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

Optimal. Leaf size=129 \[ -\frac {b^{3/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 a^{5/4} c^{5/2} \sqrt {a+b x^2}}-\frac {2 \sqrt {a+b x^2}}{3 a c (c x)^{3/2}} \]

[Out]

-2/3*(b*x^2+a)^(1/2)/a/c/(c*x)^(3/2)-1/3*b^(3/4)*(cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))^2)^(1/2)/
cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*EllipticF(sin(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))
),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/a^(5/4)/c^(5/2)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {325, 329, 220} \[ -\frac {b^{3/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{3 a^{5/4} c^{5/2} \sqrt {a+b x^2}}-\frac {2 \sqrt {a+b x^2}}{3 a c (c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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]

Rubi steps

\begin {align*} \int \frac {1}{(c x)^{5/2} \sqrt {a+b x^2}} \, dx &=-\frac {2 \sqrt {a+b x^2}}{3 a c (c x)^{3/2}}-\frac {b \int \frac {1}{\sqrt {c x} \sqrt {a+b x^2}} \, dx}{3 a c^2}\\ &=-\frac {2 \sqrt {a+b x^2}}{3 a c (c x)^{3/2}}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{3 a c^3}\\ &=-\frac {2 \sqrt {a+b x^2}}{3 a c (c x)^{3/2}}-\frac {b^{3/4} \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 )}{3 a^{5/4} c^{5/2} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 56, normalized size = 0.43 \[ -\frac {2 x \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-\frac {b x^2}{a}\right )}{3 (c x)^{5/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {c x}}{b c^{3} x^{5} + a c^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(c*x)/(b*c^3*x^5 + a*c^3*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{2} + a} \left (c x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 123, normalized size = 0.95 \[ -\frac {2 b \,x^{2}+\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, x \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )+2 a}{3 \sqrt {b \,x^{2}+a}\, \sqrt {c x}\, a \,c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b x^{2} + a} \left (c x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (c\,x\right )}^{5/2}\,\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 3.89, size = 48, normalized size = 0.37 \[ \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} c^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), b*x**2*exp_polar(I*pi)/a)/(2*sqrt(a)*c**(5/2)*x**(3/2)*gamma(1/4))

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