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

Optimal. Leaf size=125 \[ -\frac {a (-3 (A+B+C)+A+B) \tan (c+d x)}{3 d}+\frac {a (3 A+4 (B+C)) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (3 A+4 (B+C)) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a (A+B) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[Out]

1/8*a*(3*A+4*B+4*C)*arctanh(sin(d*x+c))/d-1/3*a*(-2*A-2*B-3*C)*tan(d*x+c)/d+1/8*a*(3*A+4*B+4*C)*sec(d*x+c)*tan
(d*x+c)/d+1/3*a*(A+B)*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a*A*sec(d*x+c)^3*tan(d*x+c)/d

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Rubi [A]  time = 0.25, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {3031, 3021, 2748, 3768, 3770, 3767, 8} \[ -\frac {a (-3 (A+B+C)+A+B) \tan (c+d x)}{3 d}+\frac {a (3 A+4 (B+C)) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (3 A+4 (B+C)) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a (A+B) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*A + 4*(B + C))*ArcTanh[Sin[c + d*x]])/(8*d) - (a*(A + B - 3*(A + B + C))*Tan[c + d*x])/(3*d) + (a*(3*A +
4*(B + C))*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a*(A + B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (a*A*Sec[c + d*
x]^3*Tan[c + d*x])/(4*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 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 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && 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+a \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{4} \int \left (-4 a (A+B)-a (3 A+4 (B+C)) \cos (c+d x)-4 a C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a (A+B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{12} \int (-3 a (3 A+4 (B+C))+4 a (A+B-3 (A+B+C)) \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac {a (A+B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} (a (3 A+4 (B+C))) \int \sec ^3(c+d x) \, dx-\frac {1}{3} (a (A+B-3 (A+B+C))) \int \sec ^2(c+d x) \, dx\\ &=\frac {a (3 A+4 (B+C)) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} (a (3 A+4 (B+C))) \int \sec (c+d x) \, dx+\frac {(a (A+B-3 (A+B+C))) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {a (3 A+4 (B+C)) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac {a (A+B-3 (A+B+C)) \tan (c+d x)}{3 d}+\frac {a (3 A+4 (B+C)) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (A+B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.60, size = 84, normalized size = 0.67 \[ \frac {a \left (3 (3 A+4 (B+C)) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left ((A+B) \tan ^2(c+d x)+3 (A+B+C)\right )+3 (3 A+4 (B+C)) \sec (c+d x)+6 A \sec ^3(c+d x)\right )\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*(3*A + 4*(B + C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(3*A + 4*(B + C))*Sec[c + d*x] + 6*A*Sec[c + d
*x]^3 + 8*(3*(A + B + C) + (A + B)*Tan[c + d*x]^2))))/(24*d)

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fricas [A]  time = 0.44, size = 143, normalized size = 1.14 \[ \frac {3 \, {\left (3 \, A + 4 \, B + 4 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A + 4 \, B + 4 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, A + 2 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, B + 4 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 6 \, A a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(3*A + 4*B + 4*C)*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*A + 4*B + 4*C)*a*cos(d*x + c)^4*log(-s
in(d*x + c) + 1) + 2*(8*(2*A + 2*B + 3*C)*a*cos(d*x + c)^3 + 3*(3*A + 4*B + 4*C)*a*cos(d*x + c)^2 + 8*(A + B)*
a*cos(d*x + c) + 6*A*a)*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [B]  time = 0.53, size = 254, normalized size = 2.03 \[ \frac {3 \, {\left (3 \, A a + 4 \, B a + 4 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a + 4 \, B a + 4 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 49 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 28 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 60 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 52 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 84 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 39 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 36 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*A*a + 4*B*a + 4*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a + 4*B*a + 4*C*a)*log(abs(tan(1/2
*d*x + 1/2*c) - 1)) - 2*(9*A*a*tan(1/2*d*x + 1/2*c)^7 + 12*B*a*tan(1/2*d*x + 1/2*c)^7 + 12*C*a*tan(1/2*d*x + 1
/2*c)^7 - 49*A*a*tan(1/2*d*x + 1/2*c)^5 - 28*B*a*tan(1/2*d*x + 1/2*c)^5 - 60*C*a*tan(1/2*d*x + 1/2*c)^5 + 31*A
*a*tan(1/2*d*x + 1/2*c)^3 + 52*B*a*tan(1/2*d*x + 1/2*c)^3 + 84*C*a*tan(1/2*d*x + 1/2*c)^3 - 39*A*a*tan(1/2*d*x
+ 1/2*c) - 36*B*a*tan(1/2*d*x + 1/2*c) - 36*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 0.42, size = 223, normalized size = 1.78 \[ \frac {2 a A \tan \left (d x +c \right )}{3 d}+\frac {a A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {a B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a C \tan \left (d x +c \right )}{d}+\frac {a A \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a B \tan \left (d x +c \right )}{3 d}+\frac {a B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*a*A*tan(d*x+c)/d+1/3*a*A*sec(d*x+c)^2*tan(d*x+c)/d+1/2/d*a*B*sec(d*x+c)*tan(d*x+c)+1/2/d*a*B*ln(sec(d*x+c)
+tan(d*x+c))+1/d*a*C*tan(d*x+c)+1/4*a*A*sec(d*x+c)^3*tan(d*x+c)/d+3/8*a*A*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a*A*ln
(sec(d*x+c)+tan(d*x+c))+2/3/d*a*B*tan(d*x+c)+1/3/d*a*B*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a*C*sec(d*x+c)*tan(d*x+c)
+1/2/d*a*C*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.59, size = 218, normalized size = 1.74 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a - 3 \, A a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C a \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a - 3*A*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)) - 12*B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*
C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*C*a*tan(d*x + c
))/d

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mupad [B]  time = 4.56, size = 211, normalized size = 1.69 \[ \frac {\left (-\frac {3\,A\,a}{4}-B\,a-C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {49\,A\,a}{12}+\frac {7\,B\,a}{3}+5\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {31\,A\,a}{12}-\frac {13\,B\,a}{3}-7\,C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+3\,B\,a+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,B+4\,C\right )}{2\,\left (\frac {3\,A\,a}{2}+2\,B\,a+2\,C\,a\right )}\right )\,\left (3\,A+4\,B+4\,C\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((13*A*a)/4 + 3*B*a + 3*C*a) - tan(c/2 + (d*x)/2)^7*((3*A*a)/4 + B*a + C*a) - tan(c/2 + (d
*x)/2)^3*((31*A*a)/12 + (13*B*a)/3 + 7*C*a) + tan(c/2 + (d*x)/2)^5*((49*A*a)/12 + (7*B*a)/3 + 5*C*a))/(d*(6*ta
n(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (a*atanh((
a*tan(c/2 + (d*x)/2)*(3*A + 4*B + 4*C))/(2*((3*A*a)/2 + 2*B*a + 2*C*a)))*(3*A + 4*B + 4*C))/(4*d)

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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))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

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