3.19 \(\int \sec (c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=34 \[ -\frac{a^2 \sin (c+d x)}{d}-\frac{2 a^2 \log (1-\sin (c+d x))}{d} \]

[Out]

(-2*a^2*Log[1 - Sin[c + d*x]])/d - (a^2*Sin[c + d*x])/d

________________________________________________________________________________________

Rubi [A]  time = 0.0416049, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2667, 43} \[ -\frac{a^2 \sin (c+d x)}{d}-\frac{2 a^2 \log (1-\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*Log[1 - Sin[c + d*x]])/d - (a^2*Sin[c + d*x])/d

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.019072, size = 29, normalized size = 0.85 \[ \frac{a^2 (-\sin (c+d x)-2 \log (1-\sin (c+d x)))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-2*Log[1 - Sin[c + d*x]] - Sin[c + d*x]))/d

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 53, normalized size = 1.6 \begin{align*} 2\,{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*a^2*ln(sec(d*x+c)+tan(d*x+c))-2/d*a^2*ln(cos(d*x+c))-a^2*sin(d*x+c)/d

________________________________________________________________________________________

Maxima [A]  time = 0.939936, size = 41, normalized size = 1.21 \begin{align*} -\frac{2 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + a^{2} \sin \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.64804, size = 73, normalized size = 2.15 \begin{align*} -\frac{2 \, a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + a^{2} \sin \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int 2 \sin{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int \sin ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int \sec{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.18736, size = 123, normalized size = 3.62 \begin{align*} \frac{2 \,{\left (a^{2} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 2 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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