3.5.12 \(\int (a+b \cos (c+d x)) \sec ^2(c+d x) \, dx\) [412]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2827, 3852, 8, 3855} \begin {gather*} \frac {a \tan (c+d x)}{d}+\frac {b \tanh ^{-1}(\sin (c+d x))}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3852

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 3855

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

Rubi steps

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 30, normalized size = 1.25

method	result	size
derivativedivides	\(\frac {a \tan \left (d x +c \right )+b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(30\)
default	\(\frac {a \tan \left (d x +c \right )+b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(30\)
risch	\(\frac {2 i a}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(59\)
norman	\(\frac {-\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(101\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))*sec(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 38, normalized size = 1.58 \begin {gather*} \frac {b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, a \tan \left (d x + c\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (24) = 48\).
time = 0.38, size = 60, normalized size = 2.50 \begin {gather*} \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 \, a \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+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*a*sin(d*x + c))/(d*cos(d
*x + c))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).
time = 0.46, size = 63, normalized size = 2.62 \begin {gather*} \frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.51, size = 47, normalized size = 1.96 \begin {gather*} \frac {2\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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