3.283 \(\int \frac {1}{\sqrt {a+a \tan ^2(c+d x)}} \, dx\)

Optimal. Leaf size=24 \[ \frac {\tan (c+d x)}{d \sqrt {a \sec ^2(c+d x)}} \]

[Out]

tan(d*x+c)/d/(a*sec(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {3657, 4122, 191} \[ \frac {\tan (c+d x)}{d \sqrt {a \sec ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + a*Tan[c + d*x]^2],x]

[Out]

Tan[c + d*x]/(d*Sqrt[a*Sec[c + d*x]^2])

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]

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 24, normalized size = 1.00 \[ \frac {\tan (c+d x)}{d \sqrt {a \sec ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + a*Tan[c + d*x]^2],x]

[Out]

Tan[c + d*x]/(d*Sqrt[a*Sec[c + d*x]^2])

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fricas [A]  time = 0.40, size = 38, normalized size = 1.58 \[ \frac {\sqrt {a \tan \left (d x + c\right )^{2} + a} \tan \left (d x + c\right )}{a d \tan \left (d x + c\right )^{2} + a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a*tan(d*x + c)^2 + a)*tan(d*x + c)/(a*d*tan(d*x + c)^2 + a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.37, size = 25, normalized size = 1.04 \[ \frac {\tan \left (d x +c \right )}{d \sqrt {a +a \left (\tan ^{2}\left (d x +c \right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*tan(d*x+c)/(a+a*tan(d*x+c)^2)^(1/2)

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maxima [A]  time = 1.44, size = 13, normalized size = 0.54 \[ \frac {\sin \left (d x + c\right )}{\sqrt {a} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sin(d*x + c)/(sqrt(a)*d)

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mupad [B]  time = 12.07, size = 55, normalized size = 2.29 \[ \frac {\sin \left (2\,c+2\,d\,x\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )}{4\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+3}}}{a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*tan(c + d*x)^2)^(1/2),x)

[Out]

(sin(2*c + 2*d*x)*((a*(cos(2*c + 2*d*x) + 1))/(4*cos(2*c + 2*d*x) + cos(4*c + 4*d*x) + 3))^(1/2))/(a*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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