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

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

[Out]

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

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Rubi [A]  time = 0.0265544, antiderivative size = 24, 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 = {4045, 3770} \[ \frac{A \sin (c+d x)}{d}+\frac{C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, 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 \cos (c+d x) \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \sin (c+d x)}{d}+C \int \sec (c+d x) \, dx\\ &=\frac{C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{A \sin (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.0152265, size = 35, normalized size = 1.46 \[ \frac{A \sin (c) \cos (d x)}{d}+\frac{A \cos (c) \sin (d x)}{d}+\frac{C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*ArcTanh[Sin[c + d*x]])/d + (A*Cos[d*x]*Sin[c])/d + (A*Cos[c]*Sin[d*x])/d

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Maple [A]  time = 0.043, size = 32, normalized size = 1.3 \begin{align*}{\frac{A\sin \left ( dx+c \right ) }{d}}+{\frac{C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.926645, size = 51, normalized size = 2.12 \begin{align*} \frac{C{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.50243, size = 107, normalized size = 4.46 \begin{align*} \frac{C \log \left (\sin \left (d x + c\right ) + 1\right ) - C \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, A \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21605, size = 54, normalized size = 2.25 \begin{align*} \frac{C \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - C \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 2 \, A \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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