\(\int \sin (a+b x) \tan (c+b x) \, dx\) [232]

Optimal result

Integrand size = 13, antiderivative size = 29 \[ \int \sin (a+b x) \tan (c+b x) \, dx=\frac {\text {arctanh}(\sin (c+b x)) \cos (a-c)}{b}-\frac {\sin (a+b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {4672, 2717, 3855} \[ \int \sin (a+b x) \tan (c+b x) \, dx=\frac {\cos (a-c) \text {arctanh}(\sin (b x+c))}{b}-\frac {\sin (a+b x)}{b} \]

[In]

Int[Sin[a + b*x]*Tan[c + b*x],x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3855

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

Rule 4672

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]

Rubi steps \begin{align*} \text {integral}& = \cos (a-c) \int \sec (c+b x) \, dx-\int \cos (a+b x) \, dx \\ & = \frac {\text {arctanh}(\sin (c+b x)) \cos (a-c)}{b}-\frac {\sin (a+b x)}{b} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 3.24 \[ \int \sin (a+b x) \tan (c+b x) \, dx=-\frac {2 i \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (\cos \left (\frac {b x}{2}\right ) \sin (c)+\cos (c) \sin \left (\frac {b x}{2}\right )\right )}{\cos (c) \cos \left (\frac {b x}{2}\right )-i \cos \left (\frac {b x}{2}\right ) \sin (c)}\right ) \cos (a-c)}{b}-\frac {\cos (b x) \sin (a)}{b}-\frac {\cos (a) \sin (b x)}{b} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.41

method	result	size
risch	\(\frac {i {\mathrm e}^{i \left (x b +a \right )}}{2 b}-\frac {i {\mathrm e}^{-i \left (x b +a \right )}}{2 b}+\frac {\ln \left ({\mathrm e}^{i \left (x b +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{b}-\frac {\ln \left ({\mathrm e}^{i \left (x b +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{b}\)	\(99\)

[In]

int(sin(b*x+a)*tan(b*x+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (29) = 58\).

Time = 0.26 (sec) , antiderivative size = 188, normalized size of antiderivative = 6.48 \[ \int \sin (a+b x) \tan (c+b x) \, dx=\frac {\sqrt {2} \sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1} \log \left (\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right ) - 4 \, \sin \left (b x + a\right )}{4 \, b} \]

[In]

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

[Out]

1/4*(sqrt(2)*sqrt(cos(-2*a + 2*c) + 1)*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) + cos(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)) - 4*sin(b*x + a))/b

Sympy [F]

\[ \int \sin (a+b x) \tan (c+b x) \, dx=\int \sin {\left (a + b x \right )} \tan {\left (b x + c \right )}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 131 vs. \(2 (29) = 58\).

Time = 0.39 (sec) , antiderivative size = 131, normalized size of antiderivative = 4.52 \[ \int \sin (a+b x) \tan (c+b x) \, dx=-\frac {\cos \left (-a + c\right ) \log \left (\frac {\cos \left (b x + 2 \, c\right )^{2} + \cos \left (c\right )^{2} - 2 \, \cos \left (c\right ) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} + 2 \, \cos \left (b x + 2 \, c\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}}{\cos \left (b x + 2 \, c\right )^{2} + \cos \left (c\right )^{2} + 2 \, \cos \left (c\right ) \sin \left (b x + 2 \, c\right ) + \sin \left (b x + 2 \, c\right )^{2} - 2 \, \cos \left (b x + 2 \, c\right ) \sin \left (c\right ) + \sin \left (c\right )^{2}}\right ) + 2 \, \sin \left (b x + a\right )}{2 \, b} \]

[In]

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

[Out]

-1/2*(cos(-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*sin(b*x + a))/b

Giac [F]

\[ \int \sin (a+b x) \tan (c+b x) \, dx=\int { \sin \left (b x + a\right ) \tan \left (b x + c\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 26.04 (sec) , antiderivative size = 227, normalized size of antiderivative = 7.83 \[ \int \sin (a+b x) \tan (c+b x) \, dx=-\frac {{\mathrm {e}}^{-a\,1{}\mathrm {i}-b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,b}+\frac {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,b}+\frac {\ln \left (-{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}+1\right )-\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}+1\right )\,1{}\mathrm {i}}{\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}}\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}+1\right )}{2\,b\,\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}}-\frac {\ln \left (-{\mathrm {e}}^{a\,1{}\mathrm {i}}\,{\mathrm {e}}^{b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}+1\right )+\frac {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}+1\right )\,1{}\mathrm {i}}{\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}}\,{\mathrm {e}}^{-c\,2{}\mathrm {i}}}}\right )\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}+1\right )}{2\,b\,\sqrt {{\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}}} \]

[In]

int(sin(a + b*x)*tan(c + b*x),x)

[Out]

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