3.462 \(\int \sqrt{a-a \sin ^2(e+f x)} \tan ^6(e+f x) \, dx\)
Optimal. Leaf size=120 \[ \frac{\tan ^5(e+f x) \sqrt{a \cos ^2(e+f x)}}{4 f}-\frac{5 \tan ^3(e+f x) \sqrt{a \cos ^2(e+f x)}}{8 f}-\frac{15 \tan (e+f x) \sqrt{a \cos ^2(e+f x)}}{8 f}+\frac{15 \sec (e+f x) \sqrt{a \cos ^2(e+f x)} \tanh ^{-1}(\sin (e+f x))}{8 f} \]
[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)
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Rubi [A] time = 0.128648, antiderivative size = 120, normalized size of antiderivative = 1.,
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 =
{3176, 3207, 2592, 288, 321, 206} \[ \frac{\tan ^5(e+f x) \sqrt{a \cos ^2(e+f x)}}{4 f}-\frac{5 \tan ^3(e+f x) \sqrt{a \cos ^2(e+f x)}}{8 f}-\frac{15 \tan (e+f x) \sqrt{a \cos ^2(e+f x)}}{8 f}+\frac{15 \sec (e+f x) \sqrt{a \cos ^2(e+f x)} \tanh ^{-1}(\sin (e+f x))}{8 f} \]
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 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 ^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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}
Mathematica [A] time = 0.332584, size = 75, normalized size = 0.62 \[ \frac{\sec ^5(e+f x) \sqrt{a \cos ^2(e+f x)} \left (-5 \sin (e+f x)-15 \sin (3 (e+f x))-2 \sin (5 (e+f x))+60 \cos ^4(e+f x) \tanh ^{-1}(\sin (e+f x))\right )}{32 f} \]
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 = 1.229, size = 120, normalized size = 1. \begin{align*}{\frac{a \left ( 16\,\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}+18\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -4\,\sin \left ( fx+e \right ) + \left ( 15\,\ln \left ( -1+\sin \left ( fx+e \right ) \right ) -15\,\ln \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4} \right ) }{ \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((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^6,x)
[Out]
1/16*a*(16*sin(f*x+e)*cos(f*x+e)^4+18*cos(f*x+e)^2*sin(f*x+e)-4*sin(f*x+e)+(15*ln(-1+sin(f*x+e))-15*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 = 6.59078, size = 2639, normalized size = 21.99 \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)^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 = 1.73221, size = 232, normalized size = 1.93 \begin{align*} -\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{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)^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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**6,x)
[Out]
Timed out
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Giac [B] time = 3.91622, size = 339, normalized size = 2.82 \begin{align*} -\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{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)^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