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

Optimal. Leaf size=304 \[ -\frac{c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{4 a^{3/4} b^{7/4} \sqrt{a+b x^2}}+\frac{c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{2 a^{3/4} b^{7/4} \sqrt{a+b x^2}}-\frac{c^2 \sqrt{c x} \sqrt{a+b x^2}}{2 a b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{c (c x)^{3/2}}{2 a b \sqrt{a+b x^2}}-\frac{c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}} \]

[Out]

-(c*(c*x)^(3/2))/(3*b*(a + b*x^2)^(3/2)) + (c*(c*x)^(3/2))/(2*a*b*Sqrt[a + b*x^2]) - (c^2*Sqrt[c*x]*Sqrt[a + b
*x^2])/(2*a*b^(3/2)*(Sqrt[a] + Sqrt[b]*x)) + (c^(5/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b
]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(2*a^(3/4)*b^(7/4)*Sqrt[a + b*x^2]) -
(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])/(4*a^(3/4)*b^(7/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.215871, antiderivative size = 304, 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 = {288, 290, 329, 305, 220, 1196} \[ -\frac{c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{4 a^{3/4} b^{7/4} \sqrt{a+b x^2}}+\frac{c^{5/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{2 a^{3/4} b^{7/4} \sqrt{a+b x^2}}-\frac{c^2 \sqrt{c x} \sqrt{a+b x^2}}{2 a b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{c (c x)^{3/2}}{2 a b \sqrt{a+b x^2}}-\frac{c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(5/2)/(a + b*x^2)^(5/2),x]

[Out]

-(c*(c*x)^(3/2))/(3*b*(a + b*x^2)^(3/2)) + (c*(c*x)^(3/2))/(2*a*b*Sqrt[a + b*x^2]) - (c^2*Sqrt[c*x]*Sqrt[a + b
*x^2])/(2*a*b^(3/2)*(Sqrt[a] + Sqrt[b]*x)) + (c^(5/2)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b
]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(2*a^(3/4)*b^(7/4)*Sqrt[a + b*x^2]) -
(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])/(4*a^(3/4)*b^(7/4)*Sqrt[a + b*x^2])

Rule 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 \frac{(c x)^{5/2}}{\left (a+b x^2\right )^{5/2}} \, dx &=-\frac{c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}}+\frac{c^2 \int \frac{\sqrt{c x}}{\left (a+b x^2\right )^{3/2}} \, dx}{2 b}\\ &=-\frac{c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}}+\frac{c (c x)^{3/2}}{2 a b \sqrt{a+b x^2}}-\frac{c^2 \int \frac{\sqrt{c x}}{\sqrt{a+b x^2}} \, dx}{4 a b}\\ &=-\frac{c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}}+\frac{c (c x)^{3/2}}{2 a b \sqrt{a+b x^2}}-\frac{c \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{2 a b}\\ &=-\frac{c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}}+\frac{c (c x)^{3/2}}{2 a b \sqrt{a+b x^2}}-\frac{c^2 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{2 \sqrt{a} b^{3/2}}+\frac{c^2 \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 )}{2 \sqrt{a} b^{3/2}}\\ &=-\frac{c (c x)^{3/2}}{3 b \left (a+b x^2\right )^{3/2}}+\frac{c (c x)^{3/2}}{2 a b \sqrt{a+b x^2}}-\frac{c^2 \sqrt{c x} \sqrt{a+b x^2}}{2 a b^{3/2} \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{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 )}{2 a^{3/4} b^{7/4} \sqrt{a+b x^2}}-\frac{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 )}{4 a^{3/4} b^{7/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0393135, size = 74, normalized size = 0.24 \[ \frac{2 c (c x)^{3/2} \left (\left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{3}{4},\frac{5}{2};\frac{7}{4};-\frac{b x^2}{a}\right )-a\right )}{3 a b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.03, size = 385, normalized size = 1.3 \begin{align*} -{\frac{{c}^{2}}{12\,a{b}^{2}x} \left ( 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 ){x}^{2}ab-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 ){x}^{2}ab+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}^{2}-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}^{2}-6\,{b}^{2}{x}^{4}-2\,ab{x}^{2} \right ) \sqrt{cx} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(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))*x^2*a*b-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))*x^2*a*b+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^2-3*((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^2-6*b^2*x^4-2*a*b*x^2
)/x*c^2*(c*x)^(1/2)/b^2/a/(b*x^2+a)^(3/2)

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{2} + a} \sqrt{c x} c^{2} x^{2}}{b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(c*x)*c^2*x^2/(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3), x)

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Sympy [C]  time = 40.1428, size = 44, normalized size = 0.14 \begin{align*} \frac{c^{\frac{5}{2}} x^{\frac{7}{2}} \Gamma \left (\frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{7}{4}, \frac{5}{2} \\ \frac{11}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{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)**(5/2),x)

[Out]

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

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

[Out]

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