Integrand size = 20, antiderivative size = 331 \[ \int \frac {x^{5/2}}{a+b x^2+c x^4} \, dx=-\frac {\left (-b-\sqrt {b^2-4 a c}\right )^{3/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2^{3/4} c^{3/4} \sqrt {b^2-4 a c}}+\frac {\left (-b+\sqrt {b^2-4 a c}\right )^{3/4} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2^{3/4} c^{3/4} \sqrt {b^2-4 a c}}+\frac {\left (-b-\sqrt {b^2-4 a c}\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2^{3/4} c^{3/4} \sqrt {b^2-4 a c}}-\frac {\left (-b+\sqrt {b^2-4 a c}\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2^{3/4} c^{3/4} \sqrt {b^2-4 a c}} \]
-1/2*arctan(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^(1/2))^(1/4))*(-b-(-4 *a*c+b^2)^(1/2))^(3/4)*2^(1/4)/c^(3/4)/(-4*a*c+b^2)^(1/2)+1/2*arctanh(2^(1 /4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^(1/2))^(1/4))*(-b-(-4*a*c+b^2)^(1/2)) ^(3/4)*2^(1/4)/c^(3/4)/(-4*a*c+b^2)^(1/2)+1/2*arctan(2^(1/4)*c^(1/4)*x^(1/ 2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*(-b+(-4*a*c+b^2)^(1/2))^(3/4)*2^(1/4)/c^ (3/4)/(-4*a*c+b^2)^(1/2)-1/2*arctanh(2^(1/4)*c^(1/4)*x^(1/2)/(-b+(-4*a*c+b ^2)^(1/2))^(1/4))*(-b+(-4*a*c+b^2)^(1/2))^(3/4)*2^(1/4)/c^(3/4)/(-4*a*c+b^ 2)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.15 \[ \int \frac {x^{5/2}}{a+b x^2+c x^4} \, dx=\frac {1}{2} \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {\log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^3}{b+2 c \text {$\#$1}^4}\&\right ] \]
Time = 0.48 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1435, 1710, 27, 827, 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.
| \(\displaystyle \int \frac {x^{5/2}}{a+b x^2+c x^4} \, dx\) |
| \(\Big \downarrow \) 1435 |
| \(\displaystyle 2 \int \frac {x^3}{c x^4+b x^2+a}d\sqrt {x}\) |
| \(\Big \downarrow \) 1710 |
| \(\displaystyle 2 \left (\frac {1}{2} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {2 x}{2 c x^2+b-\sqrt {b^2-4 a c}}d\sqrt {x}+\frac {1}{2} \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \int \frac {2 x}{2 c x^2+b+\sqrt {b^2-4 a c}}d\sqrt {x}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \left (\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {x}{2 c x^2+b-\sqrt {b^2-4 a c}}d\sqrt {x}+\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \int \frac {x}{2 c x^2+b+\sqrt {b^2-4 a c}}d\sqrt {x}\right )\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle 2 \left (\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {-b-\sqrt {b^2-4 a c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )+\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {\sqrt {b^2-4 a c}-b}}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle 2 \left (\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )+\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle 2 \left (\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )+\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )\right )\) |
2*((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*2^(3/4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1/4)) - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/ 4)*c^(3/4)*(-b - Sqrt[b^2 - 4*a*c])^(1/4))) + (1 - b/Sqrt[b^2 - 4*a*c])*(A rcTan[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4) *c^(3/4)*(-b + Sqrt[b^2 - 4*a*c])^(1/4)) - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x ])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2*2^(3/4)*c^(3/4)*(-b + Sqrt[b^2 - 4*a *c])^(1/4))))
3.11.64.3.1 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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) 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_)), x_Symbo l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(d^n/2)*(b/q + 1) Int[(d*x)^(m - n)/(b/2 + q/2 + c*x^n), x], x] - Simp[(d^n/2)*(b/q - 1) Int[(d*x)^(m - n)/(b/2 - q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d}, x] && EqQ[n2, 2*n] & & NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GeQ[m, n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.10 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.14
| method | result | size |
| derivativedivides | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\textit {\_R}^{6} \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}\right )}{2}\) | \(45\) |
| default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\textit {\_R}^{6} \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}\right )}{2}\) | \(45\) |
Leaf count of result is larger than twice the leaf count of optimal. 4449 vs. \(2 (253) = 506\).
Time = 0.34 (sec) , antiderivative size = 4449, normalized size of antiderivative = 13.44 \[ \int \frac {x^{5/2}}{a+b x^2+c x^4} \, dx=\text {Too large to display} \]
1/2*sqrt(sqrt(1/2)*sqrt(-(b^3 - 3*a*b*c + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2* c^5)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^6*c^6 - 12*a*b^4*c^7 + 48*a^2*b^2 *c^8 - 64*a^3*c^9)))/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)))*log(1/2*sqrt(1 /2)*(b^7 - 9*a*b^5*c + 24*a^2*b^3*c^2 - 16*a^3*b*c^3 - (b^8*c^3 - 14*a*b^6 *c^4 + 72*a^2*b^4*c^5 - 160*a^3*b^2*c^6 + 128*a^4*c^7)*sqrt((b^4 - 2*a*b^2 *c + a^2*c^2)/(b^6*c^6 - 12*a*b^4*c^7 + 48*a^2*b^2*c^8 - 64*a^3*c^9)))*sqr t(sqrt(1/2)*sqrt(-(b^3 - 3*a*b*c + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sq rt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^6*c^6 - 12*a*b^4*c^7 + 48*a^2*b^2*c^8 - 64*a^3*c^9)))/(b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)))*sqrt(-(b^3 - 3*a*b*c + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b ^6*c^6 - 12*a*b^4*c^7 + 48*a^2*b^2*c^8 - 64*a^3*c^9)))/(b^4*c^3 - 8*a*b^2* c^4 + 16*a^2*c^5)) - (a^2*b^2 - a^3*c)*sqrt(x)) - 1/2*sqrt(sqrt(1/2)*sqrt( -(b^3 - 3*a*b*c + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt((b^4 - 2*a*b^2 *c + a^2*c^2)/(b^6*c^6 - 12*a*b^4*c^7 + 48*a^2*b^2*c^8 - 64*a^3*c^9)))/(b^ 4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)))*log(-1/2*sqrt(1/2)*(b^7 - 9*a*b^5*c + 24*a^2*b^3*c^2 - 16*a^3*b*c^3 - (b^8*c^3 - 14*a*b^6*c^4 + 72*a^2*b^4*c^5 - 160*a^3*b^2*c^6 + 128*a^4*c^7)*sqrt((b^4 - 2*a*b^2*c + a^2*c^2)/(b^6*c^6 - 12*a*b^4*c^7 + 48*a^2*b^2*c^8 - 64*a^3*c^9)))*sqrt(sqrt(1/2)*sqrt(-(b^3 - 3*a*b*c + (b^4*c^3 - 8*a*b^2*c^4 + 16*a^2*c^5)*sqrt((b^4 - 2*a*b^2*c + a ^2*c^2)/(b^6*c^6 - 12*a*b^4*c^7 + 48*a^2*b^2*c^8 - 64*a^3*c^9)))/(b^4*c...
Timed out. \[ \int \frac {x^{5/2}}{a+b x^2+c x^4} \, dx=\text {Timed out} \]
\[ \int \frac {x^{5/2}}{a+b x^2+c x^4} \, dx=\int { \frac {x^{\frac {5}{2}}}{c x^{4} + b x^{2} + a} \,d x } \]
\[ \int \frac {x^{5/2}}{a+b x^2+c x^4} \, dx=\int { \frac {x^{\frac {5}{2}}}{c x^{4} + b x^{2} + a} \,d x } \]
Time = 15.29 (sec) , antiderivative size = 8093, normalized size of antiderivative = 24.45 \[ \int \frac {x^{5/2}}{a+b x^2+c x^4} \, dx=\text {Too large to display} \]
- atan(((x^(1/2)*(256*a^3*b^3*c - 768*a^4*b*c^2) + (-(b^7 + b^2*(-(4*a*c - b^2)^5)^(1/2) - 48*a^3*b*c^3 + 40*a^2*b^3*c^2 - 11*a*b^5*c - a*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(256*a^4*c^7 + b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c ^5 - 256*a^3*b^2*c^6)))^(3/4)*(32768*a^5*c^5 + x^(1/2)*(-(b^7 + b^2*(-(4*a *c - b^2)^5)^(1/2) - 48*a^3*b*c^3 + 40*a^2*b^3*c^2 - 11*a*b^5*c - a*c*(-(4 *a*c - b^2)^5)^(1/2))/(32*(256*a^4*c^7 + b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b ^4*c^5 - 256*a^3*b^2*c^6)))^(1/4)*(131072*a^5*c^6 + 8192*a^3*b^4*c^4 - 655 36*a^4*b^2*c^5) + 2048*a^3*b^4*c^3 - 16384*a^4*b^2*c^4))*(-(b^7 + b^2*(-(4 *a*c - b^2)^5)^(1/2) - 48*a^3*b*c^3 + 40*a^2*b^3*c^2 - 11*a*b^5*c - a*c*(- (4*a*c - b^2)^5)^(1/2))/(32*(256*a^4*c^7 + b^8*c^3 - 16*a*b^6*c^4 + 96*a^2 *b^4*c^5 - 256*a^3*b^2*c^6)))^(1/4)*1i + (x^(1/2)*(256*a^3*b^3*c - 768*a^4 *b*c^2) - (-(b^7 + b^2*(-(4*a*c - b^2)^5)^(1/2) - 48*a^3*b*c^3 + 40*a^2*b^ 3*c^2 - 11*a*b^5*c - a*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(256*a^4*c^7 + b^8* c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6)))^(3/4)*(32768*a^5* c^5 - x^(1/2)*(-(b^7 + b^2*(-(4*a*c - b^2)^5)^(1/2) - 48*a^3*b*c^3 + 40*a^ 2*b^3*c^2 - 11*a*b^5*c - a*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(256*a^4*c^7 + b^8*c^3 - 16*a*b^6*c^4 + 96*a^2*b^4*c^5 - 256*a^3*b^2*c^6)))^(1/4)*(131072 *a^5*c^6 + 8192*a^3*b^4*c^4 - 65536*a^4*b^2*c^5) + 2048*a^3*b^4*c^3 - 1638 4*a^4*b^2*c^4))*(-(b^7 + b^2*(-(4*a*c - b^2)^5)^(1/2) - 48*a^3*b*c^3 + 40* a^2*b^3*c^2 - 11*a*b^5*c - a*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(256*a^4*c...