3.6 \(\int \sec (c+d x) (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=40 \[ \frac{(2 A+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0254462, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {4046, 3770} \[ \frac{(2 A+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

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

Mathematica [A]  time = 0.0137766, size = 48, normalized size = 1.2 \[ \frac{A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.026, size = 59, normalized size = 1.5 \begin{align*}{\frac{A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \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)*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*A*ln(sec(d*x+c)+tan(d*x+c))+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 = 0.927594, size = 78, normalized size = 1.95 \begin{align*} \frac{{\left (2 \, A + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, A + C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \, C \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.49434, size = 192, normalized size = 4.8 \begin{align*} \frac{{\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, C \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)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/4*((2*A + C)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*A + C)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*C*si
n(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 (A + C \sec ^{2}{\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(sec(d*x+c)*(A+C*sec(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.1852, size = 81, normalized size = 2.02 \begin{align*} \frac{{\left (2 \, A + C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (2 \, A + C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \, C \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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