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

Optimal. Leaf size=101 \[ \frac {a \tan ^5(c+d x)}{5 d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

3/8*a*arctanh(sin(d*x+c))/d+a*tan(d*x+c)/d+3/8*a*sec(d*x+c)*tan(d*x+c)/d+1/4*a*sec(d*x+c)^3*tan(d*x+c)/d+2/3*a
*tan(d*x+c)^3/d+1/5*a*tan(d*x+c)^5/d

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Rubi [A]  time = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2748, 3767, 3768, 3770} \[ \frac {a \tan ^5(c+d x)}{5 d}+\frac {2 a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + (a*Tan[c + d*x])/d + (3*a*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*Sec[c + d*
x]^3*Tan[c + d*x])/(4*d) + (2*a*Tan[c + d*x]^3)/(3*d) + (a*Tan[c + d*x]^5)/(5*d)

Rule 2748

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

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Mathematica [A]  time = 0.25, size = 65, normalized size = 0.64 \[ \frac {a \left (45 \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (24 \tan ^4(c+d x)+80 \tan ^2(c+d x)+30 \sec ^3(c+d x)+45 \sec (c+d x)+120\right )\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(45*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(120 + 45*Sec[c + d*x] + 30*Sec[c + d*x]^3 + 80*Tan[c + d*x]^2 + 2
4*Tan[c + d*x]^4)))/(120*d)

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fricas [A]  time = 1.22, size = 110, normalized size = 1.09 \[ \frac {45 \, a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (64 \, a \cos \left (d x + c\right )^{4} + 45 \, a \cos \left (d x + c\right )^{3} + 32 \, a \cos \left (d x + c\right )^{2} + 30 \, a \cos \left (d x + c\right ) + 24 \, a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(45*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*a*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(64*a*cos(d*
x + c)^4 + 45*a*cos(d*x + c)^3 + 32*a*cos(d*x + c)^2 + 30*a*cos(d*x + c) + 24*a)*sin(d*x + c))/(d*cos(d*x + c)
^5)

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giac [A]  time = 1.45, size = 124, normalized size = 1.23 \[ \frac {45 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 130 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 190 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(45*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(45*a*tan(1/2*d*x
+ 1/2*c)^9 - 130*a*tan(1/2*d*x + 1/2*c)^7 + 464*a*tan(1/2*d*x + 1/2*c)^5 - 190*a*tan(1/2*d*x + 1/2*c)^3 + 195
*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 0.14, size = 112, normalized size = 1.11 \[ \frac {a \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {8 a \tan \left (d x +c \right )}{15 d}+\frac {a \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 a \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 107, normalized size = 1.06 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a - 15 \, a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*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 - 15*a*(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)))/d

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mupad [B]  time = 4.77, size = 158, normalized size = 1.56 \[ \frac {3\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d}-\frac {\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}-\frac {13\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{6}+\frac {116\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {19\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+\frac {13\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*a*atanh(tan(c/2 + (d*x)/2)))/(4*d) - ((13*a*tan(c/2 + (d*x)/2))/4 - (19*a*tan(c/2 + (d*x)/2)^3)/6 + (116*a*
tan(c/2 + (d*x)/2)^5)/15 - (13*a*tan(c/2 + (d*x)/2)^7)/6 + (3*a*tan(c/2 + (d*x)/2)^9)/4)/(d*(5*tan(c/2 + (d*x)
/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1
))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \cos {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx + \int \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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