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

Optimal. Leaf size=51 \[ -\frac{a \sin (c+d x)}{d}+\frac{a \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a \sin (c+d x) \cos (c+d x)}{2 d}+\frac{a x}{2} \]

[Out]

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

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Rubi [A]  time = 0.0818246, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {3872, 2838, 2592, 321, 206, 2635, 8} \[ -\frac{a \sin (c+d x)}{d}+\frac{a \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a \sin (c+d x) \cos (c+d x)}{2 d}+\frac{a x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.0520987, size = 54, normalized size = 1.06 \[ \frac{a (c+d x)}{2 d}-\frac{a \sin (c+d x)}{d}-\frac{a \sin (2 (c+d x))}{4 d}+\frac{a \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(c + d*x))/(2*d) + (a*ArcTanh[Sin[c + d*x]])/d - (a*Sin[c + d*x])/d - (a*Sin[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.031, size = 62, normalized size = 1.2 \begin{align*} -{\frac{a\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{ax}{2}}+{\frac{ac}{2\,d}}+{\frac{a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{a\sin \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.989886, size = 80, normalized size = 1.57 \begin{align*} \frac{{\left (2 \, d x + 2 \, c - \sin \left (2 \, d x + 2 \, c\right )\right )} a + 2 \, a{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right ) - 2 \, \sin \left (d x + c\right )\right )}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c - sin(2*d*x + 2*c))*a + 2*a*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) - 2*sin(d*x + c))
)/d

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Fricas [A]  time = 1.73897, size = 143, normalized size = 2.8 \begin{align*} \frac{a d x + a \log \left (\sin \left (d x + c\right ) + 1\right ) - a \log \left (-\sin \left (d x + c\right ) + 1\right ) -{\left (a \cos \left (d x + c\right ) + 2 \, a\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.49379, size = 119, normalized size = 2.33 \begin{align*} \frac{{\left (d x + c\right )} a + 2 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, a \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a \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}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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