\(\int \frac {1}{x^{7/2} (a+b x^2+c x^4)} \, dx\) [929]

Optimal result

Integrand size = 20, antiderivative size = 412 \[ \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx=-\frac {2}{5 a x^{5/2}}+\frac {2 b}{a^2 \sqrt {x}}+\frac {\sqrt [4]{c} \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2^{3/4} a^2 \sqrt [4]{-b-\sqrt {b^2-4 a c}}}+\frac {\sqrt [4]{c} \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2^{3/4} a^2 \sqrt [4]{-b+\sqrt {b^2-4 a c}}}-\frac {\sqrt [4]{c} \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{2^{3/4} a^2 \sqrt [4]{-b-\sqrt {b^2-4 a c}}}-\frac {\sqrt [4]{c} \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{2^{3/4} a^2 \sqrt [4]{-b+\sqrt {b^2-4 a c}}} \] Output:

-2/5/a/x^(5/2)+2*b/a^2/x^(1/2)+1/2*c^(1/4)*(b-(-2*a*c+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/(-b-(-4*a*c+b^2)^(1/2))^(1/4)+1/2*c^(1/4)*(b+(-2*a*c+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/(-b+(-4*a*c+b^2)^(1/2))^(1/4)-1/2*c^(1/4)*(b-(-2*a*c+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/(-b-(-4*a*c+b^2)^(1/2))^(1/4)-1/2*c^(1/4)*(b+(-2*a*c+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/(-b+(-4*a*c+b^2)^(1/2))^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx=\frac {-\frac {4 \left (a-5 b x^2\right )}{x^{5/2}}+5 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b^2 \log \left (\sqrt {x}-\text {$\#$1}\right )-a c \log \left (\sqrt {x}-\text {$\#$1}\right )+b c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}+2 c \text {$\#$1}^5}\&\right ]}{10 a^2} \] Input:

Integrate[1/(x^(7/2)*(a + b*x^2 + c*x^4)),x]
 

Output:

((-4*(a - 5*b*x^2))/x^(5/2) + 5*RootSum[a + b*#1^4 + c*#1^8 & , (b^2*Log[S 
qrt[x] - #1] - a*c*Log[Sqrt[x] - #1] + b*c*Log[Sqrt[x] - #1]*#1^4)/(b*#1 + 
 2*c*#1^5) & ])/(10*a^2)
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 383, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {1435, 1704, 27, 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.

\(\displaystyle \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx\)

\(\Big \downarrow \) 1435

\(\displaystyle 2 \int \frac {1}{x^3 \left (c x^4+b x^2+a\right )}d\sqrt {x}\)

\(\Big \downarrow \) 1704

\(\displaystyle 2 \left (\frac {\int -\frac {5 \left (c x^2+b\right )}{x \left (c x^4+b x^2+a\right )}d\sqrt {x}}{5 a}-\frac {1}{5 a x^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (-\frac {\int \frac {c x^2+b}{x \left (c x^4+b x^2+a\right )}d\sqrt {x}}{a}-\frac {1}{5 a x^{5/2}}\right )\)

\(\Big \downarrow \) 1828

\(\displaystyle 2 \left (-\frac {-\frac {\int \frac {x \left (b^2+c x^2 b-a c\right )}{c x^4+b x^2+a}d\sqrt {x}}{a}-\frac {b}{a \sqrt {x}}}{a}-\frac {1}{5 a x^{5/2}}\right )\)

\(\Big \downarrow \) 1834

\(\displaystyle 2 \left (-\frac {-\frac {\frac {1}{2} c \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \int \frac {2 x}{2 c x^2+b-\sqrt {b^2-4 a c}}d\sqrt {x}+\frac {1}{2} c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \int \frac {2 x}{2 c x^2+b+\sqrt {b^2-4 a c}}d\sqrt {x}}{a}-\frac {b}{a \sqrt {x}}}{a}-\frac {1}{5 a x^{5/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle 2 \left (-\frac {-\frac {c \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \int \frac {x}{2 c x^2+b-\sqrt {b^2-4 a c}}d\sqrt {x}+c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \int \frac {x}{2 c x^2+b+\sqrt {b^2-4 a c}}d\sqrt {x}}{a}-\frac {b}{a \sqrt {x}}}{a}-\frac {1}{5 a x^{5/2}}\right )\)

\(\Big \downarrow \) 827

\(\displaystyle 2 \left (-\frac {-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {-b-\sqrt {b^2-4 a c}}}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )+c \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \left (\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {\sqrt {b^2-4 a c}-b}}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )}{a}-\frac {b}{a \sqrt {x}}}{a}-\frac {1}{5 a x^{5/2}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle 2 \left (-\frac {-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )+c \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{2 \sqrt {2} \sqrt {c}}\right )}{a}-\frac {b}{a \sqrt {x}}}{a}-\frac {1}{5 a x^{5/2}}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle 2 \left (-\frac {-\frac {c \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )+c \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \left (\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{2\ 2^{3/4} c^{3/4} \sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{a}-\frac {b}{a \sqrt {x}}}{a}-\frac {1}{5 a x^{5/2}}\right )\)

Input:

Int[1/(x^(7/2)*(a + b*x^2 + c*x^4)),x]
 

Output:

2*(-1/5*1/(a*x^(5/2)) - (-(b/(a*Sqrt[x])) - (c*(b - (b^2 - 2*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)) - 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 - S 
qrt[b^2 - 4*a*c])^(1/4))) + c*(b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(ArcTa 
n[(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)/a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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]
 

rule 221

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]
 

rule 827

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]
 

rule 1435

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]
 

rule 1704

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]
 

rule 1828

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]
 

rule 1834

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]
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.20

method	result	size
risch	\(-\frac {2 \left (-5 b \,x^{2}+a \right )}{5 a^{2} x^{\frac {5}{2}}}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (b c \,\textit {\_R}^{6}+\left (-a c +b^{2}\right ) \textit {\_R}^{2}\right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{2 a^{2}}\)	\(81\)
derivativedivides	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (-b c \,\textit {\_R}^{6}+\left (a c -b^{2}\right ) \textit {\_R}^{2}\right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{2 a^{2}}-\frac {2}{5 a \,x^{\frac {5}{2}}}+\frac {2 b}{a^{2} \sqrt {x}}\)	\(84\)
default	\(-\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (-b c \,\textit {\_R}^{6}+\left (a c -b^{2}\right ) \textit {\_R}^{2}\right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{2 a^{2}}-\frac {2}{5 a \,x^{\frac {5}{2}}}+\frac {2 b}{a^{2} \sqrt {x}}\)	\(84\)

Input:

int(1/x^(7/2)/(c*x^4+b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

-2/5*(-5*b*x^2+a)/a^2/x^(5/2)+1/2/a^2*sum((b*c*_R^6+(-a*c+b^2)*_R^2)/(2*_R 
^7*c+_R^3*b)*ln(x^(1/2)-_R),_R=RootOf(_Z^8*c+_Z^4*b+a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8376 vs. \(2 (328) = 656\).

Time = 6.70 (sec) , antiderivative size = 8376, normalized size of antiderivative = 20.33 \[ \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \] Input:

integrate(1/x^(7/2)/(c*x^4+b*x^2+a),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x**(7/2)/(c*x**4+b*x**2+a),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} x^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/x^(7/2)/(c*x^4+b*x^2+a),x, algorithm="maxima")
 

Output:

2/5*(5*b/sqrt(x) - a/x^(5/2))/a^2 + integrate((b*c*x^(5/2) + (b^2 - a*c)*s 
qrt(x))/(a^2*c*x^4 + a^2*b*x^2 + a^3), x)
 

Giac [F]

\[ \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2} + a\right )} x^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/x^(7/2)/(c*x^4+b*x^2+a),x, algorithm="giac")
 

Output:

integrate(1/((c*x^4 + b*x^2 + a)*x^(7/2)), x)
 

Mupad [B] (verification not implemented)

Time = 19.64 (sec) , antiderivative size = 15149, normalized size of antiderivative = 36.77 \[ \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx=\text {Too large to display} \] Input:

int(1/(x^(7/2)*(a + b*x^2 + c*x^4)),x)
 

Output:

atan((((-(b^13 + b^8*(-(4*a*c - b^2)^5)^(1/2) + 144*a^6*b*c^6 + 115*a^2*b^ 
9*c^2 - 390*a^3*b^7*c^3 + 681*a^4*b^5*c^4 - 552*a^5*b^3*c^5 + a^4*c^4*(-(4 
*a*c - b^2)^5)^(1/2) - 17*a*b^11*c + 15*a^2*b^4*c^2*(-(4*a*c - b^2)^5)^(1/ 
2) - 10*a^3*b^2*c^3*(-(4*a*c - b^2)^5)^(1/2) - 7*a*b^6*c*(-(4*a*c - b^2)^5 
)^(1/2))/(32*(a^9*b^8 + 256*a^13*c^4 - 16*a^10*b^6*c + 96*a^11*b^4*c^2 - 2 
56*a^12*b^2*c^3)))^(3/4)*(x^(1/2)*(-(b^13 + b^8*(-(4*a*c - b^2)^5)^(1/2) + 
 144*a^6*b*c^6 + 115*a^2*b^9*c^2 - 390*a^3*b^7*c^3 + 681*a^4*b^5*c^4 - 552 
*a^5*b^3*c^5 + a^4*c^4*(-(4*a*c - b^2)^5)^(1/2) - 17*a*b^11*c + 15*a^2*b^4 
*c^2*(-(4*a*c - b^2)^5)^(1/2) - 10*a^3*b^2*c^3*(-(4*a*c - b^2)^5)^(1/2) - 
7*a*b^6*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(a^9*b^8 + 256*a^13*c^4 - 16*a^10* 
b^6*c + 96*a^11*b^4*c^2 - 256*a^12*b^2*c^3)))^(1/4)*(131072*a^28*c^9 - 409 
6*a^23*b^10*c^4 + 57344*a^24*b^8*c^5 - 299008*a^25*b^6*c^6 + 696320*a^26*b 
^4*c^7 - 655360*a^27*b^2*c^8) - 131072*a^26*b*c^9 + 2048*a^21*b^11*c^4 - 2 
8672*a^22*b^9*c^5 + 151552*a^23*b^7*c^6 - 368640*a^24*b^5*c^7 + 393216*a^2 
5*b^3*c^8) + x^(1/2)*(768*a^21*b*c^11 - 256*a^20*b^3*c^10))*(-(b^13 + b^8* 
(-(4*a*c - b^2)^5)^(1/2) + 144*a^6*b*c^6 + 115*a^2*b^9*c^2 - 390*a^3*b^7*c 
^3 + 681*a^4*b^5*c^4 - 552*a^5*b^3*c^5 + a^4*c^4*(-(4*a*c - b^2)^5)^(1/2) 
- 17*a*b^11*c + 15*a^2*b^4*c^2*(-(4*a*c - b^2)^5)^(1/2) - 10*a^3*b^2*c^3*( 
-(4*a*c - b^2)^5)^(1/2) - 7*a*b^6*c*(-(4*a*c - b^2)^5)^(1/2))/(32*(a^9*b^8 
 + 256*a^13*c^4 - 16*a^10*b^6*c + 96*a^11*b^4*c^2 - 256*a^12*b^2*c^3)))...
 

Reduce [F]

\[ \int \frac {1}{x^{7/2} \left (a+b x^2+c x^4\right )} \, dx=\int \frac {1}{x^{\frac {7}{2}} \left (c \,x^{4}+b \,x^{2}+a \right )}d x \] Input:

int(1/x^(7/2)/(c*x^4+b*x^2+a),x)
 

Output:

int(1/x^(7/2)/(c*x^4+b*x^2+a),x)