3.3.21 \(\int \frac {x^{23/2}}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=251 \[ \frac {45 \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} c^{13/4}}-\frac {45 \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} c^{13/4}}+\frac {45 \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} c^{13/4}}-\frac {45 \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} c^{13/4}}-\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}+\frac {45 \sqrt {x}}{16 c^3} \]

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Rubi [A]  time = 0.21, antiderivative size = 251, 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} \begin {gather*} -\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}+\frac {45 \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} c^{13/4}}-\frac {45 \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} c^{13/4}}+\frac {45 \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} c^{13/4}}-\frac {45 \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} c^{13/4}}-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}+\frac {45 \sqrt {x}}{16 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(23/2)/(b*x^2 + c*x^4)^3,x]

[Out]

(45*Sqrt[x])/(16*c^3) - x^(9/2)/(4*c*(b + c*x^2)^2) - (9*x^(5/2))/(16*c^2*(b + c*x^2)) + (45*b^(1/4)*ArcTan[1
- (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*c^(13/4)) - (45*b^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])
/b^(1/4)])/(32*Sqrt[2]*c^(13/4)) + (45*b^(1/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64
*Sqrt[2]*c^(13/4)) - (45*b^(1/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*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^{23/2}}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {x^{11/2}}{\left (b+c x^2\right )^3} \, dx\\ &=-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}+\frac {9 \int \frac {x^{7/2}}{\left (b+c x^2\right )^2} \, dx}{8 c}\\ &=-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}-\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}+\frac {45 \int \frac {x^{3/2}}{b+c x^2} \, dx}{32 c^2}\\ &=\frac {45 \sqrt {x}}{16 c^3}-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}-\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}-\frac {(45 b) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{32 c^3}\\ &=\frac {45 \sqrt {x}}{16 c^3}-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}-\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}-\frac {(45 b) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 c^3}\\ &=\frac {45 \sqrt {x}}{16 c^3}-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}-\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}-\frac {\left (45 \sqrt {b}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 c^3}-\frac {\left (45 \sqrt {b}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 c^3}\\ &=\frac {45 \sqrt {x}}{16 c^3}-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}-\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}-\frac {\left (45 \sqrt {b}\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 c^{7/2}}-\frac {\left (45 \sqrt {b}\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 c^{7/2}}+\frac {\left (45 \sqrt [4]{b}\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} c^{13/4}}+\frac {\left (45 \sqrt [4]{b}\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} c^{13/4}}\\ &=\frac {45 \sqrt {x}}{16 c^3}-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}-\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}+\frac {45 \sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} c^{13/4}}-\frac {45 \sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} c^{13/4}}-\frac {\left (45 \sqrt [4]{b}\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} c^{13/4}}+\frac {\left (45 \sqrt [4]{b}\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} c^{13/4}}\\ &=\frac {45 \sqrt {x}}{16 c^3}-\frac {x^{9/2}}{4 c \left (b+c x^2\right )^2}-\frac {9 x^{5/2}}{16 c^2 \left (b+c x^2\right )}+\frac {45 \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} c^{13/4}}-\frac {45 \sqrt [4]{b} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} c^{13/4}}+\frac {45 \sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} c^{13/4}}-\frac {45 \sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} c^{13/4}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 220, normalized size = 0.88 \begin {gather*} \frac {\frac {8 \sqrt [4]{c} \sqrt {x} \left (45 b^2+81 b c x^2+32 c^2 x^4\right )}{\left (b+c x^2\right )^2}+45 \sqrt {2} \sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )-45 \sqrt {2} \sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )+90 \sqrt {2} \sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )-90 \sqrt {2} \sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{128 c^{13/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(23/2)/(b*x^2 + c*x^4)^3,x]

[Out]

((8*c^(1/4)*Sqrt[x]*(45*b^2 + 81*b*c*x^2 + 32*c^2*x^4))/(b + c*x^2)^2 + 90*Sqrt[2]*b^(1/4)*ArcTan[1 - (Sqrt[2]
*c^(1/4)*Sqrt[x])/b^(1/4)] - 90*Sqrt[2]*b^(1/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)] + 45*Sqrt[2]*b^(
1/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x] - 45*Sqrt[2]*b^(1/4)*Log[Sqrt[b] + Sqrt[2]*b^(
1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(128*c^(13/4))

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IntegrateAlgebraic [A]  time = 0.55, size = 324, normalized size = 1.29 \begin {gather*} \frac {-\frac {45 b^{5/4} x^2 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{16 \sqrt {2} c^{9/4}}+\left (\frac {45 b^{5/4} x^2}{16 \sqrt {2} c^{9/4}}+\frac {45 b^{9/4}}{32 \sqrt {2} c^{13/4}}+\frac {45 \sqrt [4]{b} x^4}{32 \sqrt {2} c^{5/4}}\right ) \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )-\frac {45 b^{9/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} c^{13/4}}+\frac {45 b^2 \sqrt {x}}{16 c^3}-\frac {45 \sqrt [4]{b} x^4 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} c^{5/4}}+\frac {81 b x^{5/2}}{16 c^2}+\frac {2 x^{9/2}}{c}}{\left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(23/2)/(b*x^2 + c*x^4)^3,x]

[Out]

((45*b^2*Sqrt[x])/(16*c^3) + (81*b*x^(5/2))/(16*c^2) + (2*x^(9/2))/c + ((45*b^(9/4))/(32*Sqrt[2]*c^(13/4)) + (
45*b^(5/4)*x^2)/(16*Sqrt[2]*c^(9/4)) + (45*b^(1/4)*x^4)/(32*Sqrt[2]*c^(5/4)))*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sq
rt[2]*b^(1/4)*c^(1/4)*Sqrt[x])] - (45*b^(9/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)]
)/(32*Sqrt[2]*c^(13/4)) - (45*b^(5/4)*x^2*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(1
6*Sqrt[2]*c^(9/4)) - (45*b^(1/4)*x^4*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqr
t[2]*c^(5/4)))/(b + c*x^2)^2

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fricas [A]  time = 1.00, size = 247, normalized size = 0.98 \begin {gather*} -\frac {180 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {b}{c^{13}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {c^{6} \sqrt {-\frac {b}{c^{13}}} + x} c^{10} \left (-\frac {b}{c^{13}}\right )^{\frac {3}{4}} - c^{10} \sqrt {x} \left (-\frac {b}{c^{13}}\right )^{\frac {3}{4}}}{b}\right ) + 45 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {b}{c^{13}}\right )^{\frac {1}{4}} \log \left (45 \, c^{3} \left (-\frac {b}{c^{13}}\right )^{\frac {1}{4}} + 45 \, \sqrt {x}\right ) - 45 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )} \left (-\frac {b}{c^{13}}\right )^{\frac {1}{4}} \log \left (-45 \, c^{3} \left (-\frac {b}{c^{13}}\right )^{\frac {1}{4}} + 45 \, \sqrt {x}\right ) - 4 \, {\left (32 \, c^{2} x^{4} + 81 \, b c x^{2} + 45 \, b^{2}\right )} \sqrt {x}}{64 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

-1/64*(180*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(-b/c^13)^(1/4)*arctan((sqrt(c^6*sqrt(-b/c^13) + x)*c^10*(-b/c^13
)^(3/4) - c^10*sqrt(x)*(-b/c^13)^(3/4))/b) + 45*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(-b/c^13)^(1/4)*log(45*c^3*(
-b/c^13)^(1/4) + 45*sqrt(x)) - 45*(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)*(-b/c^13)^(1/4)*log(-45*c^3*(-b/c^13)^(1/4
) + 45*sqrt(x)) - 4*(32*c^2*x^4 + 81*b*c*x^2 + 45*b^2)*sqrt(x))/(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3)

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giac [A]  time = 0.18, size = 208, normalized size = 0.83 \begin {gather*} -\frac {45 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{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 )}{64 \, c^{4}} - \frac {45 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{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 )}{64 \, c^{4}} - \frac {45 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, c^{4}} + \frac {45 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, c^{4}} + \frac {2 \, \sqrt {x}}{c^{3}} + \frac {17 \, b c x^{\frac {5}{2}} + 13 \, b^{2} \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

-45/64*sqrt(2)*(b*c^3)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^4 - 45/64*sqr
t(2)*(b*c^3)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^4 - 45/128*sqrt(2)*(b*
c^3)^(1/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 + 45/128*sqrt(2)*(b*c^3)^(1/4)*log(-sqrt(2)*sq
rt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^4 + 2*sqrt(x)/c^3 + 1/16*(17*b*c*x^(5/2) + 13*b^2*sqrt(x))/((c*x^2 + b)^2
*c^3)

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maple [A]  time = 0.02, size = 178, normalized size = 0.71 \begin {gather*} \frac {17 b \,x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} c^{2}}+\frac {13 b^{2} \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} c^{3}}-\frac {45 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 c^{3}}-\frac {45 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 c^{3}}-\frac {45 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \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 c^{3}}+\frac {2 \sqrt {x}}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(23/2)/(c*x^4+b*x^2)^3,x)

[Out]

2*x^(1/2)/c^3+17/16/c^2*b/(c*x^2+b)^2*x^(5/2)+13/16/c^3*b^2/(c*x^2+b)^2*x^(1/2)-45/128/c^3*(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)))-45/64/c^3*(b/c)^(
1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)-45/64/c^3*(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 = 2.95, size = 229, normalized size = 0.91 \begin {gather*} \frac {17 \, b c x^{\frac {5}{2}} + 13 \, b^{2} \sqrt {x}}{16 \, {\left (c^{5} x^{4} + 2 \, b c^{4} x^{2} + b^{2} c^{3}\right )}} - \frac {45 \, {\left (\frac {2 \, \sqrt {2} \sqrt {b} \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} \sqrt {b} \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 {1}{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 {1}{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 )}}{128 \, c^{3}} + \frac {2 \, \sqrt {x}}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(23/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/16*(17*b*c*x^(5/2) + 13*b^2*sqrt(x))/(c^5*x^4 + 2*b*c^4*x^2 + b^2*c^3) - 45/128*(2*sqrt(2)*sqrt(b)*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)) + 2*sqrt(
2)*sqrt(b)*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^(1/4)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4) - sqrt(2)*b^(
1/4)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/c^(1/4))/c^3 + 2*sqrt(x)/c^3

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mupad [B]  time = 4.39, size = 101, normalized size = 0.40 \begin {gather*} \frac {\frac {13\,b^2\,\sqrt {x}}{16}+\frac {17\,b\,c\,x^{5/2}}{16}}{b^2\,c^3+2\,b\,c^4\,x^2+c^5\,x^4}+\frac {2\,\sqrt {x}}{c^3}-\frac {45\,{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,c^{13/4}}+\frac {{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,45{}\mathrm {i}}{32\,c^{13/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(23/2)/(b*x^2 + c*x^4)^3,x)

[Out]

((13*b^2*x^(1/2))/16 + (17*b*c*x^(5/2))/16)/(b^2*c^3 + c^5*x^4 + 2*b*c^4*x^2) + (2*x^(1/2))/c^3 - (45*(-b)^(1/
4)*atan((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(32*c^(13/4)) + ((-b)^(1/4)*atan((c^(1/4)*x^(1/2)*1i)/(-b)^(1/4))*45i)/
(32*c^(13/4))

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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(x**(23/2)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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