\(\int \frac {1}{x^{3/2} (b x^2+c x^4)} \, dx\) [324]

Optimal result

Integrand size = 19, antiderivative size = 215 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=-\frac {2}{5 b x^{5/2}}+\frac {2 c}{b^2 \sqrt {x}}-\frac {c^{5/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{9/4}}+\frac {c^{5/4} \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{9/4}}+\frac {c^{5/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{9/4}}-\frac {c^{5/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} b^{9/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1598, 331, 335, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=-\frac {c^{5/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} b^{9/4}}+\frac {c^{5/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} b^{9/4}}+\frac {c^{5/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{9/4}}-\frac {c^{5/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} b^{9/4}}+\frac {2 c}{b^2 \sqrt {x}}-\frac {2}{5 b x^{5/2}} \]

[In]

Int[1/(x^(3/2)*(b*x^2 + c*x^4)),x]

[Out]

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

Rule 210

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

Rule 303

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 331

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 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 631

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 642

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 1176

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 1179

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 1598

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {-\frac {4 \sqrt [4]{b} \left (b-5 c x^2\right )}{x^{5/2}}-5 \sqrt {2} c^{5/4} \arctan \left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-5 \sqrt {2} c^{5/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{10 b^{9/4}} \]

[In]

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

[Out]

((-4*b^(1/4)*(b - 5*c*x^2))/x^(5/2) - 5*Sqrt[2]*c^(5/4)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*
Sqrt[x])] - 5*Sqrt[2]*c^(5/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(10*b^(9/4))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.58

method	result	size
risch	\(-\frac {2 \left (-5 c \,x^{2}+b \right )}{5 b^{2} x^{\frac {5}{2}}}+\frac {c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(124\)
derivativedivides	\(-\frac {2}{5 b \,x^{\frac {5}{2}}}+\frac {2 c}{b^{2} \sqrt {x}}+\frac {c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(125\)
default	\(-\frac {2}{5 b \,x^{\frac {5}{2}}}+\frac {2 c}{b^{2} \sqrt {x}}+\frac {c \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 b^{2} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(125\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {5 \, b^{2} x^{3} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (b^{7} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {3}{4}} + c^{4} \sqrt {x}\right ) - 5 i \, b^{2} x^{3} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (i \, b^{7} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {3}{4}} + c^{4} \sqrt {x}\right ) + 5 i \, b^{2} x^{3} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-i \, b^{7} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {3}{4}} + c^{4} \sqrt {x}\right ) - 5 \, b^{2} x^{3} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-b^{7} \left (-\frac {c^{5}}{b^{9}}\right )^{\frac {3}{4}} + c^{4} \sqrt {x}\right ) + 4 \, {\left (5 \, c x^{2} - b\right )} \sqrt {x}}{10 \, b^{2} x^{3}} \]

[In]

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

[Out]

1/10*(5*b^2*x^3*(-c^5/b^9)^(1/4)*log(b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) - 5*I*b^2*x^3*(-c^5/b^9)^(1/4)*log(I*
b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) + 5*I*b^2*x^3*(-c^5/b^9)^(1/4)*log(-I*b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x))
- 5*b^2*x^3*(-c^5/b^9)^(1/4)*log(-b^7*(-c^5/b^9)^(3/4) + c^4*sqrt(x)) + 4*(5*c*x^2 - b)*sqrt(x))/(b^2*x^3)

Sympy [A] (verification not implemented)

Time = 25.62 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.65 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\begin {cases} \frac {\tilde {\infty }}{x^{\frac {9}{2}}} & \text {for}\: b = 0 \wedge c = 0 \\- \frac {2}{9 c x^{\frac {9}{2}}} & \text {for}\: b = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} & \text {for}\: c = 0 \\- \frac {2}{5 b x^{\frac {5}{2}}} + \frac {c \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{2} \sqrt [4]{- \frac {b}{c}}} - \frac {c \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 b^{2} \sqrt [4]{- \frac {b}{c}}} + \frac {c \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{b^{2} \sqrt [4]{- \frac {b}{c}}} + \frac {2 c}{b^{2} \sqrt {x}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo/x**(9/2), Eq(b, 0) & Eq(c, 0)), (-2/(9*c*x**(9/2)), Eq(b, 0)), (-2/(5*b*x**(5/2)), Eq(c, 0)), (
-2/(5*b*x**(5/2)) + c*log(sqrt(x) - (-b/c)**(1/4))/(2*b**2*(-b/c)**(1/4)) - c*log(sqrt(x) + (-b/c)**(1/4))/(2*
b**2*(-b/c)**(1/4)) + c*atan(sqrt(x)/(-b/c)**(1/4))/(b**2*(-b/c)**(1/4)) + 2*c/(b**2*sqrt(x)), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {c^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}} \sqrt {c}} - \frac {\sqrt {2} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {1}{4}} c^{\frac {3}{4}}}\right )}}{4 \, b^{2}} + \frac {2 \, {\left (5 \, c x^{2} - b\right )}}{5 \, b^{2} x^{\frac {5}{2}}} \]

[In]

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

[Out]

1/4*c^2*(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)))/(sq
rt(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))/sqr
t(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^2 + 2/5*(5*c*x^2 - b)/(b^2*x^(5/2))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3} c} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{2 \, b^{3} c} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{3} c} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, b^{3} c} + \frac {2 \, {\left (5 \, c x^{2} - b\right )}}{5 \, b^{2} x^{\frac {5}{2}}} \]

[In]

integrate(1/x^(3/2)/(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^3*c) + 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^3*c) - 1/4*sqrt(2)*(b*
c^3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) + 1/4*sqrt(2)*(b*c^3)^(3/4)*log(-sqrt(2)*s
qrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) + 2/5*(5*c*x^2 - b)/(b^2*x^(5/2))

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.31 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )} \, dx=\frac {{\left (-c\right )}^{5/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{9/4}}-\frac {{\left (-c\right )}^{5/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{b^{9/4}}-\frac {\frac {2}{5\,b}-\frac {2\,c\,x^2}{b^2}}{x^{5/2}} \]

[In]

int(1/(x^(3/2)*(b*x^2 + c*x^4)),x)

[Out]

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