3.230 \(\int \sin (a+b x) \tan ^3(c+b x) \, dx\)

Optimal. Leaf size=72 \[ \frac{\sin (a-c) \sec (b x+c)}{b}-\frac{3 \cos (a-c) \tanh ^{-1}(\sin (b x+c))}{2 b}+\frac{\cos (a-c) \tan (b x+c) \sec (b x+c)}{2 b}+\frac{\sin (a+b x)}{b} \]

[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)

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Rubi [A]  time = 0.0707118, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {4576, 4579, 2637, 3770, 2606, 8, 2611} \[ \frac{\sin (a-c) \sec (b x+c)}{b}-\frac{3 \cos (a-c) \tanh ^{-1}(\sin (b x+c))}{2 b}+\frac{\cos (a-c) \tan (b x+c) \sec (b x+c)}{2 b}+\frac{\sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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]

Rule 2611

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]

Rubi steps

\begin{align*} \int \sin (a+b x) \tan ^3(c+b x) \, dx &=\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{\tanh ^{-1}(\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) \operatorname{Subst}(\int 1 \, dx,x,\sec (c+b x))}{b}+\int \cos (a+b x) \, dx\\ &=-\frac{3 \tanh ^{-1}(\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]  time = 0.363034, size = 70, normalized size = 0.97 \[ \frac{\sec ^2(b x+c) (2 \sin (a-b x-2 c)+\sin (a+3 b x+2 c)+5 \sin (a+b x))-12 \cos (a-c) \tanh ^{-1}\left (\cos (c) \tan \left (\frac{b x}{2}\right )+\sin (c)\right )}{4 b} \]

Antiderivative was successfully verified.

[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)

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Maple [C]  time = 0.115, size = 186, normalized size = 2.6 \begin{align*}{\frac{-{\frac{i}{2}}{{\rm e}^{i \left ( bx+a \right ) }}}{b}}+{\frac{{\frac{i}{2}}{{\rm e}^{-i \left ( bx+a \right ) }}}{b}}-{\frac{{\frac{i}{2}} \left ( 3\,{{\rm e}^{i \left ( 3\,bx+5\,a+2\,c \right ) }}-{{\rm e}^{i \left ( 3\,bx+3\,a+4\,c \right ) }}+{{\rm e}^{i \left ( bx+5\,a \right ) }}-3\,{{\rm e}^{i \left ( bx+3\,a+2\,c \right ) }} \right ) }{b \left ({{\rm e}^{2\,i \left ( bx+a+c \right ) }}+{{\rm e}^{2\,ia}} \right ) ^{2}}}+{\frac{3\,\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}-i{{\rm e}^{i \left ( a-c \right ) }} \right ) \cos \left ( a-c \right ) }{2\,b}}-{\frac{3\,\ln \left ({{\rm e}^{i \left ( bx+a \right ) }}+i{{\rm e}^{i \left ( a-c \right ) }} \right ) \cos \left ( a-c \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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/b*ln(exp(I*(b*x+a))-I*exp(I*(a-c)))*
cos(a-c)-3/2/b*ln(exp(I*(b*x+a))+I*exp(I*(a-c)))*cos(a-c)

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Maxima [B]  time = 1.97578, size = 1386, normalized size = 19.25 \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)^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))

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Fricas [B]  time = 0.564859, size = 1015, normalized size = 14.1 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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 )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)