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

Optimal. Leaf size=83 \[ \frac{\cos (c+d x)}{a d}+\frac{\tan ^3(c+d x)}{3 a d}-\frac{\tan (c+d x)}{a d}-\frac{\sec ^3(c+d x)}{3 a d}+\frac{2 \sec (c+d x)}{a d}+\frac{x}{a} \]

[Out]

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

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Rubi [A]  time = 0.142375, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2839, 3473, 8, 2590, 270} \[ \frac{\cos (c+d x)}{a d}+\frac{\tan ^3(c+d x)}{3 a d}-\frac{\tan (c+d x)}{a d}-\frac{\sec ^3(c+d x)}{3 a d}+\frac{2 \sec (c+d x)}{a d}+\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2839

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

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.377016, size = 148, normalized size = 1.78 \[ \frac{11 \sin (c+d x)+6 c \sin (2 (c+d x))+6 d x \sin (2 (c+d x))-11 \sin (2 (c+d x))+3 \sin (3 (c+d x))+2 (6 c+6 d x-11) \cos (c+d x)+14 \cos (2 (c+d x))+18}{12 a d (\sin (c+d x)+1) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(18 + 2*(-11 + 6*c + 6*d*x)*Cos[c + d*x] + 14*Cos[2*(c + d*x)] + 11*Sin[c + d*x] - 11*Sin[2*(c + d*x)] + 6*c*S
in[2*(c + d*x)] + 6*d*x*Sin[2*(c + d*x)] + 3*Sin[3*(c + d*x)])/(12*a*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(1 + Sin[c + d*x]))

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Maple [A]  time = 0.081, size = 126, normalized size = 1.5 \begin{align*} -{\frac{1}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{1}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{2}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{5}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c)),x)

[Out]

-1/2/d/a/(tan(1/2*d*x+1/2*c)-1)+2/a/d/(1+tan(1/2*d*x+1/2*c)^2)+2/a/d*arctan(tan(1/2*d*x+1/2*c))-2/3/d/a/(tan(1
/2*d*x+1/2*c)+1)^3+1/d/a/(tan(1/2*d*x+1/2*c)+1)^2+5/2/a/d/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.49787, size = 319, normalized size = 3.84 \begin{align*} \frac{2 \,{\left (\frac{\frac{13 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{2 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{6 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 8}{a + \frac{2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{2 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac{3 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

2/3*((13*sin(d*x + c)/(cos(d*x + c) + 1) + 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 2*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 - 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 8)/(a + 2*a*sin(d
*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 2
*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + 3*arctan(sin(d*x + c)/(cos(d
*x + c) + 1))/a)/d

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Fricas [A]  time = 1.05527, size = 215, normalized size = 2.59 \begin{align*} \frac{3 \, d x \cos \left (d x + c\right ) + 7 \, \cos \left (d x + c\right )^{2} +{\left (3 \, d x \cos \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right ) + 1}{3 \,{\left (a d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a 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)^4/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(3*d*x*cos(d*x + c) + 7*cos(d*x + c)^2 + (3*d*x*cos(d*x + c) + 3*cos(d*x + c)^2 + 2)*sin(d*x + c) + 1)/(a*
d*cos(d*x + c)*sin(d*x + c) + a*d*cos(d*x + c))

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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(sec(d*x+c)**2*sin(d*x+c)**4/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.16784, size = 169, normalized size = 2.04 \begin{align*} \frac{\frac{6 \,{\left (d x + c\right )}}{a} - \frac{3 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )} a} + \frac{15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 36 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 17}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)/a - 3*(tan(1/2*d*x + 1/2*c)^2 - 4*tan(1/2*d*x + 1/2*c) + 5)/((tan(1/2*d*x + 1/2*c)^3 - tan(1/
2*d*x + 1/2*c)^2 + tan(1/2*d*x + 1/2*c) - 1)*a) + (15*tan(1/2*d*x + 1/2*c)^2 + 36*tan(1/2*d*x + 1/2*c) + 17)/(
a*(tan(1/2*d*x + 1/2*c) + 1)^3))/d