3.427 \(\int (a+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=135 \[ \frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a b \tan (c+d x) \sec (c+d x)}{4 d} \]

[Out]

3/4*a*b*arctanh(sin(d*x+c))/d+1/5*(4*a^2+5*b^2)*tan(d*x+c)/d+3/4*a*b*sec(d*x+c)*tan(d*x+c)/d+1/2*a*b*sec(d*x+c
)^3*tan(d*x+c)/d+1/5*a^2*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(4*a^2+5*b^2)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.11, antiderivative size = 135, 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 = {2789, 3768, 3770, 3012, 3767} \[ \frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a b \tan (c+d x) \sec (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^6,x]

[Out]

(3*a*b*ArcTanh[Sin[c + d*x]])/(4*d) + ((4*a^2 + 5*b^2)*Tan[c + d*x])/(5*d) + (3*a*b*Sec[c + d*x]*Tan[c + d*x])
/(4*d) + (a*b*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) + (a^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((4*a^2 + 5*b^2)*
Tan[c + d*x]^3)/(15*d)

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3012

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

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+b \cos (c+d x))^2 \sec ^6(c+d x) \, dx &=(2 a b) \int \sec ^5(c+d x) \, dx+\int \left (a^2+b^2 \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx\\ &=\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} (3 a b) \int \sec ^3(c+d x) \, dx+\frac {1}{5} \left (4 a^2+5 b^2\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 a b) \int \sec (c+d x) \, dx-\frac {\left (4 a^2+5 b^2\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {3 a b \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (4 a^2+5 b^2\right ) \tan (c+d x)}{5 d}+\frac {3 a b \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a b \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {\left (4 a^2+5 b^2\right ) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.56, size = 118, normalized size = 0.87 \[ \frac {a^2 \left (\frac {1}{5} \tan ^5(c+d x)+\frac {2}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac {a b \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac {3 a b \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{4 d}+\frac {b^2 \left (\frac {1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*Sec[c + d*x]^6,x]

[Out]

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

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fricas [A]  time = 1.43, size = 136, normalized size = 1.01 \[ \frac {45 \, a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (45 \, a b \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, a b \cos \left (d x + c\right ) + 4 \, {\left (4 \, a^{2} + 5 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 12 \, a^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/120*(45*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*a*b*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(45*a*b*
cos(d*x + c)^3 + 8*(4*a^2 + 5*b^2)*cos(d*x + c)^4 + 30*a*b*cos(d*x + c) + 4*(4*a^2 + 5*b^2)*cos(d*x + c)^2 + 1
2*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [B]  time = 0.61, size = 272, normalized size = 2.01 \[ \frac {45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 232 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/60*(45*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(60*a^2*tan(1/
2*d*x + 1/2*c)^9 - 75*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*b^2*tan(1/2*d*x + 1/2*c)^9 - 80*a^2*tan(1/2*d*x + 1/2*c)
^7 + 30*a*b*tan(1/2*d*x + 1/2*c)^7 - 160*b^2*tan(1/2*d*x + 1/2*c)^7 + 232*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*b^2
*tan(1/2*d*x + 1/2*c)^5 - 80*a^2*tan(1/2*d*x + 1/2*c)^3 - 30*a*b*tan(1/2*d*x + 1/2*c)^3 - 160*b^2*tan(1/2*d*x
+ 1/2*c)^3 + 60*a^2*tan(1/2*d*x + 1/2*c) + 75*a*b*tan(1/2*d*x + 1/2*c) + 60*b^2*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 = 0.10, size = 157, normalized size = 1.16 \[ \frac {8 a^{2} \tan \left (d x +c \right )}{15 d}+\frac {a^{2} \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 a^{2} \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d}+\frac {a b \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {2 b^{2} \tan \left (d x +c \right )}{3 d}+\frac {b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*sec(d*x+c)^6,x)

[Out]

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

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maxima [A]  time = 0.55, size = 132, normalized size = 0.98 \[ \frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} b^{2} - 15 \, a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/120*(8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*b
^2 - 15*a*b*(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 = 3.23, size = 221, normalized size = 1.64 \[ \frac {3\,a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}-\frac {\left (2\,a^2-\frac {5\,a\,b}{2}+2\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {8\,a^2}{3}+a\,b-\frac {16\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,a^2}{15}+\frac {20\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,a^2}{3}-a\,b-\frac {16\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a^2+\frac {5\,a\,b}{2}+2\,b^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cos(c + d*x))^2/cos(c + d*x)^6,x)

[Out]

(3*a*b*atanh(tan(c/2 + (d*x)/2)))/(2*d) - (tan(c/2 + (d*x)/2)^5*((116*a^2)/15 + (20*b^2)/3) + tan(c/2 + (d*x)/
2)^9*(2*a^2 - (5*a*b)/2 + 2*b^2) - tan(c/2 + (d*x)/2)^3*(a*b + (8*a^2)/3 + (16*b^2)/3) - tan(c/2 + (d*x)/2)^7*
((8*a^2)/3 - a*b + (16*b^2)/3) + tan(c/2 + (d*x)/2)*((5*a*b)/2 + 2*a^2 + 2*b^2))/(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(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))**2*sec(d*x+c)**6,x)

[Out]

Timed out

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