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

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Rubi [A]  time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 8

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

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*}

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Mathematica [C]  time = 0.10, 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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fricas [B]  time = 0.47, size = 315, normalized size = 7.16 \[ -\frac {4 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right )^{2} - 4 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {\sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \sin \left (b x + a\right )\right )} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} + 4 \, \cos \left (-2 \, a + 2 \, c\right ) + 4}{4 \, {\left (b \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - {\left (b \cos \left (-2 \, a + 2 \, c\right ) + b\right )} \cos \left (b x + a\right )\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin \left (b x + a\right ) \tan \left (b x + c\right )^{2}\,{d x} \]

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)

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maple [C]  time = 0.79, size = 143, normalized size = 3.25 \[ \frac {{\mathrm e}^{i \left (b x +a \right )}}{2 b}+\frac {{\mathrm e}^{-i \left (b x +a \right )}}{2 b}+\frac {{\mathrm e}^{i \left (b x +3 a \right )}+{\mathrm e}^{i \left (b x +a +2 c \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{b} \]

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

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maxima [B]  time = 0.53, size = 520, normalized size = 11.82 \[ \frac {{\left (\cos \left (3 \, b x + a + 2 \, c\right ) + \cos \left (b x + a\right )\right )} \cos \left (4 \, b x + 2 \, a + 2 \, c\right ) + {\left (3 \, \cos \left (2 \, b x + 2 \, a\right ) + 3 \, \cos \left (2 \, b x + 2 \, c\right ) + 1\right )} \cos \left (3 \, b x + a + 2 \, c\right ) + 3 \, \cos \left (2 \, b x + 2 \, a\right ) \cos \left (b x + a\right ) + 3 \, \cos \left (2 \, b x + 2 \, c\right ) \cos \left (b x + a\right ) + {\left (\cos \left (3 \, b x + a + 2 \, c\right )^{2} \sin \left (-a + c\right ) + 2 \, \cos \left (3 \, b x + a + 2 \, c\right ) \cos \left (b x + a\right ) \sin \left (-a + c\right ) + \cos \left (b x + a\right )^{2} \sin \left (-a + c\right ) + \sin \left (3 \, b x + a + 2 \, c\right )^{2} \sin \left (-a + c\right ) + 2 \, \sin \left (3 \, b x + a + 2 \, c\right ) \sin \left (b x + a\right ) \sin \left (-a + c\right ) + \sin \left (b x + a\right )^{2} \sin \left (-a + c\right )\right )} \log \left (\frac {\cos \left (b x + 2 \, c\right )^{2} + \cos \relax (c)^{2} - 2 \, \cos \relax (c) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} + 2 \, \cos \left (b x + 2 \, c\right ) \sin \relax (c) + \sin \relax (c)^{2}}{\cos \left (b x + 2 \, c\right )^{2} + \cos \relax (c)^{2} + 2 \, \cos \relax (c) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} - 2 \, \cos \left (b x + 2 \, c\right ) \sin \relax (c) + \sin \relax (c)^{2}}\right ) + {\left (\sin \left (3 \, b x + a + 2 \, c\right ) + \sin \left (b x + a\right )\right )} \sin \left (4 \, b x + 2 \, a + 2 \, c\right ) + 3 \, {\left (\sin \left (2 \, b x + 2 \, a\right ) + \sin \left (2 \, b x + 2 \, c\right )\right )} \sin \left (3 \, b x + a + 2 \, c\right ) + 3 \, \sin \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + 3 \, \sin \left (2 \, b x + 2 \, c\right ) \sin \left (b x + a\right ) + \cos \left (b x + a\right )}{2 \, {\left (b \cos \left (3 \, b x + a + 2 \, c\right )^{2} + 2 \, b \cos \left (3 \, b x + a + 2 \, c\right ) \cos \left (b x + a\right ) + b \cos \left (b x + a\right )^{2} + b \sin \left (3 \, b x + a + 2 \, c\right )^{2} + 2 \, b \sin \left (3 \, b x + a + 2 \, c\right ) \sin \left (b x + a\right ) + b \sin \left (b x + a\right )^{2}\right )}} \]

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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mupad [B]  time = 5.31, size = 294, normalized size = 6.68 \[ \frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}}{2\,b}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}}{2\,b}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}+1\right )\,1{}\mathrm {i}}{b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}\,1{}\mathrm {i}+{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}\right )}+\frac {\ln \left ({\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )\,1{}\mathrm {i}}{\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}}\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}-1\right )}{2\,b\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}}-\frac {\ln \left ({\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}-1\right )\,1{}\mathrm {i}}{\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}}\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}-1\right )}{2\,b\,\sqrt {-{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*tan(c + b*x)^2,x)

[Out]

exp(- a*1i - b*x*1i)/(2*b) + exp(a*1i + b*x*1i)/(2*b) + (exp(a*1i + b*x*1i)*(exp(a*2i - c*2i) + 1)*1i)/(b*(exp
(a*2i - c*2i)*1i + exp(a*2i + b*x*2i)*1i)) + (log(exp(a*1i)*exp(b*x*1i)*(exp(a*2i)*exp(-c*2i)*1i - 1i) - (exp(
a*2i)*exp(-c*2i)*(exp(a*2i)*exp(-c*2i) - 1)*1i)/(-exp(a*2i)*exp(-c*2i))^(1/2))*(exp(a*2i - c*2i) - 1))/(2*b*(-
exp(a*2i - c*2i))^(1/2)) - (log(exp(a*1i)*exp(b*x*1i)*(exp(a*2i)*exp(-c*2i)*1i - 1i) + (exp(a*2i)*exp(-c*2i)*(
exp(a*2i)*exp(-c*2i) - 1)*1i)/(-exp(a*2i)*exp(-c*2i))^(1/2))*(exp(a*2i - c*2i) - 1))/(2*b*(-exp(a*2i - c*2i))^
(1/2))

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

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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