Integrand size = 20, antiderivative size = 371 \[ \int \frac {1}{x^{5/2} \left (a+b x^2+c x^4\right )} \, dx=-\frac {2}{3 a x^{3/2}}+\frac {c^{3/4} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{\sqrt [4]{2} a \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{\sqrt [4]{2} a \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{\sqrt [4]{2} a \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}+\frac {c^{3/4} \left (1+\frac {b}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{\sqrt [4]{2} a \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}} \]
-2/3/a/x^(3/2)+1/2*c^(3/4)*arctan(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2) ^(1/2))^(1/4))*(1-b/(-4*a*c+b^2)^(1/2))*2^(3/4)/a/(-b-(-4*a*c+b^2)^(1/2))^ (3/4)+1/2*c^(3/4)*arctanh(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^(1/2))^ (1/4))*(1-b/(-4*a*c+b^2)^(1/2))*2^(3/4)/a/(-b-(-4*a*c+b^2)^(1/2))^(3/4)+1/ 2*c^(3/4)*arctan(2^(1/4)*c^(1/4)*x^(1/2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*(1 +b/(-4*a*c+b^2)^(1/2))*2^(3/4)/a/(-b+(-4*a*c+b^2)^(1/2))^(3/4)+1/2*c^(3/4) *arctanh(2^(1/4)*c^(1/4)*x^(1/2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*(1+b/(-4*a *c+b^2)^(1/2))*2^(3/4)/a/(-b+(-4*a*c+b^2)^(1/2))^(3/4)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x^{5/2} \left (a+b x^2+c x^4\right )} \, dx=-\frac {\frac {4}{x^{3/2}}+3 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log \left (\sqrt {x}-\text {$\#$1}\right )+c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]}{6 a} \]
-1/6*(4/x^(3/2) + 3*RootSum[a + b*#1^4 + c*#1^8 & , (b*Log[Sqrt[x] - #1] + c*Log[Sqrt[x] - #1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ])/a
Time = 0.48 (sec) , antiderivative size = 350, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1435, 1704, 27, 1752, 756, 218, 221}
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.
| \(\displaystyle \int \frac {1}{x^{5/2} \left (a+b x^2+c x^4\right )} \, dx\) |
| \(\Big \downarrow \) 1435 |
| \(\displaystyle 2 \int \frac {1}{x^2 \left (c x^4+b x^2+a\right )}d\sqrt {x}\) |
| \(\Big \downarrow \) 1704 |
| \(\displaystyle 2 \left (\frac {\int -\frac {3 \left (c x^2+b\right )}{c x^4+b x^2+a}d\sqrt {x}}{3 a}-\frac {1}{3 a x^{3/2}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \left (-\frac {\int \frac {c x^2+b}{c x^4+b x^2+a}d\sqrt {x}}{a}-\frac {1}{3 a x^{3/2}}\right )\) |
| \(\Big \downarrow \) 1752 |
| \(\displaystyle 2 \left (-\frac {\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}d\sqrt {x}+\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}d\sqrt {x}}{a}-\frac {1}{3 a x^{3/2}}\right )\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle 2 \left (-\frac {\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {-b-\sqrt {b^2-4 a c}}}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}\right )+\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {\sqrt {b^2-4 a c}-b}}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}\right )}{a}-\frac {1}{3 a x^{3/2}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle 2 \left (-\frac {\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{a}-\frac {1}{3 a x^{3/2}}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle 2 \left (-\frac {\frac {1}{2} c \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} c \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{a}-\frac {1}{3 a x^{3/2}}\right )\) |
2*(-1/3*1/(a*x^(3/2)) - ((c*(1 - b/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^(1/4)*c ^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b - Sqr t[b^2 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4))))/2 + ( c*(1 + b/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt [b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4) *c^(1/4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))))/2)/a)
3.11.69.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/d Subst[Int[x^(k*(m + 1) - 1)*(a + b *(x^(2*k)/d^2) + c*(x^(4*k)/d^4))^p, x], x, (d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && FractionQ[m] && IntegerQ[p]
Int[((d_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_ Symbol] :> Simp[(d*x)^(m + 1)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*d*(m + 1) )), x] - Simp[1/(a*d^n*(m + 1)) Int[(d*x)^(m + n)*(b*(m + n*(p + 1) + 1) + c*(m + 2*n*(p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{ a, b, c, d, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntegerQ[p]
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x _Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) I nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 , 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] || !IGtQ[n/2, 0])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.16
| method | result | size |
| risch | \(-\frac {2}{3 a \,x^{\frac {3}{2}}}-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (\textit {\_R}^{4} c +b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{2 a}\) | \(61\) |
| derivativedivides | \(-\frac {2}{3 a \,x^{\frac {3}{2}}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (-\textit {\_R}^{4} c -b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{2 a}\) | \(64\) |
| default | \(-\frac {2}{3 a \,x^{\frac {3}{2}}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (-\textit {\_R}^{4} c -b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{2 a}\) | \(64\) |
-2/3/a/x^(3/2)-1/2/a*sum((_R^4*c+b)/(2*_R^7*c+_R^3*b)*ln(x^(1/2)-_R),_R=Ro otOf(_Z^8*c+_Z^4*b+a))
Leaf count of result is larger than twice the leaf count of optimal. 5046 vs. \(2 (289) = 578\).
Time = 0.90 (sec) , antiderivative size = 5046, normalized size of antiderivative = 13.60 \[ \int \frac {1}{x^{5/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {1}{x^{5/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^{5/2} \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} x^{\frac {5}{2}}} \,d x } \]
-2/3*(3*b*sqrt(x) + a/x^(3/2))/a^2 + integrate((b*c*x^(7/2) + (b^2 - a*c)* x^(3/2))/(a^2*c*x^4 + a^2*b*x^2 + a^3), x)
\[ \int \frac {1}{x^{5/2} \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} x^{\frac {5}{2}}} \,d x } \]
Time = 17.21 (sec) , antiderivative size = 16557, normalized size of antiderivative = 44.63 \[ \int \frac {1}{x^{5/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \]
atan(((x^(1/2)*(512*a^10*c^10 - 256*a^9*b^2*c^9) - (-(b^11 + b^6*(-(4*a*c - b^2)^5)^(1/2) - 112*a^5*b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 280*a ^4*b^3*c^4 - a^3*c^3*(-(4*a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2 *(-(4*a*c - b^2)^5)^(1/2) - 5*a*b^4*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(a^7*b ^8 + 256*a^11*c^4 - 16*a^8*b^6*c + 96*a^9*b^4*c^2 - 256*a^10*b^2*c^3)))^(1 /4)*((x^(1/2)*(327680*a^15*b*c^8 + 4096*a^11*b^9*c^4 - 53248*a^12*b^7*c^5 + 249856*a^13*b^5*c^6 - 491520*a^14*b^3*c^7) + (-(b^11 + b^6*(-(4*a*c - b^ 2)^5)^(1/2) - 112*a^5*b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 280*a^4*b ^3*c^4 - a^3*c^3*(-(4*a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2*(-( 4*a*c - b^2)^5)^(1/2) - 5*a*b^4*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(a^7*b^8 + 256*a^11*c^4 - 16*a^8*b^6*c + 96*a^9*b^4*c^2 - 256*a^10*b^2*c^3)))^(1/4)* (524288*a^17*c^8 + 8192*a^13*b^8*c^4 - 106496*a^14*b^6*c^5 + 491520*a^15*b ^4*c^6 - 917504*a^16*b^2*c^7))*(-(b^11 + b^6*(-(4*a*c - b^2)^5)^(1/2) - 11 2*a^5*b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 280*a^4*b^3*c^4 - a^3*c^3 *(-(4*a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2*(-(4*a*c - b^2)^5)^ (1/2) - 5*a*b^4*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(a^7*b^8 + 256*a^11*c^4 - 16*a^8*b^6*c + 96*a^9*b^4*c^2 - 256*a^10*b^2*c^3)))^(3/4) - 4096*a^11*b*c^ 9 - 512*a^9*b^5*c^7 + 3072*a^10*b^3*c^8))*(-(b^11 + b^6*(-(4*a*c - b^2)^5) ^(1/2) - 112*a^5*b*c^5 + 86*a^2*b^7*c^2 - 231*a^3*b^5*c^3 + 280*a^4*b^3*c^ 4 - a^3*c^3*(-(4*a*c - b^2)^5)^(1/2) - 15*a*b^9*c + 6*a^2*b^2*c^2*(-(4*...