3.22 \(\int (a+a \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\)

Optimal. Leaf size=145 \[ -\frac {(6 A-5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {3 a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} a^3 x (6 A+5 C)+\frac {5 a^3 C \sin (c+d x)}{2 d}-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 a d}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]

[Out]

1/2*a^3*(6*A+5*C)*x+3*a^3*A*arctanh(sin(d*x+c))/d+5/2*a^3*C*sin(d*x+c)/d-1/3*(3*A-C)*(a^2+a^2*cos(d*x+c))^2*si
n(d*x+c)/a/d-1/6*(6*A-5*C)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d+A*(a+a*cos(d*x+c))^3*tan(d*x+c)/d

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Rubi [A]  time = 0.45, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3044, 2976, 2968, 3023, 2735, 3770} \[ -\frac {(6 A-5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}-\frac {(3 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 a d}+\frac {3 a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} a^3 x (6 A+5 C)+\frac {5 a^3 C \sin (c+d x)}{2 d}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^3}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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 2976

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*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*S
in[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3023

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

Rule 3044

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + 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))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x))^3 (3 a A-a (3 A-C) \cos (c+d x)) \sec (c+d x) \, dx}{a}\\ &=-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x))^2 \left (9 a^2 A-a^2 (6 A-5 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{3 a}\\ &=-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x)) \left (18 a^3 A+15 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int \left (18 a^4 A+\left (18 a^4 A+15 a^4 C\right ) \cos (c+d x)+15 a^4 C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac {5 a^3 C \sin (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {\int \left (18 a^4 A+3 a^4 (6 A+5 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=\frac {1}{2} a^3 (6 A+5 C) x+\frac {5 a^3 C \sin (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}+\left (3 a^3 A\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^3 (6 A+5 C) x+\frac {3 a^3 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 C \sin (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{3 a d}-\frac {(6 A-5 C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^3 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [B]  time = 1.97, size = 298, normalized size = 2.06 \[ \frac {1}{96} a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (\frac {3 (4 A+15 C) \sin (c) \cos (d x)}{d}+\frac {3 (4 A+15 C) \cos (c) \sin (d x)}{d}+\frac {12 A \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 A \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {36 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {36 A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+6 x (6 A+5 C)+\frac {9 C \sin (2 c) \cos (2 d x)}{d}+\frac {C \sin (3 c) \cos (3 d x)}{d}+\frac {9 C \cos (2 c) \sin (2 d x)}{d}+\frac {C \cos (3 c) \sin (3 d x)}{d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^2,x]

[Out]

(a^3*(1 + Cos[c + d*x])^3*Sec[(c + d*x)/2]^6*(6*(6*A + 5*C)*x - (36*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
)/d + (36*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (3*(4*A + 15*C)*Cos[d*x]*Sin[c])/d + (9*C*Cos[2*d*x]
*Sin[2*c])/d + (C*Cos[3*d*x]*Sin[3*c])/d + (3*(4*A + 15*C)*Cos[c]*Sin[d*x])/d + (9*C*Cos[2*c]*Sin[2*d*x])/d +
(C*Cos[3*c]*Sin[3*d*x])/d + (12*A*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])
) + (12*A*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/96

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fricas [A]  time = 0.53, size = 138, normalized size = 0.95 \[ \frac {3 \, {\left (6 \, A + 5 \, C\right )} a^{3} d x \cos \left (d x + c\right ) + 9 \, A a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, A a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{3} \cos \left (d x + c\right )^{3} + 9 \, C a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + 11 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

1/6*(3*(6*A + 5*C)*a^3*d*x*cos(d*x + c) + 9*A*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - 9*A*a^3*cos(d*x + c)*lo
g(-sin(d*x + c) + 1) + (2*C*a^3*cos(d*x + c)^3 + 9*C*a^3*cos(d*x + c)^2 + 2*(3*A + 11*C)*a^3*cos(d*x + c) + 6*
A*a^3)*sin(d*x + c))/(d*cos(d*x + c))

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giac [A]  time = 1.49, size = 210, normalized size = 1.45 \[ \frac {18 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (6 \, A a^{3} + 5 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="giac")

[Out]

1/6*(18*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 12*A*a^3*tan(
1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 3*(6*A*a^3 + 5*C*a^3)*(d*x + c) + 2*(6*A*a^3*tan(1/2*d*x + 1/2
*c)^5 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 +
6*A*a^3*tan(1/2*d*x + 1/2*c) + 33*C*a^3*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 = 0.32, size = 146, normalized size = 1.01 \[ \frac {a^{3} A \sin \left (d x +c \right )}{d}+\frac {C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{3 d}+\frac {11 a^{3} C \sin \left (d x +c \right )}{3 d}+3 A x \,a^{3}+\frac {3 A \,a^{3} c}{d}+\frac {3 C \,a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a^{3} C x}{2}+\frac {5 C \,a^{3} c}{2 d}+\frac {3 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)

[Out]

a^3*A*sin(d*x+c)/d+1/3/d*C*cos(d*x+c)^2*sin(d*x+c)*a^3+11/3*a^3*C*sin(d*x+c)/d+3*A*x*a^3+3/d*A*a^3*c+3/2/d*C*a
^3*cos(d*x+c)*sin(d*x+c)+5/2*a^3*C*x+5/2/d*C*a^3*c+3/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*a^3*tan(d*x+c)

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maxima [A]  time = 0.50, size = 137, normalized size = 0.94 \[ \frac {36 \, {\left (d x + c\right )} A a^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \, {\left (d x + c\right )} C a^{3} + 18 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 36 \, C a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

1/12*(36*(d*x + c)*A*a^3 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^
3 + 12*(d*x + c)*C*a^3 + 18*A*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 36
*C*a^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c))/d

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mupad [B]  time = 1.00, size = 189, normalized size = 1.30 \[ \frac {A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {11\,C\,a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {6\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {3\,C\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^3)/cos(c + d*x)^2,x)

[Out]

(A*a^3*sin(c + d*x))/d + (11*C*a^3*sin(c + d*x))/(3*d) + (6*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))
/d + (6*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (5*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2)))/d + (A*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (C*a^3*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (3*C*a^3*cos(c
+ d*x)*sin(c + d*x))/(2*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))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

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