3.783 \(\int \frac{\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=71 \[ \frac{4 \tan (c+d x)}{15 a^2 d}-\frac{2 \sec (c+d x)}{15 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2} \]

[Out]

Sec[c + d*x]/(5*d*(a + a*Sin[c + d*x])^2) - (2*Sec[c + d*x])/(15*d*(a^2 + a^2*Sin[c + d*x])) + (4*Tan[c + d*x]
)/(15*a^2*d)

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Rubi [A]  time = 0.128079, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2859, 2672, 3767, 8} \[ \frac{4 \tan (c+d x)}{15 a^2 d}-\frac{2 \sec (c+d x)}{15 d \left (a^2 \sin (c+d x)+a^2\right )}+\frac{\sec (c+d x)}{5 d (a \sin (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]*Tan[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

Sec[c + d*x]/(5*d*(a + a*Sin[c + d*x])^2) - (2*Sec[c + d*x])/(15*d*(a^2 + a^2*Sin[c + d*x])) + (4*Tan[c + d*x]
)/(15*a^2*d)

Rule 2859

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*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 2672

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)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[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] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{\sec (c+d x) \tan (c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}+\frac{2 \int \frac{\sec ^2(c+d x)}{a+a \sin (c+d x)} \, dx}{5 a}\\ &=\frac{\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}-\frac{2 \sec (c+d x)}{15 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{4 \int \sec ^2(c+d x) \, dx}{15 a^2}\\ &=\frac{\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}-\frac{2 \sec (c+d x)}{15 d \left (a^2+a^2 \sin (c+d x)\right )}-\frac{4 \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 a^2 d}\\ &=\frac{\sec (c+d x)}{5 d (a+a \sin (c+d x))^2}-\frac{2 \sec (c+d x)}{15 d \left (a^2+a^2 \sin (c+d x)\right )}+\frac{4 \tan (c+d x)}{15 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.226141, size = 82, normalized size = 1.15 \[ -\frac{\sec (c+d x) (-80 \sin (c+d x)-4 \sin (2 (c+d x))+16 \sin (3 (c+d x))-5 \cos (c+d x)+64 \cos (2 (c+d x))+\cos (3 (c+d x))-80)}{240 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[c + d*x]*Tan[c + d*x])/(a + a*Sin[c + d*x])^2,x]

[Out]

-(Sec[c + d*x]*(-80 - 5*Cos[c + d*x] + 64*Cos[2*(c + d*x)] + Cos[3*(c + d*x)] - 80*Sin[c + d*x] - 4*Sin[2*(c +
d*x)] + 16*Sin[3*(c + d*x)]))/(240*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.082, size = 100, normalized size = 1.4 \begin{align*} 4\,{\frac{1}{d{a}^{2}} \left ( -1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-1}+1/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}-1/2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}+{\frac{7}{12\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}-3/8\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}+1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)/(a+a*sin(d*x+c))^2,x)

[Out]

4/d/a^2*(-1/16/(tan(1/2*d*x+1/2*c)-1)+1/5/(tan(1/2*d*x+1/2*c)+1)^5-1/2/(tan(1/2*d*x+1/2*c)+1)^4+7/12/(tan(1/2*
d*x+1/2*c)+1)^3-3/8/(tan(1/2*d*x+1/2*c)+1)^2+1/16/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.03107, size = 275, normalized size = 3.87 \begin{align*} \frac{2 \,{\left (\frac{4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{20 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}}{15 \,{\left (a^{2} + \frac{4 \, a^{2} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{5 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 \, a^{2} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

2/15*(4*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 20*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)/((a^2 + 4*a^2*sin(d*x + c)/(cos(d*x + c) + 1) + 5*a^
2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 5*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a^2*sin(d*x + c)^5/(cos(
d*x + c) + 1)^5 - a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*d)

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Fricas [A]  time = 1.01259, size = 204, normalized size = 2.87 \begin{align*} \frac{8 \, \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, \cos \left (d x + c\right )^{2} - 3\right )} \sin \left (d x + c\right ) - 9}{15 \,{\left (a^{2} d \cos \left (d x + c\right )^{3} - 2 \, a^{2} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} d \cos \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/15*(8*cos(d*x + c)^2 + 2*(2*cos(d*x + c)^2 - 3)*sin(d*x + c) - 9)/(a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x +
c)*sin(d*x + c) - 2*a^2*d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\sin{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} + 2 \sin{\left (c + d x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(sin(c + d*x)*sec(c + d*x)**2/(sin(c + d*x)**2 + 2*sin(c + d*x) + 1), x)/a**2

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Giac [A]  time = 1.25893, size = 127, normalized size = 1.79 \begin{align*} -\frac{\frac{15}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}} - \frac{15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 50 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/60*(15/(a^2*(tan(1/2*d*x + 1/2*c) - 1)) - (15*tan(1/2*d*x + 1/2*c)^4 - 30*tan(1/2*d*x + 1/2*c)^3 - 40*tan(1
/2*d*x + 1/2*c)^2 - 50*tan(1/2*d*x + 1/2*c) - 7)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d