3.247 \(\int \cos (a+b x) \tan ^3(c+b x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 = {4579, 4576, 2638, 3770, 2606, 8, 2611} \[ \frac {3 \sin (a-c) \tanh ^{-1}(\sin (b x+c))}{2 b}+\frac {\cos (a-c) \sec (b x+c)}{b}-\frac {\sin (a-c) \tan (b x+c) \sec (b x+c)}{2 b}+\frac {\cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Tan[c + b*x]^3,x]

[Out]

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

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

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Mathematica [A]  time = 0.37, size = 70, normalized size = 0.97 \[ \frac {\sec ^2(b x+c) (2 \cos (a-b x-2 c)+\cos (a+3 b x+2 c)+5 \cos (a+b x))+12 \sin (a-c) \tanh ^{-1}\left (\cos (c) \tan \left (\frac {b x}{2}\right )+\sin (c)\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.49, size = 366, normalized size = 5.08 \[ \frac {16 \, \cos \left (b x + a\right )^{3} \cos \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) - 5\right )} \cos \left (b x + a\right ) + \frac {3 \, \sqrt {2} {\left (2 \, {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + {\left (2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (-2 \, a + 2 \, c\right )\right )} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}}}{8 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) + b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(16*cos(b*x + a)^3*cos(-2*a + 2*c) - 4*(4*cos(b*x + a)^2 + 1)*sin(b*x + a)*sin(-2*a + 2*c) - 4*(cos(-2*a +
2*c) - 5)*cos(b*x + a) + 3*sqrt(2)*(2*(cos(-2*a + 2*c)^2 - 1)*cos(b*x + a)*sin(b*x + a) + (2*cos(b*x + a)^2*c
os(-2*a + 2*c) - cos(-2*a + 2*c) + 1)*sin(-2*a + 2*c))*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))/(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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos \left (b x + a\right ) \tan \left (b x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.96, size = 181, normalized size = 2.51 \[ \frac {{\mathrm e}^{i \left (b x +a \right )}}{2 b}+\frac {{\mathrm e}^{-i \left (b x +a \right )}}{2 b}+\frac {3 \,{\mathrm e}^{i \left (3 b x +5 a +2 c \right )}+{\mathrm e}^{i \left (3 b x +3 a +4 c \right )}+{\mathrm e}^{i \left (b x +5 a \right )}+3 \,{\mathrm e}^{i \left (b x +3 a +2 c \right )}}{2 b \left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*exp(I*(b*x+a))/b+1/2/b*exp(-I*(b*x+a))+1/2/b/(exp(2*I*(b*x+a+c))+exp(2*I*a))^2*(3*exp(I*(3*b*x+5*a+2*c))+e
xp(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*sin(a-c
)+3/2*ln(exp(I*(b*x+a))+I*exp(I*(a-c)))/b*sin(a-c)

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maxima [B]  time = 0.54, size = 1027, normalized size = 14.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(cos(5*b*x + a + 4*c) + 2*cos(3*b*x + a + 2*c) + cos(b*x + a))*cos(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)*cos(5*b*x + a + 4*c) + 10*(2*
cos(3*b*x + a + 2*c) + cos(b*x + a))*cos(4*b*x + 2*a + 2*c) + 4*(2*cos(3*b*x + a + 2*c) + cos(b*x + a))*cos(4*
b*x + 4*c) + 4*(2*cos(2*b*x + 2*a) + 5*cos(2*b*x + 2*c) + 1)*cos(3*b*x + a + 2*c) + 4*cos(2*b*x + 2*a)*cos(b*x
+ a) + 10*cos(2*b*x + 2*c)*cos(b*x + a) + 3*(cos(5*b*x + a + 4*c)^2*sin(-a + c) + 4*cos(3*b*x + a + 2*c)^2*si
n(-a + c) + 4*cos(3*b*x + a + 2*c)*cos(b*x + a)*sin(-a + c) + cos(b*x + a)^2*sin(-a + c) + sin(5*b*x + a + 4*c
)^2*sin(-a + c) + 4*sin(3*b*x + a + 2*c)^2*sin(-a + c) + 4*sin(3*b*x + a + 2*c)*sin(b*x + a)*sin(-a + c) + sin
(b*x + a)^2*sin(-a + c) + 2*(2*cos(3*b*x + a + 2*c)*sin(-a + c) + cos(b*x + a)*sin(-a + c))*cos(5*b*x + a + 4*
c) + 2*(2*sin(3*b*x + a + 2*c)*sin(-a + c) + sin(b*x + a)*sin(-a + c))*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*(sin
(5*b*x + a + 4*c) + 2*sin(3*b*x + a + 2*c) + sin(b*x + a))*sin(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))*sin(5*b*x + a + 4*c) + 10*(2*sin(3*b*x + a +
2*c) + sin(b*x + a))*sin(4*b*x + 2*a + 2*c) + 4*(2*sin(3*b*x + a + 2*c) + sin(b*x + a))*sin(4*b*x + 4*c) + 4*
(2*sin(2*b*x + 2*a) + 5*sin(2*b*x + 2*c))*sin(3*b*x + a + 2*c) + 4*sin(2*b*x + 2*a)*sin(b*x + a) + 10*sin(2*b*
x + 2*c)*sin(b*x + a) + 2*cos(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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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.01 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*tan(c + b*x)^3,x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos {\left (a + b x \right )} \tan ^{3}{\left (b x + c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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