3.5.2 \(\int \frac {1}{x^{5/2} (b x^2+c x^4)^{3/2}} \, dx\) [402]

Optimal. Leaf size=350 \[ \frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}+\frac {77 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b^4 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}+\frac {77 c \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}-\frac {77 c^2 \sqrt {b x^2+c x^4}}{15 b^4 x^{3/2}}-\frac {77 c^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{15/4} \sqrt {b x^2+c x^4}}+\frac {77 c^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{30 b^{15/4} \sqrt {b x^2+c x^4}} \]

[Out]

1/b/x^(7/2)/(c*x^4+b*x^2)^(1/2)+77/15*c^(5/2)*x^(3/2)*(c*x^2+b)/b^4/(b^(1/2)+x*c^(1/2))/(c*x^4+b*x^2)^(1/2)-11
/9*(c*x^4+b*x^2)^(1/2)/b^2/x^(11/2)+77/45*c*(c*x^4+b*x^2)^(1/2)/b^3/x^(7/2)-77/15*c^2*(c*x^4+b*x^2)^(1/2)/b^4/
x^(3/2)-77/15*c^(9/4)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))
)*EllipticE(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1
/2))^2)^(1/2)/b^(15/4)/(c*x^4+b*x^2)^(1/2)+77/30*c^(9/4)*x*(cos(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))^2)^(1/2)/co
s(2*arctan(c^(1/4)*x^(1/2)/b^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x^(1/2)/b^(1/4))),1/2*2^(1/2))*(b^(1/2)+x*
c^(1/2))*((c*x^2+b)/(b^(1/2)+x*c^(1/2))^2)^(1/2)/b^(15/4)/(c*x^4+b*x^2)^(1/2)

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Rubi [A]
time = 0.28, antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2048, 2050, 2057, 335, 311, 226, 1210} \begin {gather*} \frac {77 c^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{30 b^{15/4} \sqrt {b x^2+c x^4}}-\frac {77 c^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{15/4} \sqrt {b x^2+c x^4}}+\frac {77 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b^4 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {77 c^2 \sqrt {b x^2+c x^4}}{15 b^4 x^{3/2}}+\frac {77 c \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}+\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

1/(b*x^(7/2)*Sqrt[b*x^2 + c*x^4]) + (77*c^(5/2)*x^(3/2)*(b + c*x^2))/(15*b^4*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2
+ c*x^4]) - (11*Sqrt[b*x^2 + c*x^4])/(9*b^2*x^(11/2)) + (77*c*Sqrt[b*x^2 + c*x^4])/(45*b^3*x^(7/2)) - (77*c^2*
Sqrt[b*x^2 + c*x^4])/(15*b^4*x^(3/2)) - (77*c^(9/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c
]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*b^(15/4)*Sqrt[b*x^2 + c*x^4]) + (77*c^(9/4)*x
*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)]
, 1/2])/(30*b^(15/4)*Sqrt[b*x^2 + c*x^4])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 335

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 1210

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

Rule 2048

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-c^(j - 1))*(c*x)^(m - j
+ 1)*((a*x^j + b*x^n)^(p + 1)/(a*(n - j)*(p + 1))), x] + Dist[c^j*((m + n*p + n - j + 1)/(a*(n - j)*(p + 1)))
, Int[(c*x)^(m - j)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n]
&& (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1]

Rule 2050

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j,
n] && (IntegersQ[j, n] || GtQ[c, 0]) && LtQ[m + j*p + 1, 0]

Rule 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p])), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \frac {1}{x^{5/2} \left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}+\frac {11 \int \frac {1}{x^{9/2} \sqrt {b x^2+c x^4}} \, dx}{2 b}\\ &=\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}-\frac {(77 c) \int \frac {1}{x^{5/2} \sqrt {b x^2+c x^4}} \, dx}{18 b^2}\\ &=\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}+\frac {77 c \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}+\frac {\left (77 c^2\right ) \int \frac {1}{\sqrt {x} \sqrt {b x^2+c x^4}} \, dx}{30 b^3}\\ &=\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}+\frac {77 c \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}-\frac {77 c^2 \sqrt {b x^2+c x^4}}{15 b^4 x^{3/2}}+\frac {\left (77 c^3\right ) \int \frac {x^{3/2}}{\sqrt {b x^2+c x^4}} \, dx}{30 b^4}\\ &=\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}+\frac {77 c \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}-\frac {77 c^2 \sqrt {b x^2+c x^4}}{15 b^4 x^{3/2}}+\frac {\left (77 c^3 x \sqrt {b+c x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {b+c x^2}} \, dx}{30 b^4 \sqrt {b x^2+c x^4}}\\ &=\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}+\frac {77 c \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}-\frac {77 c^2 \sqrt {b x^2+c x^4}}{15 b^4 x^{3/2}}+\frac {\left (77 c^3 x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 b^4 \sqrt {b x^2+c x^4}}\\ &=\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}+\frac {77 c \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}-\frac {77 c^2 \sqrt {b x^2+c x^4}}{15 b^4 x^{3/2}}+\frac {\left (77 c^{5/2} x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 b^{7/2} \sqrt {b x^2+c x^4}}-\frac {\left (77 c^{5/2} x \sqrt {b+c x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {b}}}{\sqrt {b+c x^4}} \, dx,x,\sqrt {x}\right )}{15 b^{7/2} \sqrt {b x^2+c x^4}}\\ &=\frac {1}{b x^{7/2} \sqrt {b x^2+c x^4}}+\frac {77 c^{5/2} x^{3/2} \left (b+c x^2\right )}{15 b^4 \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {b x^2+c x^4}}-\frac {11 \sqrt {b x^2+c x^4}}{9 b^2 x^{11/2}}+\frac {77 c \sqrt {b x^2+c x^4}}{45 b^3 x^{7/2}}-\frac {77 c^2 \sqrt {b x^2+c x^4}}{15 b^4 x^{3/2}}-\frac {77 c^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{15 b^{15/4} \sqrt {b x^2+c x^4}}+\frac {77 c^{9/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{30 b^{15/4} \sqrt {b x^2+c x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.03, size = 60, normalized size = 0.17 \begin {gather*} -\frac {2 \sqrt {1+\frac {c x^2}{b}} \, _2F_1\left (-\frac {9}{4},\frac {3}{2};-\frac {5}{4};-\frac {c x^2}{b}\right )}{9 b x^{7/2} \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(b*x^2 + c*x^4)^(3/2)),x]

[Out]

(-2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-9/4, 3/2, -5/4, -((c*x^2)/b)])/(9*b*x^(7/2)*Sqrt[x^2*(b + c*x^2)])

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Maple [A]
time = 0.13, size = 237, normalized size = 0.68

method	result	size
default	\(\frac {\left (c \,x^{2}+b \right ) \left (462 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticE \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-231 \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \EllipticF \left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) b \,c^{2} x^{4}-462 c^{3} x^{6}-308 b \,c^{2} x^{4}+44 b^{2} c \,x^{2}-20 b^{3}\right )}{90 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} x^{\frac {3}{2}} b^{4}}\)	\(237\)
risch	\(-\frac {2 \left (c \,x^{2}+b \right ) \left (93 c^{2} x^{4}-16 b c \,x^{2}+5 b^{2}\right )}{45 b^{4} x^{\frac {7}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {c^{3} \left (\frac {31 \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{c \sqrt {c \,x^{3}+b x}}-15 b \left (\frac {x^{2}}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}-\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {x c}{\sqrt {-b c}}}\, \left (-\frac {2 \sqrt {-b c}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}+\frac {\sqrt {-b c}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{15 b^{4} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(431\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/90/(c*x^4+b*x^2)^(3/2)/x^(3/2)*(c*x^2+b)*(462*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^
(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(
1/2))*b*c^2*x^4-231*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(
-x*c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))*b*c^2*x^4-462*c^3*x^6-
308*b*c^2*x^4+44*b^2*c*x^2-20*b^3)/b^4

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^(5/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 108, normalized size = 0.31 \begin {gather*} -\frac {231 \, {\left (c^{3} x^{8} + b c^{2} x^{6}\right )} \sqrt {c} {\rm weierstrassZeta}\left (-\frac {4 \, b}{c}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right )\right ) + {\left (231 \, c^{3} x^{6} + 154 \, b c^{2} x^{4} - 22 \, b^{2} c x^{2} + 10 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{45 \, {\left (b^{4} c x^{8} + b^{5} x^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

-1/45*(231*(c^3*x^8 + b*c^2*x^6)*sqrt(c)*weierstrassZeta(-4*b/c, 0, weierstrassPInverse(-4*b/c, 0, x)) + (231*
c^3*x^6 + 154*b*c^2*x^4 - 22*b^2*c*x^2 + 10*b^3)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(b^4*c*x^8 + b^5*x^6)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{\frac {5}{2}} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Integral(1/(x**(5/2)*(x**2*(b + c*x**2))**(3/2)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{x^{5/2}\,{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(b*x^2 + c*x^4)^(3/2)),x)

[Out]

int(1/(x^(5/2)*(b*x^2 + c*x^4)^(3/2)), x)

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