3.392 \(\int \frac{x^{15/2}}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=291 \[ -\frac{21 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{10 c^{11/4} \sqrt{b x^2+c x^4}}+\frac{21 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{5 c^{11/4} \sqrt{b x^2+c x^4}}-\frac{21 b x^{3/2} \left (b+c x^2\right )}{5 c^{5/2} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{7 \sqrt{x} \sqrt{b x^2+c x^4}}{5 c^2}-\frac{x^{9/2}}{c \sqrt{b x^2+c x^4}} \]

[Out]

-(x^(9/2)/(c*Sqrt[b*x^2 + c*x^4])) - (21*b*x^(3/2)*(b + c*x^2))/(5*c^(5/2)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 +
c*x^4]) + (7*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(5*c^2) + (21*b^(5/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt
[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(5*c^(11/4)*Sqrt[b*x^2 + c*x^4]) - (2
1*b^(5/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x
])/b^(1/4)], 1/2])/(10*c^(11/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.29648, antiderivative size = 291, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2022, 2024, 2032, 329, 305, 220, 1196} \[ -\frac{21 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{10 c^{11/4} \sqrt{b x^2+c x^4}}+\frac{21 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{5 c^{11/4} \sqrt{b x^2+c x^4}}-\frac{21 b x^{3/2} \left (b+c x^2\right )}{5 c^{5/2} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{7 \sqrt{x} \sqrt{b x^2+c x^4}}{5 c^2}-\frac{x^{9/2}}{c \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^(15/2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(x^(9/2)/(c*Sqrt[b*x^2 + c*x^4])) - (21*b*x^(3/2)*(b + c*x^2))/(5*c^(5/2)*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 +
c*x^4]) + (7*Sqrt[x]*Sqrt[b*x^2 + c*x^4])/(5*c^2) + (21*b^(5/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt
[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(5*c^(11/4)*Sqrt[b*x^2 + c*x^4]) - (2
1*b^(5/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[(c^(1/4)*Sqrt[x
])/b^(1/4)], 1/2])/(10*c^(11/4)*Sqrt[b*x^2 + c*x^4])

Rule 2022

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(n - j)*(p + 1)), x] - Dist[(c^n*(m + j*p - n + j + 1))/(b*(n - j)*(p + 1)), I
nt[(c*x)^(m - n)*(a*x^j + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

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 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{x^{15/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac{x^{9/2}}{c \sqrt{b x^2+c x^4}}+\frac{7 \int \frac{x^{7/2}}{\sqrt{b x^2+c x^4}} \, dx}{2 c}\\ &=-\frac{x^{9/2}}{c \sqrt{b x^2+c x^4}}+\frac{7 \sqrt{x} \sqrt{b x^2+c x^4}}{5 c^2}-\frac{(21 b) \int \frac{x^{3/2}}{\sqrt{b x^2+c x^4}} \, dx}{10 c^2}\\ &=-\frac{x^{9/2}}{c \sqrt{b x^2+c x^4}}+\frac{7 \sqrt{x} \sqrt{b x^2+c x^4}}{5 c^2}-\frac{\left (21 b x \sqrt{b+c x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{b+c x^2}} \, dx}{10 c^2 \sqrt{b x^2+c x^4}}\\ &=-\frac{x^{9/2}}{c \sqrt{b x^2+c x^4}}+\frac{7 \sqrt{x} \sqrt{b x^2+c x^4}}{5 c^2}-\frac{\left (21 b x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{5 c^2 \sqrt{b x^2+c x^4}}\\ &=-\frac{x^{9/2}}{c \sqrt{b x^2+c x^4}}+\frac{7 \sqrt{x} \sqrt{b x^2+c x^4}}{5 c^2}-\frac{\left (21 b^{3/2} x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{5 c^{5/2} \sqrt{b x^2+c x^4}}+\frac{\left (21 b^{3/2} x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{c} x^2}{\sqrt{b}}}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{5 c^{5/2} \sqrt{b x^2+c x^4}}\\ &=-\frac{x^{9/2}}{c \sqrt{b x^2+c x^4}}-\frac{21 b x^{3/2} \left (b+c x^2\right )}{5 c^{5/2} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{7 \sqrt{x} \sqrt{b x^2+c x^4}}{5 c^2}+\frac{21 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{5 c^{11/4} \sqrt{b x^2+c x^4}}-\frac{21 b^{5/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{10 c^{11/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0264773, size = 72, normalized size = 0.25 \[ \frac{2 x^{5/2} \left (7 b \sqrt{\frac{c x^2}{b}+1} \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\frac{c x^2}{b}\right )-7 b+c x^2\right )}{5 c^2 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(15/2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(2*x^(5/2)*(-7*b + c*x^2 + 7*b*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[3/4, 3/2, 7/4, -((c*x^2)/b)]))/(5*c^2*Sqr
t[x^2*(b + c*x^2)])

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Maple [A]  time = 0.187, size = 213, normalized size = 0.7 \begin{align*} -{\frac{c{x}^{2}+b}{10\,{c}^{3}}{x}^{{\frac{5}{2}}} \left ( 42\,{b}^{2}\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticE} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) -21\,{b}^{2}\sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{2}\sqrt{{\frac{-cx+\sqrt{-bc}}{\sqrt{-bc}}}}\sqrt{-{\frac{cx}{\sqrt{-bc}}}}{\it EllipticF} \left ( \sqrt{{\frac{cx+\sqrt{-bc}}{\sqrt{-bc}}}},1/2\,\sqrt{2} \right ) -4\,{c}^{2}{x}^{4}-14\,bc{x}^{2} \right ) \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(15/2)/(c*x^4+b*x^2)^(3/2),x)

[Out]

-1/10/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(42*b^2*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b
*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*c/(-b*c)^(1/2))^(1/2)*EllipticE(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2
*2^(1/2))-21*b^2*((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2)*(-x*
c/(-b*c)^(1/2))^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^(1/2),1/2*2^(1/2))-4*c^2*x^4-14*b*c*x^2)/c^3

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{15}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^(15/2)/(c*x^4 + b*x^2)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{c x^{4} + b x^{2}} x^{\frac{7}{2}}}{c^{2} x^{4} + 2 \, b c x^{2} + b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*x^(7/2)/(c^2*x^4 + 2*b*c*x^2 + b^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(15/2)/(c*x**4+b*x**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{15}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(15/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^(15/2)/(c*x^4 + b*x^2)^(3/2), x)