3.290 \(\int (-1-\tan ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=35 \[ \frac {1}{2} \tan ^{-1}\left (\frac {\tan (x)}{\sqrt {-\sec ^2(x)}}\right )-\frac {1}{2} \tan (x) \sqrt {-\sec ^2(x)} \]

[Out]

1/2*arctan(tan(x)/(-sec(x)^2)^(1/2))-1/2*(-sec(x)^2)^(1/2)*tan(x)

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3657, 4122, 195, 217, 203} \[ \frac {1}{2} \tan ^{-1}\left (\frac {\tan (x)}{\sqrt {-\sec ^2(x)}}\right )-\frac {1}{2} \tan (x) \sqrt {-\sec ^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[(-1 - Tan[x]^2)^(3/2),x]

[Out]

ArcTan[Tan[x]/Sqrt[-Sec[x]^2]]/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 203

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 \left (-1-\tan ^2(x)\right )^{3/2} \, dx &=\int \left (-\sec ^2(x)\right )^{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} \sqrt {-\sec ^2(x)} \tan (x)+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\tan (x)}{\sqrt {-\sec ^2(x)}}\right )\\ &=\frac {1}{2} \tan ^{-1}\left (\frac {\tan (x)}{\sqrt {-\sec ^2(x)}}\right )-\frac {1}{2} \sqrt {-\sec ^2(x)} \tan (x)\\ \end {align*}

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Mathematica [B]  time = 0.07, size = 72, normalized size = 2.06 \[ \frac {1}{4} \cos (x) \sqrt {-\sec ^2(x)} \left (\frac {1}{\sin (x)-1}+\frac {1}{\left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )^2}+2 \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-2 \log \left (\sin \left (\frac {x}{2}\right )+\cos \left (\frac {x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 - Tan[x]^2)^(3/2),x]

[Out]

(Cos[x]*Sqrt[-Sec[x]^2]*(2*Log[Cos[x/2] - Sin[x/2]] - 2*Log[Cos[x/2] + Sin[x/2]] + (Cos[x/2] + Sin[x/2])^(-2)
+ (-1 + Sin[x])^(-1)))/4

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fricas [C]  time = 0.45, size = 73, normalized size = 2.09 \[ \frac {{\left (-i \, e^{\left (4 i \, x\right )} - 2 i \, e^{\left (2 i \, x\right )} - i\right )} \log \left (e^{\left (i \, x\right )} + i\right ) + {\left (i \, e^{\left (4 i \, x\right )} + 2 i \, e^{\left (2 i \, x\right )} + i\right )} \log \left (e^{\left (i \, x\right )} - i\right ) - 2 \, e^{\left (3 i \, x\right )} + 2 \, e^{\left (i \, x\right )}}{2 \, {\left (e^{\left (4 i \, x\right )} + 2 \, e^{\left (2 i \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((-I*e^(4*I*x) - 2*I*e^(2*I*x) - I)*log(e^(I*x) + I) + (I*e^(4*I*x) + 2*I*e^(2*I*x) + I)*log(e^(I*x) - I)
- 2*e^(3*I*x) + 2*e^(I*x))/(e^(4*I*x) + 2*e^(2*I*x) + 1)

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giac [C]  time = 0.30, size = 29, normalized size = 0.83 \[ -\frac {1}{2} i \, \sqrt {\tan \relax (x)^{2} + 1} \tan \relax (x) + \frac {1}{2} i \, \log \left (\sqrt {\tan \relax (x)^{2} + 1} - \tan \relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.25, size = 32, normalized size = 0.91 \[ -\frac {\tan \relax (x ) \sqrt {-1-\left (\tan ^{2}\relax (x )\right )}}{2}+\frac {\arctan \left (\frac {\tan \relax (x )}{\sqrt {-1-\left (\tan ^{2}\relax (x )\right )}}\right )}{2} \]

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*arctan(tan(x)/(-1-tan(x)^2)^(1/2))

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maxima [C]  time = 0.58, size = 20, normalized size = 0.57 \[ -\frac {1}{2} \, \sqrt {-\tan \relax (x)^{2} - 1} \tan \relax (x) - \frac {1}{2} i \, \operatorname {arsinh}\left (\tan \relax (x)\right ) \]

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*I*arcsinh(tan(x))

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mupad [B]  time = 11.59, size = 31, normalized size = 0.89 \[ \frac {\mathrm {atan}\left (\frac {\mathrm {tan}\relax (x)}{\sqrt {-{\mathrm {tan}\relax (x)}^2-1}}\right )}{2}-\frac {\mathrm {tan}\relax (x)\,\sqrt {-{\mathrm {tan}\relax (x)}^2-1}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((- tan(x)^2 - 1)^(3/2),x)

[Out]

atan(tan(x)/(- tan(x)^2 - 1)^(1/2))/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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