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

Optimal. Leaf size=118 \[ \frac{31 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}+\frac{124 a^3 \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}+\frac{9 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{5 f} \]

[Out]

(124*a^3*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) + (31*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*f)
+ (9*Sec[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(5*f) - (2*Sec[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(5*a*f)

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Rubi [A]  time = 0.214287, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {2713, 2855, 2647, 2646} \[ \frac{31 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{15 f}+\frac{124 a^3 \cos (e+f x)}{15 f \sqrt{a \sin (e+f x)+a}}-\frac{2 \sec (e+f x) (a \sin (e+f x)+a)^{7/2}}{5 a f}+\frac{9 \sec (e+f x) (a \sin (e+f x)+a)^{5/2}}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(124*a^3*Cos[e + f*x])/(15*f*Sqrt[a + a*Sin[e + f*x]]) + (31*a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(15*f)
+ (9*Sec[e + f*x]*(a + a*Sin[e + f*x])^(5/2))/(5*f) - (2*Sec[e + f*x]*(a + a*Sin[e + f*x])^(7/2))/(5*a*f)

Rule 2713

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

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]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{5/2} \tan ^2(e+f x) \, dx &=-\frac{2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f}+\frac{2 \int \sec ^2(e+f x) (a+a \sin (e+f x))^{5/2} \left (\frac{7 a}{2}+a \sin (e+f x)\right ) \, dx}{5 a}\\ &=\frac{9 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{5 f}-\frac{2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f}-\frac{1}{10} (31 a) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=\frac{31 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}+\frac{9 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{5 f}-\frac{2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f}-\frac{1}{15} \left (62 a^2\right ) \int \sqrt{a+a \sin (e+f x)} \, dx\\ &=\frac{124 a^3 \cos (e+f x)}{15 f \sqrt{a+a \sin (e+f x)}}+\frac{31 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{15 f}+\frac{9 \sec (e+f x) (a+a \sin (e+f x))^{5/2}}{5 f}-\frac{2 \sec (e+f x) (a+a \sin (e+f x))^{7/2}}{5 a f}\\ \end{align*}

Mathematica [A]  time = 5.46323, size = 60, normalized size = 0.51 \[ \frac{a^2 \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} (-185 \sin (e+f x)+3 \sin (3 (e+f x))+22 \cos (2 (e+f x))+330)}{30 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*(330 + 22*Cos[2*(e + f*x)] - 185*Sin[e + f*x] + 3*Sin[3*(e + f*x)
]))/(30*f)

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Maple [A]  time = 0.412, size = 67, 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 ) ^{3}+11\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+44\,\sin \left ( fx+e \right ) -88 \right ) }{15\,f\cos \left ( fx+e \right ) }{\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)^2,x)

[Out]

-2/15*a^3*(1+sin(f*x+e))*(3*sin(f*x+e)^3+11*sin(f*x+e)^2+44*sin(f*x+e)-88)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [A]  time = 1.66292, size = 258, normalized size = 2.19 \begin{align*} -\frac{8 \,{\left (22 \, a^{\frac{5}{2}} - \frac{22 \, a^{\frac{5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{55 \, a^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{50 \, a^{\frac{5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{55 \, a^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{22 \, a^{\frac{5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{22 \, a^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{15 \, f{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 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)^2,x, algorithm="maxima")

[Out]

-8/15*(22*a^(5/2) - 22*a^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 55*a^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^
2 - 50*a^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 55*a^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 22*a^(5/
2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 22*a^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/(f*(sin(f*x + e)/(cos
(f*x + e) + 1) - 1)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2))

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Fricas [A]  time = 1.42272, size = 173, normalized size = 1.47 \begin{align*} \frac{2 \,{\left (11 \, a^{2} \cos \left (f x + e\right )^{2} + 77 \, a^{2} +{\left (3 \, a^{2} \cos \left (f x + e\right )^{2} - 47 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{15 \, f \cos \left (f x + e\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)^2,x, algorithm="fricas")

[Out]

2/15*(11*a^2*cos(f*x + e)^2 + 77*a^2 + (3*a^2*cos(f*x + e)^2 - 47*a^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)/
(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)**2,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 )^{2}\,{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)^2,x, algorithm="giac")

[Out]

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