3.754 \(\int \sec (c+d x) (a+a \sin (c+d x)) \tan (c+d x) \, dx\)

Optimal. Leaf size=27 \[ \frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d}-a x \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2838, 2606, 8, 3473} \[ \frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d}-a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

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 2838

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 36, normalized size = 1.33 \[ -\frac {a \tan ^{-1}(\tan (c+d x))}{d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.45, size = 60, normalized size = 2.22 \[ -\frac {a d x + {\left (a d x - a\right )} \cos \left (d x + c\right ) - {\left (a d x + a\right )} \sin \left (d x + c\right ) - a}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]

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="fricas")

[Out]

-(a*d*x + (a*d*x - a)*cos(d*x + c) - (a*d*x + a)*sin(d*x + c) - a)/(d*cos(d*x + c) - d*sin(d*x + c) + d)

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giac [A]  time = 0.16, size = 29, normalized size = 1.07 \[ -\frac {{\left (d x + c\right )} a + \frac {2 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

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]

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

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maple [A]  time = 0.14, size = 32, normalized size = 1.19 \[ \frac {a \left (\tan \left (d x +c \right )-d x -c \right )+\frac {a}{\cos \left (d x +c \right )}}{d} \]

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]

1/d*(a*(tan(d*x+c)-d*x-c)+a/cos(d*x+c))

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maxima [A]  time = 0.41, size = 32, normalized size = 1.19 \[ -\frac {{\left (d x + c - \tan \left (d x + c\right )\right )} a - \frac {a}{\cos \left (d x + c\right )}}{d} \]

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]

-((d*x + c - tan(d*x + c))*a - a/cos(d*x + c))/d

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mupad [B]  time = 8.98, size = 24, normalized size = 0.89 \[ -a\,x-\frac {2\,a}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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)

[Out]

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

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