3.590 \(\int (c x)^{5/2} \sqrt{a+b x^2} \, dx\)

Optimal. Leaf size=301 \[ -\frac{2 a^{9/4} c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}-\frac{4 a^2 c^2 \sqrt{c x} \sqrt{a+b x^2}}{15 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{4 a^{9/4} c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}+\frac{2 (c x)^{7/2} \sqrt{a+b x^2}}{9 c}+\frac{4 a c (c x)^{3/2} \sqrt{a+b x^2}}{45 b} \]

[Out]

(4*a*c*(c*x)^(3/2)*Sqrt[a + b*x^2])/(45*b) + (2*(c*x)^(7/2)*Sqrt[a + b*x^2])/(9*c) - (4*a^2*c^2*Sqrt[c*x]*Sqrt
[a + b*x^2])/(15*b^(3/2)*(Sqrt[a] + Sqrt[b]*x)) + (4*a^(9/4)*c^(5/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(S
qrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*b^(7/4)*Sqrt[a + b
*x^2]) - (2*a^(9/4)*c^(5/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan
[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*b^(7/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.229721, antiderivative size = 301, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {279, 321, 329, 305, 220, 1196} \[ -\frac{4 a^2 c^2 \sqrt{c x} \sqrt{a+b x^2}}{15 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 a^{9/4} c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}+\frac{4 a^{9/4} c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}+\frac{2 (c x)^{7/2} \sqrt{a+b x^2}}{9 c}+\frac{4 a c (c x)^{3/2} \sqrt{a+b x^2}}{45 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(5/2)*Sqrt[a + b*x^2],x]

[Out]

(4*a*c*(c*x)^(3/2)*Sqrt[a + b*x^2])/(45*b) + (2*(c*x)^(7/2)*Sqrt[a + b*x^2])/(9*c) - (4*a^2*c^2*Sqrt[c*x]*Sqrt
[a + b*x^2])/(15*b^(3/2)*(Sqrt[a] + Sqrt[b]*x)) + (4*a^(9/4)*c^(5/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(S
qrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*b^(7/4)*Sqrt[a + b
*x^2]) - (2*a^(9/4)*c^(5/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan
[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*b^(7/4)*Sqrt[a + b*x^2])

Rule 279

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

Rule 321

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

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 (c x)^{5/2} \sqrt{a+b x^2} \, dx &=\frac{2 (c x)^{7/2} \sqrt{a+b x^2}}{9 c}+\frac{1}{9} (2 a) \int \frac{(c x)^{5/2}}{\sqrt{a+b x^2}} \, dx\\ &=\frac{4 a c (c x)^{3/2} \sqrt{a+b x^2}}{45 b}+\frac{2 (c x)^{7/2} \sqrt{a+b x^2}}{9 c}-\frac{\left (2 a^2 c^2\right ) \int \frac{\sqrt{c x}}{\sqrt{a+b x^2}} \, dx}{15 b}\\ &=\frac{4 a c (c x)^{3/2} \sqrt{a+b x^2}}{45 b}+\frac{2 (c x)^{7/2} \sqrt{a+b x^2}}{9 c}-\frac{\left (4 a^2 c\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 b}\\ &=\frac{4 a c (c x)^{3/2} \sqrt{a+b x^2}}{45 b}+\frac{2 (c x)^{7/2} \sqrt{a+b x^2}}{9 c}-\frac{\left (4 a^{5/2} c^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 b^{3/2}}+\frac{\left (4 a^{5/2} c^2\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a} c}}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 b^{3/2}}\\ &=\frac{4 a c (c x)^{3/2} \sqrt{a+b x^2}}{45 b}+\frac{2 (c x)^{7/2} \sqrt{a+b x^2}}{9 c}-\frac{4 a^2 c^2 \sqrt{c x} \sqrt{a+b x^2}}{15 b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{4 a^{9/4} c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}-\frac{2 a^{9/4} c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 b^{7/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0459954, size = 85, normalized size = 0.28 \[ \frac{2 c (c x)^{3/2} \sqrt{a+b x^2} \left (\left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1}-a \, _2F_1\left (-\frac{1}{2},\frac{3}{4};\frac{7}{4};-\frac{b x^2}{a}\right )\right )}{9 b \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(5/2)*Sqrt[a + b*x^2],x]

[Out]

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

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Maple [A]  time = 0.02, size = 221, normalized size = 0.7 \begin{align*} -{\frac{2\,{c}^{2}}{45\,{b}^{2}x}\sqrt{cx} \left ( -5\,{b}^{3}{x}^{6}+6\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{3}-3\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){a}^{3}-7\,a{b}^{2}{x}^{4}-2\,{a}^{2}b{x}^{2} \right ){\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/2)*(b*x^2+a)^(1/2),x)

[Out]

-2/45/x*c^2*(c*x)^(1/2)/(b*x^2+a)^(1/2)/b^2*(-5*b^3*x^6+6*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b
*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/
2),1/2*2^(1/2))*a^3-3*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)
*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*a^3-7*a*b^2*x^4-2*a^
2*b*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)*(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*x^2 + a)*(c*x)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)*(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 52.4007, size = 46, normalized size = 0.15 \begin{align*} \frac{\sqrt{a} c^{\frac{5}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{2}, \frac{7}{4} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac{11}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(5/2)*(b*x**2+a)**(1/2),x)

[Out]

sqrt(a)*c**(5/2)*x**(7/2)*gamma(7/4)*hyper((-1/2, 7/4), (11/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(11/4))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(5/2)*(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*x^2 + a)*(c*x)^(5/2), x)