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

Optimal. Leaf size=50 \[ \frac{\tan ^3(c+d x)}{3 a d}-\frac{\sec ^3(c+d x)}{3 a d}+\frac{\sec (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.0876481, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2706, 2607, 30, 2606} \[ \frac{\tan ^3(c+d x)}{3 a d}-\frac{\sec ^3(c+d x)}{3 a d}+\frac{\sec (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

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 30

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

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])

Rubi steps

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

Mathematica [B]  time = 0.164187, size = 106, normalized size = 2.12 \[ \frac{8 \sin (c+d x)-5 \sin (2 (c+d x))-10 \cos (c+d x)+2 \cos (2 (c+d x))+6}{12 a d (\sin (c+d x)+1) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6 - 10*Cos[c + d*x] + 2*Cos[2*(c + d*x)] + 8*Sin[c + d*x] - 5*Sin[2*(c + d*x)])/(12*a*d*(Cos[(c + d*x)/2] - S
in[(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 = 0.047, size = 70, normalized size = 1.4 \begin{align*} 8\,{\frac{1}{da} \left ( -1/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-1}-1/12\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-3}+1/8\, \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(tan(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

8/d/a*(-1/16/(tan(1/2*d*x+1/2*c)-1)-1/12/(tan(1/2*d*x+1/2*c)+1)^3+1/8/(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 [A]  time = 1.09485, size = 122, normalized size = 2.44 \begin{align*} \frac{4 \,{\left (\frac{2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}}{3 \,{\left (a + \frac{2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*(2*sin(d*x + c)/(cos(d*x + c) + 1) + 1)/((a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1) - 2*a*sin(d*x + c)^3/(co
s(d*x + c) + 1)^3 - a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4)*d)

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Fricas [A]  time = 1.37095, size = 127, normalized size = 2.54 \begin{align*} \frac{\cos \left (d x + c\right )^{2} + 2 \, \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(tan(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/3*(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.65929, size = 92, normalized size = 1.84 \begin{align*} -\frac{\frac{3}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}} - \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5}{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(tan(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(3/(a*(tan(1/2*d*x + 1/2*c) - 1)) - (3*tan(1/2*d*x + 1/2*c)^2 + 12*tan(1/2*d*x + 1/2*c) + 5)/(a*(tan(1/2*
d*x + 1/2*c) + 1)^3))/d