\(\int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx\) [463]

Optimal result

Integrand size = 26, antiderivative size = 91 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=-\frac {3 \text {arctanh}(\sin (e+f x)) \sqrt {a \cos ^2(e+f x)} \sec (e+f x)}{2 f}+\frac {3 \sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{2 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan ^3(e+f x)}{2 f} \]

[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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3255, 3286, 2672, 294, 327, 212} \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=-\frac {3 \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \text {arctanh}(\sin (e+f x))}{2 f}+\frac {\tan ^3(e+f x) \sqrt {a \cos ^2(e+f x)}}{2 f}+\frac {3 \tan (e+f x) \sqrt {a \cos ^2(e+f x)}}{2 f} \]

[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 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*} \text {integral}& = \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 ) \text {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 ) \text {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 ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (e+f x)\right )}{2 f} \\ & = -\frac {3 \text {arctanh}(\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] (verified)

Time = 0.22 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.60 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=\frac {a (-3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)+(2+\cos (2 (e+f x))) \tan (e+f x))}{2 f \sqrt {a \cos ^2(e+f x)}} \]

[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])

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.44

method	result	size
default	\(\frac {\sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\, \left (2 \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\, \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {a}-3 \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}+2 a}{\cos \left (f x +e \right )}\right ) a \left (\cos ^{2}\left (f x +e \right )\right )+\sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{2 \cos \left (f x +e \right ) \sqrt {a}\, \sin \left (f x +e \right ) \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\)	\(131\)
risch	\(-\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (i {\mathrm e}^{6 i \left (f x +e \right )}+3 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) {\mathrm e}^{5 i \left (f x +e \right )}-3 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) {\mathrm e}^{5 i \left (f x +e \right )}+3 i {\mathrm e}^{4 i \left (f x +e \right )}+6 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) {\mathrm e}^{3 i \left (f x +e \right )}-6 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) {\mathrm e}^{3 i \left (f x +e \right )}-3 i {\mathrm e}^{2 i \left (f x +e \right )}+3 \ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) {\mathrm e}^{i \left (f x +e \right )}-3 \ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) {\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}\)	\(246\)

[In]

int((a-a*sin(f*x+e)^2)^(1/2)*tan(f*x+e)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.85 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (3 \, \cos \left (f x + e\right )^{2} \log \left (-\frac {\sin \left (f x + e\right ) + 1}{\sin \left (f x + e\right ) - 1}\right ) - 2 \, {\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right )\right )}}{4 \, f \cos \left (f x + e\right )^{3}} \]

[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)

Sympy [F]

\[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \tan ^{4}{\left (e + f x \right )}\, dx \]

[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)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 827 vs. \(2 (79) = 158\).

Time = 0.42 (sec) , antiderivative size = 827, normalized size of antiderivative = 9.09 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=\text {Too large to display} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (79) = 158\).

Time = 1.22 (sec) , antiderivative size = 197, normalized size of antiderivative = 2.16 \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=\frac {{\left (3 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 3 \, \log \left ({\left | \frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \right |}\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - \frac {4 \, {\left (3 \, {\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right ) - 8 \, \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )\right )}}{{\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} - \frac {4}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} - 4 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}\right )} \sqrt {a}}{4 \, f} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a-a \sin ^2(e+f x)} \tan ^4(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^4\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \]

[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)