3.2.98 \(\int \frac {x^{13/2}}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=217 \[ \frac {b^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{11/4}}-\frac {b^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{11/4}}-\frac {b^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{11/4}}+\frac {b^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{11/4}}-\frac {2 b x^{3/2}}{3 c^2}+\frac {2 x^{7/2}}{7 c} \]

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Rubi [A]  time = 0.23, antiderivative size = 217, 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, 321, 329, 297, 1162, 617, 204, 1165, 628} \begin {gather*} \frac {b^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{11/4}}-\frac {b^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{11/4}}-\frac {b^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{11/4}}+\frac {b^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{11/4}}-\frac {2 b x^{3/2}}{3 c^2}+\frac {2 x^{7/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(13/2)/(b*x^2 + c*x^4),x]

[Out]

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

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 297

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

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Mathematica [A]  time = 0.04, size = 89, normalized size = 0.41 \begin {gather*} \frac {2 c^{3/4} x^{3/2} \left (3 c x^2-7 b\right )+21 (-b)^{7/4} \tan ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )+21 b (-b)^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b}}\right )}{21 c^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(13/2)/(b*x^2 + c*x^4),x]

[Out]

(2*c^(3/4)*x^(3/2)*(-7*b + 3*c*x^2) + 21*(-b)^(7/4)*ArcTan[(c^(1/4)*Sqrt[x])/(-b)^(1/4)] + 21*(-b)^(3/4)*b*Arc
Tanh[(c^(1/4)*Sqrt[x])/(-b)^(1/4)])/(21*c^(11/4))

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IntegrateAlgebraic [A]  time = 0.20, size = 138, normalized size = 0.64 \begin {gather*} -\frac {b^{7/4} \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 )}{\sqrt {2} c^{11/4}}-\frac {b^{7/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{\sqrt {2} c^{11/4}}+\frac {2 \left (3 c x^{7/2}-7 b x^{3/2}\right )}{21 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(13/2)/(b*x^2 + c*x^4),x]

[Out]

(2*(-7*b*x^(3/2) + 3*c*x^(7/2)))/(21*c^2) - (b^(7/4)*ArcTan[(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[2]*
b^(1/4)))/Sqrt[x]])/(Sqrt[2]*c^(11/4)) - (b^(7/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]
*x)])/(Sqrt[2]*c^(11/4))

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fricas [A]  time = 0.87, size = 182, normalized size = 0.84 \begin {gather*} -\frac {84 \, c^{2} \left (-\frac {b^{7}}{c^{11}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b^{5} c^{3} \sqrt {x} \left (-\frac {b^{7}}{c^{11}}\right )^{\frac {1}{4}} - \sqrt {-b^{7} c^{5} \sqrt {-\frac {b^{7}}{c^{11}}} + b^{10} x} c^{3} \left (-\frac {b^{7}}{c^{11}}\right )^{\frac {1}{4}}}{b^{7}}\right ) - 21 \, c^{2} \left (-\frac {b^{7}}{c^{11}}\right )^{\frac {1}{4}} \log \left (c^{8} \left (-\frac {b^{7}}{c^{11}}\right )^{\frac {3}{4}} + b^{5} \sqrt {x}\right ) + 21 \, c^{2} \left (-\frac {b^{7}}{c^{11}}\right )^{\frac {1}{4}} \log \left (-c^{8} \left (-\frac {b^{7}}{c^{11}}\right )^{\frac {3}{4}} + b^{5} \sqrt {x}\right ) - 4 \, {\left (3 \, c x^{3} - 7 \, b x\right )} \sqrt {x}}{42 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

-1/42*(84*c^2*(-b^7/c^11)^(1/4)*arctan(-(b^5*c^3*sqrt(x)*(-b^7/c^11)^(1/4) - sqrt(-b^7*c^5*sqrt(-b^7/c^11) + b
^10*x)*c^3*(-b^7/c^11)^(1/4))/b^7) - 21*c^2*(-b^7/c^11)^(1/4)*log(c^8*(-b^7/c^11)^(3/4) + b^5*sqrt(x)) + 21*c^
2*(-b^7/c^11)^(1/4)*log(-c^8*(-b^7/c^11)^(3/4) + b^5*sqrt(x)) - 4*(3*c*x^3 - 7*b*x)*sqrt(x))/c^2

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giac [A]  time = 0.17, size = 197, normalized size = 0.91 \begin {gather*} \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{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 )}{2 \, c^{5}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{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 )}{2 \, c^{5}} - \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} b \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{5}} + \frac {\sqrt {2} \left (b c^{3}\right )^{\frac {3}{4}} b \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{4 \, c^{5}} + \frac {2 \, {\left (3 \, c^{6} x^{\frac {7}{2}} - 7 \, b c^{5} x^{\frac {3}{2}}\right )}}{21 \, c^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

1/2*sqrt(2)*(b*c^3)^(3/4)*b*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/c^5 + 1/2*sqrt(2
)*(b*c^3)^(3/4)*b*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/c^5 - 1/4*sqrt(2)*(b*c^3)
^(3/4)*b*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/c^5 + 1/4*sqrt(2)*(b*c^3)^(3/4)*b*log(-sqrt(2)*sqrt(
x)*(b/c)^(1/4) + x + sqrt(b/c))/c^5 + 2/21*(3*c^6*x^(7/2) - 7*b*c^5*x^(3/2))/c^7

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maple [A]  time = 0.01, size = 158, normalized size = 0.73 \begin {gather*} \frac {2 x^{\frac {7}{2}}}{7 c}-\frac {2 b \,x^{\frac {3}{2}}}{3 c^{2}}+\frac {\sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {\sqrt {2}\, b^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}}+\frac {\sqrt {2}\, b^{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 )}{4 \left (\frac {b}{c}\right )^{\frac {1}{4}} c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(c*x^4+b*x^2),x)

[Out]

2/7*x^(7/2)/c-2/3*b*x^(3/2)/c^2+1/4*b^2/c^3/(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)))+1/2*b^2/c^3/(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/
2)+1)+1/2*b^2/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 = 3.01, size = 198, normalized size = 0.91 \begin {gather*} \frac {b^{2} {\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 )}}{4 \, c^{2}} + \frac {2 \, {\left (3 \, c x^{\frac {7}{2}} - 7 \, b x^{\frac {3}{2}}\right )}}{21 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(13/2)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/4*b^2*(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)))/(sq
rt(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))/sqr
t(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)))/c^2 + 2/21*(3*c*x^(7/2) - 7*b*x^(3/2))/c^2

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mupad [B]  time = 0.11, size = 66, normalized size = 0.30 \begin {gather*} \frac {2\,x^{7/2}}{7\,c}-\frac {2\,b\,x^{3/2}}{3\,c^2}+\frac {{\left (-b\right )}^{7/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{c^{11/4}}+\frac {{\left (-b\right )}^{7/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,1{}\mathrm {i}}{c^{11/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(13/2)/(b*x^2 + c*x^4),x)

[Out]

(2*x^(7/2))/(7*c) - (2*b*x^(3/2))/(3*c^2) + ((-b)^(7/4)*atan((c^(1/4)*x^(1/2))/(-b)^(1/4)))/c^(11/4) + ((-b)^(
7/4)*atan((c^(1/4)*x^(1/2)*1i)/(-b)^(1/4))*1i)/c^(11/4)

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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**(13/2)/(c*x**4+b*x**2),x)

[Out]

Timed out

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