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

Optimal. Leaf size=62 \[ \frac{\tan (e+f x)}{2 f \sqrt{a \cos ^2(e+f x)}}-\frac{\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{2 f \sqrt{a \cos ^2(e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.119355, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3176, 3207, 2611, 3770} \[ \frac{\tan (e+f x)}{2 f \sqrt{a \cos ^2(e+f x)}}-\frac{\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{2 f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.0451097, size = 43, normalized size = 0.69 \[ \frac{\tan (e+f x)-\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{2 f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 1.423, size = 65, normalized size = 1.1 \begin{align*}{\frac{1}{f\cos \left ( fx+e \right ) } \left ({\frac{\sin \left ( fx+e \right ) }{2}}+{\frac{ \left ( \ln \left ( -1+\sin \left ( fx+e \right ) \right ) -\ln \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{4}} \right ){\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(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(1/2),x)

[Out]

(1/2*sin(f*x+e)+1/4*(ln(-1+sin(f*x+e))-ln(1+sin(f*x+e)))*cos(f*x+e)^2)/cos(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f

________________________________________________________________________________________

Maxima [B]  time = 1.73306, size = 711, normalized size = 11.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(sin(3*f*x + 3*e) - sin(f*x + e))*cos(4*f*x + 4*e) - (2*(2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + cos
(4*f*x + 4*e)^2 + 4*cos(2*f*x + 2*e)^2 + sin(4*f*x + 4*e)^2 + 4*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*sin(2*f*
x + 2*e)^2 + 4*cos(2*f*x + 2*e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) + (2*(2*cos(2*f
*x + 2*e) + 1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^2 + 4*cos(2*f*x + 2*e)^2 + sin(4*f*x + 4*e)^2 + 4*sin(4*f*x
+ 4*e)*sin(2*f*x + 2*e) + 4*sin(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2
- 2*sin(f*x + e) + 1) - 4*(cos(3*f*x + 3*e) - cos(f*x + e))*sin(4*f*x + 4*e) + 4*(2*cos(2*f*x + 2*e) + 1)*sin(
3*f*x + 3*e) - 8*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) + 8*cos(f*x + e)*sin(2*f*x + 2*e) - 8*cos(2*f*x + 2*e)*sin(
f*x + e) - 4*sin(f*x + e))/((2*(2*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + cos(4*f*x + 4*e)^2 + 4*cos(2*f*x +
2*e)^2 + sin(4*f*x + 4*e)^2 + 4*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 4*sin(2*f*x + 2*e)^2 + 4*cos(2*f*x + 2*e)
+ 1)*sqrt(a)*f)

________________________________________________________________________________________

Fricas [A]  time = 1.68546, size = 173, normalized size = 2.79 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left (\cos \left (f x + e\right )^{2} \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 )}}{4 \, a f \cos \left (f x + e\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(a*cos(f*x + e)^2)*(cos(f*x + e)^2*log(-(sin(f*x + e) + 1)/(sin(f*x + e) - 1)) - 2*sin(f*x + e))/(a*f
*cos(f*x + e)^3)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.45832, size = 228, normalized size = 3.68 \begin{align*} \frac{\frac{\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{\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 \,{\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 )\right )}}{{\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 )\right )}^{2} - 4\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}}{4 \, \sqrt{a} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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