3.629 \(\int \frac{(c x)^{7/2}}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=155 \[ \frac{5 c^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{12 \sqrt [4]{a} b^{9/4} \sqrt{a+b x^2}}-\frac{5 c^3 \sqrt{c x}}{6 b^2 \sqrt{a+b x^2}}-\frac{c (c x)^{5/2}}{3 b \left (a+b x^2\right )^{3/2}} \]

[Out]

-(c*(c*x)^(5/2))/(3*b*(a + b*x^2)^(3/2)) - (5*c^3*Sqrt[c*x])/(6*b^2*Sqrt[a + b*x^2]) + (5*c^(7/2)*(Sqrt[a] + S
qrt[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])/(12*a^(1/4)*b^(9/4)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.0887926, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {288, 329, 220} \[ -\frac{5 c^3 \sqrt{c x}}{6 b^2 \sqrt{a+b x^2}}+\frac{5 c^{7/2} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{12 \sqrt [4]{a} b^{9/4} \sqrt{a+b x^2}}-\frac{c (c x)^{5/2}}{3 b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(7/2)/(a + b*x^2)^(5/2),x]

[Out]

-(c*(c*x)^(5/2))/(3*b*(a + b*x^2)^(3/2)) - (5*c^3*Sqrt[c*x])/(6*b^2*Sqrt[a + b*x^2]) + (5*c^(7/2)*(Sqrt[a] + S
qrt[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])/(12*a^(1/4)*b^(9/4)*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 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{(c x)^{7/2}}{\left (a+b x^2\right )^{5/2}} \, dx &=-\frac{c (c x)^{5/2}}{3 b \left (a+b x^2\right )^{3/2}}+\frac{\left (5 c^2\right ) \int \frac{(c x)^{3/2}}{\left (a+b x^2\right )^{3/2}} \, dx}{6 b}\\ &=-\frac{c (c x)^{5/2}}{3 b \left (a+b x^2\right )^{3/2}}-\frac{5 c^3 \sqrt{c x}}{6 b^2 \sqrt{a+b x^2}}+\frac{\left (5 c^4\right ) \int \frac{1}{\sqrt{c x} \sqrt{a+b x^2}} \, dx}{12 b^2}\\ &=-\frac{c (c x)^{5/2}}{3 b \left (a+b x^2\right )^{3/2}}-\frac{5 c^3 \sqrt{c x}}{6 b^2 \sqrt{a+b x^2}}+\frac{\left (5 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{6 b^2}\\ &=-\frac{c (c x)^{5/2}}{3 b \left (a+b x^2\right )^{3/2}}-\frac{5 c^3 \sqrt{c x}}{6 b^2 \sqrt{a+b x^2}}+\frac{5 c^{7/2} \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 )}{12 \sqrt [4]{a} b^{9/4} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0510805, size = 80, normalized size = 0.52 \[ \frac{c^3 \sqrt{c x} \left (5 \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )-5 a-7 b x^2\right )}{6 b^2 \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(7/2)/(a + b*x^2)^(5/2),x]

[Out]

(c^3*Sqrt[c*x]*(-5*a - 7*b*x^2 + 5*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/
a)]))/(6*b^2*(a + b*x^2)^(3/2))

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Maple [A]  time = 0.031, size = 219, normalized size = 1.4 \begin{align*}{\frac{{c}^{3}}{12\,{b}^{3}x} \left ( 5\,\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 ) \sqrt{-ab}{x}^{2}b+5\,\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 ) \sqrt{-ab}a-14\,{b}^{2}{x}^{3}-10\,abx \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)^(7/2)/(b*x^2+a)^(5/2),x)

[Out]

1/12*(5*((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*b)^(1/2)*x^2*b+5*((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)*Elliptic
F(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*(-a*b)^(1/2)*a-14*b^2*x^3-10*a*b*x)/x*c^3*(c*x)^(1/2)/b
^3/(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{7}{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)^(7/2)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((c*x)^(7/2)/(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^{3} x^{3}}{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)^(7/2)/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(c*x)*c^3*x^3/(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 = 159.189, size = 44, normalized size = 0.28 \begin{align*} \frac{c^{\frac{7}{2}} x^{\frac{9}{2}} \Gamma \left (\frac{9}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{9}{4}, \frac{5}{2} \\ \frac{13}{4} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac{5}{2}} \Gamma \left (\frac{13}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(7/2)/(b*x**2+a)**(5/2),x)

[Out]

c**(7/2)*x**(9/2)*gamma(9/4)*hyper((9/4, 5/2), (13/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/2)*gamma(13/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c x\right )^{\frac{7}{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)^(7/2)/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate((c*x)^(7/2)/(b*x^2 + a)^(5/2), x)