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]
-3/2*arctanh(sin(f*x+e))*sec(f*x+e)*(a*cos(f*x+e)^2)^(1/2)/f+3/2*(a*cos(f*x+e)^2)^(1/2)*tan(f*x+e)/f+1/2*(a*co
s(f*x+e)^2)^(1/2)*tan(f*x+e)^3/f
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Rubi [A] time = 0.12, antiderivative size = 91, normalized size of antiderivative = 1.00,
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 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])
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 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 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
]])
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*}
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Mathematica [A] time = 0.19, size = 55, normalized size = 0.60 \[ \frac {a \left ((\cos (2 (e+f x))+2) \tan (e+f x)-3 \cos (e+f x) \tanh ^{-1}(\sin (e+f x))\right )}{2 f \sqrt {a \cos ^2(e+f x)}} \]
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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fricas [A] time = 0.47, size = 77, normalized size = 0.85 \[ -\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (3 \, \cos \left (f x + e\right )^{2} \log \left (-\frac {\sin \left (f x + e\right ) + 1}{\sin \left (f x + e\right ) - 1}\right ) - 2 \, {\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right )\right )}}{4 \, f \cos \left (f x + e\right )^{3}} \]
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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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
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]
Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)2/f*sqrt(a)*(-1/2*(3*(tan((f*x+exp(1))/2)+1/tan((f*x+exp(1))/2))^2*sign(tan((f*x+exp(1))/2)^4-1)-8*sign(ta
n((f*x+exp(1))/2)^4-1))/((tan((f*x+exp(1))/2)+1/tan((f*x+exp(1))/2))^3-4*(tan((f*x+exp(1))/2)+1/tan((f*x+exp(1
))/2)))+3/8*sign(tan((f*x+exp(1))/2)^4-1)*ln(abs(tan((f*x+exp(1))/2)+2+1/tan((f*x+exp(1))/2)))-3/8*sign(tan((f
*x+exp(1))/2)^4-1)*ln(abs(tan((f*x+exp(1))/2)-2+1/tan((f*x+exp(1))/2))))
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maple [A] time = 1.85, size = 84, normalized size = 0.92 \[ -\frac {a \left (-4 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-2 \sin \left (f x +e \right )+\left (-3 \ln \left (\sin \left (f x +e \right )-1\right )+3 \ln \left (1+\sin \left (f x +e \right )\right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )\right )}{4 \cos \left (f x +e \right ) \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f} \]
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(sin(f*x+e)-1)+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 = 0.54, size = 827, normalized size = 9.09 \[ \text {result too large to display} \]
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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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (e+f\,x\right )}^4\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(e + f*x)^4*(a - a*sin(e + f*x)^2)^(1/2),x)
[Out]
int(tan(e + f*x)^4*(a - a*sin(e + f*x)^2)^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{4}{\left (e + f x \right )}\, dx \]
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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