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

Optimal. Leaf size=160 \[ -\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac {a^3 (7 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {(4 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {1}{2} a^3 x (2 A+7 C)+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 a d}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d} \]

[Out]

1/2*a^3*(2*A+7*C)*x+1/2*a^3*(7*A+2*C)*arctanh(sin(d*x+c))/d-5/2*a^3*(A-C)*sin(d*x+c)/d-1/2*(4*A-C)*(a^3+a^3*co
s(d*x+c))*sin(d*x+c)/d+3/2*A*(a^2+a^2*cos(d*x+c))^2*tan(d*x+c)/a/d+1/2*A*(a+a*cos(d*x+c))^3*sec(d*x+c)*tan(d*x
+c)/d

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Rubi [A]  time = 0.48, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3044, 2975, 2976, 2968, 3023, 2735, 3770} \[ -\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac {a^3 (7 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {(4 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}+\frac {3 A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{2 a d}+\frac {1}{2} a^3 x (2 A+7 C)+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(2*A + 7*C)*x)/2 + (a^3*(7*A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) - (5*a^3*(A - C)*Sin[c + d*x])/(2*d) - (
(4*A - C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/(2*d) + (3*A*(a^2 + a^2*Cos[c + d*x])^2*Tan[c + d*x])/(2*a*d)
+ (A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(2*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 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 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 ^3(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x))^3 (3 a A-2 a (A-C) \cos (c+d x)) \sec ^2(c+d x) \, dx}{2 a}\\ &=\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x))^2 \left (a^2 (7 A+2 C)-2 a^2 (4 A-C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac {(4 A-C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int (a+a \cos (c+d x)) \left (2 a^3 (7 A+2 C)-10 a^3 (A-C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{4 a}\\ &=-\frac {(4 A-C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int \left (2 a^4 (7 A+2 C)+\left (-10 a^4 (A-C)+2 a^4 (7 A+2 C)\right ) \cos (c+d x)-10 a^4 (A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{4 a}\\ &=-\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}-\frac {(4 A-C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {\int \left (2 a^4 (7 A+2 C)+2 a^4 (2 A+7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{4 a}\\ &=\frac {1}{2} a^3 (2 A+7 C) x-\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}-\frac {(4 A-C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a^3 (7 A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^3 (2 A+7 C) x+\frac {a^3 (7 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {5 a^3 (A-C) \sin (c+d x)}{2 d}-\frac {(4 A-C) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {3 A \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{2 a d}+\frac {A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 1.98, size = 214, normalized size = 1.34 \[ \frac {a^3 \left (12 A \tan (c+d x)+\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}-14 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+14 A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 A c+4 A d x+12 C \sin (c+d x)+C \sin (2 (c+d x))-4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+14 c C+14 C d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(4*A*c + 14*c*C + 4*A*d*x + 14*C*d*x - 14*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 4*C*Log[Cos[(c + d
*x)/2] - Sin[(c + d*x)/2]] + 14*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 4*C*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + A/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - A/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 12*C*Sin
[c + d*x] + C*Sin[2*(c + d*x)] + 12*A*Tan[c + d*x]))/(4*d)

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fricas [A]  time = 1.16, size = 148, normalized size = 0.92 \[ \frac {2 \, {\left (2 \, A + 7 \, C\right )} a^{3} d x \cos \left (d x + c\right )^{2} + {\left (7 \, A + 2 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (7 \, A + 2 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{3} \cos \left (d x + c\right )^{3} + 6 \, C a^{3} \cos \left (d x + c\right )^{2} + 6 \, A a^{3} \cos \left (d x + c\right ) + A a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

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)^3,x, algorithm="fricas")

[Out]

1/4*(2*(2*A + 7*C)*a^3*d*x*cos(d*x + c)^2 + (7*A + 2*C)*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (7*A + 2*C)
*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(C*a^3*cos(d*x + c)^3 + 6*C*a^3*cos(d*x + c)^2 + 6*A*a^3*cos(d*
x + c) + A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

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giac [A]  time = 2.84, size = 230, normalized size = 1.44 \[ \frac {{\left (2 \, A a^{3} + 7 \, C a^{3}\right )} {\left (d x + c\right )} + {\left (7 \, A a^{3} + 2 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (7 \, A a^{3} + 2 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 7 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1\right )}^{2}}}{2 \, d} \]

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)^3,x, algorithm="giac")

[Out]

1/2*((2*A*a^3 + 7*C*a^3)*(d*x + c) + (7*A*a^3 + 2*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (7*A*a^3 + 2*C*a
^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 5*C*a^3*tan(1/2*d*x + 1/2*c)^7 +
3*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 9*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 9*C*a^3*tan
(1/2*d*x + 1/2*c)^3 - 7*A*a^3*tan(1/2*d*x + 1/2*c) - 7*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1
)^2)/d

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maple [A]  time = 0.32, size = 151, normalized size = 0.94 \[ A x \,a^{3}+\frac {A \,a^{3} c}{d}+\frac {C \,a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {7 a^{3} C x}{2}+\frac {7 C \,a^{3} c}{2 d}+\frac {7 A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {3 a^{3} C \sin \left (d x +c \right )}{d}+\frac {3 A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

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)^3,x)

[Out]

A*x*a^3+1/d*A*a^3*c+1/2/d*C*a^3*cos(d*x+c)*sin(d*x+c)+7/2*a^3*C*x+7/2/d*C*a^3*c+7/2/d*A*a^3*ln(sec(d*x+c)+tan(
d*x+c))+3*a^3*C*sin(d*x+c)/d+3/d*A*a^3*tan(d*x+c)+1/2/d*A*a^3*sec(d*x+c)*tan(d*x+c)+1/d*C*a^3*ln(sec(d*x+c)+ta
n(d*x+c))

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maxima [A]  time = 0.37, size = 175, normalized size = 1.09 \[ \frac {4 \, {\left (d x + c\right )} A a^{3} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \, {\left (d x + c\right )} C a^{3} - A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{4 \, d} \]

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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.04, size = 207, normalized size = 1.29 \[ \frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

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)^3,x)

[Out]

(3*C*a^3*sin(c + d*x))/d + (2*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*A*a^3*atanh(sin(c/2 +
(d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^3*atanh(sin
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*A*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (A*a^3*sin(c + d*x))/(2*d*c
os(c + d*x)^2) + (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)**3,x)

[Out]

Timed out

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