\(\int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx\) [464]

Optimal result

Integrand size = 26, antiderivative size = 57 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {\text {arctanh}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{f}-\frac {\sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{f} \]

[Out]

arctanh(sin(f*x+e))*sec(f*x+e)*(a*cos(f*x+e)^2)^(1/2)/f-(a*cos(f*x+e)^2)^(1/2)*tan(f*x+e)/f

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3255, 3286, 2672, 327, 212} \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} \text {arctanh}(\sin (e+f x))}{f}-\frac {\tan (e+f x) \sqrt {a \cos ^2(e+f x)}}{f} \]

[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 212

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

Rule 327

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 2672

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 3255

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 3286

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
]])

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {arctanh}(\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] (verified)

Time = 0.05 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.70 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {\sqrt {a \cos ^2(e+f x)} \sec (e+f x) (\text {arctanh}(\sin (e+f x))-\sin (e+f x))}{f} \]

[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

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.68

method	result	size
default	\(-\frac {\cos \left (f x +e \right ) \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\, \left (\sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}-\ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}+2 a}{\cos \left (f x +e \right )}\right ) a \right )}{\sqrt {a}\, \sin \left (f x +e \right ) \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\)	\(96\)
risch	\(\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}-\frac {\ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\)	\(238\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.96 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (\log \left (-\frac {\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) + 2 \, \sin \left (f x + e\right )\right )}}{2 \, f \cos \left (f x + e\right )} \]

[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))

Sympy [F]

\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{2}{\left (e + f x \right )}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\frac {\sqrt {a} {\left (\log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 2 \, \sin \left (f x + e\right )\right )}}{2 \, f} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (53) = 106\).

Time = 0.58 (sec) , antiderivative size = 127, normalized size of antiderivative = 2.23 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=-\frac {{\left (\log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - \frac {4 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}{\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}\right )} \sqrt {a}}{2 \, f} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^2(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^2\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \]

[In]

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

[Out]

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