3.625 \(\int \frac {1}{\sqrt {c x} (a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=126 \[ \frac {\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{2 a^{5/4} \sqrt [4]{b} \sqrt {c} \sqrt {a+b x^2}}+\frac {\sqrt {c x}}{a c \sqrt {a+b x^2}} \]

[Out]

(c*x)^(1/2)/a/c/(b*x^2+a)^(1/2)+1/2*(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)/b^(1/4)/c^(1/2)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 126, 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 = {290, 329, 220} \[ \frac {\left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{2 a^{5/4} \sqrt [4]{b} \sqrt {c} \sqrt {a+b x^2}}+\frac {\sqrt {c x}}{a c \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[c*x]*(a + b*x^2)^(3/2)),x]

[Out]

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

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 59, normalized size = 0.47 \[ \frac {x \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^2}{a}\right )+x}{a \sqrt {c x} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[c*x]*(a + b*x^2)^(3/2)),x]

[Out]

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

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {c x}}{b^{2} c x^{5} + 2 \, a b c x^{3} + a^{2} c x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(1/2)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {c x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(1/2)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/2)*sqrt(c*x)), x)

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maple [A]  time = 0.02, size = 114, normalized size = 0.90 \[ \frac {2 b x +\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 \sqrt {b \,x^{2}+a}\, \sqrt {c x}\, a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {c x}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(1/2)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/2)*sqrt(c*x)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {c\,x}\,{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 1.80, size = 44, normalized size = 0.35 \[ \frac {\sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \sqrt {c} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(1/2)/(b*x**2+a)**(3/2),x)

[Out]

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

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