\,f^4,z,k\right )\right )-\frac {e\,x^2}{2\,b}-\frac {f\,x^3}{3\,b}-\frac {d\,x}{b} \]
symsum(log(- (a^4*f^3 + a^2*b^2*c^2*d + a^3*b*d*e^2 - a^3*b*d^2*f - 2*a^3* b*c*e*f)/b^2 - root(256*b^7*z^4 + 256*b^6*c*z^3 - 64*a*b^4*d*f*z^2 - 32*a* b^4*e^2*z^2 + 96*b^5*c^2*z^2 - 32*a*b^3*c*d*f*z + 16*a^2*b^2*e*f^2*z + 16* a*b^3*d^2*e*z - 16*a*b^3*c*e^2*z + 16*b^4*c^3*z - 4*a^2*b*d*e^2*f + 4*a^2* b*c*e*f^2 - 4*a*b^2*c^2*d*f + 4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 - 2*a*b^2* c^2*e^2 + a^2*b*e^4 + b^3*c^4 - a*b^2*d^4 - a^3*f^4, z, k)*(root(256*b^7*z ^4 + 256*b^6*c*z^3 - 64*a*b^4*d*f*z^2 - 32*a*b^4*e^2*z^2 + 96*b^5*c^2*z^2 - 32*a*b^3*c*d*f*z + 16*a^2*b^2*e*f^2*z + 16*a*b^3*d^2*e*z - 16*a*b^3*c*e^ 2*z + 16*b^4*c^3*z - 4*a^2*b*d*e^2*f + 4*a^2*b*c*e*f^2 - 4*a*b^2*c^2*d*f + 4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 - 2*a*b^2*c^2*e^2 + a^2*b*e^4 + b^3*c^4 - a*b^2*d^4 - a^3*f^4, z, k)*(16*a^2*b^2*d - 16*a^2*b^2*e*x) + (8*a^2*b^3 *c*d - 8*a^3*b^2*e*f)/b^2 + (x*(4*a^3*b*f^2 + 4*a^2*b^2*d^2 - 8*a^2*b^2*c* e))/b) - (x*(a^3*e^3 + a^3*c*f^2 - 2*a^3*d*e*f + a^2*b*c*d^2 - a^2*b*c^2*e ))/b)*root(256*b^7*z^4 + 256*b^6*c*z^3 - 64*a*b^4*d*f*z^2 - 32*a*b^4*e^2*z ^2 + 96*b^5*c^2*z^2 - 32*a*b^3*c*d*f*z + 16*a^2*b^2*e*f^2*z + 16*a*b^3*d^2 *e*z - 16*a*b^3*c*e^2*z + 16*b^4*c^3*z - 4*a^2*b*d*e^2*f + 4*a^2*b*c*e*f^2 - 4*a*b^2*c^2*d*f + 4*a*b^2*c*d^2*e + 2*a^2*b*d^2*f^2 - 2*a*b^2*c^2*e^2 + a^2*b*e^4 + b^3*c^4 - a*b^2*d^4 - a^3*f^4, z, k), k, 1, 4) - (e*x^2)/(2*b ) - (f*x^3)/(3*b) - (d*x)/b