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

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

[Out]

(-2*a^5*Cos[e + f*x]^5)/(5*f*(a + a*Sin[e + f*x])^(5/2)) + (8*a^4*Cos[e + f*x]^3)/(3*f*(a + a*Sin[e + f*x])^(3
/2)) - (12*a^3*Cos[e + f*x])/(f*Sqrt[a + a*Sin[e + f*x]]) - (8*a^2*Sec[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/f +
(2*a*Sec[e + f*x]^3*(a + a*Sin[e + f*x])^(3/2))/(3*f)

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Rubi [A]  time = 0.979098, 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 = {2714, 2647, 2646, 4401, 2673, 2878, 2855} \[ -\frac{16 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}-\frac{64 a^3 \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{46 a^2 \sec (e+f x) \sqrt{a \sin (e+f x)+a}}{3 f}-\frac{2 a \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{5 f}-\frac{4 \sec ^3(e+f x) (a \sin (e+f x)+a)^{7/2}}{a f}+\frac{26 \sec ^3(e+f x) (a \sin (e+f x)+a)^{5/2}}{3 f}-\frac{2 a \sec ^3(e+f x) (a \sin (e+f x)+a)^{3/2}}{3 f} \]

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 2714

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 2647

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] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2646

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 4401

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

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((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 - 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 2878

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 2855

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]

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*}

Mathematica [A]  time = 5.45924, size = 112, normalized size = 0.74 \[ \frac{a^2 \sqrt{a (\sin (e+f x)+1)} (1488 \sin (e+f x)+16 \sin (3 (e+f x))+204 \cos (2 (e+f x))-3 \cos (4 (e+f x))-1225)}{60 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

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 = 0.404, size = 87, normalized size = 0.6 \begin{align*}{\frac{2\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) \left ( 3\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}+8\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}+48\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}-192\,\sin \left ( fx+e \right ) +128 \right ) }{ \left ( -15+15\,\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

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)

[Out]

2/15*a^3*(1+sin(f*x+e))/(-1+sin(f*x+e))*(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]  time = 1.76375, size = 374, normalized size = 2.48 \begin{align*} \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{align*}

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 = 1.43246, size = 246, normalized size = 1.63 \begin{align*} \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{align*}

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)]  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))**(5/2)*tan(f*x+e)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}} \tan \left (f x + e\right )^{4}\,{d x} \end{align*}

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]

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