3.2.5 \(\int \frac {1}{(\frac {b c}{d}+b x^2) \sqrt {c+d x^2}} \, dx\)

Optimal. Leaf size=20 \[ \frac {d x}{b c \sqrt {c+d x^2}} \]

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Rubi [A]  time = 0.00, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {21, 191} \begin {gather*} \frac {d x}{b c \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x)/(b*c*Sqrt[c + d*x^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin {align*} \int \frac {1}{\left (\frac {b c}{d}+b x^2\right ) \sqrt {c+d x^2}} \, dx &=\frac {d \int \frac {1}{\left (c+d x^2\right )^{3/2}} \, dx}{b}\\ &=\frac {d x}{b c \sqrt {c+d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} \frac {d x}{b c \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x)/(b*c*Sqrt[c + d*x^2])

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IntegrateAlgebraic [A]  time = 0.08, size = 20, normalized size = 1.00 \begin {gather*} \frac {d x}{b c \sqrt {c+d x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(((b*c)/d + b*x^2)*Sqrt[c + d*x^2]),x]

[Out]

(d*x)/(b*c*Sqrt[c + d*x^2])

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fricas [A]  time = 0.87, size = 27, normalized size = 1.35 \begin {gather*} \frac {\sqrt {d x^{2} + c} d x}{b c d x^{2} + b c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(d*x^2 + c)*d*x/(b*c*d*x^2 + b*c^2)

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giac [A]  time = 0.60, size = 18, normalized size = 0.90 \begin {gather*} \frac {d x}{\sqrt {d x^{2} + c} b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*x/(sqrt(d*x^2 + c)*b*c)

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maple [A]  time = 0.00, size = 19, normalized size = 0.95 \begin {gather*} \frac {d x}{\sqrt {d \,x^{2}+c}\, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.77, size = 18, normalized size = 0.90 \begin {gather*} \frac {d\,x}{b\,c\,\sqrt {d\,x^2+c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(d*x)/(b*c*(c + d*x^2)^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {d \int \frac {1}{c \sqrt {c + d x^{2}} + d x^{2} \sqrt {c + d x^{2}}}\, dx}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*Integral(1/(c*sqrt(c + d*x**2) + d*x**2*sqrt(c + d*x**2)), x)/b

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