3.636 \(\int \frac{1}{(c x)^{7/2} (a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=362 \[ -\frac{77 b^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{20 a^{15/4} c^{7/2} \sqrt{a+b x^2}}-\frac{77 b^{3/2} \sqrt{c x} \sqrt{a+b x^2}}{10 a^4 c^4 \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{77 b^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{10 a^{15/4} c^{7/2} \sqrt{a+b x^2}}+\frac{77 b \sqrt{a+b x^2}}{10 a^4 c^3 \sqrt{c x}}-\frac{77 \sqrt{a+b x^2}}{30 a^3 c (c x)^{5/2}}+\frac{11}{6 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}+\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}} \]

[Out]

1/(3*a*c*(c*x)^(5/2)*(a + b*x^2)^(3/2)) + 11/(6*a^2*c*(c*x)^(5/2)*Sqrt[a + b*x^2]) - (77*Sqrt[a + b*x^2])/(30*
a^3*c*(c*x)^(5/2)) + (77*b*Sqrt[a + b*x^2])/(10*a^4*c^3*Sqrt[c*x]) - (77*b^(3/2)*Sqrt[c*x]*Sqrt[a + b*x^2])/(1
0*a^4*c^4*(Sqrt[a] + Sqrt[b]*x)) + (77*b^(5/4)*(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])/(10*a^(15/4)*c^(7/2)*Sqrt[a + b*x^2]) - (77*
b^(5/4)*(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])/(20*a^(15/4)*c^(7/2)*Sqrt[a + b*x^2])

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Rubi [A]  time = 0.295031, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {290, 325, 329, 305, 220, 1196} \[ -\frac{77 b^{3/2} \sqrt{c x} \sqrt{a+b x^2}}{10 a^4 c^4 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{77 b^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{20 a^{15/4} c^{7/2} \sqrt{a+b x^2}}+\frac{77 b^{5/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{10 a^{15/4} c^{7/2} \sqrt{a+b x^2}}+\frac{77 b \sqrt{a+b x^2}}{10 a^4 c^3 \sqrt{c x}}-\frac{77 \sqrt{a+b x^2}}{30 a^3 c (c x)^{5/2}}+\frac{11}{6 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}+\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(7/2)*(a + b*x^2)^(5/2)),x]

[Out]

1/(3*a*c*(c*x)^(5/2)*(a + b*x^2)^(3/2)) + 11/(6*a^2*c*(c*x)^(5/2)*Sqrt[a + b*x^2]) - (77*Sqrt[a + b*x^2])/(30*
a^3*c*(c*x)^(5/2)) + (77*b*Sqrt[a + b*x^2])/(10*a^4*c^3*Sqrt[c*x]) - (77*b^(3/2)*Sqrt[c*x]*Sqrt[a + b*x^2])/(1
0*a^4*c^4*(Sqrt[a] + Sqrt[b]*x)) + (77*b^(5/4)*(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])/(10*a^(15/4)*c^(7/2)*Sqrt[a + b*x^2]) - (77*
b^(5/4)*(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])/(20*a^(15/4)*c^(7/2)*Sqrt[a + b*x^2])

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -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{1}{(c x)^{7/2} \left (a+b x^2\right )^{5/2}} \, dx &=\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}}+\frac{11 \int \frac{1}{(c x)^{7/2} \left (a+b x^2\right )^{3/2}} \, dx}{6 a}\\ &=\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}}+\frac{11}{6 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}+\frac{77 \int \frac{1}{(c x)^{7/2} \sqrt{a+b x^2}} \, dx}{12 a^2}\\ &=\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}}+\frac{11}{6 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}-\frac{77 \sqrt{a+b x^2}}{30 a^3 c (c x)^{5/2}}-\frac{(77 b) \int \frac{1}{(c x)^{3/2} \sqrt{a+b x^2}} \, dx}{20 a^3 c^2}\\ &=\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}}+\frac{11}{6 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}-\frac{77 \sqrt{a+b x^2}}{30 a^3 c (c x)^{5/2}}+\frac{77 b \sqrt{a+b x^2}}{10 a^4 c^3 \sqrt{c x}}-\frac{\left (77 b^2\right ) \int \frac{\sqrt{c x}}{\sqrt{a+b x^2}} \, dx}{20 a^4 c^4}\\ &=\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}}+\frac{11}{6 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}-\frac{77 \sqrt{a+b x^2}}{30 a^3 c (c x)^{5/2}}+\frac{77 b \sqrt{a+b x^2}}{10 a^4 c^3 \sqrt{c x}}-\frac{\left (77 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{10 a^4 c^5}\\ &=\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}}+\frac{11}{6 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}-\frac{77 \sqrt{a+b x^2}}{30 a^3 c (c x)^{5/2}}+\frac{77 b \sqrt{a+b x^2}}{10 a^4 c^3 \sqrt{c x}}-\frac{\left (77 b^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{10 a^{7/2} c^4}+\frac{\left (77 b^{3/2}\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 )}{10 a^{7/2} c^4}\\ &=\frac{1}{3 a c (c x)^{5/2} \left (a+b x^2\right )^{3/2}}+\frac{11}{6 a^2 c (c x)^{5/2} \sqrt{a+b x^2}}-\frac{77 \sqrt{a+b x^2}}{30 a^3 c (c x)^{5/2}}+\frac{77 b \sqrt{a+b x^2}}{10 a^4 c^3 \sqrt{c x}}-\frac{77 b^{3/2} \sqrt{c x} \sqrt{a+b x^2}}{10 a^4 c^4 \left (\sqrt{a}+\sqrt{b} x\right )}+\frac{77 b^{5/4} \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 )}{10 a^{15/4} c^{7/2} \sqrt{a+b x^2}}-\frac{77 b^{5/4} \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 )}{20 a^{15/4} c^{7/2} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0129009, size = 59, normalized size = 0.16 \[ -\frac{2 x \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (-\frac{5}{4},\frac{5}{2};-\frac{1}{4};-\frac{b x^2}{a}\right )}{5 a^2 (c x)^{7/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(7/2)*(a + b*x^2)^(5/2)),x]

[Out]

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

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Maple [A]  time = 0.019, size = 410, normalized size = 1.1 \begin{align*} -{\frac{1}{60\,{x}^{2}{a}^{4}{c}^{3}} \left ( 462\,\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}^{4}a{b}^{2}-231\,\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}^{4}a{b}^{2}+462\,\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 ) \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}{x}^{2}{a}^{2}b-231\,\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{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}{x}^{2}{a}^{2}b-462\,{b}^{3}{x}^{6}-770\,a{b}^{2}{x}^{4}-264\,{a}^{2}b{x}^{2}+24\,{a}^{3} \right ){\frac{1}{\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(1/(c*x)^(7/2)/(b*x^2+a)^(5/2),x)

[Out]

-1/60*(462*((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^4*a*b^2-231*((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^4*a*b^2+462*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))*((b*x+(-a*b)^(1/2
))/(-a*b)^(1/2))^(1/2)*x^2*a^2*b-231*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))*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*x^2*a
^2*b-462*b^3*x^6-770*a*b^2*x^4-264*a^2*b*x^2+24*a^3)/x^2/a^4/c^3/(c*x)^(1/2)/(b*x^2+a)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{5}{2}} \left (c x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(5/2)*(c*x)^(7/2)), 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}}{b^{3} c^{4} x^{10} + 3 \, a b^{2} c^{4} x^{8} + 3 \, a^{2} b c^{4} x^{6} + a^{3} c^{4} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(c*x)/(b^3*c^4*x^10 + 3*a*b^2*c^4*x^8 + 3*a^2*b*c^4*x^6 + a^3*c^4*x^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(7/2)/(b*x**2+a)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a\right )}^{\frac{5}{2}} \left (c x\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(5/2)*(c*x)^(7/2)), x)