3.19 \(\int (a+a \cos (c+d x))^2 \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=34 \[ \frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}+a^2 x \]

[Out]

a^2*x+2*a^2*arctanh(sin(d*x+c))/d+a^2*tan(d*x+c)/d

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Rubi [A]  time = 0.06, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2757, 3770, 3767, 8} \[ \frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \tanh ^{-1}(\sin (c+d x))}{d}+a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*x + (2*a^2*ArcTanh[Sin[c + d*x]])/d + (a^2*Tan[c + d*x])/d

Rule 8

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

Rule 2757

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(a + b*sin[e + f*x])^m*(d*sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] &
& IGtQ[m, 0] && RationalQ[n]

Rule 3767

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.82 \[ a^2 \left (\frac {\tan (c+d x)}{d}+\frac {2 \tanh ^{-1}(\sin (c+d x))}{d}+x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*(x + (2*ArcTanh[Sin[c + d*x]])/d + Tan[c + d*x]/d)

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fricas [B]  time = 1.61, size = 76, normalized size = 2.24 \[ \frac {a^{2} d x \cos \left (d x + c\right ) + a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + a^{2} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.58, size = 79, normalized size = 2.32 \[ \frac {{\left (d x + c\right )} a^{2} + 2 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.11, size = 50, normalized size = 1.47 \[ a^{2} x +\frac {2 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} \tan \left (d x +c \right )}{d}+\frac {a^{2} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.99, size = 49, normalized size = 1.44 \[ \frac {{\left (d x + c\right )} a^{2} + a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + a^{2} \tan \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.39, size = 56, normalized size = 1.65 \[ a^2\,x+\frac {4\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {2\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + (4*a^2*atanh(tan(c/2 + (d*x)/2)))/d - (2*a^2*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d*x)/2)^2 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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