3.3.27 \(\int \frac {x^{11/2}}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=239 \[ -\frac {21 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {21 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {21 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {21 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {7 \sqrt {x}}{16 b^2 \left (b+c x^2\right )}+\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2} \]

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Rubi [A]  time = 0.19, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {1584, 290, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {7 \sqrt {x}}{16 b^2 \left (b+c x^2\right )}-\frac {21 \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {21 \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {21 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {21 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(11/2)/(b*x^2 + c*x^4)^3,x]

[Out]

Sqrt[x]/(4*b*(b + c*x^2)^2) + (7*Sqrt[x])/(16*b^2*(b + c*x^2)) - (21*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1
/4)])/(32*Sqrt[2]*b^(11/4)*c^(1/4)) + (21*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(11/4)*
c^(1/4)) - (21*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(11/4)*c^(1/4)) + (21
*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(11/4)*c^(1/4))

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 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^{11/2}}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {1}{\sqrt {x} \left (b+c x^2\right )^3} \, dx\\ &=\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2}+\frac {7 \int \frac {1}{\sqrt {x} \left (b+c x^2\right )^2} \, dx}{8 b}\\ &=\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2}+\frac {7 \sqrt {x}}{16 b^2 \left (b+c x^2\right )}+\frac {21 \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{32 b^2}\\ &=\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2}+\frac {7 \sqrt {x}}{16 b^2 \left (b+c x^2\right )}+\frac {21 \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^2}\\ &=\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2}+\frac {7 \sqrt {x}}{16 b^2 \left (b+c x^2\right )}+\frac {21 \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{5/2}}+\frac {21 \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{5/2}}\\ &=\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2}+\frac {7 \sqrt {x}}{16 b^2 \left (b+c x^2\right )}+\frac {21 \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^{5/2} \sqrt {c}}+\frac {21 \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^{5/2} \sqrt {c}}-\frac {21 \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^{11/4} \sqrt [4]{c}}-\frac {21 \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^{11/4} \sqrt [4]{c}}\\ &=\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2}+\frac {7 \sqrt {x}}{16 b^2 \left (b+c x^2\right )}-\frac {21 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {21 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {21 \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^{11/4} \sqrt [4]{c}}-\frac {21 \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^{11/4} \sqrt [4]{c}}\\ &=\frac {\sqrt {x}}{4 b \left (b+c x^2\right )^2}+\frac {7 \sqrt {x}}{16 b^2 \left (b+c x^2\right )}-\frac {21 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {21 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{11/4} \sqrt [4]{c}}-\frac {21 \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {21 \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{11/4} \sqrt [4]{c}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 220, normalized size = 0.92 \begin {gather*} \frac {\frac {32 b^{7/4} \sqrt {x}}{\left (b+c x^2\right )^2}+\frac {56 b^{3/4} \sqrt {x}}{b+c x^2}-\frac {21 \sqrt {2} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{\sqrt [4]{c}}+\frac {21 \sqrt {2} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{\sqrt [4]{c}}-\frac {42 \sqrt {2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt [4]{c}}+\frac {42 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt [4]{c}}}{128 b^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(11/2)/(b*x^2 + c*x^4)^3,x]

[Out]

((32*b^(7/4)*Sqrt[x])/(b + c*x^2)^2 + (56*b^(3/4)*Sqrt[x])/(b + c*x^2) - (42*Sqrt[2]*ArcTan[1 - (Sqrt[2]*c^(1/
4)*Sqrt[x])/b^(1/4)])/c^(1/4) + (42*Sqrt[2]*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/c^(1/4) - (21*Sqrt[
2]*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4) + (21*Sqrt[2]*Log[Sqrt[b] + Sqrt[2]*b^(
1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/c^(1/4))/(128*b^(11/4))

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IntegrateAlgebraic [A]  time = 0.27, size = 149, normalized size = 0.62 \begin {gather*} -\frac {21 \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^{11/4} \sqrt [4]{c}}+\frac {21 \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{11/4} \sqrt [4]{c}}+\frac {\sqrt {x} \left (11 b+7 c x^2\right )}{16 b^2 \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(11/2)/(b*x^2 + c*x^4)^3,x]

[Out]

(Sqrt[x]*(11*b + 7*c*x^2))/(16*b^2*(b + c*x^2)^2) - (21*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[
2]*b^(1/4)))/Sqrt[x]])/(32*Sqrt[2]*b^(11/4)*c^(1/4)) + (21*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b]
+ Sqrt[c]*x)])/(32*Sqrt[2]*b^(11/4)*c^(1/4))

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fricas [A]  time = 1.10, size = 241, normalized size = 1.01 \begin {gather*} \frac {84 \, {\left (b^{2} c^{2} x^{4} + 2 \, b^{3} c x^{2} + b^{4}\right )} \left (-\frac {1}{b^{11} c}\right )^{\frac {1}{4}} \arctan \left (\sqrt {b^{6} \sqrt {-\frac {1}{b^{11} c}} + x} b^{8} c \left (-\frac {1}{b^{11} c}\right )^{\frac {3}{4}} - b^{8} c \sqrt {x} \left (-\frac {1}{b^{11} c}\right )^{\frac {3}{4}}\right ) + 21 \, {\left (b^{2} c^{2} x^{4} + 2 \, b^{3} c x^{2} + b^{4}\right )} \left (-\frac {1}{b^{11} c}\right )^{\frac {1}{4}} \log \left (b^{3} \left (-\frac {1}{b^{11} c}\right )^{\frac {1}{4}} + \sqrt {x}\right ) - 21 \, {\left (b^{2} c^{2} x^{4} + 2 \, b^{3} c x^{2} + b^{4}\right )} \left (-\frac {1}{b^{11} c}\right )^{\frac {1}{4}} \log \left (-b^{3} \left (-\frac {1}{b^{11} c}\right )^{\frac {1}{4}} + \sqrt {x}\right ) + 4 \, {\left (7 \, c x^{2} + 11 \, b\right )} \sqrt {x}}{64 \, {\left (b^{2} c^{2} x^{4} + 2 \, b^{3} c x^{2} + b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/64*(84*(b^2*c^2*x^4 + 2*b^3*c*x^2 + b^4)*(-1/(b^11*c))^(1/4)*arctan(sqrt(b^6*sqrt(-1/(b^11*c)) + x)*b^8*c*(-
1/(b^11*c))^(3/4) - b^8*c*sqrt(x)*(-1/(b^11*c))^(3/4)) + 21*(b^2*c^2*x^4 + 2*b^3*c*x^2 + b^4)*(-1/(b^11*c))^(1
/4)*log(b^3*(-1/(b^11*c))^(1/4) + sqrt(x)) - 21*(b^2*c^2*x^4 + 2*b^3*c*x^2 + b^4)*(-1/(b^11*c))^(1/4)*log(-b^3
*(-1/(b^11*c))^(1/4) + sqrt(x)) + 4*(7*c*x^2 + 11*b)*sqrt(x))/(b^2*c^2*x^4 + 2*b^3*c*x^2 + b^4)

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giac [A]  time = 0.21, size = 209, normalized size = 0.87 \begin {gather*} \frac {21 \, \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 \, b^{3} c} + \frac {21 \, \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 \, b^{3} c} + \frac {21 \, \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 \, b^{3} c} - \frac {21 \, \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 \, b^{3} c} + \frac {7 \, c x^{\frac {5}{2}} + 11 \, b \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

21/64*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))/(b^3*c) + 21/64*
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))/(b^3*c) + 21/128*sqrt
(2)*(b*c^3)^(1/4)*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) - 21/128*sqrt(2)*(b*c^3)^(1/4)*log(
-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/(b^3*c) + 1/16*(7*c*x^(5/2) + 11*b*sqrt(x))/((c*x^2 + b)^2*b^2)

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maple [A]  time = 0.01, size = 166, normalized size = 0.69 \begin {gather*} \frac {\sqrt {x}}{4 \left (c \,x^{2}+b \right )^{2} b}+\frac {7 \sqrt {x}}{16 \left (c \,x^{2}+b \right ) b^{2}}+\frac {21 \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 b^{3}}+\frac {21 \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 b^{3}}+\frac {21 \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 b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)/(c*x^4+b*x^2)^3,x)

[Out]

1/4*x^(1/2)/b/(c*x^2+b)^2+7/16*x^(1/2)/b^2/(c*x^2+b)+21/128/b^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)))+21/64/b^3*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)
/(b/c)^(1/4)*x^(1/2)+1)+21/64/b^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 = 2.97, size = 217, normalized size = 0.91 \begin {gather*} \frac {7 \, c x^{\frac {5}{2}} + 11 \, b \sqrt {x}}{16 \, {\left (b^{2} c^{2} x^{4} + 2 \, b^{3} c x^{2} + b^{4}\right )}} + \frac {21 \, {\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 {b} \sqrt {\sqrt {b} \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 {b} \sqrt {\sqrt {b} \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 {3}{4}} c^{\frac {1}{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 {3}{4}} c^{\frac {1}{4}}}\right )}}{128 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/16*(7*c*x^(5/2) + 11*b*sqrt(x))/(b^2*c^2*x^4 + 2*b^3*c*x^2 + b^4) + 21/128*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sq
rt(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)*
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)*log(sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)) - sqrt(2)*lo
g(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + sqrt(b))/(b^(3/4)*c^(1/4)))/b^2

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mupad [B]  time = 4.29, size = 86, normalized size = 0.36 \begin {gather*} \frac {\frac {11\,\sqrt {x}}{16\,b}+\frac {7\,c\,x^{5/2}}{16\,b^2}}{b^2+2\,b\,c\,x^2+c^2\,x^4}-\frac {21\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,{\left (-b\right )}^{11/4}\,c^{1/4}}-\frac {21\,\mathrm {atanh}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{32\,{\left (-b\right )}^{11/4}\,c^{1/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)/(b*x^2 + c*x^4)^3,x)

[Out]

((11*x^(1/2))/(16*b) + (7*c*x^(5/2))/(16*b^2))/(b^2 + c^2*x^4 + 2*b*c*x^2) - (21*atan((c^(1/4)*x^(1/2))/(-b)^(
1/4)))/(32*(-b)^(11/4)*c^(1/4)) - (21*atanh((c^(1/4)*x^(1/2))/(-b)^(1/4)))/(32*(-b)^(11/4)*c^(1/4))

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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**(11/2)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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