3.327 \(\int \frac {x^{19/2}}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=243 \[ -\frac {9 b^{5/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{13/4}}-\frac {9 b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{13/4}}-\frac {9 b \sqrt {x}}{2 c^3}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}+\frac {9 x^{5/2}}{10 c^2} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1584, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {9 b^{5/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{8 \sqrt {2} c^{13/4}}-\frac {9 b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} c^{13/4}}-\frac {9 b \sqrt {x}}{2 c^3}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}+\frac {9 x^{5/2}}{10 c^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(19/2)/(b*x^2 + c*x^4)^2,x]

[Out]

(-9*b*Sqrt[x])/(2*c^3) + (9*x^(5/2))/(10*c^2) - x^(9/2)/(2*c*(b + c*x^2)) - (9*b^(5/4)*ArcTan[1 - (Sqrt[2]*c^(
1/4)*Sqrt[x])/b^(1/4)])/(4*Sqrt[2]*c^(13/4)) + (9*b^(5/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(4*Sq
rt[2]*c^(13/4)) - (9*b^(5/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(13/4))
+ (9*b^(5/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(8*Sqrt[2]*c^(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 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 288

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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 {x^{19/2}}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^{11/2}}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}+\frac {9 \int \frac {x^{7/2}}{b+c x^2} \, dx}{4 c}\\ &=\frac {9 x^{5/2}}{10 c^2}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}-\frac {(9 b) \int \frac {x^{3/2}}{b+c x^2} \, dx}{4 c^2}\\ &=-\frac {9 b \sqrt {x}}{2 c^3}+\frac {9 x^{5/2}}{10 c^2}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}+\frac {\left (9 b^2\right ) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{4 c^3}\\ &=-\frac {9 b \sqrt {x}}{2 c^3}+\frac {9 x^{5/2}}{10 c^2}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}+\frac {\left (9 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{2 c^3}\\ &=-\frac {9 b \sqrt {x}}{2 c^3}+\frac {9 x^{5/2}}{10 c^2}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}+\frac {\left (9 b^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^3}+\frac {\left (9 b^{3/2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{4 c^3}\\ &=-\frac {9 b \sqrt {x}}{2 c^3}+\frac {9 x^{5/2}}{10 c^2}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}+\frac {\left (9 b^{3/2}\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 )}{8 c^{7/2}}+\frac {\left (9 b^{3/2}\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 )}{8 c^{7/2}}-\frac {\left (9 b^{5/4}\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 )}{8 \sqrt {2} c^{13/4}}-\frac {\left (9 b^{5/4}\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 )}{8 \sqrt {2} c^{13/4}}\\ &=-\frac {9 b \sqrt {x}}{2 c^3}+\frac {9 x^{5/2}}{10 c^2}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}-\frac {9 b^{5/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{13/4}}+\frac {\left (9 b^{5/4}\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 )}{4 \sqrt {2} c^{13/4}}-\frac {\left (9 b^{5/4}\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 )}{4 \sqrt {2} c^{13/4}}\\ &=-\frac {9 b \sqrt {x}}{2 c^3}+\frac {9 x^{5/2}}{10 c^2}-\frac {x^{9/2}}{2 c \left (b+c x^2\right )}-\frac {9 b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} c^{13/4}}-\frac {9 b^{5/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{13/4}}+\frac {9 b^{5/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{8 \sqrt {2} c^{13/4}}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 220, normalized size = 0.91 \[ \frac {-45 \sqrt {2} b^{5/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+45 \sqrt {2} b^{5/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-90 \sqrt {2} b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )+90 \sqrt {2} b^{5/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )+\frac {8 \sqrt [4]{c} \sqrt {x} \left (-45 b^2-36 b c x^2+4 c^2 x^4\right )}{b+c x^2}}{80 c^{13/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(19/2)/(b*x^2 + c*x^4)^2,x]

[Out]

((8*c^(1/4)*Sqrt[x]*(-45*b^2 - 36*b*c*x^2 + 4*c^2*x^4))/(b + c*x^2) - 90*Sqrt[2]*b^(5/4)*ArcTan[1 - (Sqrt[2]*c
^(1/4)*Sqrt[x])/b^(1/4)] + 90*Sqrt[2]*b^(5/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] - 45*Sqrt[2]*b^(5/
4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] + 45*Sqrt[2]*b^(5/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/
4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(80*c^(13/4))

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fricas [A]  time = 0.83, size = 227, normalized size = 0.93 \[ \frac {180 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b c^{10} \sqrt {x} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {3}{4}} - \sqrt {c^{6} \sqrt {-\frac {b^{5}}{c^{13}}} + b^{2} x} c^{10} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {3}{4}}}{b^{5}}\right ) + 45 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} \log \left (9 \, c^{3} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} + 9 \, b \sqrt {x}\right ) - 45 \, {\left (c^{4} x^{2} + b c^{3}\right )} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} \log \left (-9 \, c^{3} \left (-\frac {b^{5}}{c^{13}}\right )^{\frac {1}{4}} + 9 \, b \sqrt {x}\right ) + 4 \, {\left (4 \, c^{2} x^{4} - 36 \, b c x^{2} - 45 \, b^{2}\right )} \sqrt {x}}{40 \, {\left (c^{4} x^{2} + b c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

1/40*(180*(c^4*x^2 + b*c^3)*(-b^5/c^13)^(1/4)*arctan(-(b*c^10*sqrt(x)*(-b^5/c^13)^(3/4) - sqrt(c^6*sqrt(-b^5/c
^13) + b^2*x)*c^10*(-b^5/c^13)^(3/4))/b^5) + 45*(c^4*x^2 + b*c^3)*(-b^5/c^13)^(1/4)*log(9*c^3*(-b^5/c^13)^(1/4
) + 9*b*sqrt(x)) - 45*(c^4*x^2 + b*c^3)*(-b^5/c^13)^(1/4)*log(-9*c^3*(-b^5/c^13)^(1/4) + 9*b*sqrt(x)) + 4*(4*c
^2*x^4 - 36*b*c*x^2 - 45*b^2)*sqrt(x))/(c^4*x^2 + b*c^3)

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giac [A]  time = 0.20, size = 216, normalized size = 0.89 \[ \frac {9 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} b \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{4}} + \frac {9 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} b \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{8 \, c^{4}} + \frac {9 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} b \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{4}} - \frac {9 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} b \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{16 \, c^{4}} - \frac {b^{2} \sqrt {x}}{2 \, {\left (c x^{2} + b\right )} c^{3}} + \frac {2 \, {\left (c^{8} x^{\frac {5}{2}} - 10 \, b c^{7} \sqrt {x}\right )}}{5 \, c^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

9/8*sqrt(2)*(b*c^3)^(1/4)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^4 + 9/8*sqrt(2
)*(b*c^3)^(1/4)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^4 + 9/16*sqrt(2)*(b*c^3
)^(1/4)*b*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 - 9/16*sqrt(2)*(b*c^3)^(1/4)*b*log(-sqrt(2)*sqr
t(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 - 1/2*b^2*sqrt(x)/((c*x^2 + b)*c^3) + 2/5*(c^8*x^(5/2) - 10*b*c^7*sqrt(x
))/c^10

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maple [A]  time = 0.01, size = 172, normalized size = 0.71 \[ \frac {2 x^{\frac {5}{2}}}{5 c^{2}}-\frac {b^{2} \sqrt {x}}{2 \left (c \,x^{2}+b \right ) c^{3}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{8 c^{3}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, b \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{8 c^{3}}+\frac {9 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, b \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{16 c^{3}}-\frac {4 b \sqrt {x}}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(19/2)/(c*x^4+b*x^2)^2,x)

[Out]

2/5*x^(5/2)/c^2-4*b*x^(1/2)/c^3-1/2/c^3*b^2*x^(1/2)/(c*x^2+b)+9/16/c^3*b*(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)))+9/8/c^3*b*(b/c)^(1/4)*2^(1/2)*arcta
n(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+9/8/c^3*b*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)-1)

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maxima [A]  time = 3.02, size = 217, normalized size = 0.89 \[ -\frac {b^{2} \sqrt {x}}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {2 \, {\left (c x^{\frac {5}{2}} - 10 \, b \sqrt {x}\right )}}{5 \, c^{3}} + \frac {9 \, {\left (\frac {2 \, \sqrt {2} b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} b^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} b^{\frac {5}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}} - \frac {\sqrt {2} b^{\frac {5}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{c^{\frac {1}{4}}}\right )}}{16 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(19/2)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

-1/2*b^2*sqrt(x)/(c^4*x^2 + b*c^3) + 2/5*(c*x^(5/2) - 10*b*sqrt(x))/c^3 + 9/16*(2*sqrt(2)*b^(3/2)*arctan(1/2*s
qrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/sqrt(sqrt(b)*sqrt(c)) + 2*sqrt(2)*
b^(3/2)*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(2)*b^(5/4)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4) - sqrt(2)*b^(5/4
)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4))/c^3

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mupad [B]  time = 0.10, size = 92, normalized size = 0.38 \[ \frac {2\,x^{5/2}}{5\,c^2}-\frac {b^2\,\sqrt {x}}{2\,\left (c^4\,x^2+b\,c^3\right )}-\frac {4\,b\,\sqrt {x}}{c^3}-\frac {9\,{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{4\,c^{13/4}}+\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,9{}\mathrm {i}}{4\,c^{13/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(19/2)/(b*x^2 + c*x^4)^2,x)

[Out]

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

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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(x**(19/2)/(c*x**4+b*x**2)**2,x)

[Out]

Timed out

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