3.223 \(\int (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=47 \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d}+\frac {C \tan (c+d x)}{d} \]

[Out]

1/2*B*arctanh(sin(d*x+c))/d+C*tan(d*x+c)/d+1/2*B*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.08, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3010, 2748, 3768, 3770, 3767, 8} \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d}+\frac {C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(B*ArcTanh[Sin[c + d*x]])/(2*d) + (C*Tan[c + d*x])/d + (B*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 8

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3010

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x
_Symbol] :> Dist[1/b, Int[(b*Sin[e + f*x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x
]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\int (B+C \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=B \int \sec ^3(c+d x) \, dx+C \int \sec ^2(c+d x) \, dx\\ &=\frac {B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} B \int \sec (c+d x) \, dx-\frac {C \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {C \tan (c+d x)}{d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 47, normalized size = 1.00 \[ \frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d}+\frac {C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(B*ArcTanh[Sin[c + d*x]])/(2*d) + (C*Tan[c + d*x])/d + (B*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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fricas [A]  time = 0.42, size = 74, normalized size = 1.57 \[ \frac {B \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C \cos \left (d x + c\right ) + B\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/4*(B*cos(d*x + c)^2*log(sin(d*x + c) + 1) - B*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*cos(d*x + c) +
B)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [B]  time = 0.51, size = 105, normalized size = 2.23 \[ \frac {B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/2*(B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - B*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(B*tan(1/2*d*x + 1/2*c)^3
- 2*C*tan(1/2*d*x + 1/2*c)^3 + B*tan(1/2*d*x + 1/2*c) + 2*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1
)^2)/d

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maple [A]  time = 0.24, size = 51, normalized size = 1.09 \[ \frac {C \tan \left (d x +c \right )}{d}+\frac {B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

C*tan(d*x+c)/d+1/2*B*sec(d*x+c)*tan(d*x+c)/d+1/2/d*B*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.54, size = 58, normalized size = 1.23 \[ -\frac {B {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, C \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

-1/4*(B*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 4*C*tan(d*x +
c))/d

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mupad [B]  time = 1.62, size = 81, normalized size = 1.72 \[ \frac {\left (B-2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*cos(c + d*x) + C*cos(c + d*x)^2)/cos(c + d*x)^4,x)

[Out]

(tan(c/2 + (d*x)/2)*(B + 2*C) + tan(c/2 + (d*x)/2)^3*(B - 2*C))/(d*(tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2
)^2 + 1)) + (B*atanh(tan(c/2 + (d*x)/2)))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (B + C \cos {\left (c + d x \right )}\right ) \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Integral((B + C*cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**4, x)

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