3.37 \(\int \sec ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx\)

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

[Out]

a*arctanh(sin(d*x+c))/d+b*sec(d*x+c)/d

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Rubi [A]  time = 0.05, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3090, 3770, 2606, 8} \[ \frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (b*Sec[c + d*x])/d

Rule 8

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> Int[ExpandTrig[cos[c + d*x]^m*(a*cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] &&
IntegerQ[m] && IGtQ[n, 0]

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 ^2(c+d x) (a \cos (c+d x)+b \sin (c+d x)) \, dx &=\int (a \sec (c+d x)+b \sec (c+d x) \tan (c+d x)) \, dx\\ &=a \int \sec (c+d x) \, dx+b \int \sec (c+d x) \tan (c+d x) \, dx\\ &=\frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \operatorname {Subst}(\int 1 \, dx,x,\sec (c+d x))}{d}\\ &=\frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \sec (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \[ \frac {a \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a*Cos[c + d*x] + b*Sin[c + d*x]),x]

[Out]

(a*ArcTanh[Sin[c + d*x]])/d + (b*Sec[c + d*x])/d

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fricas [B]  time = 0.72, size = 54, normalized size = 2.25 \[ \frac {a \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - a \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, b}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(a*cos(d*x + c)*log(sin(d*x + c) + 1) - a*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*b)/(d*cos(d*x + c))

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giac [B]  time = 0.23, size = 54, normalized size = 2.25 \[ \frac {a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, b}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="giac")

[Out]

(a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*b/(tan(1/2*d*x + 1/2*c)^2 - 1
))/d

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maple [A]  time = 1.77, size = 34, normalized size = 1.42 \[ \frac {a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b}{d \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

1/d*a*ln(sec(d*x+c)+tan(d*x+c))+1/d*b/cos(d*x+c)

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maxima [A]  time = 0.32, size = 40, normalized size = 1.67 \[ \frac {a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, b}{\cos \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*(a*cos(d*x+c)+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/2*(a*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*b/cos(d*x + c))/d

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mupad [B]  time = 0.41, size = 38, normalized size = 1.58 \[ \frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,b}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*cos(c + d*x) + b*sin(c + d*x))/cos(c + d*x)^2,x)

[Out]

(2*a*atanh(tan(c/2 + (d*x)/2)))/d - (2*b)/(d*(tan(c/2 + (d*x)/2)^2 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(a*cos(d*x+c)+b*sin(d*x+c)),x)

[Out]

Integral((a*cos(c + d*x) + b*sin(c + d*x))*sec(c + d*x)**2, x)

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