3.39 \(\int \sec (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

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

[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)

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Rubi [A]  time = 0.0428867, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {4047, 3767, 8, 12, 3768, 3770} \[ \frac{B \tan (c+d x)}{d}+\frac{C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, 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 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.0173433, size = 47, normalized size = 1. \[ \frac{B \tan (c+d x)}{d}+\frac{C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.023, size = 51, normalized size = 1.1 \begin{align*}{\frac{B\tan \left ( dx+c \right ) }{d}}+{\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*tan(d*x+c)/d+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))

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Maxima [A]  time = 0.93785, size = 78, normalized size = 1.66 \begin{align*} -\frac{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 )} - 4 \, B \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.494562, size = 198, normalized size = 4.21 \begin{align*} \frac{C \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - C \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, B \cos \left (d x + c\right ) + C\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (B + C \sec{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.22851, size = 144, normalized size = 3.06 \begin{align*} \frac{C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - C \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (2 \, 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 )^{3} - 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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