\(\int \sin (a+b x) \tan ^3(c+b x) \, dx\) [230]

Optimal result

Integrand size = 15, antiderivative size = 72 \[ \int \sin (a+b x) \tan ^3(c+b x) \, dx=-\frac {3 \text {arctanh}(\sin (c+b x)) \cos (a-c)}{2 b}+\frac {\sec (c+b x) \sin (a-c)}{b}+\frac {\sin (a+b x)}{b}+\frac {\cos (a-c) \sec (c+b x) \tan (c+b x)}{2 b} \]

[Out]

-3/2*arctanh(sin(b*x+c))*cos(a-c)/b+sec(b*x+c)*sin(a-c)/b+sin(b*x+a)/b+1/2*cos(a-c)*sec(b*x+c)*tan(b*x+c)/b

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {4672, 4675, 2717, 3855, 2686, 8, 2691} \[ \int \sin (a+b x) \tan ^3(c+b x) \, dx=-\frac {3 \cos (a-c) \text {arctanh}(\sin (b x+c))}{2 b}+\frac {\sin (a-c) \sec (b x+c)}{b}+\frac {\cos (a-c) \tan (b x+c) \sec (b x+c)}{2 b}+\frac {\sin (a+b x)}{b} \]

[In]

Int[Sin[a + b*x]*Tan[c + b*x]^3,x]

[Out]

(-3*ArcTanh[Sin[c + b*x]]*Cos[a - c])/(2*b) + (Sec[c + b*x]*Sin[a - c])/b + Sin[a + b*x]/b + (Cos[a - c]*Sec[c
+ b*x]*Tan[c + b*x])/(2*b)

Rule 8

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

Rule 2686

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 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

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]

Rule 4675

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*} \text {integral}& = \cos (a-c) \int \sec (c+b x) \tan ^2(c+b x) \, dx-\int \cos (a+b x) \tan ^2(c+b x) \, dx \\ & = \frac {\cos (a-c) \sec (c+b x) \tan (c+b x)}{2 b}-\frac {1}{2} \cos (a-c) \int \sec (c+b x) \, dx+\sin (a-c) \int \sec (c+b x) \tan (c+b x) \, dx-\int \sin (a+b x) \tan (c+b x) \, dx \\ & = -\frac {\text {arctanh}(\sin (c+b x)) \cos (a-c)}{2 b}+\frac {\cos (a-c) \sec (c+b x) \tan (c+b x)}{2 b}-\cos (a-c) \int \sec (c+b x) \, dx+\frac {\sin (a-c) \text {Subst}(\int 1 \, dx,x,\sec (c+b x))}{b}+\int \cos (a+b x) \, dx \\ & = -\frac {3 \text {arctanh}(\sin (c+b x)) \cos (a-c)}{2 b}+\frac {\sec (c+b x) \sin (a-c)}{b}+\frac {\sin (a+b x)}{b}+\frac {\cos (a-c) \sec (c+b x) \tan (c+b x)}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97 \[ \int \sin (a+b x) \tan ^3(c+b x) \, dx=\frac {-12 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {b x}{2}\right )\right ) \cos (a-c)+\sec ^2(c+b x) (2 \sin (a-2 c-b x)+5 \sin (a+b x)+\sin (a+2 c+3 b x))}{4 b} \]

[In]

Integrate[Sin[a + b*x]*Tan[c + b*x]^3,x]

[Out]

(-12*ArcTanh[Sin[c] + Cos[c]*Tan[(b*x)/2]]*Cos[a - c] + Sec[c + b*x]^2*(2*Sin[a - 2*c - b*x] + 5*Sin[a + b*x]
+ Sin[a + 2*c + 3*b*x]))/(4*b)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.58

method	result	size
risch	\(-\frac {i {\mathrm e}^{i \left (x b +a \right )}}{2 b}+\frac {i {\mathrm e}^{-i \left (x b +a \right )}}{2 b}-\frac {i \left (3 \,{\mathrm e}^{i \left (3 x b +5 a +2 c \right )}-{\mathrm e}^{i \left (3 x b +3 a +4 c \right )}+{\mathrm e}^{i \left (x b +5 a \right )}-3 \,{\mathrm e}^{i \left (x b +3 a +2 c \right )}\right )}{2 b \left ({\mathrm e}^{2 i \left (x b +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{2 b}\)	\(186\)

[In]

int(sin(b*x+a)*tan(b*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

-1/2*I*exp(I*(b*x+a))/b+1/2*I/b*exp(-I*(b*x+a))-1/2*I/b/(exp(2*I*(b*x+a+c))+exp(2*I*a))^2*(3*exp(I*(3*b*x+5*a+
2*c))-exp(I*(3*b*x+3*a+4*c))+exp(I*(b*x+5*a))-3*exp(I*(b*x+3*a+2*c)))+3/2*ln(exp(I*(b*x+a))-I*exp(I*(a-c)))/b*
cos(a-c)-3/2*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. 376 vs. \(2 (68) = 136\).

Time = 0.28 (sec) , antiderivative size = 376, normalized size of antiderivative = 5.22 \[ \int \sin (a+b x) \tan ^3(c+b x) \, dx=-\frac {\frac {3 \, \sqrt {2} {\left (2 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 2 \, {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} + \cos \left (-2 \, a + 2 \, c\right )\right )} \cos \left (b x + a\right )^{2} + \cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - 4 \, {\left (4 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 3 \, \cos \left (-2 \, a + 2 \, c\right ) + 5\right )} \sin \left (b x + a\right ) - 4 \, {\left (4 \, \cos \left (b x + a\right )^{3} - 5 \, \cos \left (b x + a\right )\right )} \sin \left (-2 \, a + 2 \, c\right )}{8 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) + b\right )}} \]

[In]

integrate(sin(b*x+a)*tan(b*x+c)^3,x, algorithm="fricas")

[Out]

-1/8*(3*sqrt(2)*(2*(cos(-2*a + 2*c) + 1)*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - 2*(cos(-2*a + 2*c)^2 + co
s(-2*a + 2*c))*cos(b*x + a)^2 + cos(-2*a + 2*c)^2 - 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))/sqrt(cos(-2*a + 2*c) + 1) - 4*(4*cos(b*x + a)^2*cos(-2*a + 2*c) - 3
*cos(-2*a + 2*c) + 5)*sin(b*x + a) - 4*(4*cos(b*x + a)^3 - 5*cos(b*x + a))*sin(-2*a + 2*c))/(2*b*cos(b*x + a)^
2*cos(-2*a + 2*c) - 2*b*cos(b*x + a)*sin(b*x + a)*sin(-2*a + 2*c) - b*cos(-2*a + 2*c) + b)

Sympy [F]

\[ \int \sin (a+b x) \tan ^3(c+b x) \, dx=\int \sin {\left (a + b x \right )} \tan ^{3}{\left (b x + c \right )}\, dx \]

[In]

integrate(sin(b*x+a)*tan(b*x+c)**3,x)

[Out]

Integral(sin(a + b*x)*tan(b*x + c)**3, x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1027 vs. \(2 (68) = 136\).

Time = 0.42 (sec) , antiderivative size = 1027, normalized size of antiderivative = 14.26 \[ \int \sin (a+b x) \tan ^3(c+b x) \, dx=\text {Too large to display} \]

[In]

integrate(sin(b*x+a)*tan(b*x+c)^3,x, algorithm="maxima")

[Out]

-1/4*(2*(sin(5*b*x + a + 4*c) + 2*sin(3*b*x + a + 2*c) + sin(b*x + a))*cos(6*b*x + 2*a + 4*c) - 2*(5*sin(4*b*x
+ 2*a + 2*c) - 2*sin(4*b*x + 4*c) + 2*sin(2*b*x + 2*a) - 5*sin(2*b*x + 2*c))*cos(5*b*x + a + 4*c) + 10*(2*sin
(3*b*x + a + 2*c) + sin(b*x + a))*cos(4*b*x + 2*a + 2*c) - 4*(2*sin(3*b*x + a + 2*c) + sin(b*x + a))*cos(4*b*x
+ 4*c) - 4*(2*sin(2*b*x + 2*a) - 5*sin(2*b*x + 2*c))*cos(3*b*x + a + 2*c) - 3*(cos(5*b*x + a + 4*c)^2*cos(-a
+ c) + 4*cos(3*b*x + a + 2*c)^2*cos(-a + c) + 4*cos(3*b*x + a + 2*c)*cos(b*x + a)*cos(-a + c) + cos(b*x + a)^2
*cos(-a + c) + cos(-a + c)*sin(5*b*x + a + 4*c)^2 + 4*cos(-a + c)*sin(3*b*x + a + 2*c)^2 + 4*cos(-a + c)*sin(3
*b*x + a + 2*c)*sin(b*x + a) + cos(-a + c)*sin(b*x + a)^2 + 2*(2*cos(3*b*x + a + 2*c)*cos(-a + c) + cos(b*x +
a)*cos(-a + c))*cos(5*b*x + a + 4*c) + 2*(2*cos(-a + c)*sin(3*b*x + a + 2*c) + cos(-a + c)*sin(b*x + a))*sin(5
*b*x + a + 4*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*(cos(5*b*x + a + 4*c) + 2*cos(3*b*x + a + 2*c) + cos(b*x + a))*sin(6*b*x + 2*a +
4*c) + 2*(5*cos(4*b*x + 2*a + 2*c) - 2*cos(4*b*x + 4*c) + 2*cos(2*b*x + 2*a) - 5*cos(2*b*x + 2*c) - 1)*sin(5*
b*x + a + 4*c) - 10*(2*cos(3*b*x + a + 2*c) + cos(b*x + a))*sin(4*b*x + 2*a + 2*c) + 4*(2*cos(3*b*x + a + 2*c)
+ cos(b*x + a))*sin(4*b*x + 4*c) + 4*(2*cos(2*b*x + 2*a) - 5*cos(2*b*x + 2*c) - 1)*sin(3*b*x + a + 2*c) - 4*c
os(b*x + a)*sin(2*b*x + 2*a) + 10*cos(b*x + a)*sin(2*b*x + 2*c) + 4*cos(2*b*x + 2*a)*sin(b*x + a) - 10*cos(2*b
*x + 2*c)*sin(b*x + a) - 2*sin(b*x + a))/(b*cos(5*b*x + a + 4*c)^2 + 4*b*cos(3*b*x + a + 2*c)^2 + 4*b*cos(3*b*
x + a + 2*c)*cos(b*x + a) + b*cos(b*x + a)^2 + b*sin(5*b*x + a + 4*c)^2 + 4*b*sin(3*b*x + a + 2*c)^2 + 4*b*sin
(3*b*x + a + 2*c)*sin(b*x + a) + b*sin(b*x + a)^2 + 2*(2*b*cos(3*b*x + a + 2*c) + b*cos(b*x + a))*cos(5*b*x +
a + 4*c) + 2*(2*b*sin(3*b*x + a + 2*c) + b*sin(b*x + a))*sin(5*b*x + a + 4*c))

Giac [F]

\[ \int \sin (a+b x) \tan ^3(c+b x) \, dx=\int { \sin \left (b x + a\right ) \tan \left (b x + c\right )^{3} \,d x } \]

[In]

integrate(sin(b*x+a)*tan(b*x+c)^3,x, algorithm="giac")

[Out]

integrate(sin(b*x + a)*tan(b*x + c)^3, x)

Mupad [F(-1)]

Timed out. \[ \int \sin (a+b x) \tan ^3(c+b x) \, dx=\text {Hanged} \]

[In]

int(sin(a + b*x)*tan(c + b*x)^3,x)

[Out]

\text{Hanged}