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

Optimal. Leaf size=143 \[ \frac{4 c^{7/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{7 \sqrt [4]{b} \sqrt{b x^2+c x^4}}-\frac{4 c \sqrt{b x^2+c x^4}}{7 x^{5/2}}-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}} \]

[Out]

(-4*c*Sqrt[b*x^2 + c*x^4])/(7*x^(5/2)) - (2*(b*x^2 + c*x^4)^(3/2))/(7*x^(13/2)) + (4*c^(7/4)*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])/(7*b^(1/
4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.18338, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2020, 2032, 329, 220} \[ \frac{4 c^{7/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{7 \sqrt [4]{b} \sqrt{b x^2+c x^4}}-\frac{4 c \sqrt{b x^2+c x^4}}{7 x^{5/2}}-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(15/2),x]

[Out]

(-4*c*Sqrt[b*x^2 + c*x^4])/(7*x^(5/2)) - (2*(b*x^2 + c*x^4)^(3/2))/(7*x^(13/2)) + (4*c^(7/4)*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])/(7*b^(1/
4)*Sqrt[b*x^2 + c*x^4])

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(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] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*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{\left (b x^2+c x^4\right )^{3/2}}{x^{15/2}} \, dx &=-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac{1}{7} (6 c) \int \frac{\sqrt{b x^2+c x^4}}{x^{7/2}} \, dx\\ &=-\frac{4 c \sqrt{b x^2+c x^4}}{7 x^{5/2}}-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac{1}{7} \left (4 c^2\right ) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx\\ &=-\frac{4 c \sqrt{b x^2+c x^4}}{7 x^{5/2}}-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac{\left (4 c^2 x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{7 \sqrt{b x^2+c x^4}}\\ &=-\frac{4 c \sqrt{b x^2+c x^4}}{7 x^{5/2}}-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac{\left (8 c^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 \sqrt{b x^2+c x^4}}\\ &=-\frac{4 c \sqrt{b x^2+c x^4}}{7 x^{5/2}}-\frac{2 \left (b x^2+c x^4\right )^{3/2}}{7 x^{13/2}}+\frac{4 c^{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 )}{7 \sqrt [4]{b} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0166989, size = 58, normalized size = 0.41 \[ -\frac{2 b \sqrt{x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac{7}{4},-\frac{3}{2};-\frac{3}{4};-\frac{c x^2}{b}\right )}{7 x^{9/2} \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*Sqrt[x^2*(b + c*x^2)]*Hypergeometric2F1[-7/4, -3/2, -3/4, -((c*x^2)/b)])/(7*x^(9/2)*Sqrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.185, size = 140, normalized size = 1. \begin{align*}{\frac{2}{7\, \left ( c{x}^{2}+b \right ) ^{2}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 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 ) \sqrt{-bc}{x}^{3}c-3\,{c}^{2}{x}^{4}-4\,bc{x}^{2}-{b}^{2} \right ){x}^{-{\frac{13}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(c*x^4+b*x^2)^(3/2)/x^(13/2)/(c*x^2+b)^2*(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))*(-b*c)^(1/2)*x^3*c-3*c^2*x^4-4*b*c*x^2-b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(15/2), 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}}{\left (c x^{2} + b\right )}}{x^{\frac{11}{2}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*x^4 + b*x^2)*(c*x^2 + b)/x^(11/2), 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((c*x**4+b*x**2)**(3/2)/x**(15/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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