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

Optimal. Leaf size=37 \[ \frac {\sec ^3(c+d x)}{3 a d}-\frac {\tan ^3(c+d x)}{3 a d} \]

[Out]

1/3*sec(d*x+c)^3/a/d-1/3*tan(d*x+c)^3/a/d

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Rubi [A]  time = 0.09, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2839, 2606, 30, 2607} \[ \frac {\sec ^3(c+d x)}{3 a d}-\frac {\tan ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [B]  time = 0.14, size = 104, normalized size = 2.81 \[ \frac {-2 \sin (c+d x)+\frac {1}{2} \sin (2 (c+d x))+\cos (c+d x)+\cos (2 (c+d x))-3}{6 a d (\sin (c+d x)+1) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3 + Cos[c + d*x] + Cos[2*(c + d*x)] - 2*Sin[c + d*x] + Sin[2*(c + d*x)]/2)/(6*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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fricas [A]  time = 0.43, size = 47, normalized size = 1.27 \[ -\frac {\cos \left (d x + c\right )^{2} - \sin \left (d x + c\right ) - 2}{3 \, {\left (a d \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )\right )}} \]

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)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 57, normalized size = 1.54 \[ -\frac {\frac {3}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{3}}}{6 \, d} \]

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)),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 + 1)/(a*(tan(1/2*d*x + 1/2*c) + 1)^3))/d

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maple [B]  time = 0.28, size = 70, normalized size = 1.89 \[ \frac {-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {4}{8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8}}{d a} \]

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)),x)

[Out]

4/d/a*(-1/8/(tan(1/2*d*x+1/2*c)-1)+1/6/(tan(1/2*d*x+1/2*c)+1)^3-1/4/(tan(1/2*d*x+1/2*c)+1)^2+1/8/(tan(1/2*d*x+
1/2*c)+1))

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maxima [B]  time = 0.32, size = 110, normalized size = 2.97 \[ \frac {2 \, {\left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{3 \, {\left (a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {2 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}\right )} d} \]

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)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.91, size = 60, normalized size = 1.62 \[ -\frac {2\,\left (3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{3\,a\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

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)),x)

[Out]

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

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