3.3.33 \(\int \frac {1}{\sqrt {x} (b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=279 \[ \frac {285 c^{11/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}+\frac {285 c^{11/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{23/4}}-\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {285 c}{112 b^4 x^{7/2}}-\frac {285}{176 b^3 x^{11/2}}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2} \]

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Rubi [A]  time = 0.26, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1584, 290, 325, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {285 c^{11/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}+\frac {285 c^{11/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{23/4}}+\frac {285 c}{112 b^4 x^{7/2}}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}-\frac {285}{176 b^3 x^{11/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(b*x^2 + c*x^4)^3),x]

[Out]

-285/(176*b^3*x^(11/2)) + (285*c)/(112*b^4*x^(7/2)) - (95*c^2)/(16*b^5*x^(3/2)) + 1/(4*b*x^(11/2)*(b + c*x^2)^
2) + 19/(16*b^2*x^(11/2)*(b + c*x^2)) + (285*c^(11/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[
2]*b^(23/4)) - (285*c^(11/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(23/4)) + (285*c^(11
/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(23/4)) - (285*c^(11/4)*Log[Sqrt
[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(23/4))

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 211

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

Rule 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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}{\sqrt {x} \left (b x^2+c x^4\right )^3} \, dx &=\int \frac {1}{x^{13/2} \left (b+c x^2\right )^3} \, dx\\ &=\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19 \int \frac {1}{x^{13/2} \left (b+c x^2\right )^2} \, dx}{8 b}\\ &=\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}+\frac {285 \int \frac {1}{x^{13/2} \left (b+c x^2\right )} \, dx}{32 b^2}\\ &=-\frac {285}{176 b^3 x^{11/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}-\frac {(285 c) \int \frac {1}{x^{9/2} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=-\frac {285}{176 b^3 x^{11/2}}+\frac {285 c}{112 b^4 x^{7/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}+\frac {\left (285 c^2\right ) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{32 b^4}\\ &=-\frac {285}{176 b^3 x^{11/2}}+\frac {285 c}{112 b^4 x^{7/2}}-\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}-\frac {\left (285 c^3\right ) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{32 b^5}\\ &=-\frac {285}{176 b^3 x^{11/2}}+\frac {285 c}{112 b^4 x^{7/2}}-\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}-\frac {\left (285 c^3\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^5}\\ &=-\frac {285}{176 b^3 x^{11/2}}+\frac {285 c}{112 b^4 x^{7/2}}-\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}-\frac {\left (285 c^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{11/2}}-\frac {\left (285 c^3\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{11/2}}\\ &=-\frac {285}{176 b^3 x^{11/2}}+\frac {285 c}{112 b^4 x^{7/2}}-\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}-\frac {\left (285 c^{5/2}\right ) \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 )}{64 b^{11/2}}-\frac {\left (285 c^{5/2}\right ) \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 )}{64 b^{11/2}}+\frac {\left (285 c^{11/4}\right ) \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 )}{64 \sqrt {2} b^{23/4}}+\frac {\left (285 c^{11/4}\right ) \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 )}{64 \sqrt {2} b^{23/4}}\\ &=-\frac {285}{176 b^3 x^{11/2}}+\frac {285 c}{112 b^4 x^{7/2}}-\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}+\frac {285 c^{11/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}-\frac {\left (285 c^{11/4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}+\frac {\left (285 c^{11/4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}\\ &=-\frac {285}{176 b^3 x^{11/2}}+\frac {285 c}{112 b^4 x^{7/2}}-\frac {95 c^2}{16 b^5 x^{3/2}}+\frac {1}{4 b x^{11/2} \left (b+c x^2\right )^2}+\frac {19}{16 b^2 x^{11/2} \left (b+c x^2\right )}+\frac {285 c^{11/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{23/4}}+\frac {285 c^{11/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{23/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.10 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {11}{4},3;-\frac {7}{4};-\frac {c x^2}{b}\right )}{11 b^3 x^{11/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Hypergeometric2F1[-11/4, 3, -7/4, -((c*x^2)/b)])/(11*b^3*x^(11/2))

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IntegrateAlgebraic [A]  time = 0.49, size = 182, normalized size = 0.65 \begin {gather*} \frac {285 c^{11/4} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2} \sqrt [4]{c}}-\frac {\sqrt [4]{c} x}{\sqrt {2} \sqrt [4]{b}}}{\sqrt {x}}\right )}{32 \sqrt {2} b^{23/4}}-\frac {285 c^{11/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{23/4}}+\frac {-224 b^4+608 b^3 c x^2-3040 b^2 c^2 x^4-11495 b c^3 x^6-7315 c^4 x^8}{1232 b^5 x^{11/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-224*b^4 + 608*b^3*c*x^2 - 3040*b^2*c^2*x^4 - 11495*b*c^3*x^6 - 7315*c^4*x^8)/(1232*b^5*x^(11/2)*(b + c*x^2)^
2) + (285*c^(11/4)*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[2]*b^(1/4)))/Sqrt[x]])/(32*Sqrt[2]*b^
(23/4)) - (285*c^(11/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqrt[2]*b^(23/4)
)

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fricas [A]  time = 2.08, size = 311, normalized size = 1.11 \begin {gather*} -\frac {87780 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b^{17} c^{3} \sqrt {x} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {3}{4}} - \sqrt {b^{12} \sqrt {-\frac {c^{11}}{b^{23}}} + c^{6} x} b^{17} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {3}{4}}}{c^{11}}\right ) + 21945 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} \log \left (285 \, b^{6} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} + 285 \, c^{3} \sqrt {x}\right ) - 21945 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} \log \left (-285 \, b^{6} \left (-\frac {c^{11}}{b^{23}}\right )^{\frac {1}{4}} + 285 \, c^{3} \sqrt {x}\right ) + 4 \, {\left (7315 \, c^{4} x^{8} + 11495 \, b c^{3} x^{6} + 3040 \, b^{2} c^{2} x^{4} - 608 \, b^{3} c x^{2} + 224 \, b^{4}\right )} \sqrt {x}}{4928 \, {\left (b^{5} c^{2} x^{10} + 2 \, b^{6} c x^{8} + b^{7} x^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4928*(87780*(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)*(-c^11/b^23)^(1/4)*arctan(-(b^17*c^3*sqrt(x)*(-c^11/b^23
)^(3/4) - sqrt(b^12*sqrt(-c^11/b^23) + c^6*x)*b^17*(-c^11/b^23)^(3/4))/c^11) + 21945*(b^5*c^2*x^10 + 2*b^6*c*x
^8 + b^7*x^6)*(-c^11/b^23)^(1/4)*log(285*b^6*(-c^11/b^23)^(1/4) + 285*c^3*sqrt(x)) - 21945*(b^5*c^2*x^10 + 2*b
^6*c*x^8 + b^7*x^6)*(-c^11/b^23)^(1/4)*log(-285*b^6*(-c^11/b^23)^(1/4) + 285*c^3*sqrt(x)) + 4*(7315*c^4*x^8 +
11495*b*c^3*x^6 + 3040*b^2*c^2*x^4 - 608*b^3*c*x^2 + 224*b^4)*sqrt(x))/(b^5*c^2*x^10 + 2*b^6*c*x^8 + b^7*x^6)

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giac [A]  time = 0.17, size = 243, normalized size = 0.87 \begin {gather*} -\frac {285 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c^{2} \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 )}{64 \, b^{6}} - \frac {285 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c^{2} \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 )}{64 \, b^{6}} - \frac {285 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c^{2} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{6}} + \frac {285 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c^{2} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{6}} - \frac {31 \, c^{4} x^{\frac {5}{2}} + 35 \, b c^{3} \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{5}} - \frac {2 \, {\left (154 \, c^{2} x^{4} - 33 \, b c x^{2} + 7 \, b^{2}\right )}}{77 \, b^{5} x^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-285/64*sqrt(2)*(b*c^3)^(1/4)*c^2*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/b^6 - 285/
64*sqrt(2)*(b*c^3)^(1/4)*c^2*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^6 - 285/128*
sqrt(2)*(b*c^3)^(1/4)*c^2*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^6 + 285/128*sqrt(2)*(b*c^3)^(1/4)
*c^2*log(-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^6 - 1/16*(31*c^4*x^(5/2) + 35*b*c^3*sqrt(x))/((c*x^2
+ b)^2*b^5) - 2/77*(154*c^2*x^4 - 33*b*c*x^2 + 7*b^2)/(b^5*x^(11/2))

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maple [A]  time = 0.02, size = 209, normalized size = 0.75 \begin {gather*} -\frac {31 c^{4} x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{5}}-\frac {35 c^{3} \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} b^{4}}-\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 b^{6}}-\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{3} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 b^{6}}-\frac {285 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{3} \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 )}{128 b^{6}}-\frac {4 c^{2}}{b^{5} x^{\frac {3}{2}}}+\frac {6 c}{7 b^{4} x^{\frac {7}{2}}}-\frac {2}{11 b^{3} x^{\frac {11}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-31/16*c^4/b^5/(c*x^2+b)^2*x^(5/2)-35/16*c^3/b^4/(c*x^2+b)^2*x^(1/2)-285/128*c^3/b^6*(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)))-285/64*c^3/b^6*(b/c)^(1
/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-285/64*c^3/b^6*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4
)*x^(1/2)-1)-2/11/b^3/x^(11/2)-4*c^2/b^5/x^(3/2)+6/7*c/b^4/x^(7/2)

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maxima [A]  time = 3.06, size = 257, normalized size = 0.92 \begin {gather*} -\frac {7315 \, c^{4} x^{8} + 11495 \, b c^{3} x^{6} + 3040 \, b^{2} c^{2} x^{4} - 608 \, b^{3} c x^{2} + 224 \, b^{4}}{1232 \, {\left (b^{5} c^{2} x^{\frac {19}{2}} + 2 \, b^{6} c x^{\frac {15}{2}} + b^{7} x^{\frac {11}{2}}\right )}} - \frac {285 \, {\left (\frac {2 \, \sqrt {2} c^{3} \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 {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} c^{3} \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 {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} c^{\frac {11}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} c^{\frac {11}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{128 \, b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1232*(7315*c^4*x^8 + 11495*b*c^3*x^6 + 3040*b^2*c^2*x^4 - 608*b^3*c*x^2 + 224*b^4)/(b^5*c^2*x^(19/2) + 2*b^
6*c*x^(15/2) + b^7*x^(11/2)) - 285/128*(2*sqrt(2)*c^3*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*
sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*c^3*arctan(-1/2*sqrt(2)*(sqrt(2)*b
^(1/4)*c^(1/4) - 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*c^(11/4)*
log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/b^(3/4) - sqrt(2)*c^(11/4)*log(-sqrt(2)*b^(1/4)*c^(
1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/b^(3/4))/b^5

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mupad [B]  time = 4.39, size = 121, normalized size = 0.43 \begin {gather*} \frac {285\,{\left (-c\right )}^{11/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{23/4}}-\frac {\frac {2}{11\,b}-\frac {38\,c\,x^2}{77\,b^2}+\frac {190\,c^2\,x^4}{77\,b^3}+\frac {1045\,c^3\,x^6}{112\,b^4}+\frac {95\,c^4\,x^8}{16\,b^5}}{b^2\,x^{11/2}+c^2\,x^{19/2}+2\,b\,c\,x^{15/2}}+\frac {285\,{\left (-c\right )}^{11/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{23/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(1/2)*(b*x^2 + c*x^4)^3),x)

[Out]

(285*(-c)^(11/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4)))/(32*b^(23/4)) - (2/(11*b) - (38*c*x^2)/(77*b^2) + (190*c^
2*x^4)/(77*b^3) + (1045*c^3*x^6)/(112*b^4) + (95*c^4*x^8)/(16*b^5))/(b^2*x^(11/2) + c^2*x^(19/2) + 2*b*c*x^(15
/2)) + (285*(-c)^(11/4)*atanh(((-c)^(1/4)*x^(1/2))/b^(1/4)))/(32*b^(23/4))

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sympy [F(-1)]  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/(c*x**4+b*x**2)**3/x**(1/2),x)

[Out]

Timed out

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