3.3.48 \(\int \cos (a+b x) \tan ^2(c+b x) \, dx\) [248]

Optimal. Leaf size=46 \[ \frac {\tanh ^{-1}(\sin (c+b x)) \cos (a-c)}{b}-\frac {\sec (c+b x) \sin (a-c)}{b}-\frac {\sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4675, 4672, 2717, 3855, 2686, 8} \begin {gather*} -\frac {\sin (a-c) \sec (b x+c)}{b}+\frac {\cos (a-c) \tanh ^{-1}(\sin (b x+c))}{b}-\frac {\sin (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Tan[c + b*x]^2,x]

[Out]

(ArcTanh[Sin[c + b*x]]*Cos[a - c])/b - (Sec[c + b*x]*Sin[a - c])/b - Sin[a + b*x]/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 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*} \int \cos (a+b x) \tan ^2(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\\ &=\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 {\tanh ^{-1}(\sin (c+b x)) \cos (a-c)}{b}-\frac {\sec (c+b x) \sin (a-c)}{b}-\frac {\sin (a+b x)}{b}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.10, size = 111, normalized size = 2.41 \begin {gather*} -\frac {2 i \text {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 {\sec (c+b x) \sin (a-c)}{b}-\frac {\cos (a) \sin (b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Tan[c + b*x]^2,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 - (Sec[c + b*x]*Sin[a - c])/b - (Cos[a]*Sin[b*x])/b

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Maple [C] Result contains complex when optimal does not.
time = 0.13, size = 149, normalized size = 3.24

method	result	size
risch	\(\frac {i {\mathrm e}^{i \left (b x +a \right )}}{2 b}-\frac {i {\mathrm e}^{-i \left (b x +a \right )}}{2 b}-\frac {i \left (-{\mathrm e}^{i \left (b x +3 a \right )}+{\mathrm e}^{i \left (b x +a +2 c \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a +c \right )}+{\mathrm e}^{2 i a}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i {\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i {\mathrm e}^{i \left (a -c \right )}\right ) \cos \left (a -c \right )}{b}\)	\(149\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*tan(b*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*I*exp(I*(b*x+a))/b-1/2*I/b*exp(-I*(b*x+a))-I/b/(exp(2*I*(b*x+a+c))+exp(2*I*a))*(-exp(I*(b*x+3*a))+exp(I*(b
*x+a+2*c)))-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)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (46) = 92\).
time = 0.55, size = 526, normalized size = 11.43 \begin {gather*} \frac {{\left (\sin \left (3 \, b x + a + 2 \, c\right ) + \sin \left (b x + a\right )\right )} \cos \left (4 \, b x + 2 \, a + 2 \, c\right ) - 3 \, {\left (\sin \left (2 \, b x + 2 \, a\right ) - \sin \left (2 \, b x + 2 \, c\right )\right )} \cos \left (3 \, b x + a + 2 \, c\right ) - {\left (\cos \left (3 \, b x + a + 2 \, c\right )^{2} \cos \left (-a + c\right ) + 2 \, \cos \left (3 \, b x + a + 2 \, c\right ) \cos \left (b x + a\right ) \cos \left (-a + c\right ) + \cos \left (b x + a\right )^{2} \cos \left (-a + c\right ) + \cos \left (-a + c\right ) \sin \left (3 \, b x + a + 2 \, c\right )^{2} + 2 \, \cos \left (-a + c\right ) \sin \left (3 \, b x + a + 2 \, c\right ) \sin \left (b x + a\right ) + \cos \left (-a + c\right ) \sin \left (b x + a\right )^{2}\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 ) - {\left (\cos \left (3 \, b x + a + 2 \, c\right ) + \cos \left (b x + a\right )\right )} \sin \left (4 \, b x + 2 \, a + 2 \, c\right ) + {\left (3 \, \cos \left (2 \, b x + 2 \, a\right ) - 3 \, \cos \left (2 \, b x + 2 \, c\right ) - 1\right )} \sin \left (3 \, b x + a + 2 \, c\right ) - 3 \, \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 3 \, \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, c\right ) + 3 \, \cos \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) - 3 \, \cos \left (2 \, b x + 2 \, c\right ) \sin \left (b x + a\right ) - \sin \left (b x + a\right )}{2 \, {\left (b \cos \left (3 \, b x + a + 2 \, c\right )^{2} + 2 \, b \cos \left (3 \, b x + a + 2 \, c\right ) \cos \left (b x + a\right ) + b \cos \left (b x + a\right )^{2} + b \sin \left (3 \, b x + a + 2 \, c\right )^{2} + 2 \, b \sin \left (3 \, b x + a + 2 \, c\right ) \sin \left (b x + a\right ) + b \sin \left (b x + a\right )^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*tan(b*x+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (46) = 92\).
time = 2.66, size = 316, normalized size = 6.87 \begin {gather*} \frac {4 \, {\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + \frac {\sqrt {2} {\left ({\left (\cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - {\left (\cos \left (-2 \, a + 2 \, c\right )^{2} + 2 \, \cos \left (-2 \, a + 2 \, c\right ) + 1\right )} \cos \left (b x + a\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}} + 4 \, {\left (\cos \left (b x + a\right )^{2} - 2\right )} \sin \left (-2 \, a + 2 \, c\right )}{4 \, {\left (b \sin \left (b x + a\right ) \sin \left (-2 \, a + 2 \, c\right ) - {\left (b \cos \left (-2 \, a + 2 \, c\right ) + b\right )} \cos \left (b x + a\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*tan(b*x+c)^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \cos {\left (a + b x \right )} \tan ^{2}{\left (b x + c \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*tan(b*x+c)**2,x)

[Out]

Integral(cos(a + b*x)*tan(b*x + c)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*tan(b*x+c)^2,x, algorithm="giac")

[Out]

integrate(cos(b*x + a)*tan(b*x + c)^2, x)

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Mupad [B]
time = 5.24, size = 285, normalized size = 6.20 \begin {gather*} -\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 {{\mathrm {e}}^{a\,1{}\mathrm {i}+b\,x\,1{}\mathrm {i}}\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}-1\right )}{b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}-c\,2{}\mathrm {i}}\,1{}\mathrm {i}+{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}\right )}+\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}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*tan(c + b*x)^2,x)

[Out]

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

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