\(\int \cos (a+b x) \tan ^3(c+b x) \, dx\) [247]
Optimal result
Integrand size = 15, antiderivative size = 72 \[
\int \cos (a+b x) \tan ^3(c+b x) \, dx=\frac {\cos (a+b x)}{b}+\frac {\cos (a-c) \sec (c+b x)}{b}+\frac {3 \text {arctanh}(\sin (c+b x)) \sin (a-c)}{2 b}-\frac {\sec (c+b x) \sin (a-c) \tan (c+b x)}{2 b}
\]
[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
Rubi [A] (verified)
Time = 0.10 (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 = {4675, 4672, 2718, 3855, 2686,
8, 2691} \[
\int \cos (a+b x) \tan ^3(c+b x) \, dx=\frac {3 \sin (a-c) \text {arctanh}(\sin (b x+c))}{2 b}+\frac {\cos (a-c) \sec (b x+c)}{b}-\frac {\sin (a-c) \tan (b x+c) \sec (b x+c)}{2 b}+\frac {\cos (a+b x)}{b}
\]
[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 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 2718
Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[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}& = -\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 {\text {arctanh}(\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) \text {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 \text {arctanh}(\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*}
Mathematica [A] (verified)
Time = 0.45 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.97
\[
\int \cos (a+b x) \tan ^3(c+b x) \, dx=\frac {(2 \cos (a-2 c-b x)+5 \cos (a+b x)+\cos (a+2 c+3 b x)) \sec ^2(c+b x)+12 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {b x}{2}\right )\right ) \sin (a-c)}{4 b}
\]
[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)
Maple [C] (verified)
Result contains complex when optimal does not.
Time = 0.34 (sec) , antiderivative size = 181, normalized size of antiderivative = 2.51
| | |
method | result | size |
| | |
risch |
\(\frac {{\mathrm e}^{i \left (x b +a \right )}}{2 b}+\frac {{\mathrm e}^{-i \left (x b +a \right )}}{2 b}+\frac {3 \,{\mathrm e}^{i \left (3 x b +5 a +2 c \right )}+{\mathrm e}^{i \left (3 x b +3 a +4 c \right )}+{\mathrm e}^{i \left (x b +5 a \right )}+3 \,{\mathrm e}^{i \left (x b +3 a +2 c \right )}}{2 b \left ({\mathrm e}^{2 i \left (x b +a +c \right )}+{\mathrm e}^{2 i a}\right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}+\frac {3 \ln \left ({\mathrm e}^{i \left (x b +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \sin \left (a -c \right )}{2 b}\) |
\(181\) |
| | |
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[In]
int(cos(b*x+a)*tan(b*x+c)^3,x,method=_RETURNVERBOSE)
[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)
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 366 vs. \(2 (68) = 136\).
Time = 0.28 (sec) , antiderivative size = 366, normalized size of antiderivative = 5.08
\[
\int \cos (a+b x) \tan ^3(c+b x) \, dx=\frac {16 \, \cos \left (b x + a\right )^{3} \cos \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - 4 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) - 5\right )} \cos \left (b x + a\right ) + \frac {3 \, \sqrt {2} {\left (2 \, {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} - 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + {\left (2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (-2 \, a + 2 \, c\right )\right )} \log \left (-\frac {2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) + \frac {2 \, \sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) + \cos \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right )\right )}}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}} - \cos \left (-2 \, a + 2 \, c\right ) - 3}{2 \, \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - \cos \left (-2 \, a + 2 \, c\right ) + 1}\right )}{\sqrt {\cos \left (-2 \, a + 2 \, c\right ) + 1}}}{8 \, {\left (2 \, b \cos \left (b x + a\right )^{2} \cos \left (-2 \, a + 2 \, c\right ) - 2 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - b \cos \left (-2 \, a + 2 \, c\right ) + b\right )}}
\]
[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)
Sympy [F]
\[
\int \cos (a+b x) \tan ^3(c+b x) \, dx=\int \cos {\left (a + b x \right )} \tan ^{3}{\left (b x + c \right )}\, dx
\]
[In]
integrate(cos(b*x+a)*tan(b*x+c)**3,x)
[Out]
Integral(cos(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 = 0.39 (sec) , antiderivative size = 1027, normalized size of antiderivative = 14.26
\[
\int \cos (a+b x) \tan ^3(c+b x) \, dx=\text {Too large to display}
\]
[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))
Giac [F]
\[
\int \cos (a+b x) \tan ^3(c+b x) \, dx=\int { \cos \left (b x + a\right ) \tan \left (b x + c\right )^{3} \,d x }
\]
[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)
Mupad [F(-1)]
Timed out. \[
\int \cos (a+b x) \tan ^3(c+b x) \, dx=\text {Hanged}
\]
[In]
int(cos(a + b*x)*tan(c + b*x)^3,x)
[Out]
\text{Hanged}