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]

1/2*(2*A+C)*arctanh(sin(d*x+c))/d+1/2*C*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.00, 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 3770

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

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]

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*}

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Mathematica [A]  time = 0.02, size = 48, normalized size = 1.20 \[ \frac {A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {C \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {C \tan (c+d x) \sec (c+d x)}{2 d} \]

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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fricas [A]  time = 0.63, size = 72, normalized size = 1.80 \[ \frac {{\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, C \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

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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giac [A]  time = 0.24, size = 60, normalized size = 1.50 \[ \frac {{\left (2 \, A + C\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (2 \, A + C\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, C \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

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

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maple [A]  time = 0.90, size = 59, normalized size = 1.48 \[ \frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

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

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maxima [A]  time = 0.34, size = 58, normalized size = 1.45 \[ \frac {{\left (2 \, A + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + C\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, C \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \]

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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mupad [B]  time = 2.41, size = 41, normalized size = 1.02 \[ \frac {\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A+\frac {C}{2}\right )}{d}-\frac {C\,\sin \left (c+d\,x\right )}{2\,d\,\left ({\sin \left (c+d\,x\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}\, dx \]

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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