3.32 \(\int (a+a \cos (c+d x))^3 \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=114 \[ \frac{a^3 \tan ^5(c+d x)}{5 d}+\frac{5 a^3 \tan ^3(c+d x)}{3 d}+\frac{4 a^3 \tan (c+d x)}{d}+\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{13 a^3 \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

(13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a^3*Tan[c + d*x])/d + (13*a^3*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (3*
a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^3*Tan[c + d*x]^3)/(3*d) + (a^3*Tan[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.127425, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2757, 3768, 3770, 3767} \[ \frac{a^3 \tan ^5(c+d x)}{5 d}+\frac{5 a^3 \tan ^3(c+d x)}{3 d}+\frac{4 a^3 \tan (c+d x)}{d}+\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{13 a^3 \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*Sec[c + d*x]^6,x]

[Out]

(13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (4*a^3*Tan[c + d*x])/d + (13*a^3*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (3*
a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^3*Tan[c + d*x]^3)/(3*d) + (a^3*Tan[c + d*x]^5)/(5*d)

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 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]

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]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x))^3 \sec ^6(c+d x) \, dx &=\int \left (a^3 \sec ^3(c+d x)+3 a^3 \sec ^4(c+d x)+3 a^3 \sec ^5(c+d x)+a^3 \sec ^6(c+d x)\right ) \, dx\\ &=a^3 \int \sec ^3(c+d x) \, dx+a^3 \int \sec ^6(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^4(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^5(c+d x) \, dx\\ &=\frac{a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{2} a^3 \int \sec (c+d x) \, dx+\frac{1}{4} \left (9 a^3\right ) \int \sec ^3(c+d x) \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (1+2 x^2+x^4\right ) \, dx,x,-\tan (c+d x)\right )}{d}-\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{4 a^3 \tan (c+d x)}{d}+\frac{13 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{5 a^3 \tan ^3(c+d x)}{3 d}+\frac{a^3 \tan ^5(c+d x)}{5 d}+\frac{1}{8} \left (9 a^3\right ) \int \sec (c+d x) \, dx\\ &=\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{4 a^3 \tan (c+d x)}{d}+\frac{13 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{3 a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{5 a^3 \tan ^3(c+d x)}{3 d}+\frac{a^3 \tan ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [B]  time = 1.31052, size = 487, normalized size = 4.27 \[ -\frac{a^3 \sec (c) \sec ^5(c+d x) \left (1440 \sin (2 c+d x)-1500 \sin (c+2 d x)-1500 \sin (3 c+2 d x)-3040 \sin (2 c+3 d x)-390 \sin (3 c+4 d x)-390 \sin (5 c+4 d x)-608 \sin (4 c+5 d x)+975 \cos (2 c+3 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+975 \cos (4 c+3 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+195 \cos (4 c+5 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+195 \cos (6 c+5 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+1950 \cos (d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+1950 \cos (2 c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-975 \cos (2 c+3 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-975 \cos (4 c+3 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-195 \cos (4 c+5 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-195 \cos (6 c+5 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-4640 \sin (d x)\right )}{3840 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*Sec[c + d*x]^6,x]

[Out]

-(a^3*Sec[c]*Sec[c + d*x]^5*(975*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 975*Cos[4*c + 3*d
*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 195*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] +
195*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 1950*Cos[d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 1950*Cos[2*c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 975*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*
x)/2]] - 975*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 195*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x
)/2] + Sin[(c + d*x)/2]] - 195*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 4640*Sin[d*x] + 144
0*Sin[2*c + d*x] - 1500*Sin[c + 2*d*x] - 1500*Sin[3*c + 2*d*x] - 3040*Sin[2*c + 3*d*x] - 390*Sin[3*c + 4*d*x]
- 390*Sin[5*c + 4*d*x] - 608*Sin[4*c + 5*d*x]))/(3840*d)

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Maple [A]  time = 0.086, size = 124, normalized size = 1.1 \begin{align*}{\frac{13\,{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{13\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{38\,{a}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{19\,{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{3\,{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)^3*sec(d*x+c)^6,x)

[Out]

13/8/d*a^3*sec(d*x+c)*tan(d*x+c)+13/8/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+38/15*a^3*tan(d*x+c)/d+19/15/d*a^3*tan(d
*x+c)*sec(d*x+c)^2+3/4/d*a^3*tan(d*x+c)*sec(d*x+c)^3+1/5/d*a^3*tan(d*x+c)*sec(d*x+c)^4

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Maxima [A]  time = 1.1471, size = 242, normalized size = 2.12 \begin{align*} \frac{16 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{3} - 45 \, a^{3}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, a^{3}{\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 )}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))
*a^3 - 45*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x +
c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + lo
g(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 1.69043, size = 329, normalized size = 2.89 \begin{align*} \frac{195 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 195 \, a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (304 \, a^{3} \cos \left (d x + c\right )^{4} + 195 \, a^{3} \cos \left (d x + c\right )^{3} + 152 \, a^{3} \cos \left (d x + c\right )^{2} + 90 \, a^{3} \cos \left (d x + c\right ) + 24 \, a^{3}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(195*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 195*a^3*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(304*a
^3*cos(d*x + c)^4 + 195*a^3*cos(d*x + c)^3 + 152*a^3*cos(d*x + c)^2 + 90*a^3*cos(d*x + c) + 24*a^3)*sin(d*x +
c))/(d*cos(d*x + c)^5)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**3*sec(d*x+c)**6,x)

[Out]

Timed out

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Giac [A]  time = 1.36579, size = 186, normalized size = 1.63 \begin{align*} \frac{195 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 195 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (195 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 910 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1664 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1330 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 765 \, a^{3} \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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(195*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 195*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(195*a^3*ta
n(1/2*d*x + 1/2*c)^9 - 910*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*a^3*tan(1/2*d*x + 1/2*c)^5 - 1330*a^3*tan(1/2*d*x
+ 1/2*c)^3 + 765*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d