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

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

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 34, 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 = {3788, 8, 4045, 3770} \[ \frac {a^2 \sin (c+d x)}{d}+\frac {a^2 \tanh ^{-1}(\sin (c+d x))}{d}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 3770

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

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 47, normalized size = 1.38 \[ \frac {a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \sin (c) \cos (d x)}{d}+\frac {a^2 \cos (c) \sin (d x)}{d}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 53, normalized size = 1.56 \[ \frac {4 \, a^{2} d x + a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a^2*d*x + a^2*log(sin(d*x + c) + 1) - a^2*log(-sin(d*x + c) + 1) + 2*a^2*sin(d*x + c))/d

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giac [B]  time = 0.41, size = 79, normalized size = 2.32 \[ \frac {2 \, {\left (d x + c\right )} a^{2} + a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(d*x + c)*a^2 + a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*a^2*tan
(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 = 0.54, size = 51, normalized size = 1.50 \[ 2 a^{2} x +\frac {a^{2} \sin \left (d x +c \right )}{d}+\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{2} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.83, size = 52, normalized size = 1.53 \[ \frac {4 \, {\left (d x + c\right )} a^{2} + a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, a^{2} \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(d*x + c)*a^2 + a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*a^2*sin(d*x + c))/d

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mupad [B]  time = 0.70, size = 33, normalized size = 0.97 \[ 2\,a^2\,x+\frac {a^2\,\left (2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+\sin \left (c+d\,x\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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