3.326 \(\int \frac {1}{x^{7/2} (b x^2+c x^4)} \, dx\)

Optimal. Leaf size=230 \[ -\frac {c^{9/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{13/4}}+\frac {c^{9/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{13/4}}+\frac {c^{9/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{13/4}}-\frac {c^{9/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{13/4}}-\frac {2 c^2}{b^3 \sqrt {x}}+\frac {2 c}{5 b^2 x^{5/2}}-\frac {2}{9 b x^{9/2}} \]

[Out]

-2/9/b/x^(9/2)+2/5*c/b^2/x^(5/2)+1/2*c^(9/4)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(13/4)*2^(1/2)-1/2*c^
(9/4)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(13/4)*2^(1/2)-1/4*c^(9/4)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1
/4)*2^(1/2)*x^(1/2))/b^(13/4)*2^(1/2)+1/4*c^(9/4)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(13/
4)*2^(1/2)-2*c^2/b^3/x^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1584, 325, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac {2 c^2}{b^3 \sqrt {x}}-\frac {c^{9/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{13/4}}+\frac {c^{9/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{13/4}}+\frac {c^{9/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{13/4}}-\frac {c^{9/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{13/4}}+\frac {2 c}{5 b^2 x^{5/2}}-\frac {2}{9 b x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(7/2)*(b*x^2 + c*x^4)),x]

[Out]

-2/(9*b*x^(9/2)) + (2*c)/(5*b^2*x^(5/2)) - (2*c^2)/(b^3*Sqrt[x]) + (c^(9/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x
])/b^(1/4)])/(Sqrt[2]*b^(13/4)) - (c^(9/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*b^(13/4)) -
(c^(9/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(13/4)) + (c^(9/4)*Log[Sqrt
[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*b^(13/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {1}{x^{7/2} \left (b x^2+c x^4\right )} \, dx &=\int \frac {1}{x^{11/2} \left (b+c x^2\right )} \, dx\\ &=-\frac {2}{9 b x^{9/2}}-\frac {c \int \frac {1}{x^{7/2} \left (b+c x^2\right )} \, dx}{b}\\ &=-\frac {2}{9 b x^{9/2}}+\frac {2 c}{5 b^2 x^{5/2}}+\frac {c^2 \int \frac {1}{x^{3/2} \left (b+c x^2\right )} \, dx}{b^2}\\ &=-\frac {2}{9 b x^{9/2}}+\frac {2 c}{5 b^2 x^{5/2}}-\frac {2 c^2}{b^3 \sqrt {x}}-\frac {c^3 \int \frac {\sqrt {x}}{b+c x^2} \, dx}{b^3}\\ &=-\frac {2}{9 b x^{9/2}}+\frac {2 c}{5 b^2 x^{5/2}}-\frac {2 c^2}{b^3 \sqrt {x}}-\frac {\left (2 c^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=-\frac {2}{9 b x^{9/2}}+\frac {2 c}{5 b^2 x^{5/2}}-\frac {2 c^2}{b^3 \sqrt {x}}+\frac {c^{5/2} \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{b^3}-\frac {c^{5/2} \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=-\frac {2}{9 b x^{9/2}}+\frac {2 c}{5 b^2 x^{5/2}}-\frac {2 c^2}{b^3 \sqrt {x}}-\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{2 b^3}-\frac {c^2 \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{2 b^3}-\frac {c^{9/4} \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2} b^{13/4}}-\frac {c^{9/4} \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{2 \sqrt {2} b^{13/4}}\\ &=-\frac {2}{9 b x^{9/2}}+\frac {2 c}{5 b^2 x^{5/2}}-\frac {2 c^2}{b^3 \sqrt {x}}-\frac {c^{9/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{13/4}}+\frac {c^{9/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{13/4}}-\frac {c^{9/4} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{13/4}}+\frac {c^{9/4} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{13/4}}\\ &=-\frac {2}{9 b x^{9/2}}+\frac {2 c}{5 b^2 x^{5/2}}-\frac {2 c^2}{b^3 \sqrt {x}}+\frac {c^{9/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{13/4}}-\frac {c^{9/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{13/4}}-\frac {c^{9/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{13/4}}+\frac {c^{9/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{13/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 29, normalized size = 0.13 \[ -\frac {2 \, _2F_1\left (-\frac {9}{4},1;-\frac {5}{4};-\frac {c x^2}{b}\right )}{9 b x^{9/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Hypergeometric2F1[-9/4, 1, -5/4, -((c*x^2)/b)])/(9*b*x^(9/2))

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 204, normalized size = 0.89 \[ \frac {180 \, b^{3} x^{5} \left (-\frac {c^{9}}{b^{13}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b^{3} c^{7} \sqrt {x} \left (-\frac {c^{9}}{b^{13}}\right )^{\frac {1}{4}} - \sqrt {-b^{7} c^{9} \sqrt {-\frac {c^{9}}{b^{13}}} + c^{14} x} b^{3} \left (-\frac {c^{9}}{b^{13}}\right )^{\frac {1}{4}}}{c^{9}}\right ) - 45 \, b^{3} x^{5} \left (-\frac {c^{9}}{b^{13}}\right )^{\frac {1}{4}} \log \left (b^{10} \left (-\frac {c^{9}}{b^{13}}\right )^{\frac {3}{4}} + c^{7} \sqrt {x}\right ) + 45 \, b^{3} x^{5} \left (-\frac {c^{9}}{b^{13}}\right )^{\frac {1}{4}} \log \left (-b^{10} \left (-\frac {c^{9}}{b^{13}}\right )^{\frac {3}{4}} + c^{7} \sqrt {x}\right ) - 4 \, {\left (45 \, c^{2} x^{4} - 9 \, b c x^{2} + 5 \, b^{2}\right )} \sqrt {x}}{90 \, b^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90*(180*b^3*x^5*(-c^9/b^13)^(1/4)*arctan(-(b^3*c^7*sqrt(x)*(-c^9/b^13)^(1/4) - sqrt(-b^7*c^9*sqrt(-c^9/b^13)
+ c^14*x)*b^3*(-c^9/b^13)^(1/4))/c^9) - 45*b^3*x^5*(-c^9/b^13)^(1/4)*log(b^10*(-c^9/b^13)^(3/4) + c^7*sqrt(x)
) + 45*b^3*x^5*(-c^9/b^13)^(1/4)*log(-b^10*(-c^9/b^13)^(3/4) + c^7*sqrt(x)) - 4*(45*c^2*x^4 - 9*b*c*x^2 + 5*b^
2)*sqrt(x))/(b^3*x^5)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 199, normalized size = 0.87 \[ -\frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{4}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{4}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{4}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{4}} - \frac {2 \, {\left (45 \, c^{2} x^{4} - 9 \, b c x^{2} + 5 \, b^{2}\right )}}{45 \, b^{3} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*(b*c^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/b^4 - 1/2*sqrt(2)
*(b*c^3)^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^4 + 1/4*sqrt(2)*(b*c^3)^(3
/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^4 - 1/4*sqrt(2)*(b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c
)^(1/4) + x + sqrt(b/c))/b^4 - 2/45*(45*c^2*x^4 - 9*b*c*x^2 + 5*b^2)/(b^3*x^(9/2))

________________________________________________________________________________________

maple [A]  time = 0.01, size = 169, normalized size = 0.73 \[ -\frac {\sqrt {2}\, c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {\sqrt {2}\, c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {\sqrt {2}\, c^{2} \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} b^{3}}-\frac {2 c^{2}}{b^{3} \sqrt {x}}+\frac {2 c}{5 b^{2} x^{\frac {5}{2}}}-\frac {2}{9 b \,x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c^2/b^3/(b/c)^(1/4)*2^(1/2)*ln((x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x+(b/c)^(1/4)*2^(1/2)*x^(1/2)
+(b/c)^(1/2)))-1/2*c^2/b^3/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-1/2*c^2/b^3/(b/c)^(1/4)*2
^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)-2/9/b/x^(9/2)-2*c^2/b^3/x^(1/2)+2/5*c/b^2/x^(5/2)

________________________________________________________________________________________

maxima [A]  time = 3.09, size = 209, normalized size = 0.91 \[ -\frac {c^{3} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{4 \, b^{3}} - \frac {2 \, {\left (45 \, c^{2} x^{4} - 9 \, b c x^{2} + 5 \, b^{2}\right )}}{45 \, b^{3} x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*c^3*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(s
qrt(sqrt(b)*sqrt(c))*sqrt(c)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sq
rt(sqrt(b)*sqrt(c)))/(sqrt(sqrt(b)*sqrt(c))*sqrt(c)) - sqrt(2)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x
+ sqrt(b))/(b^(1/4)*c^(3/4)) + sqrt(2)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(1/4)*c
^(3/4)))/b^3 - 2/45*(45*c^2*x^4 - 9*b*c*x^2 + 5*b^2)/(b^3*x^(9/2))

________________________________________________________________________________________

mupad [B]  time = 4.47, size = 77, normalized size = 0.33 \[ \frac {{\left (-c\right )}^{9/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{13/4}}-\frac {{\left (-c\right )}^{9/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{13/4}}-\frac {\frac {2}{9\,b}-\frac {2\,c\,x^2}{5\,b^2}+\frac {2\,c^2\,x^4}{b^3}}{x^{9/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(7/2)*(b*x^2 + c*x^4)),x)

[Out]

((-c)^(9/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4)))/b^(13/4) - ((-c)^(9/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4)))/b^
(13/4) - (2/(9*b) - (2*c*x^2)/(5*b^2) + (2*c^2*x^4)/b^3)/x^(9/2)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________