3.1.99 \(\int (a+a \sin (e+f x))^{5/2} \tan ^4(e+f x) \, dx\) [99]

Optimal. Leaf size=151 \[ -\frac {2 a^5 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^{5/2}}+\frac {8 a^4 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^{3/2}}-\frac {12 a^3 \cos (e+f x)}{f \sqrt {a+a \sin (e+f x)}}-\frac {8 a^2 \sec (e+f x) \sqrt {a+a \sin (e+f x)}}{f}+\frac {2 a \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f} \]

[Out]

-2/5*a^5*cos(f*x+e)^5/f/(a+a*sin(f*x+e))^(5/2)+8/3*a^4*cos(f*x+e)^3/f/(a+a*sin(f*x+e))^(3/2)+2/3*a*sec(f*x+e)^
3*(a+a*sin(f*x+e))^(3/2)/f-12*a^3*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-8*a^2*sec(f*x+e)*(a+a*sin(f*x+e))^(1/2)/
f

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Rubi [A]
time = 0.67, antiderivative size = 208, normalized size of antiderivative = 1.38, number of steps used = 10, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2793, 2726, 2725, 4486, 2752, 2957, 2934} \begin {gather*} -\frac {64 a^3 \cos (e+f x)}{15 f \sqrt {a \sin (e+f x)+a}}-\frac {16 a^2 \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{15 f}-\frac {46 a^2 \sec (e+f x) \sqrt {a \sin (e+f x)+a}}{3 f}-\frac {2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}-\frac {4 \sec ^3(e+f x) (a \sin (e+f x)+a)^{7/2}}{a f}+\frac {26 \sec ^3(e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f}-\frac {2 a \sec ^3(e+f x) (a \sin (e+f x)+a)^{3/2}}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)*Tan[e + f*x]^4,x]

[Out]

(-64*a^3*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) - (16*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*f)
- (46*a^2*Sec[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(3*f) - (2*a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(5*f)
- (2*a*Sec[e + f*x]^3*(a + a*Sin[e + f*x])^(3/2))/(3*f) + (26*Sec[e + f*x]^3*(a + a*Sin[e + f*x])^(5/2))/(3*f)
- (4*Sec[e + f*x]^3*(a + a*Sin[e + f*x])^(7/2))/(a*f)

Rule 2725

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

Rule 2726

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

Rule 2752

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 - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

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 2957

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

Rule 4486

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

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{5/2} \tan ^4(e+f x) \, dx &=\int (a+a \sin (e+f x))^{5/2} \, dx-\int \sec ^4(e+f x) (a+a \sin (e+f x))^{5/2} \left (1-2 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac {2 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}+\frac {1}{5} (8 a) \int (a+a \sin (e+f x))^{3/2} \, dx-\int \left (\sec ^4(e+f x) (a (1+\sin (e+f x)))^{5/2}-2 \sec ^2(e+f x) (a (1+\sin (e+f x)))^{5/2} \tan ^2(e+f x)\right ) \, dx\\ &=-\frac {16 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}+2 \int \sec ^2(e+f x) (a (1+\sin (e+f x)))^{5/2} \tan ^2(e+f x) \, dx+\frac {1}{15} \left (32 a^2\right ) \int \sqrt {a+a \sin (e+f x)} \, dx-\int \sec ^4(e+f x) (a (1+\sin (e+f x)))^{5/2} \, dx\\ &=-\frac {64 a^3 \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac {2 a \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}-\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{7/2}}{a f}+\frac {4 \int \sec ^4(e+f x) (a+a \sin (e+f x))^{5/2} \left (\frac {7 a}{2}+3 a \sin (e+f x)\right ) \, dx}{a}\\ &=-\frac {64 a^3 \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {2 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac {2 a \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}+\frac {26 \sec ^3(e+f x) (a+a \sin (e+f x))^{5/2}}{3 f}-\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{7/2}}{a f}-\frac {1}{3} (23 a) \int \sec ^2(e+f x) (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac {64 a^3 \cos (e+f x)}{15 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{15 f}-\frac {46 a^2 \sec (e+f x) \sqrt {a+a \sin (e+f x)}}{3 f}-\frac {2 a \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{5 f}-\frac {2 a \sec ^3(e+f x) (a+a \sin (e+f x))^{3/2}}{3 f}+\frac {26 \sec ^3(e+f x) (a+a \sin (e+f x))^{5/2}}{3 f}-\frac {4 \sec ^3(e+f x) (a+a \sin (e+f x))^{7/2}}{a f}\\ \end {align*}

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Mathematica [A]
time = 5.31, size = 112, normalized size = 0.74 \begin {gather*} \frac {a^2 \sqrt {a (1+\sin (e+f x))} (-1225+204 \cos (2 (e+f x))-3 \cos (4 (e+f x))+1488 \sin (e+f x)+16 \sin (3 (e+f x)))}{60 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)*Tan[e + f*x]^4,x]

[Out]

(a^2*Sqrt[a*(1 + Sin[e + f*x])]*(-1225 + 204*Cos[2*(e + f*x)] - 3*Cos[4*(e + f*x)] + 1488*Sin[e + f*x] + 16*Si
n[3*(e + f*x)]))/(60*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]
time = 1.56, size = 87, normalized size = 0.58

method	result	size
default	\(\frac {2 a^{3} \left (1+\sin \left (f x +e \right )\right ) \left (3 \left (\sin ^{4}\left (f x +e \right )\right )+8 \left (\sin ^{3}\left (f x +e \right )\right )+48 \left (\sin ^{2}\left (f x +e \right )\right )-192 \sin \left (f x +e \right )+128\right )}{15 \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^4,x,method=_RETURNVERBOSE)

[Out]

2/15*a^3*(1+sin(f*x+e))/(sin(f*x+e)-1)*(3*sin(f*x+e)^4+8*sin(f*x+e)^3+48*sin(f*x+e)^2-192*sin(f*x+e)+128)/cos(
f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 301 vs. \(2 (145) = 290\).
time = 0.55, size = 301, normalized size = 1.99 \begin {gather*} \frac {32 \, {\left (8 \, a^{\frac {5}{2}} - \frac {24 \, a^{\frac {5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {44 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {68 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {75 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {68 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {44 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {24 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {8 \, a^{\frac {5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )}}{15 \, f {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^4,x, algorithm="maxima")

[Out]

32/15*(8*a^(5/2) - 24*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 44*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
- 68*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 75*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 68*a^(5/2
)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 44*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 24*a^(5/2)*sin(f*x +
e)^7/(cos(f*x + e) + 1)^7 + 8*a^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)/(f*(3*sin(f*x + e)/(cos(f*x + e) +
1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)*(sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 1)^(5/2))

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Fricas [A]
time = 0.36, size = 106, normalized size = 0.70 \begin {gather*} \frac {2 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{4} - 54 \, a^{2} \cos \left (f x + e\right )^{2} + 179 \, a^{2} - 8 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + 23 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{15 \, {\left (f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - f \cos \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*tan(f*x+e)^4,x, algorithm="fricas")

[Out]

2/15*(3*a^2*cos(f*x + e)^4 - 54*a^2*cos(f*x + e)^2 + 179*a^2 - 8*(a^2*cos(f*x + e)^2 + 23*a^2)*sin(f*x + e))*s
qrt(a*sin(f*x + e) + a)/(f*cos(f*x + e)*sin(f*x + e) - f*cos(f*x + e))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(5/2)*tan(f*x+e)**4,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 1580 vs. \(2 (145) = 290\).
time = 162.75, size = 1580, normalized size = 10.46 \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))^(5/2)*tan(f*x+e)^4,x, algorithm="giac")

[Out]

-1/122880*sqrt(2)*(2560*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^16 - 79560*pi*a
^2*floor(-1/8*(pi - 2*f*x - 2*e)/pi + 1/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^
13 - 79560*pi*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(tan(-1/8*pi + 1/4*f*x + 1/4*e))*tan(-1/8*pi + 1/4*f*
x + 1/4*e)^13 - 9945*(pi - 2*f*x - 2*e)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)
^13 + 79560*a^2*arctan(1/tan(-1/8*pi + 1/4*f*x + 1/4*e))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4
*f*x + 1/4*e)^13 - 225280*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^14 - 397800*p
i*a^2*floor(-1/8*(pi - 2*f*x - 2*e)/pi + 1/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*
e)^11 - 397800*pi*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(tan(-1/8*pi + 1/4*f*x + 1/4*e))*tan(-1/8*pi + 1/
4*f*x + 1/4*e)^11 - 49725*(pi - 2*f*x - 2*e)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1
/4*e)^11 + 397800*a^2*arctan(1/tan(-1/8*pi + 1/4*f*x + 1/4*e))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi
+ 1/4*f*x + 1/4*e)^11 - 4352000*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^12 - 7
95600*pi*a^2*floor(-1/8*(pi - 2*f*x - 2*e)/pi + 1/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x
+ 1/4*e)^9 - 795600*pi*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(tan(-1/8*pi + 1/4*f*x + 1/4*e))*tan(-1/8*p
i + 1/4*f*x + 1/4*e)^9 - 99450*(pi - 2*f*x - 2*e)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*
x + 1/4*e)^9 + 795600*a^2*arctan(1/tan(-1/8*pi + 1/4*f*x + 1/4*e))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/
8*pi + 1/4*f*x + 1/4*e)^9 - 10096640*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^10
- 795600*pi*a^2*floor(-1/8*(pi - 2*f*x - 2*e)/pi + 1/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4
*f*x + 1/4*e)^7 - 795600*pi*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(tan(-1/8*pi + 1/4*f*x + 1/4*e))*tan(-1
/8*pi + 1/4*f*x + 1/4*e)^7 - 99450*(pi - 2*f*x - 2*e)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/
4*f*x + 1/4*e)^7 + 795600*a^2*arctan(1/tan(-1/8*pi + 1/4*f*x + 1/4*e))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan
(-1/8*pi + 1/4*f*x + 1/4*e)^7 - 18236416*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e
)^8 - 397800*pi*a^2*floor(-1/8*(pi - 2*f*x - 2*e)/pi + 1/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi +
1/4*f*x + 1/4*e)^5 - 397800*pi*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(tan(-1/8*pi + 1/4*f*x + 1/4*e))*tan
(-1/8*pi + 1/4*f*x + 1/4*e)^5 - 49725*(pi - 2*f*x - 2*e)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi +
1/4*f*x + 1/4*e)^5 + 397800*a^2*arctan(1/tan(-1/8*pi + 1/4*f*x + 1/4*e))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*
tan(-1/8*pi + 1/4*f*x + 1/4*e)^5 - 10096640*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/
4*e)^6 - 79560*pi*a^2*floor(-1/8*(pi - 2*f*x - 2*e)/pi + 1/2)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi
+ 1/4*f*x + 1/4*e)^3 - 79560*pi*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(tan(-1/8*pi + 1/4*f*x + 1/4*e))*ta
n(-1/8*pi + 1/4*f*x + 1/4*e)^3 - 9945*(pi - 2*f*x - 2*e)*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi +
1/4*f*x + 1/4*e)^3 + 79560*a^2*arctan(1/tan(-1/8*pi + 1/4*f*x + 1/4*e))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*t
an(-1/8*pi + 1/4*f*x + 1/4*e)^3 - 4352000*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*
e)^4 - 225280*a^2*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*tan(-1/8*pi + 1/4*f*x + 1/4*e)^2 + 2560*a^2*sgn(cos(-1/4
*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/((tan(-1/8*pi + 1/4*f*x + 1/4*e)^13 + 5*tan(-1/8*pi + 1/4*f*x + 1/4*e)^11 + 1
0*tan(-1/8*pi + 1/4*f*x + 1/4*e)^9 + 10*tan(-1/8*pi + 1/4*f*x + 1/4*e)^7 + 5*tan(-1/8*pi + 1/4*f*x + 1/4*e)^5
+ tan(-1/8*pi + 1/4*f*x + 1/4*e)^3)*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 )}^4\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2} \,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))^(5/2),x)

[Out]

int(tan(e + f*x)^4*(a + a*sin(e + f*x))^(5/2), x)

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