3.224 \(\int (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=63 \[ \frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d}+\frac{C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Tan[c + d*x])/d + (C*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (B*Tan[c + d*x]^3
)/(3*d)

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Rubi [A]  time = 0.0800781, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3010, 2748, 3767, 3768, 3770} \[ \frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d}+\frac{C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Tan[c + d*x])/d + (C*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (B*Tan[c + d*x]^3
)/(3*d)

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 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 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 ^5(c+d x) \, dx &=\int (B+C \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=B \int \sec ^4(c+d x) \, dx+C \int \sec ^3(c+d x) \, dx\\ &=\frac{C \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} C \int \sec (c+d x) \, dx-\frac{B \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{B \tan (c+d x)}{d}+\frac{C \sec (c+d x) \tan (c+d x)}{2 d}+\frac{B \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.151043, size = 60, normalized size = 0.95 \[ \frac{B \left (\frac{1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac{C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(C*ArcTanh[Sin[c + d*x]])/(2*d) + (C*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (B*(Tan[c + d*x] + Tan[c + d*x]^3/3))/
d

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Maple [A]  time = 0.039, size = 72, normalized size = 1.1 \begin{align*}{\frac{C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

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)^5,x)

[Out]

1/2*C*sec(d*x+c)*tan(d*x+c)/d+1/2/d*C*ln(sec(d*x+c)+tan(d*x+c))+2/3*B*tan(d*x+c)/d+1/3/d*B*tan(d*x+c)*sec(d*x+
c)^2

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Maxima [A]  time = 1.33846, size = 95, normalized size = 1.51 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B - 3 \, C{\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 )}}{12 \, d} \end{align*}

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)^5,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B - 3*C*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1)
+ log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 1.67288, size = 236, normalized size = 3.75 \begin{align*} \frac{3 \, C \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (4 \, B \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + 2 \, B\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

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)^5,x, algorithm="fricas")

[Out]

1/12*(3*C*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*C*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(4*B*cos(d*x +
c)^2 + 3*C*cos(d*x + c) + 2*B)*sin(d*x + c))/(d*cos(d*x + c)^3)

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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((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.32756, size = 165, normalized size = 2.62 \begin{align*} \frac{3 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, 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 )}^{3}}}{6 \, d} \end{align*}

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)^5,x, algorithm="giac")

[Out]

1/6*(3*C*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*C*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*tan(1/2*d*x + 1/
2*c)^5 - 3*C*tan(1/2*d*x + 1/2*c)^5 - 4*B*tan(1/2*d*x + 1/2*c)^3 + 6*B*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d