3.4.17 \(\int \frac {x^{11/2}}{b x^2+c x^4} \, dx\) [317]

Optimal. Leaf size=215 \[ -\frac {2 b \sqrt {x}}{c^2}+\frac {2 x^{5/2}}{5 c}-\frac {b^{5/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {b^{5/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}-\frac {b^{5/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {b^{5/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}} \]

[Out]

2/5*x^(5/2)/c-1/2*b^(5/4)*arctan(1-c^(1/4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(9/4)*2^(1/2)+1/2*b^(5/4)*arctan(1+c^(1/
4)*2^(1/2)*x^(1/2)/b^(1/4))/c^(9/4)*2^(1/2)-1/4*b^(5/4)*ln(b^(1/2)+x*c^(1/2)-b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/
c^(9/4)*2^(1/2)+1/4*b^(5/4)*ln(b^(1/2)+x*c^(1/2)+b^(1/4)*c^(1/4)*2^(1/2)*x^(1/2))/c^(9/4)*2^(1/2)-2*b*x^(1/2)/
c^2

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Rubi [A]
time = 0.13, antiderivative size = 215, 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 = {1598, 327, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {b^{5/4} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{\sqrt {2} c^{9/4}}+\frac {b^{5/4} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} c^{9/4}}-\frac {b^{5/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}+\frac {b^{5/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{2 \sqrt {2} c^{9/4}}-\frac {2 b \sqrt {x}}{c^2}+\frac {2 x^{5/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(11/2)/(b*x^2 + c*x^4),x]

[Out]

(-2*b*Sqrt[x])/c^2 + (2*x^(5/2))/(5*c) - (b^(5/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(9
/4)) + (b^(5/4)*ArcTan[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(Sqrt[2]*c^(9/4)) - (b^(5/4)*Log[Sqrt[b] - Sqrt
[2]*b^(1/4)*c^(1/4)*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(9/4)) + (b^(5/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)
*Sqrt[x] + Sqrt[c]*x])/(2*Sqrt[2]*c^(9/4))

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 217

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 327

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 335

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 631

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 642

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 1176

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 1179

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 1598

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

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Mathematica [A]
time = 0.14, size = 128, normalized size = 0.60 \begin {gather*} \frac {4 \sqrt [4]{c} \sqrt {x} \left (-5 b+c x^2\right )-5 \sqrt {2} b^{5/4} \tan ^{-1}\left (\frac {\sqrt {b}-\sqrt {c} x}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}\right )+5 \sqrt {2} b^{5/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{10 c^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(11/2)/(b*x^2 + c*x^4),x]

[Out]

(4*c^(1/4)*Sqrt[x]*(-5*b + c*x^2) - 5*Sqrt[2]*b^(5/4)*ArcTan[(Sqrt[b] - Sqrt[c]*x)/(Sqrt[2]*b^(1/4)*c^(1/4)*Sq
rt[x])] + 5*Sqrt[2]*b^(5/4)*ArcTanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(10*c^(9/4))

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Maple [A]
time = 0.12, size = 125, normalized size = 0.58

method	result	size
derivativedivides	\(-\frac {2 \left (-\frac {c \,x^{\frac {5}{2}}}{5}+b \sqrt {x}\right )}{c^{2}}+\frac {b \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2}}\)	\(125\)
default	\(-\frac {2 \left (-\frac {c \,x^{\frac {5}{2}}}{5}+b \sqrt {x}\right )}{c^{2}}+\frac {b \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )\right )}{4 c^{2}}\)	\(125\)
risch	\(-\frac {2 \left (-c \,x^{2}+5 b \right ) \sqrt {x}}{5 c^{2}}+\frac {b \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {b}{c}}}\right )}{4 c^{2}}+\frac {b \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{2 c^{2}}+\frac {b \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{2 c^{2}}\)	\(153\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)

[Out]

-2/c^2*(-1/5*c*x^(5/2)+b*x^(1/2))+1/4*b/c^2*(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)))+2*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+2*arctan(2^(1/2)/(b/c)^
(1/4)*x^(1/2)-1))

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Maxima [A]
time = 0.52, size = 194, normalized size = 0.90 \begin {gather*} \frac {2 \, {\left (c x^{\frac {5}{2}} - 5 \, b \sqrt {x}\right )}}{5 \, c^{2}} + \frac {\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}}}}{4 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(c*x^(5/2) - 5*b*sqrt(x))/c^2 + 1/4*(2*sqrt(2)*b^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqr
t(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^2

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Fricas [A]
time = 0.35, size = 170, normalized size = 0.79 \begin {gather*} \frac {20 \, c^{2} \left (-\frac {b^{5}}{c^{9}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b c^{7} \sqrt {x} \left (-\frac {b^{5}}{c^{9}}\right )^{\frac {3}{4}} - \sqrt {c^{4} \sqrt {-\frac {b^{5}}{c^{9}}} + b^{2} x} c^{7} \left (-\frac {b^{5}}{c^{9}}\right )^{\frac {3}{4}}}{b^{5}}\right ) + 5 \, c^{2} \left (-\frac {b^{5}}{c^{9}}\right )^{\frac {1}{4}} \log \left (c^{2} \left (-\frac {b^{5}}{c^{9}}\right )^{\frac {1}{4}} + b \sqrt {x}\right ) - 5 \, c^{2} \left (-\frac {b^{5}}{c^{9}}\right )^{\frac {1}{4}} \log \left (-c^{2} \left (-\frac {b^{5}}{c^{9}}\right )^{\frac {1}{4}} + b \sqrt {x}\right ) + 4 \, {\left (c x^{2} - 5 \, b\right )} \sqrt {x}}{10 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(20*c^2*(-b^5/c^9)^(1/4)*arctan(-(b*c^7*sqrt(x)*(-b^5/c^9)^(3/4) - sqrt(c^4*sqrt(-b^5/c^9) + b^2*x)*c^7*(
-b^5/c^9)^(3/4))/b^5) + 5*c^2*(-b^5/c^9)^(1/4)*log(c^2*(-b^5/c^9)^(1/4) + b*sqrt(x)) - 5*c^2*(-b^5/c^9)^(1/4)*
log(-c^2*(-b^5/c^9)^(1/4) + b*sqrt(x)) + 4*(c*x^2 - 5*b)*sqrt(x))/c^2

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Sympy [A]
time = 109.38, size = 136, normalized size = 0.63 \begin {gather*} \begin {cases} \tilde {\infty } x^{\frac {5}{2}} & \text {for}\: b = 0 \wedge c = 0 \\\frac {2 x^{\frac {9}{2}}}{9 b} & \text {for}\: c = 0 \\\frac {2 x^{\frac {5}{2}}}{5 c} & \text {for}\: b = 0 \\- \frac {2 b \sqrt {x}}{c^{2}} - \frac {b \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} - \sqrt [4]{- \frac {b}{c}} \right )}}{2 c^{2}} + \frac {b \sqrt [4]{- \frac {b}{c}} \log {\left (\sqrt {x} + \sqrt [4]{- \frac {b}{c}} \right )}}{2 c^{2}} + \frac {b \sqrt [4]{- \frac {b}{c}} \operatorname {atan}{\left (\frac {\sqrt {x}}{\sqrt [4]{- \frac {b}{c}}} \right )}}{c^{2}} + \frac {2 x^{\frac {5}{2}}}{5 c} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**(5/2), Eq(b, 0) & Eq(c, 0)), (2*x**(9/2)/(9*b), Eq(c, 0)), (2*x**(5/2)/(5*c), Eq(b, 0)), (-2
*b*sqrt(x)/c**2 - b*(-b/c)**(1/4)*log(sqrt(x) - (-b/c)**(1/4))/(2*c**2) + b*(-b/c)**(1/4)*log(sqrt(x) + (-b/c)
**(1/4))/(2*c**2) + b*(-b/c)**(1/4)*atan(sqrt(x)/(-b/c)**(1/4))/c**2 + 2*x**(5/2)/(5*c), True))

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Giac [A]
time = 4.99, size = 196, normalized size = 0.91 \begin {gather*} \frac {\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 )}{2 \, c^{3}} + \frac {\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 )}{2 \, c^{3}} + \frac {\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 )}{4 \, c^{3}} - \frac {\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 )}{4 \, c^{3}} + \frac {2 \, {\left (c^{4} x^{\frac {5}{2}} - 5 \, b c^{3} \sqrt {x}\right )}}{5 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.49, size = 67, normalized size = 0.31 \begin {gather*} \frac {2\,x^{5/2}}{5\,c}-\frac {2\,b\,\sqrt {x}}{c^2}-\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}}{{\left (-b\right )}^{1/4}}\right )}{c^{9/4}}+\frac {{\left (-b\right )}^{5/4}\,\mathrm {atan}\left (\frac {c^{1/4}\,\sqrt {x}\,1{}\mathrm {i}}{{\left (-b\right )}^{1/4}}\right )\,1{}\mathrm {i}}{c^{9/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(11/2)/(b*x^2 + c*x^4),x)

[Out]

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

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