3.368 \(\int \frac{(b x^2+c x^4)^{3/2}}{x^{5/2}} \, dx\)

Optimal. Leaf size=290 \[ \frac{4 b^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}-\frac{8 b^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}+\frac{8 b^2 x^{3/2} \left (b+c x^2\right )}{15 \sqrt{c} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{4}{15} b \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}} \]

[Out]

(8*b^2*x^(3/2)*(b + c*x^2))/(15*Sqrt[c]*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) + (4*b*Sqrt[x]*Sqrt[b*x^2 +
c*x^4])/15 + (2*(b*x^2 + c*x^4)^(3/2))/(9*x^(3/2)) - (8*b^(9/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqr
t[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[b*x^2 + c*x^4]) + (
4*b^(9/4)*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])/(15*c^(3/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.298659, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2021, 2032, 329, 305, 220, 1196} \[ \frac{4 b^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}-\frac{8 b^{9/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 c^{3/4} \sqrt{b x^2+c x^4}}+\frac{8 b^2 x^{3/2} \left (b+c x^2\right )}{15 \sqrt{c} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{4}{15} b \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(5/2),x]

[Out]

(8*b^2*x^(3/2)*(b + c*x^2))/(15*Sqrt[c]*(Sqrt[b] + Sqrt[c]*x)*Sqrt[b*x^2 + c*x^4]) + (4*b*Sqrt[x]*Sqrt[b*x^2 +
c*x^4])/15 + (2*(b*x^2 + c*x^4)^(3/2))/(9*x^(3/2)) - (8*b^(9/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqr
t[b] + Sqrt[c]*x)^2]*EllipticE[2*ArcTan[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(15*c^(3/4)*Sqrt[b*x^2 + c*x^4]) + (
4*b^(9/4)*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])/(15*c^(3/4)*Sqrt[b*x^2 + c*x^4])

Rule 2021

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 + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 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{\left (b x^2+c x^4\right )^{3/2}}{x^{5/2}} \, dx &=\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}+\frac{1}{3} (2 b) \int \frac{\sqrt{b x^2+c x^4}}{\sqrt{x}} \, dx\\ &=\frac{4}{15} b \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}+\frac{1}{15} \left (4 b^2\right ) \int \frac{x^{3/2}}{\sqrt{b x^2+c x^4}} \, dx\\ &=\frac{4}{15} b \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}+\frac{\left (4 b^2 x \sqrt{b+c x^2}\right ) \int \frac{\sqrt{x}}{\sqrt{b+c x^2}} \, dx}{15 \sqrt{b x^2+c x^4}}\\ &=\frac{4}{15} b \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}+\frac{\left (8 b^2 x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{15 \sqrt{b x^2+c x^4}}\\ &=\frac{4}{15} b \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}+\frac{\left (8 b^{5/2} x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{15 \sqrt{c} \sqrt{b x^2+c x^4}}-\frac{\left (8 b^{5/2} 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 )}{15 \sqrt{c} \sqrt{b x^2+c x^4}}\\ &=\frac{8 b^2 x^{3/2} \left (b+c x^2\right )}{15 \sqrt{c} \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{b x^2+c x^4}}+\frac{4}{15} b \sqrt{x} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{9 x^{3/2}}-\frac{8 b^{9/4} 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 )}{15 c^{3/4} \sqrt{b x^2+c x^4}}+\frac{4 b^{9/4} 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 )}{15 c^{3/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0168486, size = 58, normalized size = 0.2 \[ \frac{2 b \sqrt{x} \sqrt{x^2 \left (b+c x^2\right )} \, _2F_1\left (-\frac{3}{2},\frac{3}{4};\frac{7}{4};-\frac{c x^2}{b}\right )}{3 \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2)/x^(5/2),x]

[Out]

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

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Maple [A]  time = 0.188, size = 226, normalized size = 0.8 \begin{align*}{\frac{2}{45\, \left ( c{x}^{2}+b \right ) ^{2}c} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( 5\,{c}^{3}{x}^{6}+12\,{b}^{3}\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 ) -6\,{b}^{3}\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 ) +16\,b{c}^{2}{x}^{4}+11\,{b}^{2}c{x}^{2} \right ){x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2)^(3/2)/x^(5/2),x)

[Out]

2/45*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2/c*(5*c^3*x^6+12*b^3*((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))-6*b^3*((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))+16*b*c^2*x^4+
11*b^2*c*x^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{5}{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^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(5/2), 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 )}}{\sqrt{x}}, 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^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac{3}{2}}}{x^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2)**(3/2)/x**(5/2),x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**(5/2), x)

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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{5}{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^(5/2),x, algorithm="giac")

[Out]

integrate((c*x^4 + b*x^2)^(3/2)/x^(5/2), x)