3.4.38 \(\int \frac {1}{x^{3/2} (b x^2+c x^4)^2} \, dx\) [338]

Optimal. Leaf size=258 \[ -\frac {13}{18 b^2 x^{9/2}}+\frac {13 c}{10 b^3 x^{5/2}}-\frac {13 c^2}{2 b^4 \sqrt {x}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}+\frac {13 c^{9/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}-\frac {13 c^{9/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}-\frac {13 c^{9/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{17/4}}+\frac {13 c^{9/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{17/4}} \]

[Out]

-13/18/b^2/x^(9/2)+13/10*c/b^3/x^(5/2)+1/2/b/x^(9/2)/(c*x^2+b)+13/8*c^(9/4)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b
^(1/4))/b^(17/4)*2^(1/2)-13/8*c^(9/4)*arctan(1+c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/b^(17/4)*2^(1/2)-13/16*c^(9/4)
*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(17/4)*2^(1/2)+13/16*c^(9/4)*ln(b^(1/2)+x*c^(1/2)+b^(
1/4)*c^(1/4)*2^(1/2)*x^(1/2))/b^(17/4)*2^(1/2)-13/2*c^2/b^4/x^(1/2)

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Rubi [A]
time = 0.15, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1598, 296, 331, 335, 303, 1176, 631, 210, 1179, 642} \begin {gather*} \frac {13 c^{9/4} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}-\frac {13 c^{9/4} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} b^{17/4}}-\frac {13 c^{9/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{17/4}}+\frac {13 c^{9/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} b^{17/4}}-\frac {13 c^2}{2 b^4 \sqrt {x}}+\frac {13 c}{10 b^3 x^{5/2}}-\frac {13}{18 b^2 x^{9/2}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(b*x^2 + c*x^4)^2),x]

[Out]

-13/(18*b^2*x^(9/2)) + (13*c)/(10*b^3*x^(5/2)) - (13*c^2)/(2*b^4*Sqrt[x]) + 1/(2*b*x^(9/2)*(b + c*x^2)) + (13*
c^(9/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(17/4)) - (13*c^(9/4)*ArcTan[1 + (Sqrt[2]*
c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*b^(17/4)) - (13*c^(9/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] +
Sqrt[c]*x])/(8*Sqrt[2]*b^(17/4)) + (13*c^(9/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*
Sqrt[2]*b^(17/4))

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 296

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

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*} \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^2} \, dx &=\int \frac {1}{x^{11/2} \left (b+c x^2\right )^2} \, dx\\ &=\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}+\frac {13 \int \frac {1}{x^{11/2} \left (b+c x^2\right )} \, dx}{4 b}\\ &=-\frac {13}{18 b^2 x^{9/2}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}-\frac {(13 c) \int \frac {1}{x^{7/2} \left (b+c x^2\right )} \, dx}{4 b^2}\\ &=-\frac {13}{18 b^2 x^{9/2}}+\frac {13 c}{10 b^3 x^{5/2}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}+\frac {\left (13 c^2\right ) \int \frac {1}{x^{3/2} \left (b+c x^2\right )} \, dx}{4 b^3}\\ &=-\frac {13}{18 b^2 x^{9/2}}+\frac {13 c}{10 b^3 x^{5/2}}-\frac {13 c^2}{2 b^4 \sqrt {x}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}-\frac {\left (13 c^3\right ) \int \frac {\sqrt {x}}{b+c x^2} \, dx}{4 b^4}\\ &=-\frac {13}{18 b^2 x^{9/2}}+\frac {13 c}{10 b^3 x^{5/2}}-\frac {13 c^2}{2 b^4 \sqrt {x}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}-\frac {\left (13 c^3\right ) \text {Subst}\left (\int \frac {x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 b^4}\\ &=-\frac {13}{18 b^2 x^{9/2}}+\frac {13 c}{10 b^3 x^{5/2}}-\frac {13 c^2}{2 b^4 \sqrt {x}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}+\frac {\left (13 c^{5/2}\right ) \text {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^4}-\frac {\left (13 c^{5/2}\right ) \text {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 b^4}\\ &=-\frac {13}{18 b^2 x^{9/2}}+\frac {13 c}{10 b^3 x^{5/2}}-\frac {13 c^2}{2 b^4 \sqrt {x}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}-\frac {\left (13 c^2\right ) \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 )}{8 b^4}-\frac {\left (13 c^2\right ) \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 )}{8 b^4}-\frac {\left (13 c^{9/4}\right ) \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 )}{8 \sqrt {2} b^{17/4}}-\frac {\left (13 c^{9/4}\right ) \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 )}{8 \sqrt {2} b^{17/4}}\\ &=-\frac {13}{18 b^2 x^{9/2}}+\frac {13 c}{10 b^3 x^{5/2}}-\frac {13 c^2}{2 b^4 \sqrt {x}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}-\frac {13 c^{9/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{17/4}}+\frac {13 c^{9/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{17/4}}-\frac {\left (13 c^{9/4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}+\frac {\left (13 c^{9/4}\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}\\ &=-\frac {13}{18 b^2 x^{9/2}}+\frac {13 c}{10 b^3 x^{5/2}}-\frac {13 c^2}{2 b^4 \sqrt {x}}+\frac {1}{2 b x^{9/2} \left (b+c x^2\right )}+\frac {13 c^{9/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}-\frac {13 c^{9/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} b^{17/4}}-\frac {13 c^{9/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{17/4}}+\frac {13 c^{9/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} b^{17/4}}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 160, normalized size = 0.62 \begin {gather*} \frac {-\frac {4 \sqrt [4]{b} \left (20 b^3-52 b^2 c x^2+468 b c^2 x^4+585 c^3 x^6\right )}{x^{9/2} \left (b+c x^2\right )}+585 \sqrt {2} c^{9/4} \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+585 \sqrt {2} c^{9/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{360 b^{17/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-4*b^(1/4)*(20*b^3 - 52*b^2*c*x^2 + 468*b*c^2*x^4 + 585*c^3*x^6))/(x^(9/2)*(b + c*x^2)) + 585*Sqrt[2]*c^(9/4
)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] + 585*Sqrt[2]*c^(9/4)*ArcTanh[(Sqrt[2]*b^(1/
4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(360*b^(17/4))

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Maple [A]
time = 0.14, size = 158, normalized size = 0.61

method	result	size
derivativedivides	\(-\frac {2 c^{3} \left (\frac {x^{\frac {3}{2}}}{4 c \,x^{2}+4 b}+\frac {13 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{4}}-\frac {2}{9 b^{2} x^{\frac {9}{2}}}-\frac {6 c^{2}}{b^{4} \sqrt {x}}+\frac {4 c}{5 b^{3} x^{\frac {5}{2}}}\)	\(158\)
default	\(-\frac {2 c^{3} \left (\frac {x^{\frac {3}{2}}}{4 c \,x^{2}+4 b}+\frac {13 \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{32 c \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{b^{4}}-\frac {2}{9 b^{2} x^{\frac {9}{2}}}-\frac {6 c^{2}}{b^{4} \sqrt {x}}+\frac {4 c}{5 b^{3} x^{\frac {5}{2}}}\)	\(158\)
risch	\(-\frac {2 \left (135 c^{2} x^{4}-18 b c \,x^{2}+5 b^{2}\right )}{45 b^{4} x^{\frac {9}{2}}}-\frac {c^{3} x^{\frac {3}{2}}}{2 b^{4} \left (c \,x^{2}+b \right )}-\frac {13 c^{2} \sqrt {2}\, \ln \left (\frac {x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )}{16 b^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}-\frac {13 c^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 b^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}-\frac {13 c^{2} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 b^{4} \left (\frac {b}{c}\right )^{\frac {1}{4}}}\)	\(190\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c^3/b^4*(1/4*x^(3/2)/(c*x^2+b)+13/32/c/(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)))-2/9/b^2/x^(9/2)-6*c^2/b^4/x^(1/2)+4/5*c/b^3/x^(5/2)

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Maxima [A]
time = 0.51, size = 232, normalized size = 0.90 \begin {gather*} -\frac {585 \, c^{3} x^{6} + 468 \, b c^{2} x^{4} - 52 \, b^{2} c x^{2} + 20 \, b^{3}}{90 \, {\left (b^{4} c x^{\frac {13}{2}} + b^{5} x^{\frac {9}{2}}\right )}} - \frac {13 \, 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 )}}{16 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/90*(585*c^3*x^6 + 468*b*c^2*x^4 - 52*b^2*c*x^2 + 20*b^3)/(b^4*c*x^(13/2) + b^5*x^(9/2)) - 13/16*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)))/(sqrt(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))/sqrt(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^4

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Fricas [A]
time = 0.38, size = 262, normalized size = 1.02 \begin {gather*} \frac {2340 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )} \left (-\frac {c^{9}}{b^{17}}\right )^{\frac {1}{4}} \arctan \left (-\frac {2197 \, b^{4} c^{7} \sqrt {x} \left (-\frac {c^{9}}{b^{17}}\right )^{\frac {1}{4}} - \sqrt {-4826809 \, b^{9} c^{9} \sqrt {-\frac {c^{9}}{b^{17}}} + 4826809 \, c^{14} x} b^{4} \left (-\frac {c^{9}}{b^{17}}\right )^{\frac {1}{4}}}{2197 \, c^{9}}\right ) - 585 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )} \left (-\frac {c^{9}}{b^{17}}\right )^{\frac {1}{4}} \log \left (2197 \, b^{13} \left (-\frac {c^{9}}{b^{17}}\right )^{\frac {3}{4}} + 2197 \, c^{7} \sqrt {x}\right ) + 585 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )} \left (-\frac {c^{9}}{b^{17}}\right )^{\frac {1}{4}} \log \left (-2197 \, b^{13} \left (-\frac {c^{9}}{b^{17}}\right )^{\frac {3}{4}} + 2197 \, c^{7} \sqrt {x}\right ) - 4 \, {\left (585 \, c^{3} x^{6} + 468 \, b c^{2} x^{4} - 52 \, b^{2} c x^{2} + 20 \, b^{3}\right )} \sqrt {x}}{360 \, {\left (b^{4} c x^{7} + b^{5} x^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/360*(2340*(b^4*c*x^7 + b^5*x^5)*(-c^9/b^17)^(1/4)*arctan(-1/2197*(2197*b^4*c^7*sqrt(x)*(-c^9/b^17)^(1/4) - s
qrt(-4826809*b^9*c^9*sqrt(-c^9/b^17) + 4826809*c^14*x)*b^4*(-c^9/b^17)^(1/4))/c^9) - 585*(b^4*c*x^7 + b^5*x^5)
*(-c^9/b^17)^(1/4)*log(2197*b^13*(-c^9/b^17)^(3/4) + 2197*c^7*sqrt(x)) + 585*(b^4*c*x^7 + b^5*x^5)*(-c^9/b^17)
^(1/4)*log(-2197*b^13*(-c^9/b^17)^(3/4) + 2197*c^7*sqrt(x)) - 4*(585*c^3*x^6 + 468*b*c^2*x^4 - 52*b^2*c*x^2 +
20*b^3)*sqrt(x))/(b^4*c*x^7 + b^5*x^5)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 3.92, size = 219, normalized size = 0.85 \begin {gather*} -\frac {c^{3} x^{\frac {3}{2}}}{2 \, {\left (c x^{2} + b\right )} b^{4}} - \frac {13 \, \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 )}{8 \, b^{5}} - \frac {13 \, \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 )}{8 \, b^{5}} + \frac {13 \, \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 )}{16 \, b^{5}} - \frac {13 \, \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 )}{16 \, b^{5}} - \frac {2 \, {\left (135 \, c^{2} x^{4} - 18 \, b c x^{2} + 5 \, b^{2}\right )}}{45 \, b^{4} x^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^3*x^(3/2)/((c*x^2 + b)*b^4) - 13/8*sqrt(2)*(b*c^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sq
rt(x))/(b/c)^(1/4))/b^5 - 13/8*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^5 + 13/16*sqrt(2)*(b*c^3)^(3/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 - 13/16*sqrt(
2)*(b*c^3)^(3/4)*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 - 2/45*(135*c^2*x^4 - 18*b*c*x^2 + 5*b^
2)/(b^4*x^(9/2))

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Mupad [B]
time = 4.37, size = 99, normalized size = 0.38 \begin {gather*} \frac {13\,{\left (-c\right )}^{9/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{4\,b^{17/4}}-\frac {13\,{\left (-c\right )}^{9/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{4\,b^{17/4}}-\frac {\frac {2}{9\,b}-\frac {26\,c\,x^2}{45\,b^2}+\frac {26\,c^2\,x^4}{5\,b^3}+\frac {13\,c^3\,x^6}{2\,b^4}}{b\,x^{9/2}+c\,x^{13/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(3/2)*(b*x^2 + c*x^4)^2),x)

[Out]

(13*(-c)^(9/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4)))/(4*b^(17/4)) - (13*(-c)^(9/4)*atan(((-c)^(1/4)*x^(1/2))/b^
(1/4)))/(4*b^(17/4)) - (2/(9*b) - (26*c*x^2)/(45*b^2) + (26*c^2*x^4)/(5*b^3) + (13*c^3*x^6)/(2*b^4))/(b*x^(9/2
) + c*x^(13/2))

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