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

Optimal. Leaf size=173 \[ -\frac{4 b^{11/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{77 c^{5/4} \sqrt{b x^2+c x^4}}+\frac{8 b^2 \sqrt{b x^2+c x^4}}{77 c \sqrt{x}}+\frac{12}{77} b x^{3/2} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt{x}} \]

[Out]

(8*b^2*Sqrt[b*x^2 + c*x^4])/(77*c*Sqrt[x]) + (12*b*x^(3/2)*Sqrt[b*x^2 + c*x^4])/77 + (2*(b*x^2 + c*x^4)^(3/2))
/(11*Sqrt[x]) - (4*b^(11/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcT
an[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(77*c^(5/4)*Sqrt[b*x^2 + c*x^4])

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Rubi [A]  time = 0.235517, antiderivative size = 173, 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 = {2021, 2024, 2032, 329, 220} \[ -\frac{4 b^{11/4} x \left (\sqrt{b}+\sqrt{c} x\right ) \sqrt{\frac{b+c x^2}{\left (\sqrt{b}+\sqrt{c} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{c} \sqrt{x}}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{77 c^{5/4} \sqrt{b x^2+c x^4}}+\frac{8 b^2 \sqrt{b x^2+c x^4}}{77 c \sqrt{x}}+\frac{12}{77} b x^{3/2} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^(3/2),x]

[Out]

(8*b^2*Sqrt[b*x^2 + c*x^4])/(77*c*Sqrt[x]) + (12*b*x^(3/2)*Sqrt[b*x^2 + c*x^4])/77 + (2*(b*x^2 + c*x^4)^(3/2))
/(11*Sqrt[x]) - (4*b^(11/4)*x*(Sqrt[b] + Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcT
an[(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(77*c^(5/4)*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 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{\left (b x^2+c x^4\right )^{3/2}}{x^{3/2}} \, dx &=\frac{2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt{x}}+\frac{1}{11} (6 b) \int \sqrt{x} \sqrt{b x^2+c x^4} \, dx\\ &=\frac{12}{77} b x^{3/2} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt{x}}+\frac{1}{77} \left (12 b^2\right ) \int \frac{x^{5/2}}{\sqrt{b x^2+c x^4}} \, dx\\ &=\frac{8 b^2 \sqrt{b x^2+c x^4}}{77 c \sqrt{x}}+\frac{12}{77} b x^{3/2} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt{x}}-\frac{\left (4 b^3\right ) \int \frac{\sqrt{x}}{\sqrt{b x^2+c x^4}} \, dx}{77 c}\\ &=\frac{8 b^2 \sqrt{b x^2+c x^4}}{77 c \sqrt{x}}+\frac{12}{77} b x^{3/2} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt{x}}-\frac{\left (4 b^3 x \sqrt{b+c x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{b+c x^2}} \, dx}{77 c \sqrt{b x^2+c x^4}}\\ &=\frac{8 b^2 \sqrt{b x^2+c x^4}}{77 c \sqrt{x}}+\frac{12}{77} b x^{3/2} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt{x}}-\frac{\left (8 b^3 x \sqrt{b+c x^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b+c x^4}} \, dx,x,\sqrt{x}\right )}{77 c \sqrt{b x^2+c x^4}}\\ &=\frac{8 b^2 \sqrt{b x^2+c x^4}}{77 c \sqrt{x}}+\frac{12}{77} b x^{3/2} \sqrt{b x^2+c x^4}+\frac{2 \left (b x^2+c x^4\right )^{3/2}}{11 \sqrt{x}}-\frac{4 b^{11/4} 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 )}{77 c^{5/4} \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0390194, size = 90, normalized size = 0.52 \[ \frac{2 \sqrt{x^2 \left (b+c x^2\right )} \left (\left (b+c x^2\right )^2 \sqrt{\frac{c x^2}{b}+1}-b^2 \, _2F_1\left (-\frac{3}{2},\frac{1}{4};\frac{5}{4};-\frac{c x^2}{b}\right )\right )}{11 c \sqrt{x} \sqrt{\frac{c x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x^2*(b + c*x^2)]*((b + c*x^2)^2*Sqrt[1 + (c*x^2)/b] - b^2*Hypergeometric2F1[-3/2, 1/4, 5/4, -((c*x^2)/
b)]))/(11*c*Sqrt[x]*Sqrt[1 + (c*x^2)/b])

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Maple [A]  time = 0.183, size = 157, normalized size = 0.9 \begin{align*} -{\frac{2}{77\, \left ( c{x}^{2}+b \right ) ^{2}{c}^{2}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{{\frac{3}{2}}} \left ( -7\,{x}^{7}{c}^{4}+2\,{b}^{3}\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 ) -20\,{x}^{5}b{c}^{3}-17\,{b}^{2}{c}^{2}{x}^{3}-4\,x{b}^{3}c \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^(3/2),x)

[Out]

-2/77*(c*x^4+b*x^2)^(3/2)/x^(7/2)/(c*x^2+b)^2*(-7*x^7*c^4+2*b^3*(-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))-20*x^5*b*c^3-17*b^2*c^2*x^3-4*x*b^3*c)/c^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{3}{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^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\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^(3/2),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{3}{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**(3/2),x)

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**(3/2), 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{3}{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^(3/2),x, algorithm="giac")

[Out]

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