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

Optimal. Leaf size=73 \[ \frac{2 \tan ^5(c+d x)}{5 a^2 d}+\frac{\tan ^3(c+d x)}{3 a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{2 \sec ^3(c+d x)}{3 a^2 d} \]

[Out]

(2*Sec[c + d*x]^3)/(3*a^2*d) - (2*Sec[c + d*x]^5)/(5*a^2*d) + Tan[c + d*x]^3/(3*a^2*d) + (2*Tan[c + d*x]^5)/(5
*a^2*d)

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Rubi [A]  time = 0.189761, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2711, 2607, 14, 2606, 30} \[ \frac{2 \tan ^5(c+d x)}{5 a^2 d}+\frac{\tan ^3(c+d x)}{3 a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{2 \sec ^3(c+d x)}{3 a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sec[c + d*x]^3)/(3*a^2*d) - (2*Sec[c + d*x]^5)/(5*a^2*d) + Tan[c + d*x]^3/(3*a^2*d) + (2*Tan[c + d*x]^5)/(5
*a^2*d)

Rule 2711

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[a^(2*
m), Int[ExpandIntegrand[(g*Tan[e + f*x])^p/Sec[e + f*x]^m, (a*Sec[e + f*x] - b*Tan[e + f*x])^(-m), x], x], x]
/; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[m, 0]

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\tan ^2(c+d x)}{(a+a \sin (c+d x))^2} \, dx &=\frac{\int \left (a^2 \sec ^4(c+d x) \tan ^2(c+d x)-2 a^2 \sec ^3(c+d x) \tan ^3(c+d x)+a^2 \sec ^2(c+d x) \tan ^4(c+d x)\right ) \, dx}{a^4}\\ &=\frac{\int \sec ^4(c+d x) \tan ^2(c+d x) \, dx}{a^2}+\frac{\int \sec ^2(c+d x) \tan ^4(c+d x) \, dx}{a^2}-\frac{2 \int \sec ^3(c+d x) \tan ^3(c+d x) \, dx}{a^2}\\ &=\frac{\operatorname{Subst}\left (\int x^4 \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac{\operatorname{Subst}\left (\int x^2 \left (1+x^2\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac{2 \operatorname{Subst}\left (\int x^2 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac{\tan ^5(c+d x)}{5 a^2 d}+\frac{\operatorname{Subst}\left (\int \left (x^2+x^4\right ) \, dx,x,\tan (c+d x)\right )}{a^2 d}-\frac{2 \operatorname{Subst}\left (\int \left (-x^2+x^4\right ) \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=\frac{2 \sec ^3(c+d x)}{3 a^2 d}-\frac{2 \sec ^5(c+d x)}{5 a^2 d}+\frac{\tan ^3(c+d x)}{3 a^2 d}+\frac{2 \tan ^5(c+d x)}{5 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.236509, size = 86, normalized size = 1.18 \[ -\frac{\sec (c+d x) \left (-35 \sin (c+d x)+11 \sin (2 (c+d x))+\sin (3 (c+d x))+\frac{55}{4} \cos (c+d x)+4 \cos (2 (c+d x))-\frac{11}{4} \cos (3 (c+d x))-20\right )}{60 a^2 d (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]*(-20 + (55*Cos[c + d*x])/4 + 4*Cos[2*(c + d*x)] - (11*Cos[3*(c + d*x)])/4 - 35*Sin[c + d*x] + 1
1*Sin[2*(c + d*x)] + Sin[3*(c + d*x)]))/(60*a^2*d*(1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.089, size = 100, normalized size = 1.4 \begin{align*} 8\,{\frac{1}{d{a}^{2}} \left ( -1/32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-1}-1/10\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-5}+1/4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}-{\frac{5}{24\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}+1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-2}+1/32\, \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)^2/(a+a*sin(d*x+c))^2,x)

[Out]

8/d/a^2*(-1/32/(tan(1/2*d*x+1/2*c)-1)-1/10/(tan(1/2*d*x+1/2*c)+1)^5+1/4/(tan(1/2*d*x+1/2*c)+1)^4-5/24/(tan(1/2
*d*x+1/2*c)+1)^3+1/16/(tan(1/2*d*x+1/2*c)+1)^2+1/32/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.02728, size = 248, normalized size = 3.4 \begin{align*} \frac{8 \,{\left (\frac{4 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{5 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 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)^2/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

8/15*(4*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 5*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 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 = 1.05302, size = 198, normalized size = 2.71 \begin{align*} \frac{2 \, \cos \left (d x + c\right )^{2} +{\left (\cos \left (d x + c\right )^{2} - 9\right )} \sin \left (d x + c\right ) - 6}{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)^2/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/15*(2*cos(d*x + c)^2 + (cos(d*x + c)^2 - 9)*sin(d*x + c) - 6)/(a^2*d*cos(d*x + c)^3 - 2*a^2*d*cos(d*x + c)*s
in(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 ^{2}{\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)**2/(a+a*sin(d*x+c))**2,x)

[Out]

Integral(sin(c + d*x)**2*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 = 1.21631, size = 127, normalized size = 1.74 \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} + 90 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 80 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 70 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 17}{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)^2/(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 + 90*tan(1/2*d*x + 1/2*c)^3 + 80*tan(1
/2*d*x + 1/2*c)^2 + 70*tan(1/2*d*x + 1/2*c) + 17)/(a^2*(tan(1/2*d*x + 1/2*c) + 1)^5))/d