3.464 \(\int \sqrt{a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx\)

Optimal. Leaf size=57 \[ \frac{\sec (e+f x) \sqrt{a \cos ^2(e+f x)} \tanh ^{-1}(\sin (e+f x))}{f}-\frac{\tan (e+f x) \sqrt{a \cos ^2(e+f x)}}{f} \]

[Out]

(ArcTanh[Sin[e + f*x]]*Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x])/f - (Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x])/f

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Rubi [A]  time = 0.103233, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3176, 3207, 2592, 321, 206} \[ \frac{\sec (e+f x) \sqrt{a \cos ^2(e+f x)} \tanh ^{-1}(\sin (e+f x))}{f}-\frac{\tan (e+f x) \sqrt{a \cos ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^2,x]

[Out]

(ArcTanh[Sin[e + f*x]]*Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x])/f - (Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x])/f

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

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

Rubi steps

\begin{align*} \int \sqrt{a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx &=\int \sqrt{a \cos ^2(e+f x)} \tan ^2(e+f x) \, dx\\ &=\left (\sqrt{a \cos ^2(e+f x)} \sec (e+f x)\right ) \int \sin (e+f x) \tan (e+f x) \, dx\\ &=\frac{\left (\sqrt{a \cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac{\sqrt{a \cos ^2(e+f x)} \tan (e+f x)}{f}+\frac{\left (\sqrt{a \cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\tanh ^{-1}(\sin (e+f x)) \sqrt{a \cos ^2(e+f x)} \sec (e+f x)}{f}-\frac{\sqrt{a \cos ^2(e+f x)} \tan (e+f x)}{f}\\ \end{align*}

Mathematica [A]  time = 0.0461756, size = 40, normalized size = 0.7 \[ \frac{\sec (e+f x) \sqrt{a \cos ^2(e+f x)} \left (\tanh ^{-1}(\sin (e+f x))-\sin (e+f x)\right )}{f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^2,x]

[Out]

(Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x]*(ArcTanh[Sin[e + f*x]] - Sin[e + f*x]))/f

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Maple [A]  time = 1.321, size = 54, normalized size = 1. \begin{align*} -{\frac{a\cos \left ( fx+e \right ) \left ( 2\,\sin \left ( fx+e \right ) +\ln \left ( -1+\sin \left ( fx+e \right ) \right ) -\ln \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) }{2\,f}{\frac{1}{\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x)

[Out]

-1/2*a*cos(f*x+e)*(2*sin(f*x+e)+ln(-1+sin(f*x+e))-ln(1+sin(f*x+e)))/(a*cos(f*x+e)^2)^(1/2)/f

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Maxima [A]  time = 1.69868, size = 99, normalized size = 1.74 \begin{align*} \frac{\sqrt{a}{\left (\log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 2 \, \sin \left (f x + e\right )\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="maxima")

[Out]

1/2*sqrt(a)*(log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - log(cos(f*x + e)^2 + sin(f*x + e)^2 -
2*sin(f*x + e) + 1) - 2*sin(f*x + e))/f

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Fricas [A]  time = 1.69902, size = 147, normalized size = 2.58 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left (\log \left (-\frac{\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) + 2 \, \sin \left (f x + e\right )\right )}}{2 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="fricas")

[Out]

-1/2*sqrt(a*cos(f*x + e)^2)*(log(-(sin(f*x + e) - 1)/(sin(f*x + e) + 1)) + 2*sin(f*x + e))/(f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{- a \left (\sin{\left (e + f x \right )} - 1\right ) \left (\sin{\left (e + f x \right )} + 1\right )} \tan ^{2}{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sin(f*x+e)**2)**(1/2)*tan(f*x+e)**2,x)

[Out]

Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*tan(e + f*x)**2, x)

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Giac [B]  time = 1.37759, size = 184, normalized size = 3.23 \begin{align*} -\frac{{\left (\log \left ({\left | \frac{1}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \right |}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) - \log \left ({\left | \frac{1}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \right |}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) - \frac{4 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}{\frac{1}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}\right )} \sqrt{a}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^2,x, algorithm="giac")

[Out]

-1/2*(log(abs(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e) + 2))*sgn(tan(1/2*f*x + 1/2*e)^4 - 1) - log(abs(1/
tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e) - 2))*sgn(tan(1/2*f*x + 1/2*e)^4 - 1) - 4*sgn(tan(1/2*f*x + 1/2*e)
^4 - 1)/(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e)))*sqrt(a)/f