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

Optimal. Leaf size=24 \[ \frac{B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{C \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0180718, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {3770, 3767, 8} \[ \frac{B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3770

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

Rubi steps

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

Mathematica [A]  time = 0.0117249, size = 24, normalized size = 1. \[ \frac{B \tanh ^{-1}(\sin (c+d x))}{d}+\frac{C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 32, normalized size = 1.3 \begin{align*}{\frac{B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{C\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.922269, size = 42, normalized size = 1.75 \begin{align*} \frac{B \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac{C \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*log(sec(d*x + c) + tan(d*x + c))/d + C*tan(d*x + c)/d

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Fricas [B]  time = 0.48849, size = 162, normalized size = 6.75 \begin{align*} \frac{B \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - B \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, C \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.21941, size = 77, normalized size = 3.21 \begin{align*} \frac{B{\left (\log \left ({\left | \frac{1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) + 2 \right |}\right ) - \log \left ({\left | \frac{1}{\sin \left (d x + c\right )} + \sin \left (d x + c\right ) - 2 \right |}\right )\right )}}{4 \, d} + \frac{C \tan \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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