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

Optimal. Leaf size=44 \[ \frac {\sin (c+d x)}{a d}-\frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3872, 2839, 2637, 2635, 8} \[ \frac {\sin (c+d x)}{a d}-\frac {\sin (c+d x) \cos (c+d x)}{2 a d}-\frac {x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-x/(2*a) + Sin[c + d*x]/(a*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*a*d)

Rule 8

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

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 68, normalized size = 1.55 \[ -\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (-4 \sin (c+d x)+\sin (2 (c+d x))-c+\tan \left (\frac {c}{2}\right )+2 d x\right )}{2 a d (\sec (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(Cos[(c + d*x)/2]^2*Sec[c + d*x]*(-c + 2*d*x - 4*Sin[c + d*x] + Sin[2*(c + d*x)] + Tan[c/2]))/(a*d*(1 + S
ec[c + d*x]))

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fricas [A]  time = 1.69, size = 27, normalized size = 0.61 \[ -\frac {d x + {\left (\cos \left (d x + c\right ) - 2\right )} \sin \left (d x + c\right )}{2 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 58, normalized size = 1.32 \[ -\frac {\frac {d x + c}{a} - \frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.33, size = 85, normalized size = 1.93 \[ \frac {3 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.45, size = 112, normalized size = 2.55 \[ \frac {\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.96, size = 30, normalized size = 0.68 \[ -\frac {\sin \left (2\,c+2\,d\,x\right )-4\,\sin \left (c+d\,x\right )+2\,d\,x}{4\,a\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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