3.320 \(\int \frac{x^{5/2}}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=192 \[ \frac{\log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{b} c^{3/4}}-\frac{\log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{b} c^{3/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} \sqrt [4]{b} c^{3/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} \sqrt [4]{b} c^{3/4}} \]

[Out]

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

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Rubi [A]  time = 0.142039, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {1584, 329, 297, 1162, 617, 204, 1165, 628} \[ \frac{\log \left (-\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{b} c^{3/4}}-\frac{\log \left (\sqrt{2} \sqrt [4]{b} \sqrt [4]{c} \sqrt{x}+\sqrt{b}+\sqrt{c} x\right )}{2 \sqrt{2} \sqrt [4]{b} c^{3/4}}-\frac{\tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )}{\sqrt{2} \sqrt [4]{b} c^{3/4}}+\frac{\tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}+1\right )}{\sqrt{2} \sqrt [4]{b} c^{3/4}} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(b*x^2 + c*x^4),x]

[Out]

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

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]

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 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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0250279, size = 54, normalized size = 0.28 \[ \frac{b \left (\tan ^{-1}\left (\frac{b \sqrt [4]{c} \sqrt{x}}{(-b)^{5/4}}\right )+\tanh ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{-b}}\right )\right )}{(-b)^{5/4} c^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(b*x^2 + c*x^4),x]

[Out]

(b*(ArcTan[(b*c^(1/4)*Sqrt[x])/(-b)^(5/4)] + ArcTanh[(c^(1/4)*Sqrt[x])/(-b)^(1/4)]))/((-b)^(5/4)*c^(3/4))

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Maple [A]  time = 0.049, size = 132, normalized size = 0.7 \begin{align*}{\frac{\sqrt{2}}{4\,c}\ln \left ({ \left ( x-\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) \left ( x+\sqrt [4]{{\frac{b}{c}}}\sqrt{x}\sqrt{2}+\sqrt{{\frac{b}{c}}} \right ) ^{-1}} \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{\sqrt{2}}{2\,c}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}+{\frac{\sqrt{2}}{2\,c}\arctan \left ({\sqrt{2}\sqrt{x}{\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}}-1 \right ){\frac{1}{\sqrt [4]{{\frac{b}{c}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(c*x^4+b*x^2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33029, size = 332, normalized size = 1.73 \begin{align*} -2 \, \left (-\frac{1}{b c^{3}}\right )^{\frac{1}{4}} \arctan \left (\sqrt{-b c \sqrt{-\frac{1}{b c^{3}}} + x} c \left (-\frac{1}{b c^{3}}\right )^{\frac{1}{4}} - c \sqrt{x} \left (-\frac{1}{b c^{3}}\right )^{\frac{1}{4}}\right ) + \frac{1}{2} \, \left (-\frac{1}{b c^{3}}\right )^{\frac{1}{4}} \log \left (b c^{2} \left (-\frac{1}{b c^{3}}\right )^{\frac{3}{4}} + \sqrt{x}\right ) - \frac{1}{2} \, \left (-\frac{1}{b c^{3}}\right )^{\frac{1}{4}} \log \left (-b c^{2} \left (-\frac{1}{b c^{3}}\right )^{\frac{3}{4}} + \sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 72.7057, size = 170, normalized size = 0.89 \begin{align*} \begin{cases} \frac{\tilde{\infty }}{\sqrt{x}} & \text{for}\: b = 0 \wedge c = 0 \\\frac{2 x^{\frac{3}{2}}}{3 b} & \text{for}\: c = 0 \\- \frac{2}{c \sqrt{x}} & \text{for}\: b = 0 \\- \frac{\left (-1\right )^{\frac{3}{4}} \log{\left (- \sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac{1}{c}} + \sqrt{x} \right )}}{2 \sqrt [4]{b} c^{2} \left (\frac{1}{c}\right )^{\frac{5}{4}}} + \frac{\left (-1\right )^{\frac{3}{4}} \log{\left (\sqrt [4]{-1} \sqrt [4]{b} \sqrt [4]{\frac{1}{c}} + \sqrt{x} \right )}}{2 \sqrt [4]{b} c^{2} \left (\frac{1}{c}\right )^{\frac{5}{4}}} + \frac{\left (-1\right )^{\frac{3}{4}} \operatorname{atan}{\left (\frac{\left (-1\right )^{\frac{3}{4}} \sqrt{x}}{\sqrt [4]{b} \sqrt [4]{\frac{1}{c}}} \right )}}{\sqrt [4]{b} c^{2} \left (\frac{1}{c}\right )^{\frac{5}{4}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(c*x**4+b*x**2),x)

[Out]

Piecewise((zoo/sqrt(x), Eq(b, 0) & Eq(c, 0)), (2*x**(3/2)/(3*b), Eq(c, 0)), (-2/(c*sqrt(x)), Eq(b, 0)), (-(-1)
**(3/4)*log(-(-1)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(1/4)*c**2*(1/c)**(5/4)) + (-1)**(3/4)*log((-1
)**(1/4)*b**(1/4)*(1/c)**(1/4) + sqrt(x))/(2*b**(1/4)*c**2*(1/c)**(5/4)) + (-1)**(3/4)*atan((-1)**(3/4)*sqrt(x
)/(b**(1/4)*(1/c)**(1/4)))/(b**(1/4)*c**2*(1/c)**(5/4)), True))

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Giac [A]  time = 1.13058, size = 246, normalized size = 1.28 \begin{align*} \frac{\sqrt{2} \left (b c^{3}\right )^{\frac{3}{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 )}{2 \, b c^{3}} + \frac{\sqrt{2} \left (b c^{3}\right )^{\frac{3}{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 )}{2 \, b c^{3}} - \frac{\sqrt{2} \left (b c^{3}\right )^{\frac{3}{4}} \log \left (\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{4 \, b c^{3}} + \frac{\sqrt{2} \left (b c^{3}\right )^{\frac{3}{4}} \log \left (-\sqrt{2} \sqrt{x} \left (\frac{b}{c}\right )^{\frac{1}{4}} + x + \sqrt{\frac{b}{c}}\right )}{4 \, b c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

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