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

Optimal. Leaf size=91 \[ \frac{\tan ^3(e+f x)}{4 f \sqrt{a \cos ^2(e+f x)}}-\frac{3 \tan (e+f x)}{8 f \sqrt{a \cos ^2(e+f x)}}+\frac{3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 f \sqrt{a \cos ^2(e+f x)}} \]

[Out]

(3*ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(8*f*Sqrt[a*Cos[e + f*x]^2]) - (3*Tan[e + f*x])/(8*f*Sqrt[a*Cos[e + f*x
]^2]) + Tan[e + f*x]^3/(4*f*Sqrt[a*Cos[e + f*x]^2])

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Rubi [A]  time = 0.138509, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3176, 3207, 2611, 3770} \[ \frac{\tan ^3(e+f x)}{4 f \sqrt{a \cos ^2(e+f x)}}-\frac{3 \tan (e+f x)}{8 f \sqrt{a \cos ^2(e+f x)}}+\frac{3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(8*f*Sqrt[a*Cos[e + f*x]^2]) - (3*Tan[e + f*x])/(8*f*Sqrt[a*Cos[e + f*x
]^2]) + Tan[e + f*x]^3/(4*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 ^4(e+f x)}{\sqrt{a-a \sin ^2(e+f x)}} \, dx &=\int \frac{\tan ^4(e+f x)}{\sqrt{a \cos ^2(e+f x)}} \, dx\\ &=\frac{\cos (e+f x) \int \sec (e+f x) \tan ^4(e+f x) \, dx}{\sqrt{a \cos ^2(e+f x)}}\\ &=\frac{\tan ^3(e+f x)}{4 f \sqrt{a \cos ^2(e+f x)}}-\frac{(3 \cos (e+f x)) \int \sec (e+f x) \tan ^2(e+f x) \, dx}{4 \sqrt{a \cos ^2(e+f x)}}\\ &=-\frac{3 \tan (e+f x)}{8 f \sqrt{a \cos ^2(e+f x)}}+\frac{\tan ^3(e+f x)}{4 f \sqrt{a \cos ^2(e+f x)}}+\frac{(3 \cos (e+f x)) \int \sec (e+f x) \, dx}{8 \sqrt{a \cos ^2(e+f x)}}\\ &=\frac{3 \tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{8 f \sqrt{a \cos ^2(e+f x)}}-\frac{3 \tan (e+f x)}{8 f \sqrt{a \cos ^2(e+f x)}}+\frac{\tan ^3(e+f x)}{4 f \sqrt{a \cos ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 0.129723, size = 66, normalized size = 0.73 \[ \frac{\tan (e+f x) \left (8 \tan ^2(e+f x)-6 \sec ^2(e+f x)+3\right )+3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*ArcTanh[Sin[e + f*x]]*Cos[e + f*x] + Tan[e + f*x]*(3 - 6*Sec[e + f*x]^2 + 8*Tan[e + f*x]^2))/(8*f*Sqrt[a*Co
s[e + f*x]^2])

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Maple [A]  time = 1.333, size = 103, normalized size = 1.1 \begin{align*}{\frac{10\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -4\,\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 ) ^{4}}{ \left ( 16+16\,\sin \left ( fx+e \right ) \right ) \left ( -1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) 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(tan(f*x+e)^4/(a-a*sin(f*x+e)^2)^(1/2),x)

[Out]

1/16*(10*cos(f*x+e)^2*sin(f*x+e)-4*sin(f*x+e)+(-3*ln(1+sin(f*x+e))+3*ln(-1+sin(f*x+e)))*cos(f*x+e)^4)/(1+sin(f
*x+e))/(-1+sin(f*x+e))/cos(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f

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Maxima [B]  time = 2.68279, size = 2049, normalized size = 22.52 \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)^4/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/16*(4*(5*sin(7*f*x + 7*e) - 3*sin(5*f*x + 5*e) + 3*sin(3*f*x + 3*e) - 5*sin(f*x + e))*cos(8*f*x + 8*e) - 40
*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) - 16*(3*sin(5*f*x + 5*e) - 3*
sin(3*f*x + 3*e) + 5*sin(f*x + e))*cos(6*f*x + 6*e) + 24*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*cos(5*f*x +
5*e) + 24*(3*sin(3*f*x + 3*e) - 5*sin(f*x + e))*cos(4*f*x + 4*e) - 3*(2*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x + 4
*e) + 4*cos(2*f*x + 2*e) + 1)*cos(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*
e) + 1)*cos(6*f*x + 6*e) + 16*cos(6*f*x + 6*e)^2 + 12*(4*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 36*cos(4*f*x
+ 4*e)^2 + 16*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(8*f*x
+ 8*e) + sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) + 16*sin(6*f*x +
6*e)^2 + 36*sin(4*f*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*sin(2*f*x + 2*e)^2 + 8*cos(2*f*x +
2*e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) + 3*(2*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x +
4*e) + 4*cos(2*f*x + 2*e) + 1)*cos(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*cos(4*f*x + 4*e) + 4*cos(2*f*x +
2*e) + 1)*cos(6*f*x + 6*e) + 16*cos(6*f*x + 6*e)^2 + 12*(4*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 36*cos(4*f
*x + 4*e)^2 + 16*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(8*f
*x + 8*e) + sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) + 16*sin(6*f*x
+ 6*e)^2 + 36*sin(4*f*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*sin(2*f*x + 2*e)^2 + 8*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*(5*cos(7*f*x + 7*e) - 3*cos(5*f*x +
5*e) + 3*cos(3*f*x + 3*e) - 5*cos(f*x + e))*sin(8*f*x + 8*e) + 20*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x + 4*e) + 4
*cos(2*f*x + 2*e) + 1)*sin(7*f*x + 7*e) + 16*(3*cos(5*f*x + 5*e) - 3*cos(3*f*x + 3*e) + 5*cos(f*x + e))*sin(6*
f*x + 6*e) - 12*(6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*sin(5*f*x + 5*e) - 24*(3*cos(3*f*x + 3*e) - 5*co
s(f*x + e))*sin(4*f*x + 4*e) + 12*(4*cos(2*f*x + 2*e) + 1)*sin(3*f*x + 3*e) - 48*cos(3*f*x + 3*e)*sin(2*f*x +
2*e) + 80*cos(f*x + e)*sin(2*f*x + 2*e) - 80*cos(2*f*x + 2*e)*sin(f*x + e) - 20*sin(f*x + e))/((2*(4*cos(6*f*x
+ 6*e) + 6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*cos(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*cos(4*f*x
+ 4*e) + 4*cos(2*f*x + 2*e) + 1)*cos(6*f*x + 6*e) + 16*cos(6*f*x + 6*e)^2 + 12*(4*cos(2*f*x + 2*e) + 1)*cos(4*
f*x + 4*e) + 36*cos(4*f*x + 4*e)^2 + 16*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*si
n(2*f*x + 2*e))*sin(8*f*x + 8*e) + sin(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(6*f*x
+ 6*e) + 16*sin(6*f*x + 6*e)^2 + 36*sin(4*f*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*sin(2*f*x
+ 2*e)^2 + 8*cos(2*f*x + 2*e) + 1)*sqrt(a)*f)

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Fricas [A]  time = 1.75814, size = 208, normalized size = 2.29 \begin{align*} -\frac{{\left (3 \, \cos \left (f x + e\right )^{4} \log \left (-\frac{\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) + 2 \,{\left (5 \, \cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right )\right )} \sqrt{a \cos \left (f x + e\right )^{2}}}{16 \, a f \cos \left (f x + e\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/16*(3*cos(f*x + e)^4*log(-(sin(f*x + e) - 1)/(sin(f*x + e) + 1)) + 2*(5*cos(f*x + e)^2 - 2)*sin(f*x + e))*s
qrt(a*cos(f*x + e)^2)/(a*f*cos(f*x + e)^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{4}{\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)**4/(a-a*sin(f*x+e)**2)**(1/2),x)

[Out]

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

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Giac [B]  time = 2.14093, size = 271, normalized size = 2.98 \begin{align*} -\frac{\frac{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{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 )}^{3} - \frac{20}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )} - 20 \, \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 )}^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}}{16 \, \sqrt{a} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^4/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

-1/16*(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(a
bs(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))^3 - 20/tan(1/2*f*x + 1/2*e) - 20*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)^2*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)))/(sqrt(a)*f)