3.231 \(\int \sin (a+b x) \tan ^2(c+b x) \, dx\)

Optimal. Leaf size=44 \[ \frac{\sin (a-c) \tanh ^{-1}(\sin (b x+c))}{b}+\frac{\cos (a-c) \sec (b x+c)}{b}+\frac{\cos (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.035963, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4576, 4579, 2638, 3770, 2606, 8} \[ \frac{\sin (a-c) \tanh ^{-1}(\sin (b x+c))}{b}+\frac{\cos (a-c) \sec (b x+c)}{b}+\frac{\cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Tan[c + b*x]^2,x]

[Out]

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

Rule 4576

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

Rule 4579

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

Rule 2638

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

Rule 3770

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

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]

Rubi steps

\begin{align*} \int \sin (a+b x) \tan ^2(c+b x) \, dx &=\cos (a-c) \int \sec (c+b x) \tan (c+b x) \, dx-\int \cos (a+b x) \tan (c+b x) \, dx\\ &=\frac{\cos (a-c) \operatorname{Subst}(\int 1 \, dx,x,\sec (c+b x))}{b}+\sin (a-c) \int \sec (c+b x) \, dx-\int \sin (a+b x) \, dx\\ &=\frac{\cos (a+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 = 0.10031, size = 109, normalized size = 2.48 \[ \frac{\cos (a-c) \sec (b x+c)}{b}-\frac{2 i \sin (a-c) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\sin (c) \cos \left (\frac{b x}{2}\right )+\cos (c) \sin \left (\frac{b x}{2}\right )\right )}{\cos (c) \cos \left (\frac{b x}{2}\right )-i \sin (c) \cos \left (\frac{b x}{2}\right )}\right )}{b}-\frac{\sin (a) \sin (b x)}{b}+\frac{\cos (a) \cos (b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.083, size = 143, normalized size = 3.3 \begin{align*}{\frac{{{\rm e}^{i \left ( bx+a \right ) }}}{2\,b}}+{\frac{{{\rm e}^{-i \left ( bx+a \right ) }}}{2\,b}}+{\frac{{{\rm e}^{i \left ( bx+3\,a \right ) }}+{{\rm e}^{i \left ( bx+a+2\,c \right ) }}}{b \left ({{\rm e}^{2\,i \left ( bx+a+c \right ) }}+{{\rm e}^{2\,ia}} \right ) }}-{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-i{{\rm e}^{i \left ( a-c \right ) }} \right ) \sin \left ( a-c \right ) }{b}}+{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+i{{\rm e}^{i \left ( a-c \right ) }} \right ) \sin \left ( a-c \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*exp(I*(b*x+a))/b+1/2/b*exp(-I*(b*x+a))+1/b/(exp(2*I*(b*x+a+c))+exp(2*I*a))*(exp(I*(b*x+3*a))+exp(I*(b*x+a+
2*c)))-1/b*ln(exp(I*(b*x+a))-I*exp(I*(a-c)))*sin(a-c)+1/b*ln(exp(I*(b*x+a))+I*exp(I*(a-c)))*sin(a-c)

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Maxima [B]  time = 1.99207, size = 702, normalized size = 15.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.538988, size = 857, normalized size = 19.48 \begin{align*} -\frac{4 \,{\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right )^{2} - 4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac{\sqrt{2}{\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) +{\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \sin \left (b x + a\right )\right )} \log \left (-\frac{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac{2 \, \sqrt{2}{\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt{\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right )}{\sqrt{\cos \left (-2 \, a + 2 \, c\right ) + 1}} + 4 \, \cos \left (-2 \, a + 2 \, c\right ) + 4}{4 \,{\left (b \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) -{\left (b \cos \left (-2 \, a + 2 \, c\right ) + b\right )} \cos \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(4*(cos(-2*a + 2*c) + 1)*cos(b*x + a)^2 - 4*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) + sqrt(2)*((cos(-2*
a + 2*c) + 1)*cos(b*x + a)*sin(-2*a + 2*c) + (cos(-2*a + 2*c)^2 - 1)*sin(b*x + a))*log(-(2*cos(b*x + a)^2*cos(
-2*a + 2*c) - 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) + 2*sqrt(2)*((cos(-2*a + 2*c) + 1)*sin(b*x + a) + co
s(b*x + a)*sin(-2*a + 2*c))/sqrt(cos(-2*a + 2*c) + 1) - cos(-2*a + 2*c) - 3)/(2*cos(b*x + a)^2*cos(-2*a + 2*c)
- 2*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - cos(-2*a + 2*c) + 1))/sqrt(cos(-2*a + 2*c) + 1) + 4*cos(-2*a
+ 2*c) + 4)/(b*sin(b*x + a)*sin(-2*a + 2*c) - (b*cos(-2*a + 2*c) + b)*cos(b*x + a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)*tan(b*x + c)**2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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