3.391 \(\int \frac{x^{17/2}}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=174 \[ \frac{15 b^{7/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{14 c^{13/4} \sqrt{b x^2+c x^4}}+\frac{9 x^{3/2} \sqrt{b x^2+c x^4}}{7 c^2}-\frac{15 b \sqrt{b x^2+c x^4}}{7 c^3 \sqrt{x}}-\frac{x^{11/2}}{c \sqrt{b x^2+c x^4}} \]

[Out]

-(x^(11/2)/(c*Sqrt[b*x^2 + c*x^4])) - (15*b*Sqrt[b*x^2 + c*x^4])/(7*c^3*Sqrt[x]) + (9*x^(3/2)*Sqrt[b*x^2 + c*x
^4])/(7*c^2) + (15*b^(7/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTa
n[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(14*c^(13/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.242447, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2022, 2024, 2032, 329, 220} \[ \frac{15 b^{7/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{14 c^{13/4} \sqrt{b x^2+c x^4}}+\frac{9 x^{3/2} \sqrt{b x^2+c x^4}}{7 c^2}-\frac{15 b \sqrt{b x^2+c x^4}}{7 c^3 \sqrt{x}}-\frac{x^{11/2}}{c \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^(17/2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-(x^(11/2)/(c*Sqrt[b*x^2 + c*x^4])) - (15*b*Sqrt[b*x^2 + c*x^4])/(7*c^3*Sqrt[x]) + (9*x^(3/2)*Sqrt[b*x^2 + c*x
^4])/(7*c^2) + (15*b^(7/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTa
n[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(14*c^(13/4)*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 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]

Rubi steps

\begin{align*} \int \frac{x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=-\frac{x^{11/2}}{c \sqrt{b x^2+c x^4}}+\frac{9 \int \frac{x^{9/2}}{\sqrt{b x^2+c x^4}} \, dx}{2 c}\\ &=-\frac{x^{11/2}}{c \sqrt{b x^2+c x^4}}+\frac{9 x^{3/2} \sqrt{b x^2+c x^4}}{7 c^2}-\frac{(45 b) \int \frac{x^{5/2}}{\sqrt{b x^2+c x^4}} \, dx}{14 c^2}\\ &=-\frac{x^{11/2}}{c \sqrt{b x^2+c x^4}}-\frac{15 b \sqrt{b x^2+c x^4}}{7 c^3 \sqrt{x}}+\frac{9 x^{3/2} \sqrt{b x^2+c x^4}}{7 c^2}+\frac{\left (15 b^2\right ) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{14 c^3}\\ &=-\frac{x^{11/2}}{c \sqrt{b x^2+c x^4}}-\frac{15 b \sqrt{b x^2+c x^4}}{7 c^3 \sqrt{x}}+\frac{9 x^{3/2} \sqrt{b x^2+c x^4}}{7 c^2}+\frac{\left (15 b^2 x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{14 c^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{x^{11/2}}{c \sqrt{b x^2+c x^4}}-\frac{15 b \sqrt{b x^2+c x^4}}{7 c^3 \sqrt{x}}+\frac{9 x^{3/2} \sqrt{b x^2+c x^4}}{7 c^2}+\frac{\left (15 b^2 x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{7 c^3 \sqrt{b x^2+c x^4}}\\ &=-\frac{x^{11/2}}{c \sqrt{b x^2+c x^4}}-\frac{15 b \sqrt{b x^2+c x^4}}{7 c^3 \sqrt{x}}+\frac{9 x^{3/2} \sqrt{b x^2+c x^4}}{7 c^2}+\frac{15 b^{7/4} 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 )}{14 c^{13/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0292939, size = 86, normalized size = 0.49 \[ \frac{x^{3/2} \left (15 b^2 \sqrt{\frac{c x^2}{b}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{c x^2}{b}\right )-15 b^2-6 b c x^2+2 c^2 x^4\right )}{7 c^3 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(17/2)/(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x^(3/2)*(-15*b^2 - 6*b*c*x^2 + 2*c^2*x^4 + 15*b^2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((c*x
^2)/b)]))/(7*c^3*Sqrt[x^2*(b + c*x^2)])

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Maple [A]  time = 0.191, size = 144, normalized size = 0.8 \begin{align*}{\frac{c{x}^{2}+b}{14\,{c}^{4}}{x}^{{\frac{5}{2}}} \left ( 15\,{b}^{2}\sqrt{-bc}\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}^{3}{x}^{5}-12\,b{c}^{2}{x}^{3}-30\,{b}^{2}cx \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^(17/2)/(c*x^4+b*x^2)^(3/2),x)

[Out]

1/14/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(15*b^2*(-b*c)^(1/2)*((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^3*x^5-12*b*c^2*x^3-30*b^2*c*x)/c^4

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{\frac{17}{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^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^(17/2)/(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{9}{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^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^4 + b*x^2)*x^(9/2)/(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**(17/2)/(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{17}{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^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

integrate(x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)