3.20 \(\int (a+a \cos (c+d x))^2 \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=54 \[ \frac {2 a^2 \tan (c+d x)}{d}+\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

3/2*a^2*arctanh(sin(d*x+c))/d+2*a^2*tan(d*x+c)/d+1/2*a^2*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.08, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2757, 3770, 3767, 8, 3768} \[ \frac {2 a^2 \tan (c+d x)}{d}+\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3,x]

[Out]

(3*a^2*ArcTanh[Sin[c + d*x]])/(2*d) + (2*a^2*Tan[c + d*x])/d + (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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 ^3(c+d x) \, dx &=\int \left (a^2 \sec (c+d x)+2 a^2 \sec ^2(c+d x)+a^2 \sec ^3(c+d x)\right ) \, dx\\ &=a^2 \int \sec (c+d x) \, dx+a^2 \int \sec ^3(c+d x) \, dx+\left (2 a^2\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} a^2 \int \sec (c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{d}\\ &=\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 54, normalized size = 1.00 \[ \frac {2 a^2 \tan (c+d x)}{d}+\frac {3 a^2 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3,x]

[Out]

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

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fricas [A]  time = 1.34, size = 83, normalized size = 1.54 \[ \frac {3 \, a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

1/4*(3*a^2*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 3*a^2*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(4*a^2*cos(d
*x + c) + a^2)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [A]  time = 0.76, size = 90, normalized size = 1.67 \[ \frac {3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="giac")

[Out]

1/2*(3*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*a^2*tan(1/2*d*
x + 1/2*c)^3 - 5*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

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maple [A]  time = 0.13, size = 58, normalized size = 1.07 \[ \frac {3 a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} \tan \left (d x +c \right )}{d}+\frac {a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x)

[Out]

3/2/d*a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a^2*tan(d*x+c)/d+1/2*a^2*sec(d*x+c)*tan(d*x+c)/d

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maxima [A]  time = 0.45, size = 88, normalized size = 1.63 \[ -\frac {a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 8 \, a^{2} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/4*(a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 2*a^2*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 8*a^2*tan(d*x + c))/d

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mupad [B]  time = 0.71, size = 83, normalized size = 1.54 \[ \frac {3\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-5\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*cos(c + d*x))^2/cos(c + d*x)^3,x)

[Out]

(3*a^2*atanh(tan(c/2 + (d*x)/2)))/d - (3*a^2*tan(c/2 + (d*x)/2)^3 - 5*a^2*tan(c/2 + (d*x)/2))/(d*(tan(c/2 + (d
*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^2 + 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**2*sec(d*x+c)**3,x)

[Out]

a**2*(Integral(2*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(se
c(c + d*x)**3, x))

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