3.287 \(\int (1+\tan ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=22 \[ \frac {1}{2} \tan (x) \sqrt {\sec ^2(x)}+\frac {1}{2} \sinh ^{-1}(\tan (x)) \]

[Out]

1/2*arcsinh(tan(x))+1/2*(sec(x)^2)^(1/2)*tan(x)

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Rubi [A]  time = 0.02, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3657, 4122, 195, 215} \[ \frac {1}{2} \tan (x) \sqrt {\sec ^2(x)}+\frac {1}{2} \sinh ^{-1}(\tan (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Tan[x]^2)^(3/2),x]

[Out]

ArcSinh[Tan[x]]/2 + (Sqrt[Sec[x]^2]*Tan[x])/2

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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

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Mathematica [B]  time = 0.07, size = 52, normalized size = 2.36 \[ \frac {1}{2} \cos (x) \sqrt {\sec ^2(x)} \left (\tan (x) \sec (x)-\log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Tan[x]^2)^(3/2),x]

[Out]

(Cos[x]*Sqrt[Sec[x]^2]*(-Log[Cos[x/2] - Sin[x/2]] + Log[Cos[x/2] + Sin[x/2]] + Sec[x]*Tan[x]))/2

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fricas [B]  time = 0.39, size = 72, normalized size = 3.27 \[ \frac {1}{2} \, \sqrt {\tan \relax (x)^{2} + 1} \tan \relax (x) + \frac {1}{4} \, \log \left (\frac {\tan \relax (x)^{2} + \sqrt {\tan \relax (x)^{2} + 1} \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) - \frac {1}{4} \, \log \left (\frac {\tan \relax (x)^{2} - \sqrt {\tan \relax (x)^{2} + 1} \tan \relax (x) + 1}{\tan \relax (x)^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(x)^2)^(3/2),x, algorithm="fricas")

[Out]

1/2*sqrt(tan(x)^2 + 1)*tan(x) + 1/4*log((tan(x)^2 + sqrt(tan(x)^2 + 1)*tan(x) + 1)/(tan(x)^2 + 1)) - 1/4*log((
tan(x)^2 - sqrt(tan(x)^2 + 1)*tan(x) + 1)/(tan(x)^2 + 1))

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giac [A]  time = 0.31, size = 29, normalized size = 1.32 \[ \frac {1}{2} \, \sqrt {\tan \relax (x)^{2} + 1} \tan \relax (x) - \frac {1}{2} \, \log \left (\sqrt {\tan \relax (x)^{2} + 1} - \tan \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(x)^2)^(3/2),x, algorithm="giac")

[Out]

1/2*sqrt(tan(x)^2 + 1)*tan(x) - 1/2*log(sqrt(tan(x)^2 + 1) - tan(x))

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maple [A]  time = 0.14, size = 19, normalized size = 0.86 \[ \frac {\tan \relax (x ) \sqrt {1+\tan ^{2}\relax (x )}}{2}+\frac {\arcsinh \left (\tan \relax (x )\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+tan(x)^2)^(3/2),x)

[Out]

1/2*tan(x)*(1+tan(x)^2)^(1/2)+1/2*arcsinh(tan(x))

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maxima [A]  time = 0.55, size = 18, normalized size = 0.82 \[ \frac {1}{2} \, \sqrt {\tan \relax (x)^{2} + 1} \tan \relax (x) + \frac {1}{2} \, \operatorname {arsinh}\left (\tan \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(x)^2)^(3/2),x, algorithm="maxima")

[Out]

1/2*sqrt(tan(x)^2 + 1)*tan(x) + 1/2*arcsinh(tan(x))

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mupad [B]  time = 0.11, size = 18, normalized size = 0.82 \[ \frac {\mathrm {asinh}\left (\mathrm {tan}\relax (x)\right )}{2}+\frac {\mathrm {tan}\relax (x)\,\sqrt {{\mathrm {tan}\relax (x)}^2+1}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tan(x)^2 + 1)^(3/2),x)

[Out]

asinh(tan(x))/2 + (tan(x)*(tan(x)^2 + 1)^(1/2))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{2}{\relax (x )} + 1\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+tan(x)**2)**(3/2),x)

[Out]

Integral((tan(x)**2 + 1)**(3/2), x)

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