3.241 \(\int (a+a \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=113 \[ \frac {a^2 (5 B+6 C) \tan (c+d x)}{3 d}+\frac {a^2 (2 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d} \]

[Out]

1/2*a^2*(2*B+3*C)*arctanh(sin(d*x+c))/d+1/3*a^2*(5*B+6*C)*tan(d*x+c)/d+1/6*a^2*(4*B+3*C)*sec(d*x+c)*tan(d*x+c)
/d+1/3*B*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.36, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3029, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac {a^2 (5 B+6 C) \tan (c+d x)}{3 d}+\frac {a^2 (2 B+3 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{6 d}+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(a^2*(2*B + 3*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a^2*(5*B + 6*C)*Tan[c + d*x])/(3*d) + (a^2*(4*B + 3*C)*Sec[c
+ d*x]*Tan[c + d*x])/(6*d) + (B*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

Rule 8

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

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 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^2*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d
, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

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

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Mathematica [B]  time = 5.89, size = 451, normalized size = 3.99 \[ \frac {a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\frac {4 (5 B+6 C) \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 (5 B+6 C) \sin \left (\frac {d x}{2}\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {(7 B+3 C) \cos \left (\frac {c}{2}\right )-(5 B+3 C) \sin \left (\frac {c}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {(5 B+3 C) \sin \left (\frac {c}{2}\right )+(7 B+3 C) \cos \left (\frac {c}{2}\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-6 (2 B+3 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 (2 B+3 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 B \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2 B \sin \left (\frac {d x}{2}\right )}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(-6*(2*B + 3*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 6*(2*B
+ 3*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*B*Sin[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2]
- Sin[(c + d*x)/2])^3) + ((7*B + 3*C)*Cos[c/2] - (5*B + 3*C)*Sin[c/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/
2] - Sin[(c + d*x)/2])^2) + (4*(5*B + 6*C)*Sin[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d
*x)/2])) + (2*B*Sin[(d*x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) - ((7*B + 3*C)*C
os[c/2] + (5*B + 3*C)*Sin[c/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(5*B + 6*
C)*Sin[(d*x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(48*d)

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fricas [A]  time = 0.45, size = 125, normalized size = 1.11 \[ \frac {3 \, {\left (2 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (5 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/12*(3*(2*B + 3*C)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*B + 3*C)*a^2*cos(d*x + c)^3*log(-sin(d*x +
c) + 1) + 2*(2*(5*B + 6*C)*a^2*cos(d*x + c)^2 + 3*(2*B + C)*a^2*cos(d*x + c) + 2*B*a^2)*sin(d*x + c))/(d*cos(
d*x + c)^3)

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giac [A]  time = 0.47, size = 178, normalized size = 1.58 \[ \frac {3 \, {\left (2 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/6*(3*(2*B*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*B*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) - 2*(6*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 16*B*a^2*tan(1/2*d*x + 1/2*c
)^3 - 24*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 18*B*a^2*tan(1/2*d*x + 1/2*c) + 15*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/
2*d*x + 1/2*c)^2 - 1)^3)/d

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maple [A]  time = 0.36, size = 141, normalized size = 1.25 \[ \frac {3 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {5 a^{2} B \tan \left (d x +c \right )}{3 d}+\frac {2 a^{2} C \tan \left (d x +c \right )}{d}+\frac {a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

3/2/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+5/3*a^2*B*tan(d*x+c)/d+2/d*a^2*C*tan(d*x+c)+a^2*B*sec(d*x+c)*tan(d*x+c)/
d+1/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a^2*C*sec(d*x+c)*tan(d*x+c)+1/3*a^2*B*sec(d*x+c)^2*tan(d*x+c)/d

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maxima [A]  time = 0.48, size = 174, normalized size = 1.54 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 6 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{2} \tan \left (d x + c\right ) + 24 \, C a^{2} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 - 6*B*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)) - 3*C*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)) + 6*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*B*a^2*tan(d*x + c) + 24*C*a
^2*tan(d*x + c))/d

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mupad [B]  time = 2.89, size = 145, normalized size = 1.28 \[ \frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B+\frac {3\,C}{2}\right )}{d}-\frac {\left (2\,B\,a^2+3\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,B\,a^2}{3}-8\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,B\,a^2+5\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

(2*a^2*atanh(tan(c/2 + (d*x)/2))*(B + (3*C)/2))/d - (tan(c/2 + (d*x)/2)*(6*B*a^2 + 5*C*a^2) + tan(c/2 + (d*x)/
2)^5*(2*B*a^2 + 3*C*a^2) - tan(c/2 + (d*x)/2)^3*((16*B*a^2)/3 + 8*C*a^2))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c
/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 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+a*cos(d*x+c))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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