3.463 \(\int \sqrt{a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx\)

Optimal. Leaf size=91 \[ \frac{\tan ^3(e+f x) \sqrt{a \cos ^2(e+f x)}}{2 f}+\frac{3 \tan (e+f x) \sqrt{a \cos ^2(e+f x)}}{2 f}-\frac{3 \sec (e+f x) \sqrt{a \cos ^2(e+f x)} \tanh ^{-1}(\sin (e+f x))}{2 f} \]

[Out]

(-3*ArcTanh[Sin[e + f*x]]*Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x])/(2*f) + (3*Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x])
/(2*f) + (Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x]^3)/(2*f)

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Rubi [A]  time = 0.123761, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3176, 3207, 2592, 288, 321, 206} \[ \frac{\tan ^3(e+f x) \sqrt{a \cos ^2(e+f x)}}{2 f}+\frac{3 \tan (e+f x) \sqrt{a \cos ^2(e+f x)}}{2 f}-\frac{3 \sec (e+f x) \sqrt{a \cos ^2(e+f x)} \tanh ^{-1}(\sin (e+f x))}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^4,x]

[Out]

(-3*ArcTanh[Sin[e + f*x]]*Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x])/(2*f) + (3*Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x])
/(2*f) + (Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x]^3)/(2*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 ^4(e+f x) \, dx &=\int \sqrt{a \cos ^2(e+f x)} \tan ^4(e+f x) \, dx\\ &=\left (\sqrt{a \cos ^2(e+f x)} \sec (e+f x)\right ) \int \sin (e+f x) \tan ^3(e+f x) \, dx\\ &=\frac{\left (\sqrt{a \cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\sqrt{a \cos ^2(e+f x)} \tan ^3(e+f x)}{2 f}-\frac{\left (3 \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 )}{2 f}\\ &=\frac{3 \sqrt{a \cos ^2(e+f x)} \tan (e+f x)}{2 f}+\frac{\sqrt{a \cos ^2(e+f x)} \tan ^3(e+f x)}{2 f}-\frac{\left (3 \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 )}{2 f}\\ &=-\frac{3 \tanh ^{-1}(\sin (e+f x)) \sqrt{a \cos ^2(e+f x)} \sec (e+f x)}{2 f}+\frac{3 \sqrt{a \cos ^2(e+f x)} \tan (e+f x)}{2 f}+\frac{\sqrt{a \cos ^2(e+f x)} \tan ^3(e+f x)}{2 f}\\ \end{align*}

Mathematica [A]  time = 0.195319, size = 55, normalized size = 0.6 \[ \frac{a \left ((\cos (2 (e+f x))+2) \tan (e+f x)-3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))\right )}{2 f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^4,x]

[Out]

(a*(-3*ArcTanh[Sin[e + f*x]]*Cos[e + f*x] + (2 + Cos[2*(e + f*x)])*Tan[e + f*x]))/(2*f*Sqrt[a*Cos[e + f*x]^2])

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Maple [A]  time = 1.273, size = 84, normalized size = 0.9 \begin{align*}{\frac{a \left ( 4\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +2\,\sin \left ( fx+e \right ) + \left ( -3\,\ln \left ( 1+\sin \left ( fx+e \right ) \right ) +3\,\ln \left ( -1+\sin \left ( fx+e \right ) \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{4\,f\cos \left ( fx+e \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((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^4,x)

[Out]

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

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Maxima [B]  time = 1.91776, size = 1116, normalized size = 12.26 \begin{align*} \text{result too large to display} \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)^4,x, algorithm="maxima")

[Out]

-1/4*(2*(sin(5*f*x + 5*e) + 2*sin(3*f*x + 3*e) + sin(f*x + e))*cos(6*f*x + 6*e) - 6*(sin(4*f*x + 4*e) - sin(2*
f*x + 2*e))*cos(5*f*x + 5*e) + 6*(2*sin(3*f*x + 3*e) + sin(f*x + e))*cos(4*f*x + 4*e) + 3*(2*(2*cos(3*f*x + 3*
e) + cos(f*x + e))*cos(5*f*x + 5*e) + cos(5*f*x + 5*e)^2 + 4*cos(3*f*x + 3*e)^2 + 4*cos(3*f*x + 3*e)*cos(f*x +
e) + cos(f*x + e)^2 + 2*(2*sin(3*f*x + 3*e) + sin(f*x + e))*sin(5*f*x + 5*e) + sin(5*f*x + 5*e)^2 + 4*sin(3*f
*x + 3*e)^2 + 4*sin(3*f*x + 3*e)*sin(f*x + e) + sin(f*x + e)^2)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*
x + e) + 1) - 3*(2*(2*cos(3*f*x + 3*e) + cos(f*x + e))*cos(5*f*x + 5*e) + cos(5*f*x + 5*e)^2 + 4*cos(3*f*x + 3
*e)^2 + 4*cos(3*f*x + 3*e)*cos(f*x + e) + cos(f*x + e)^2 + 2*(2*sin(3*f*x + 3*e) + sin(f*x + e))*sin(5*f*x + 5
*e) + sin(5*f*x + 5*e)^2 + 4*sin(3*f*x + 3*e)^2 + 4*sin(3*f*x + 3*e)*sin(f*x + e) + sin(f*x + e)^2)*log(cos(f*
x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 2*(cos(5*f*x + 5*e) + 2*cos(3*f*x + 3*e) + cos(f*x + e))*sin
(6*f*x + 6*e) + 2*(3*cos(4*f*x + 4*e) - 3*cos(2*f*x + 2*e) - 1)*sin(5*f*x + 5*e) - 6*(2*cos(3*f*x + 3*e) + cos
(f*x + e))*sin(4*f*x + 4*e) - 4*(3*cos(2*f*x + 2*e) + 1)*sin(3*f*x + 3*e) + 12*cos(3*f*x + 3*e)*sin(2*f*x + 2*
e) + 6*cos(f*x + e)*sin(2*f*x + 2*e) - 6*cos(2*f*x + 2*e)*sin(f*x + e) - 2*sin(f*x + e))*sqrt(a)/((2*(2*cos(3*
f*x + 3*e) + cos(f*x + e))*cos(5*f*x + 5*e) + cos(5*f*x + 5*e)^2 + 4*cos(3*f*x + 3*e)^2 + 4*cos(3*f*x + 3*e)*c
os(f*x + e) + cos(f*x + e)^2 + 2*(2*sin(3*f*x + 3*e) + sin(f*x + e))*sin(5*f*x + 5*e) + sin(5*f*x + 5*e)^2 + 4
*sin(3*f*x + 3*e)^2 + 4*sin(3*f*x + 3*e)*sin(f*x + e) + sin(f*x + e)^2)*f)

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Fricas [A]  time = 1.6846, size = 204, normalized size = 2.24 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left (3 \, \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 \,{\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right )\right )}}{4 \, f \cos \left (f x + e\right )^{3}} \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)^4,x, algorithm="fricas")

[Out]

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

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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 ^{4}{\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)**4,x)

[Out]

Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*tan(e + f*x)**4, x)

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Giac [B]  time = 1.98275, size = 285, normalized size = 3.13 \begin{align*} \frac{{\left (3 \, \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 ) - 3 \, \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 (3 \,{\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} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) - 8 \, \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )\right )}}{{\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 )}^{3} - \frac{4}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} - 4 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}\right )} \sqrt{a}}{4 \, 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)^4,x, algorithm="giac")

[Out]

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