3.214 \(\int \sec ^2(c+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=34 \[ \frac{\sin (a-c) \tanh ^{-1}(\sin (b x+c))}{b}+\frac{\cos (a-c) \sec (b x+c)}{b} \]

[Out]

(Cos[a - c]*Sec[c + b*x])/b + (ArcTanh[Sin[c + b*x]]*Sin[a - c])/b

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Rubi [A]  time = 0.0268219, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {4580, 2606, 8, 3770} \[ \frac{\sin (a-c) \tanh ^{-1}(\sin (b x+c))}{b}+\frac{\cos (a-c) \sec (b x+c)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + b*x]^2*Sin[a + b*x],x]

[Out]

(Cos[a - c]*Sec[c + b*x])/b + (ArcTanh[Sin[c + b*x]]*Sin[a - c])/b

Rule 4580

Int[Sec[w_]^(n_.)*Sin[v_], x_Symbol] :> Dist[Cos[v - w], Int[Tan[w]*Sec[w]^(n - 1), x], x] + Dist[Sin[v - w],
Int[Sec[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]

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 8

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [C]  time = 0.0914404, size = 88, normalized size = 2.59 \[ \frac{\cos (a-c) \sec (b x+c)}{b}-\frac{2 i \sin (a-c) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\sin (c) \cos \left (\frac{b x}{2}\right )+\cos (c) \sin \left (\frac{b x}{2}\right )\right )}{\cos (c) \cos \left (\frac{b x}{2}\right )-i \sin (c) \cos \left (\frac{b x}{2}\right )}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + b*x]^2*Sin[a + b*x],x]

[Out]

(Cos[a - c]*Sec[c + b*x])/b - ((2*I)*ArcTan[((I*Cos[c] + Sin[c])*(Cos[(b*x)/2]*Sin[c] + Cos[c]*Sin[(b*x)/2]))/
(Cos[c]*Cos[(b*x)/2] - I*Cos[(b*x)/2]*Sin[c])]*Sin[a - c])/b

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Maple [B]  time = 0.424, size = 888, normalized size = 26.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*sin(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*sin(c)^2)/(cos(a)*cos(c)*tan(1/2
*b*x+1/2*a)^2+sin(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*sin(c)-2*tan(1/2*b*x+1/2*a)*sin(a
)*cos(c)-cos(a)*cos(c)-sin(a)*sin(c))*tan(1/2*b*x+1/2*a)*cos(a)*sin(c)+8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*si
n(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*sin(c)^2)/(cos(a)*cos(c)*tan(1/2*b*x+1/2*a)^2+sin(a)*sin(c)*tan(1/2*b*x+
1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*sin(c)-2*tan(1/2*b*x+1/2*a)*sin(a)*cos(c)-cos(a)*cos(c)-sin(a)*sin(c))*ta
n(1/2*b*x+1/2*a)*sin(a)*cos(c)+8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*sin(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*si
n(c)^2)/(cos(a)*cos(c)*tan(1/2*b*x+1/2*a)^2+sin(a)*sin(c)*tan(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*sin
(c)-2*tan(1/2*b*x+1/2*a)*sin(a)*cos(c)-cos(a)*cos(c)-sin(a)*sin(c))*cos(a)*cos(c)+8/b/(-4*cos(a)^2*cos(c)^2-4*
cos(a)^2*sin(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*sin(c)^2)/(cos(a)*cos(c)*tan(1/2*b*x+1/2*a)^2+sin(a)*sin(c)*t
an(1/2*b*x+1/2*a)^2+2*tan(1/2*b*x+1/2*a)*cos(a)*sin(c)-2*tan(1/2*b*x+1/2*a)*sin(a)*cos(c)-cos(a)*cos(c)-sin(a)
*sin(c))*sin(a)*sin(c)-8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*sin(c)^2-4*cos(c)^2*sin(a)^2-4*sin(a)^2*sin(c)^2)/
(-cos(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2)*arctan(1/2*(2*(cos(a)*cos(c)+
sin(a)*sin(c))*tan(1/2*b*x+1/2*a)+2*cos(a)*sin(c)-2*sin(a)*cos(c))/(-cos(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c
)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2))*cos(a)*sin(c)+8/b/(-4*cos(a)^2*cos(c)^2-4*cos(a)^2*sin(c)^2-4*cos(c)^2*
sin(a)^2-4*sin(a)^2*sin(c)^2)/(-cos(a)^2*cos(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2)
*arctan(1/2*(2*(cos(a)*cos(c)+sin(a)*sin(c))*tan(1/2*b*x+1/2*a)+2*cos(a)*sin(c)-2*sin(a)*cos(c))/(-cos(a)^2*co
s(c)^2-cos(a)^2*sin(c)^2-cos(c)^2*sin(a)^2-sin(a)^2*sin(c)^2)^(1/2))*sin(a)*cos(c)

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Maxima [B]  time = 2.04218, size = 522, normalized size = 15.35 \begin{align*} \frac{2 \,{\left (\cos \left (b x + 2 \, a\right ) + \cos \left (b x + 2 \, c\right )\right )} \cos \left (2 \, b x + a + 2 \, c\right ) + 2 \, \cos \left (b x + 2 \, a\right ) \cos \left (a\right ) + 2 \, \cos \left (b x + 2 \, c\right ) \cos \left (a\right ) +{\left (\cos \left (2 \, b x + a + 2 \, c\right )^{2} \sin \left (-a + c\right ) + 2 \, \cos \left (2 \, b x + a + 2 \, c\right ) \cos \left (a\right ) \sin \left (-a + c\right ) + \sin \left (2 \, b x + a + 2 \, c\right )^{2} \sin \left (-a + c\right ) + 2 \, \sin \left (2 \, b x + a + 2 \, c\right ) \sin \left (a\right ) \sin \left (-a + c\right ) +{\left (\cos \left (a\right )^{2} + \sin \left (a\right )^{2}\right )} \sin \left (-a + c\right )\right )} \log \left (\frac{\cos \left (b x + 2 \, c\right )^{2} + \cos \left (c\right )^{2} - 2 \, \cos \left (c\right ) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} + 2 \, \cos \left (b x + 2 \, c\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}}{\cos \left (b x + 2 \, c\right )^{2} + \cos \left (c\right )^{2} + 2 \, \cos \left (c\right ) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} - 2 \, \cos \left (b x + 2 \, c\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}}\right ) + 2 \,{\left (\sin \left (b x + 2 \, a\right ) + \sin \left (b x + 2 \, c\right )\right )} \sin \left (2 \, b x + a + 2 \, c\right ) + 2 \, \sin \left (b x + 2 \, a\right ) \sin \left (a\right ) + 2 \, \sin \left (b x + 2 \, c\right ) \sin \left (a\right )}{2 \,{\left (b \cos \left (2 \, b x + a + 2 \, c\right )^{2} + 2 \, b \cos \left (2 \, b x + a + 2 \, c\right ) \cos \left (a\right ) + b \sin \left (2 \, b x + a + 2 \, c\right )^{2} + 2 \, b \sin \left (2 \, b x + a + 2 \, c\right ) \sin \left (a\right ) +{\left (\cos \left (a\right )^{2} + \sin \left (a\right )^{2}\right )} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(cos(b*x + 2*a) + cos(b*x + 2*c))*cos(2*b*x + a + 2*c) + 2*cos(b*x + 2*a)*cos(a) + 2*cos(b*x + 2*c)*cos
(a) + (cos(2*b*x + a + 2*c)^2*sin(-a + c) + 2*cos(2*b*x + a + 2*c)*cos(a)*sin(-a + c) + sin(2*b*x + a + 2*c)^2
*sin(-a + c) + 2*sin(2*b*x + a + 2*c)*sin(a)*sin(-a + c) + (cos(a)^2 + sin(a)^2)*sin(-a + c))*log((cos(b*x + 2
*c)^2 + cos(c)^2 - 2*cos(c)*sin(b*x + 2*c) + sin(b*x + 2*c)^2 + 2*cos(b*x + 2*c)*sin(c) + sin(c)^2)/(cos(b*x +
2*c)^2 + cos(c)^2 + 2*cos(c)*sin(b*x + 2*c) + sin(b*x + 2*c)^2 - 2*cos(b*x + 2*c)*sin(c) + sin(c)^2)) + 2*(si
n(b*x + 2*a) + sin(b*x + 2*c))*sin(2*b*x + a + 2*c) + 2*sin(b*x + 2*a)*sin(a) + 2*sin(b*x + 2*c)*sin(a))/(b*co
s(2*b*x + a + 2*c)^2 + 2*b*cos(2*b*x + a + 2*c)*cos(a) + b*sin(2*b*x + a + 2*c)^2 + 2*b*sin(2*b*x + a + 2*c)*s
in(a) + (cos(a)^2 + sin(a)^2)*b)

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Fricas [B]  time = 0.503744, size = 186, normalized size = 5.47 \begin{align*} -\frac{\cos \left (b x + c\right ) \log \left (\sin \left (b x + c\right ) + 1\right ) \sin \left (-a + c\right ) - \cos \left (b x + c\right ) \log \left (-\sin \left (b x + c\right ) + 1\right ) \sin \left (-a + c\right ) - 2 \, \cos \left (-a + c\right )}{2 \, b \cos \left (b x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.17763, size = 335, normalized size = 9.85 \begin{align*} \frac{2 \,{\left (\frac{{\left (\tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right ) - \tan \left (\frac{1}{2} \, a\right ) \tan \left (\frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, a\right ) - \tan \left (\frac{1}{2} \, c\right )\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{\tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, a\right )^{2} + \tan \left (\frac{1}{2} \, c\right )^{2} + 1} - \frac{{\left (\tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right ) - \tan \left (\frac{1}{2} \, a\right ) \tan \left (\frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, a\right ) - \tan \left (\frac{1}{2} \, c\right )\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{\tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, a\right )^{2} + \tan \left (\frac{1}{2} \, c\right )^{2} + 1} - \frac{\tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right )^{2} - \tan \left (\frac{1}{2} \, a\right )^{2} + 4 \, \tan \left (\frac{1}{2} \, a\right ) \tan \left (\frac{1}{2} \, c\right ) - \tan \left (\frac{1}{2} \, c\right )^{2} + 1}{{\left (\tan \left (\frac{1}{2} \, a\right )^{2} \tan \left (\frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, a\right )^{2} + \tan \left (\frac{1}{2} \, c\right )^{2} + 1\right )}{\left (\tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, c\right )^{2} - 1\right )}}\right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 + tan(1/2*a) - tan(1/2*c))*log(abs(tan(1/2*b*x + 1/2*c)
+ 1))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1) - (tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*ta
n(1/2*c)^2 + tan(1/2*a) - tan(1/2*c))*log(abs(tan(1/2*b*x + 1/2*c) - 1))/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*
a)^2 + tan(1/2*c)^2 + 1) - (tan(1/2*a)^2*tan(1/2*c)^2 - tan(1/2*a)^2 + 4*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2
+ 1)/((tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 + 1)*(tan(1/2*b*x + 1/2*c)^2 - 1)))/b