Integrand size = 20, antiderivative size = 573 \[ \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=-\frac {5 b^2-18 a c}{2 a^2 \left (b^2-4 a c\right ) \sqrt {x}}+\frac {b^2-2 a c+b c x^2}{2 a \left (b^2-4 a c\right ) \sqrt {x} \left (a+b x^2+c x^4\right )}+\frac {\sqrt [4]{c} \left (5 b^3-28 a b c-\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-b-\sqrt {b^2-4 a c}}}-\frac {\sqrt [4]{c} \left (5 b^3-28 a b c+\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-b+\sqrt {b^2-4 a c}}}-\frac {\sqrt [4]{c} \left (5 b^3-28 a b c-\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-b-\sqrt {b^2-4 a c}}}+\frac {\sqrt [4]{c} \left (5 b^3-28 a b c+\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{4\ 2^{3/4} a^2 \left (b^2-4 a c\right )^{3/2} \sqrt [4]{-b+\sqrt {b^2-4 a c}}} \] Output:
-1/2*(-18*a*c+5*b^2)/a^2/(-4*a*c+b^2)/x^(1/2)+1/2*(b*c*x^2-2*a*c+b^2)/a/(- 4*a*c+b^2)/x^(1/2)/(c*x^4+b*x^2+a)+1/8*c^(1/4)*(5*b^3-28*a*b*c-(-18*a*c+5* b^2)*(-4*a*c+b^2)^(1/2))*arctan(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^( 1/2))^(1/4))*2^(1/4)/a^2/(-4*a*c+b^2)^(3/2)/(-b-(-4*a*c+b^2)^(1/2))^(1/4)- 1/8*c^(1/4)*(5*b^3-28*a*b*c+(-18*a*c+5*b^2)*(-4*a*c+b^2)^(1/2))*arctan(2^( 1/4)*c^(1/4)*x^(1/2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*2^(1/4)/a^2/(-4*a*c+b^ 2)^(3/2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4)-1/8*c^(1/4)*(5*b^3-28*a*b*c-(-18*a* c+5*b^2)*(-4*a*c+b^2)^(1/2))*arctanh(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b ^2)^(1/2))^(1/4))*2^(1/4)/a^2/(-4*a*c+b^2)^(3/2)/(-b-(-4*a*c+b^2)^(1/2))^( 1/4)+1/8*c^(1/4)*(5*b^3-28*a*b*c+(-18*a*c+5*b^2)*(-4*a*c+b^2)^(1/2))*arcta nh(2^(1/4)*c^(1/4)*x^(1/2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*2^(1/4)/a^2/(-4* a*c+b^2)^(3/2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.49 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.46 \[ \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=-\frac {-\frac {4 \left (16 a^2 c-5 b^2 x^2 \left (b+c x^2\right )+a \left (-4 b^2+19 b c x^2+18 c^2 x^4\right )\right )}{\left (b^2-4 a c\right ) \sqrt {x} \left (a+b x^2+c x^4\right )}+4 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log \left (\sqrt {x}-\text {$\#$1}\right )+c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ]+\frac {\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b^3 \log \left (\sqrt {x}-\text {$\#$1}\right )-7 a b c \log \left (\sqrt {x}-\text {$\#$1}\right )+b^2 c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4-2 a c^2 \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ]}{b^2-4 a c}}{8 a^2} \] Input:
Integrate[1/(x^(3/2)*(a + b*x^2 + c*x^4)^2),x]
Output:
-1/8*((-4*(16*a^2*c - 5*b^2*x^2*(b + c*x^2) + a*(-4*b^2 + 19*b*c*x^2 + 18* c^2*x^4)))/((b^2 - 4*a*c)*Sqrt[x]*(a + b*x^2 + c*x^4)) + 4*RootSum[a + b*# 1^4 + c*#1^8 & , (b*Log[Sqrt[x] - #1] + c*Log[Sqrt[x] - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ] + RootSum[a + b*#1^4 + c*#1^8 & , (b^3*Log[Sqrt[x] - #1] - 7 *a*b*c*Log[Sqrt[x] - #1] + b^2*c*Log[Sqrt[x] - #1]*#1^4 - 2*a*c^2*Log[Sqrt [x] - #1]*#1^4)/(b*#1 + 2*c*#1^5) & ]/(b^2 - 4*a*c))/a^2
Time = 1.47 (sec) , antiderivative size = 487, normalized size of antiderivative = 0.85, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1435, 1702, 25, 1828, 1834, 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.
| \(\displaystyle \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 1435 |
| \(\displaystyle 2 \int \frac {1}{x \left (c x^4+b x^2+a\right )^2}d\sqrt {x}\) |
| \(\Big \downarrow \) 1702 |
| \(\displaystyle 2 \left (\frac {-2 a c+b^2+b c x^2}{4 a \sqrt {x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int -\frac {5 b^2+5 c x^2 b-18 a c}{x \left (c x^4+b x^2+a\right )}d\sqrt {x}}{4 a \left (b^2-4 a c\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle 2 \left (\frac {\int \frac {5 b^2+5 c x^2 b-18 a c}{x \left (c x^4+b x^2+a\right )}d\sqrt {x}}{4 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{4 a \sqrt {x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1828 |
| \(\displaystyle 2 \left (\frac {-\frac {\int \frac {x \left (c \left (5 b^2-18 a c\right ) x^2+b \left (5 b^2-23 a c\right )\right )}{c x^4+b x^2+a}d\sqrt {x}}{a}-\frac {5 b^2-18 a c}{a \sqrt {x}}}{4 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{4 a \sqrt {x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 1834 |
| \(\displaystyle 2 \left (\frac {-\frac {\frac {1}{2} c \left (-\frac {28 a b c}{\sqrt {b^2-4 a c}}+\frac {5 b^3}{\sqrt {b^2-4 a c}}-18 a c+5 b^2\right ) \int \frac {2 x}{2 c x^2+b-\sqrt {b^2-4 a c}}d\sqrt {x}-\frac {c \left (-\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}-28 a b c+5 b^3\right ) \int \frac {2 x}{2 c x^2+b+\sqrt {b^2-4 a c}}d\sqrt {x}}{2 \sqrt {b^2-4 a c}}}{a}-\frac {5 b^2-18 a c}{a \sqrt {x}}}{4 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{4 a \sqrt {x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 2 \left (\frac {-\frac {c \left (-\frac {28 a b c}{\sqrt {b^2-4 a c}}+\frac {5 b^3}{\sqrt {b^2-4 a c}}-18 a c+5 b^2\right ) \int \frac {x}{2 c x^2+b-\sqrt {b^2-4 a c}}d\sqrt {x}-\frac {c \left (-\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}-28 a b c+5 b^3\right ) \int \frac {x}{2 c x^2+b+\sqrt {b^2-4 a c}}d\sqrt {x}}{\sqrt {b^2-4 a c}}}{a}-\frac {5 b^2-18 a c}{a \sqrt {x}}}{4 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{4 a \sqrt {x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle 2 \left (\frac {-\frac {c \left (-\frac {28 a b c}{\sqrt {b^2-4 a c}}+\frac {5 b^3}{\sqrt {b^2-4 a c}}-18 a c+5 b^2\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {\sqrt {b^2-4 a c}-b}}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )-\frac {c \left (-\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}-28 a b c+5 b^3\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {-b-\sqrt {b^2-4 a c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )}{\sqrt {b^2-4 a c}}}{a}-\frac {5 b^2-18 a c}{a \sqrt {x}}}{4 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{4 a \sqrt {x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle 2 \left (\frac {-\frac {c \left (-\frac {28 a b c}{\sqrt {b^2-4 a c}}+\frac {5 b^3}{\sqrt {b^2-4 a c}}-18 a c+5 b^2\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )-\frac {c \left (-\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}-28 a b c+5 b^3\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )}{\sqrt {b^2-4 a c}}}{a}-\frac {5 b^2-18 a c}{a \sqrt {x}}}{4 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{4 a \sqrt {x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle 2 \left (\frac {-\frac {c \left (-\frac {28 a b c}{\sqrt {b^2-4 a c}}+\frac {5 b^3}{\sqrt {b^2-4 a c}}-18 a c+5 b^2\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )-\frac {c \left (-\left (5 b^2-18 a c\right ) \sqrt {b^2-4 a c}-28 a b c+5 b^3\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt {b^2-4 a c}}}{a}-\frac {5 b^2-18 a c}{a \sqrt {x}}}{4 a \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{4 a \sqrt {x} \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}\right )\) |
Input:
Int[1/(x^(3/2)*(a + b*x^2 + c*x^4)^2),x]
Output:
2*((b^2 - 2*a*c + b*c*x^2)/(4*a*(b^2 - 4*a*c)*Sqrt[x]*(a + b*x^2 + c*x^4)) + (-((5*b^2 - 18*a*c)/(a*Sqrt[x])) - (-((c*(5*b^3 - 28*a*b*c - (5*b^2 - 1 8*a*c)*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)) - Arc Tanh[(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))))/Sqrt[b^2 - 4*a*c]) + c*(5*b^2 - 18*a*c + (5*b^3)/Sqrt[b^2 - 4*a*c] - (28*a*b*c)/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))))/a)/(4*a*(b^2 - 4*a*c)))
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_))^(p_), x _Symbol] :> Simp[(-(d*x)^(m + 1))*(b^2 - 2*a*c + b*c*x^n)*((a + b*x^n + c*x ^(2*n))^(p + 1)/(a*d*n*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/(a*n*(p + 1)*(b ^2 - 4*a*c)) Int[(d*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[b^2*(m + n* (p + 1) + 1) - 2*a*c*(m + 2*n*(p + 1) + 1) + b*c*(m + n*(2*p + 3) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c , 0] && IGtQ[n, 0] && ILtQ[p, -1]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + ( c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[d*(f*x)^(m + 1)*((a + b*x^n + c*x^ (2*n))^(p + 1)/(a*f*(m + 1))), x] + Simp[1/(a*f^n*(m + 1)) Int[(f*x)^(m + n)*(a + b*x^n + c*x^(2*n))^p*Simp[a*e*(m + 1) - b*d*(m + n*(p + 1) + 1) - c*d*(m + 2*n*(p + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x ] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && LtQ[m, -1] && Int egerQ[p]
Int[(((f_.)*(x_))^(m_.)*((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[(f*x)^m/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[(f*x)^m/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ [{a, b, c, d, e, f, m}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n , 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.24 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.30
| method | result | size |
| derivativedivides | \(-\frac {2}{a^{2} \sqrt {x}}-\frac {2 \left (\frac {\frac {c \left (2 a c -b^{2}\right ) x^{\frac {7}{2}}}{16 a c -4 b^{2}}+\frac {b \left (3 a c -b^{2}\right ) x^{\frac {3}{2}}}{16 a c -4 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (c \left (18 a c -5 b^{2}\right ) \textit {\_R}^{6}+b \left (23 a c -5 b^{2}\right ) \textit {\_R}^{2}\right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{64 a c -16 b^{2}}\right )}{a^{2}}\) | \(172\) |
| default | \(-\frac {2}{a^{2} \sqrt {x}}-\frac {2 \left (\frac {\frac {c \left (2 a c -b^{2}\right ) x^{\frac {7}{2}}}{16 a c -4 b^{2}}+\frac {b \left (3 a c -b^{2}\right ) x^{\frac {3}{2}}}{16 a c -4 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (c \left (18 a c -5 b^{2}\right ) \textit {\_R}^{6}+b \left (23 a c -5 b^{2}\right ) \textit {\_R}^{2}\right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{64 a c -16 b^{2}}\right )}{a^{2}}\) | \(172\) |
| risch | \(-\frac {2}{a^{2} \sqrt {x}}-\frac {\frac {\frac {2 c \left (2 a c -b^{2}\right ) x^{\frac {7}{2}}}{16 a c -4 b^{2}}+\frac {2 b \left (3 a c -b^{2}\right ) x^{\frac {3}{2}}}{16 a c -4 b^{2}}}{c \,x^{4}+b \,x^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (c \left (18 a c -5 b^{2}\right ) \textit {\_R}^{6}+b \left (23 a c -5 b^{2}\right ) \textit {\_R}^{2}\right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{32 a c -8 b^{2}}}{a^{2}}\) | \(173\) |
Input:
int(1/x^(3/2)/(c*x^4+b*x^2+a)^2,x,method=_RETURNVERBOSE)
Output:
-2/a^2/x^(1/2)-2/a^2*((1/4*c*(2*a*c-b^2)/(4*a*c-b^2)*x^(7/2)+1/4*b*(3*a*c- b^2)/(4*a*c-b^2)*x^(3/2))/(c*x^4+b*x^2+a)+1/16/(4*a*c-b^2)*sum((c*(18*a*c- 5*b^2)*_R^6+b*(23*a*c-5*b^2)*_R^2)/(2*_R^7*c+_R^3*b)*ln(x^(1/2)-_R),_R=Roo tOf(_Z^8*c+_Z^4*b+a)))
Leaf count of result is larger than twice the leaf count of optimal. 16004 vs. \(2 (469) = 938\).
Time = 80.48 (sec) , antiderivative size = 16004, normalized size of antiderivative = 27.93 \[ \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:
integrate(1/x^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \] Input:
integrate(1/x**(3/2)/(c*x**4+b*x**2+a)**2,x)
Output:
Timed out
\[ \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )}^{2} x^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/x^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm="maxima")
Output:
-1/2*((5*b^2*c - 18*a*c^2)*x^(7/2) + (5*b^3 - 19*a*b*c)*x^(3/2) + 4*(a*b^2 - 4*a^2*c)/sqrt(x))/(a^3*b^2 - 4*a^4*c + (a^2*b^2*c - 4*a^3*c^2)*x^4 + (a ^2*b^3 - 4*a^3*b*c)*x^2) - integrate(1/4*((5*b^2*c - 18*a*c^2)*x^(5/2) + ( 5*b^3 - 23*a*b*c)*sqrt(x))/(a^3*b^2 - 4*a^4*c + (a^2*b^2*c - 4*a^3*c^2)*x^ 4 + (a^2*b^3 - 4*a^3*b*c)*x^2), x)
\[ \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )}^{2} x^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/x^(3/2)/(c*x^4+b*x^2+a)^2,x, algorithm="giac")
Output:
integrate(1/((c*x^4 + b*x^2 + a)^2*x^(3/2)), x)
Time = 25.43 (sec) , antiderivative size = 31145, normalized size of antiderivative = 54.35 \[ \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \] Input:
int(1/(x^(3/2)*(a + b*x^2 + c*x^4)^2),x)
Output:
atan(((x^(1/2)*(602332119171072*a^31*b*c^21 - 54080000*a^20*b^23*c^10 + 26 04992000*a^21*b^21*c^11 - 57034444800*a^22*b^19*c^12 + 749118545920*a^23*b ^17*c^13 - 6557747642368*a^24*b^15*c^14 + 40169229778944*a^25*b^13*c^15 - 175670703423488*a^26*b^11*c^16 + 548447002296320*a^27*b^9*c^17 - 119782124 8143360*a^28*b^7*c^18 + 1742819580444672*a^29*b^5*c^19 - 1520311317037056* a^30*b^3*c^20) + (-(625*b^25 - 625*b^10*(-(4*a*c - b^2)^15)^(1/2) + 310542 3360*a^12*b*c^12 + 638475*a^2*b^21*c^2 - 8264990*a^3*b^19*c^3 + 71483001*a ^4*b^17*c^4 - 434478624*a^5*b^15*c^5 + 1898983360*a^6*b^13*c^6 - 599668992 0*a^7*b^11*c^7 + 13524825600*a^8*b^9*c^8 - 21122310144*a^9*b^7*c^9 + 21483 012096*a^10*b^5*c^10 - 12575047680*a^11*b^3*c^11 + 26244*a^5*c^5*(-(4*a*c - b^2)^15)^(1/2) - 29625*a*b^23*c - 68475*a^2*b^6*c^2*(-(4*a*c - b^2)^15)^ (1/2) + 181990*a^3*b^4*c^3*(-(4*a*c - b^2)^15)^(1/2) - 171801*a^4*b^2*c^4* (-(4*a*c - b^2)^15)^(1/2) + 10875*a*b^8*c*(-(4*a*c - b^2)^15)^(1/2))/(8192 *(a^9*b^24 + 16777216*a^21*c^12 - 48*a^10*b^22*c + 1056*a^11*b^20*c^2 - 14 080*a^12*b^18*c^3 + 126720*a^13*b^16*c^4 - 811008*a^14*b^14*c^5 + 3784704* a^15*b^12*c^6 - 12976128*a^16*b^10*c^7 + 32440320*a^17*b^8*c^8 - 57671680* a^18*b^6*c^9 + 69206016*a^19*b^4*c^10 - 50331648*a^20*b^2*c^11)))^(3/4)*(3 2768000*a^21*b^34*c^4 - 25649407252758528*a^38*c^21 - 2123366400*a^22*b^32 *c^5 + 64398295040*a^23*b^30*c^6 - 1213399564288*a^24*b^28*c^7 + 158983630 35648*a^25*b^26*c^8 - 153599583715328*a^26*b^24*c^9 + 1132021560639488*...
\[ \int \frac {1}{x^{3/2} \left (a+b x^2+c x^4\right )^2} \, dx=\int \frac {1}{x^{\frac {3}{2}} \left (c \,x^{4}+b \,x^{2}+a \right )^{2}}d x \] Input:
int(1/x^(3/2)/(c*x^4+b*x^2+a)^2,x)
Output:
int(1/x^(3/2)/(c*x^4+b*x^2+a)^2,x)