3.2.3 \(\int \frac {\tan ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx\) [103]

Optimal. Leaf size=150 \[ -\frac {67 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{64 \sqrt {2} \sqrt {a} f}-\frac {\sec (e+f x) (53+127 \sin (e+f x))}{192 f \sqrt {a+a \sin (e+f x)}}+\frac {a \sin (e+f x) \tan (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}+\frac {\tan ^3(e+f x)}{3 f \sqrt {a+a \sin (e+f x)}} \]

[Out]

-67/128*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/f*2^(1/2)/a^(1/2)-1/192*sec(f*x+e)*(53+
127*sin(f*x+e))/f/(a+a*sin(f*x+e))^(1/2)+1/24*a*sin(f*x+e)*tan(f*x+e)/f/(a+a*sin(f*x+e))^(3/2)+1/3*tan(f*x+e)^
3/f/(a+a*sin(f*x+e))^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.62, antiderivative size = 241, normalized size of antiderivative = 1.61, number of steps used = 17, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {2793, 2728, 212, 4486, 2766, 2760, 2729, 2956, 2934} \begin {gather*} \frac {61 a \cos (e+f x)}{64 f (a \sin (e+f x)+a)^{3/2}}+\frac {7 \sec ^3(e+f x) \sqrt {a \sin (e+f x)+a}}{12 a f}-\frac {5 \sec ^3(e+f x)}{6 f \sqrt {a \sin (e+f x)+a}}-\frac {61 \sec (e+f x)}{48 f \sqrt {a \sin (e+f x)+a}}+\frac {7 a \sec (e+f x)}{24 f (a \sin (e+f x)+a)^{3/2}}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f}+\frac {61 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{64 \sqrt {2} \sqrt {a} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(61*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(64*Sqrt[2]*Sqrt[a]*f) - (Sqrt[2]*ArcT
anh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f) + (61*a*Cos[e + f*x])/(64*f*(a + a
*Sin[e + f*x])^(3/2)) + (7*a*Sec[e + f*x])/(24*f*(a + a*Sin[e + f*x])^(3/2)) - (61*Sec[e + f*x])/(48*f*Sqrt[a
+ a*Sin[e + f*x]]) - (5*Sec[e + f*x]^3)/(6*f*Sqrt[a + a*Sin[e + f*x]]) + (7*Sec[e + f*x]^3*Sqrt[a + a*Sin[e +
f*x]])/(12*a*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 2728

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, b*(C
os[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[c + d*x]*((a + b*Sin[c + d*x])^n/(a*d
*(2*n + 1))), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1))), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rule 2766

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*((
g*Cos[e + f*x])^(p + 1)/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[a*((2*p + 1)/(2*g^2*(p + 1))), In
t[(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0
] && LtQ[p, -1] && IntegerQ[2*p]

Rule 2793

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^4, x_Symbol] :> Int[(a + b*Sin[e + f*x
])^m, x] - Int[(a + b*Sin[e + f*x])^m*((1 - 2*Sin[e + f*x]^2)/Cos[e + f*x]^4), x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[m - 1/2]

Rule 2934

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p +
1))), x] + Dist[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1))), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*
x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2956

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(2*m + p + 1))), x] - Dist[1/(a^
2*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2*m + p + 1)*Sin[e + f*x]), x]
, x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && LeQ[m, -2^(-1)] && NeQ[2*m + p + 1, 0]

Rule 4486

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \frac {\tan ^4(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx &=\int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx-\int \frac {\sec ^4(e+f x) \left (1-2 \sin ^2(e+f x)\right )}{\sqrt {a+a \sin (e+f x)}} \, dx\\ &=-\frac {2 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{f}-\int \left (\frac {\sec ^4(e+f x)}{\sqrt {a (1+\sin (e+f x))}}-\frac {2 \sec ^2(e+f x) \tan ^2(e+f x)}{\sqrt {a (1+\sin (e+f x))}}\right ) \, dx\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}+2 \int \frac {\sec ^2(e+f x) \tan ^2(e+f x)}{\sqrt {a (1+\sin (e+f x))}} \, dx-\int \frac {\sec ^4(e+f x)}{\sqrt {a (1+\sin (e+f x))}} \, dx\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}-\frac {5 \sec ^3(e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}+\frac {\int \sec ^4(e+f x) \sqrt {a+a \sin (e+f x)} \left (-\frac {a}{2}+4 a \sin (e+f x)\right ) \, dx}{2 a^2}-\frac {1}{6} (7 a) \int \frac {\sec ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}+\frac {7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac {5 \sec ^3(e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}+\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{12 a f}-\frac {13}{24} \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx-\frac {35}{48} \int \frac {\sec ^2(e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}+\frac {7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac {61 \sec (e+f x)}{48 f \sqrt {a+a \sin (e+f x)}}-\frac {5 \sec ^3(e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}+\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{12 a f}-\frac {1}{16} (13 a) \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx-\frac {1}{32} (35 a) \int \frac {1}{(a+a \sin (e+f x))^{3/2}} \, dx\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}+\frac {61 a \cos (e+f x)}{64 f (a+a \sin (e+f x))^{3/2}}+\frac {7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac {61 \sec (e+f x)}{48 f \sqrt {a+a \sin (e+f x)}}-\frac {5 \sec ^3(e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}+\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{12 a f}-\frac {13}{64} \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx-\frac {35}{128} \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx\\ &=-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}+\frac {61 a \cos (e+f x)}{64 f (a+a \sin (e+f x))^{3/2}}+\frac {7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac {61 \sec (e+f x)}{48 f \sqrt {a+a \sin (e+f x)}}-\frac {5 \sec ^3(e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}+\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{12 a f}+\frac {13 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{32 f}+\frac {35 \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{64 f}\\ &=\frac {61 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{64 \sqrt {2} \sqrt {a} f}-\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}+\frac {61 a \cos (e+f x)}{64 f (a+a \sin (e+f x))^{3/2}}+\frac {7 a \sec (e+f x)}{24 f (a+a \sin (e+f x))^{3/2}}-\frac {61 \sec (e+f x)}{48 f \sqrt {a+a \sin (e+f x)}}-\frac {5 \sec ^3(e+f x)}{6 f \sqrt {a+a \sin (e+f x)}}+\frac {7 \sec ^3(e+f x) \sqrt {a+a \sin (e+f x)}}{12 a f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.47, size = 118, normalized size = 0.79 \begin {gather*} \frac {(804+804 i) (-1)^{3/4} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )-\sec ^3(e+f x) (90+122 \cos (2 (e+f x))-41 \sin (e+f x)+183 \sin (3 (e+f x)))}{768 f \sqrt {a (1+\sin (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((804 + 804*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2]) - Sec[e + f*x]^3*(90 + 122*Cos[2*(e + f*x)] - 41*Sin[e + f*x] + 183*Sin[3*(e + f*x)]))/(768*f*Sqrt[a
*(1 + Sin[e + f*x])])

________________________________________________________________________________________

Maple [A]
time = 2.34, size = 231, normalized size = 1.54

method	result	size
default	\(\frac {366 a^{\frac {7}{2}} \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\left (402 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a^{2}-112 a^{\frac {7}{2}}\right ) \sin \left (f x +e \right )+\left (-201 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a^{2}+122 a^{\frac {7}{2}}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+402 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} a^{2}-16 a^{\frac {7}{2}}}{384 a^{\frac {7}{2}} \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(366*a^(7/2)*sin(f*x+e)*cos(f*x+e)^2+(402*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*(a
-a*sin(f*x+e))^(3/2)*a^2-112*a^(7/2))*sin(f*x+e)+(-201*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1
/2))*(a-a*sin(f*x+e))^(3/2)*a^2+122*a^(7/2))*cos(f*x+e)^2+402*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/
2)/a^(1/2))*(a-a*sin(f*x+e))^(3/2)*a^2-16*a^(7/2))/a^(7/2)/(sin(f*x+e)-1)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f
*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(f*x + e)^4/sqrt(a*sin(f*x + e) + a), x)

________________________________________________________________________________________

Fricas [A]
time = 0.37, size = 250, normalized size = 1.67 \begin {gather*} \frac {201 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + \cos \left (f x + e\right )^{3}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (61 \, \cos \left (f x + e\right )^{2} + {\left (183 \, \cos \left (f x + e\right )^{2} - 56\right )} \sin \left (f x + e\right ) - 8\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{768 \, {\left (a f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) + a f \cos \left (f x + e\right )^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(201*sqrt(2)*(cos(f*x + e)^3*sin(f*x + e) + cos(f*x + e)^3)*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*s
qrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*
sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) - 4*(61*cos(f*x + e
)^2 + (183*cos(f*x + e)^2 - 56)*sin(f*x + e) - 8)*sqrt(a*sin(f*x + e) + a))/(a*f*cos(f*x + e)^3*sin(f*x + e) +
a*f*cos(f*x + e)^3)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{4}{\left (e + f x \right )}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**4/(a+a*sin(f*x+e))**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 42.65, size = 229, normalized size = 1.53 \begin {gather*} -\frac {\frac {201 \, \sqrt {2} \log \left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {201 \, \sqrt {2} \log \left (-\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {6 \, \sqrt {2} {\left (21 \, \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 19 \, \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} \sqrt {a} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {16 \, \sqrt {2} {\left (15 \, \sqrt {a} \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - \sqrt {a}\right )}}{a \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (\frac {3}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}}{768 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(201*sqrt(2)*log(sin(3/4*pi + 1/2*f*x + 1/2*e) + 1)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) - 201
*sqrt(2)*log(-sin(3/4*pi + 1/2*f*x + 1/2*e) + 1)/(sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + 6*sqrt(2)*(21
*sin(3/4*pi + 1/2*f*x + 1/2*e)^3 - 19*sin(3/4*pi + 1/2*f*x + 1/2*e))/((sin(3/4*pi + 1/2*f*x + 1/2*e)^2 - 1)^2*
sqrt(a)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))) + 16*sqrt(2)*(15*sqrt(a)*sin(3/4*pi + 1/2*f*x + 1/2*e)^2 - sqrt(a
))/(a*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(3/4*pi + 1/2*f*x + 1/2*e)^3))/f

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{\sqrt {a+a\,\sin \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________