3.213 \(\int \sec (c+b x) \sin (a+b x) \, dx\)

Optimal. Leaf size=27 \[ x \sin (a-c)-\frac{\cos (a-c) \log (\cos (b x+c))}{b} \]

[Out]

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

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Rubi [A]  time = 0.0175518, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4580, 3475, 8} \[ x \sin (a-c)-\frac{\cos (a-c) \log (\cos (b x+c))}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + b*x]*Sin[a + b*x],x]

[Out]

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

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 3475

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

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.142382, size = 27, normalized size = 1. \[ x \sin (a-c)-\frac{\cos (a-c) \log (\cos (b x+c))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.201, size = 563, normalized size = 20.9 \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)*sin(b*x+a),x)

[Out]

1/b/(cos(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)/(sin(a)*cos(c)-cos(a)*sin(c))*ln
(-tan(b*x+a)*cos(a)*sin(c)+tan(b*x+a)*sin(a)*cos(c)+cos(a)*cos(c)+sin(a)*sin(c))*cos(a)^2*cos(c)*sin(c)-1/b/(c
os(a)^2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)/(sin(a)*cos(c)-cos(a)*sin(c))*ln(-tan(
b*x+a)*cos(a)*sin(c)+tan(b*x+a)*sin(a)*cos(c)+cos(a)*cos(c)+sin(a)*sin(c))*cos(a)*cos(c)^2*sin(a)+1/b/(cos(a)^
2*cos(c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)/(sin(a)*cos(c)-cos(a)*sin(c))*ln(-tan(b*x+a)
*cos(a)*sin(c)+tan(b*x+a)*sin(a)*cos(c)+cos(a)*cos(c)+sin(a)*sin(c))*cos(a)*sin(a)*sin(c)^2-1/b/(cos(a)^2*cos(
c)^2+cos(a)^2*sin(c)^2+cos(c)^2*sin(a)^2+sin(a)^2*sin(c)^2)/(sin(a)*cos(c)-cos(a)*sin(c))*ln(-tan(b*x+a)*cos(a
)*sin(c)+tan(b*x+a)*sin(a)*cos(c)+cos(a)*cos(c)+sin(a)*sin(c))*cos(c)*sin(a)^2*sin(c)+1/2/b/(cos(c)^2+sin(c)^2
)/(cos(a)^2+sin(a)^2)*ln(1+tan(b*x+a)^2)*cos(a)*cos(c)+1/2/b/(cos(c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*ln(1+tan(
b*x+a)^2)*sin(a)*sin(c)-1/b/(cos(c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*cos(a)*sin(c)*arctan(tan(b*x+a))+1/b/(cos(
c)^2+sin(c)^2)/(cos(a)^2+sin(a)^2)*cos(c)*sin(a)*arctan(tan(b*x+a))

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Maxima [B]  time = 1.25903, size = 99, normalized size = 3.67 \begin{align*} -\frac{2 \, b x \sin \left (-a + c\right ) + \cos \left (-a + c\right ) \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, c\right ) + \cos \left (2 \, c\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, c\right ) + \sin \left (2 \, c\right )^{2}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.49795, size = 74, normalized size = 2.74 \begin{align*} -\frac{b x \sin \left (-a + c\right ) + \cos \left (-a + c\right ) \log \left (-\cos \left (b x + c\right )\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 169.994, size = 435, normalized size = 16.11 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((0, Eq(b, 0)), (-x, Eq(c, pi/2)), (x, Eq(c, -pi/2)), (-2*b*x*tan(c/2)/(b*tan(c/2)**2 + b) - log(tan(
b*x/2)**2 + 1)*tan(c/2)**2/(b*tan(c/2)**2 + b) + log(tan(b*x/2)**2 + 1)/(b*tan(c/2)**2 + b) + log(tan(b*x/2) -
tan(c/2)/(tan(c/2) - 1) - 1/(tan(c/2) - 1))*tan(c/2)**2/(b*tan(c/2)**2 + b) - log(tan(b*x/2) - tan(c/2)/(tan(
c/2) - 1) - 1/(tan(c/2) - 1))/(b*tan(c/2)**2 + b) + log(tan(b*x/2) + tan(c/2)/(tan(c/2) + 1) - 1/(tan(c/2) + 1
))*tan(c/2)**2/(b*tan(c/2)**2 + b) - log(tan(b*x/2) + tan(c/2)/(tan(c/2) + 1) - 1/(tan(c/2) + 1))/(b*tan(c/2)*
*2 + b), True))*cos(a) + Piecewise((x/cos(c), Eq(b, 0)), (-log(sin(b*x))/b, Eq(c, pi/2)), (log(sin(b*x))/b, Eq
(c, -pi/2)), (-b*x*tan(c/2)**2/(b*tan(c/2)**2 + b) + b*x/(b*tan(c/2)**2 + b) + 2*log(tan(b*x/2)**2 + 1)*tan(c/
2)/(b*tan(c/2)**2 + b) - 2*log(tan(b*x/2) - tan(c/2)/(tan(c/2) - 1) - 1/(tan(c/2) - 1))*tan(c/2)/(b*tan(c/2)**
2 + b) - 2*log(tan(b*x/2) + tan(c/2)/(tan(c/2) + 1) - 1/(tan(c/2) + 1))*tan(c/2)/(b*tan(c/2)**2 + b), True))*s
in(a)

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Giac [B]  time = 1.16072, size = 213, normalized size = 7.89 \begin{align*} \frac{\frac{4 \,{\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 )}{\left (b x + c\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 )^{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\right )} \log \left (\tan \left (b x + c\right )^{2} + 1\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}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(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))*(b*x + c)/(tan(1/2*a)^2*t
an(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)*log(tan(b*x + c)^2 + 1)/(tan(1/2*a)^2*tan(1/2*c)^2 + tan(1/2*a)^2 + tan(1/2*c)^2 +
1))/b