3.5.62 \(\int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx\) [462]

Optimal. Leaf size=120 \[ \frac {15 \tanh ^{-1}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{8 f}-\frac {15 \sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{8 f}-\frac {5 \sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f} \]

[Out]

15/8*arctanh(sin(f*x+e))*sec(f*x+e)*(a*cos(f*x+e)^2)^(1/2)/f-15/8*(a*cos(f*x+e)^2)^(1/2)*tan(f*x+e)/f-5/8*(a*c
os(f*x+e)^2)^(1/2)*tan(f*x+e)^3/f+1/4*(a*cos(f*x+e)^2)^(1/2)*tan(f*x+e)^5/f

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Rubi [A]
time = 0.09, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3255, 3286, 2672, 294, 327, 212} \begin {gather*} \frac {\tan ^5(e+f x) \sqrt {a \cos ^2(e+f x)}}{4 f}-\frac {5 \tan ^3(e+f x) \sqrt {a \cos ^2(e+f x)}}{8 f}-\frac {15 \tan (e+f x) \sqrt {a \cos ^2(e+f x)}}{8 f}+\frac {15 \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \tanh ^{-1}(\sin (e+f x))}{8 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^6,x]

[Out]

(15*ArcTanh[Sin[e + f*x]]*Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x])/(8*f) - (15*Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x]
)/(8*f) - (5*Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x]^3)/(8*f) + (Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x]^5)/(4*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 294

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 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*} \int \sqrt {a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx &=\int \sqrt {a \cos ^2(e+f x)} \tan ^6(e+f x) \, dx\\ &=\left (\sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \int \sin (e+f x) \tan ^5(e+f x) \, dx\\ &=\frac {\left (\sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^6}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f}-\frac {\left (5 \sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 f}\\ &=-\frac {5 \sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f}+\frac {\left (15 \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 )}{8 f}\\ &=-\frac {15 \sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{8 f}-\frac {5 \sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f}+\frac {\left (15 \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 )}{8 f}\\ &=\frac {15 \tanh ^{-1}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{8 f}-\frac {15 \sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{8 f}-\frac {5 \sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{8 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^5(e+f x)}{4 f}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 75, normalized size = 0.62 \begin {gather*} \frac {\sqrt {a \cos ^2(e+f x)} \sec ^5(e+f x) \left (60 \tanh ^{-1}(\sin (e+f x)) \cos ^4(e+f x)-5 \sin (e+f x)-15 \sin (3 (e+f x))-2 \sin (5 (e+f x))\right )}{32 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - a*Sin[e + f*x]^2]*Tan[e + f*x]^6,x]

[Out]

(Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x]^5*(60*ArcTanh[Sin[e + f*x]]*Cos[e + f*x]^4 - 5*Sin[e + f*x] - 15*Sin[3*(e
+ f*x)] - 2*Sin[5*(e + f*x)]))/(32*f)

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Maple [A]
time = 7.11, size = 120, normalized size = 1.00

method	result	size
default	\(\frac {a \left (16 \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+18 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-4 \sin \left (f x +e \right )+\left (15 \ln \left (\sin \left (f x +e \right )-1\right )-15 \ln \left (1+\sin \left (f x +e \right )\right )\right ) \left (\cos ^{4}\left (f x +e \right )\right )\right )}{16 \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\)	\(120\)
risch	\(\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) f}+\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (9 \,{\mathrm e}^{8 i \left (f x +e \right )}+{\mathrm e}^{6 i \left (f x +e \right )}-{\mathrm e}^{4 i \left (f x +e \right )}-9 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{4 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}+\frac {15 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{8 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {15 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{8 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\)	\(327\)

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)^6,x,method=_RETURNVERBOSE)

[Out]

1/16*a*(16*cos(f*x+e)^4*sin(f*x+e)+18*cos(f*x+e)^2*sin(f*x+e)-4*sin(f*x+e)+(15*ln(sin(f*x+e)-1)-15*ln(1+sin(f*
x+e)))*cos(f*x+e)^4)/(1+sin(f*x+e))/(sin(f*x+e)-1)/cos(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 2144 vs. \(2 (113) = 226\).
time = 2.12, size = 2144, normalized size = 17.87 \begin {gather*} \text {Too large to display} \end {gather*}

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)^6,x, algorithm="maxima")

[Out]

1/16*(8*(sin(9*f*x + 9*e) + 4*sin(7*f*x + 7*e) + 6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + sin(f*x + e))*cos(1
0*f*x + 10*e) - 20*(3*sin(8*f*x + 8*e) + sin(6*f*x + 6*e) - sin(4*f*x + 4*e) - 3*sin(2*f*x + 2*e))*cos(9*f*x +
9*e) + 60*(4*sin(7*f*x + 7*e) + 6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + sin(f*x + e))*cos(8*f*x + 8*e) - 80
*(sin(6*f*x + 6*e) - sin(4*f*x + 4*e) - 3*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) + 20*(6*sin(5*f*x + 5*e) + 4*sin(
3*f*x + 3*e) + sin(f*x + e))*cos(6*f*x + 6*e) + 120*(sin(4*f*x + 4*e) + 3*sin(2*f*x + 2*e))*cos(5*f*x + 5*e) -
20*(4*sin(3*f*x + 3*e) + sin(f*x + e))*cos(4*f*x + 4*e) + 15*(2*(4*cos(7*f*x + 7*e) + 6*cos(5*f*x + 5*e) + 4*
cos(3*f*x + 3*e) + cos(f*x + e))*cos(9*f*x + 9*e) + cos(9*f*x + 9*e)^2 + 8*(6*cos(5*f*x + 5*e) + 4*cos(3*f*x +
3*e) + cos(f*x + e))*cos(7*f*x + 7*e) + 16*cos(7*f*x + 7*e)^2 + 12*(4*cos(3*f*x + 3*e) + cos(f*x + e))*cos(5*
f*x + 5*e) + 36*cos(5*f*x + 5*e)^2 + 16*cos(3*f*x + 3*e)^2 + 8*cos(3*f*x + 3*e)*cos(f*x + e) + cos(f*x + e)^2
+ 2*(4*sin(7*f*x + 7*e) + 6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(9*f*x + 9*e) + sin(9*f*x
+ 9*e)^2 + 8*(6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(7*f*x + 7*e) + 16*sin(7*f*x + 7*e)^
2 + 12*(4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(5*f*x + 5*e) + 36*sin(5*f*x + 5*e)^2 + 16*sin(3*f*x + 3*e)^2 +
8*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) -
15*(2*(4*cos(7*f*x + 7*e) + 6*cos(5*f*x + 5*e) + 4*cos(3*f*x + 3*e) + cos(f*x + e))*cos(9*f*x + 9*e) + cos(9*f
*x + 9*e)^2 + 8*(6*cos(5*f*x + 5*e) + 4*cos(3*f*x + 3*e) + cos(f*x + e))*cos(7*f*x + 7*e) + 16*cos(7*f*x + 7*e
)^2 + 12*(4*cos(3*f*x + 3*e) + cos(f*x + e))*cos(5*f*x + 5*e) + 36*cos(5*f*x + 5*e)^2 + 16*cos(3*f*x + 3*e)^2
+ 8*cos(3*f*x + 3*e)*cos(f*x + e) + cos(f*x + e)^2 + 2*(4*sin(7*f*x + 7*e) + 6*sin(5*f*x + 5*e) + 4*sin(3*f*x
+ 3*e) + sin(f*x + e))*sin(9*f*x + 9*e) + sin(9*f*x + 9*e)^2 + 8*(6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + si
n(f*x + e))*sin(7*f*x + 7*e) + 16*sin(7*f*x + 7*e)^2 + 12*(4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(5*f*x + 5*e)
+ 36*sin(5*f*x + 5*e)^2 + 16*sin(3*f*x + 3*e)^2 + 8*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) - 8*(cos(9*f*x + 9*e) + 4*cos(7*f*x + 7*e) + 6*cos(5*f*x + 5*
e) + 4*cos(3*f*x + 3*e) + cos(f*x + e))*sin(10*f*x + 10*e) + 4*(15*cos(8*f*x + 8*e) + 5*cos(6*f*x + 6*e) - 5*c
os(4*f*x + 4*e) - 15*cos(2*f*x + 2*e) - 2)*sin(9*f*x + 9*e) - 60*(4*cos(7*f*x + 7*e) + 6*cos(5*f*x + 5*e) + 4*
cos(3*f*x + 3*e) + cos(f*x + e))*sin(8*f*x + 8*e) + 16*(5*cos(6*f*x + 6*e) - 5*cos(4*f*x + 4*e) - 15*cos(2*f*x
+ 2*e) - 2)*sin(7*f*x + 7*e) - 20*(6*cos(5*f*x + 5*e) + 4*cos(3*f*x + 3*e) + cos(f*x + e))*sin(6*f*x + 6*e) -
24*(5*cos(4*f*x + 4*e) + 15*cos(2*f*x + 2*e) + 2)*sin(5*f*x + 5*e) + 20*(4*cos(3*f*x + 3*e) + cos(f*x + e))*s
in(4*f*x + 4*e) - 16*(15*cos(2*f*x + 2*e) + 2)*sin(3*f*x + 3*e) + 240*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) + 60*c
os(f*x + e)*sin(2*f*x + 2*e) - 60*cos(2*f*x + 2*e)*sin(f*x + e) - 8*sin(f*x + e))*sqrt(a)/((2*(4*cos(7*f*x + 7
*e) + 6*cos(5*f*x + 5*e) + 4*cos(3*f*x + 3*e) + cos(f*x + e))*cos(9*f*x + 9*e) + cos(9*f*x + 9*e)^2 + 8*(6*cos
(5*f*x + 5*e) + 4*cos(3*f*x + 3*e) + cos(f*x + e))*cos(7*f*x + 7*e) + 16*cos(7*f*x + 7*e)^2 + 12*(4*cos(3*f*x
+ 3*e) + cos(f*x + e))*cos(5*f*x + 5*e) + 36*cos(5*f*x + 5*e)^2 + 16*cos(3*f*x + 3*e)^2 + 8*cos(3*f*x + 3*e)*c
os(f*x + e) + cos(f*x + e)^2 + 2*(4*sin(7*f*x + 7*e) + 6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + sin(f*x + e))
*sin(9*f*x + 9*e) + sin(9*f*x + 9*e)^2 + 8*(6*sin(5*f*x + 5*e) + 4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(7*f*x
+ 7*e) + 16*sin(7*f*x + 7*e)^2 + 12*(4*sin(3*f*x + 3*e) + sin(f*x + e))*sin(5*f*x + 5*e) + 36*sin(5*f*x + 5*e)
^2 + 16*sin(3*f*x + 3*e)^2 + 8*sin(3*f*x + 3*e)*sin(f*x + e) + sin(f*x + e)^2)*f)

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Fricas [A]
time = 0.41, size = 87, normalized size = 0.72 \begin {gather*} -\frac {{\left (15 \, \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 (8 \, \cos \left (f x + e\right )^{4} + 9 \, \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 \, f \cos \left (f x + e\right )^{5}} \end {gather*}

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)^6,x, algorithm="fricas")

[Out]

-1/16*(15*cos(f*x + e)^4*log(-(sin(f*x + e) - 1)/(sin(f*x + e) + 1)) + 2*(8*cos(f*x + e)^4 + 9*cos(f*x + e)^2
- 2)*sin(f*x + e))*sqrt(a*cos(f*x + e)^2)/(f*cos(f*x + e)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{6}{\left (e + f x \right )}\, dx \end {gather*}

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)**6,x)

[Out]

Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*tan(e + f*x)**6, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 251 vs. \(2 (113) = 226\).
time = 1.97, size = 251, normalized size = 2.09 \begin {gather*} -\frac {{\left (15 \, \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 ) - 15 \, \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 {32 \, \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 )} - \frac {4 \, {\left (7 \, {\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} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 36 \, {\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 )} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\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}}\right )} \sqrt {a}}{16 \, f} \end {gather*}

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)^6,x, algorithm="giac")

[Out]

-1/16*(15*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) - 15*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) - 32*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)) - 4*(7*(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x +
1/2*e))^3*sgn(tan(1/2*f*x + 1/2*e)^4 - 1) - 36*(1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e))*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))^2 - 4)^2)*sqrt(a)/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {tan}\left (e+f\,x\right )}^6\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(e + f*x)^6*(a - a*sin(e + f*x)^2)^(1/2),x)

[Out]

int(tan(e + f*x)^6*(a - a*sin(e + f*x)^2)^(1/2), x)

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