3.34 \(\int (a+a \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=198 \[ -\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac {2 a^4 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {(22 A+3 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (2 A+13 C)+\frac {(8 A+3 C) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}+\frac {2 a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

[Out]

1/2*a^4*(2*A+13*C)*x+2*a^4*(3*A+2*C)*arctanh(sin(d*x+c))/d-5/2*a^4*(2*A-C)*sin(d*x+c)/d-1/6*(22*A+3*C)*(a^4+a^
4*cos(d*x+c))*sin(d*x+c)/d+1/3*(8*A+3*C)*(a^2+a^2*cos(d*x+c))^2*tan(d*x+c)/d+2/3*a*A*(a+a*cos(d*x+c))^3*sec(d*
x+c)*tan(d*x+c)/d+1/3*A*(a+a*cos(d*x+c))^4*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.69, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}+\frac {2 a^4 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {(22 A+3 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac {(8 A+3 C) \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}+\frac {1}{2} a^4 x (2 A+13 C)+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^4}{3 d}+\frac {2 a A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(a^4*(2*A + 13*C)*x)/2 + (2*a^4*(3*A + 2*C)*ArcTanh[Sin[c + d*x]])/d - (5*a^4*(2*A - C)*Sin[c + d*x])/(2*d) -
((22*A + 3*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(6*d) + ((8*A + 3*C)*(a^2 + a^2*Cos[c + d*x])^2*Tan[c + d
*x])/(3*d) + (2*a*A*(a + a*Cos[c + d*x])^3*Sec[c + d*x]*Tan[c + d*x])/(3*d) + (A*(a + a*Cos[c + d*x])^4*Sec[c
+ d*x]^2*Tan[c + d*x])/(3*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))^4 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^4 (4 a A-a (2 A-3 C) \cos (c+d x)) \sec ^3(c+d x) \, dx}{3 a}\\ &=\frac {2 a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^3 \left (2 a^2 (8 A+3 C)-6 a^2 (2 A-C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a}\\ &=\frac {(8 A+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^2 \left (12 a^3 (3 A+2 C)-2 a^3 (22 A+3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a}\\ &=-\frac {(22 A+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {(8 A+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x)) \left (24 a^4 (3 A+2 C)-30 a^4 (2 A-C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac {(22 A+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {(8 A+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (24 a^5 (3 A+2 C)+\left (-30 a^5 (2 A-C)+24 a^5 (3 A+2 C)\right ) \cos (c+d x)-30 a^5 (2 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=-\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}-\frac {(22 A+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {(8 A+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (24 a^5 (3 A+2 C)+6 a^5 (2 A+13 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{12 a}\\ &=\frac {1}{2} a^4 (2 A+13 C) x-\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}-\frac {(22 A+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {(8 A+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\left (2 a^4 (3 A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^4 (2 A+13 C) x+\frac {2 a^4 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {5 a^4 (2 A-C) \sin (c+d x)}{2 d}-\frac {(22 A+3 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {(8 A+3 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \tan (c+d x)}{3 d}+\frac {2 a A (a+a \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^4 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 6.22, size = 386, normalized size = 1.95 \[ a^4 \left (\frac {(2 A+13 C) (c+d x)}{2 d}+\frac {20 A \sin \left (\frac {1}{2} (c+d x)\right )+3 C \sin \left (\frac {1}{2} (c+d x)\right )}{3 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {20 A \sin \left (\frac {1}{2} (c+d x)\right )+3 C \sin \left (\frac {1}{2} (c+d x)\right )}{3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2 (3 A+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (3 A+2 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {13 A}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {13 A}{12 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {A \sin \left (\frac {1}{2} (c+d x)\right )}{6 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {4 C \sin (c+d x)}{d}+\frac {C \sin (2 (c+d x))}{4 d}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^4*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

a^4*(((2*A + 13*C)*(c + d*x))/(2*d) - (2*(3*A + 2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(3*A + 2
*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (13*A)/(12*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (A*S
in[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) + (A*Sin[(c + d*x)/2])/(6*d*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^3) - (13*A)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (20*A*Sin[(c + d*x)/2] + 3*C*S
in[(c + d*x)/2])/(3*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (20*A*Sin[(c + d*x)/2] + 3*C*Sin[(c + d*x)/2])/
(3*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (4*C*Sin[c + d*x])/d + (C*Sin[2*(c + d*x)])/(4*d))

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fricas [A]  time = 0.79, size = 170, normalized size = 0.86 \[ \frac {3 \, {\left (2 \, A + 13 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 6 \, {\left (3 \, A + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \, {\left (3 \, A + 2 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (3 \, C a^{4} \cos \left (d x + c\right )^{4} + 24 \, C a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (20 \, A + 3 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 12 \, A a^{4} \cos \left (d x + c\right ) + 2 \, A a^{4}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/6*(3*(2*A + 13*C)*a^4*d*x*cos(d*x + c)^3 + 6*(3*A + 2*C)*a^4*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 6*(3*A +
2*C)*a^4*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (3*C*a^4*cos(d*x + c)^4 + 24*C*a^4*cos(d*x + c)^3 + 2*(20*A
+ 3*C)*a^4*cos(d*x + c)^2 + 12*A*a^4*cos(d*x + c) + 2*A*a^4)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.62, size = 248, normalized size = 1.25 \[ \frac {3 \, {\left (2 \, A a^{4} + 13 \, C a^{4}\right )} {\left (d x + c\right )} + 12 \, {\left (3 \, A a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, {\left (3 \, A a^{4} + 2 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {6 \, {\left (7 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}} - \frac {4 \, {\left (15 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 38 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/6*(3*(2*A*a^4 + 13*C*a^4)*(d*x + c) + 12*(3*A*a^4 + 2*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*(3*A*a^
4 + 2*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(7*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*C*a^4*tan(1/2*d*x + 1/
2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 4*(15*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 3
8*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 6*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 27*A*a^4*tan(1/2*d*x + 1/2*c) + 3*C*a^4*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 = 0.41, size = 189, normalized size = 0.95 \[ A \,a^{4} x +\frac {A \,a^{4} c}{d}+\frac {a^{4} C \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {13 a^{4} C x}{2}+\frac {13 a^{4} C c}{2 d}+\frac {6 A \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 a^{4} C \sin \left (d x +c \right )}{d}+\frac {20 A \,a^{4} \tan \left (d x +c \right )}{3 d}+\frac {2 A \,a^{4} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {4 a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,a^{4} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{4} C \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

A*a^4*x+1/d*A*a^4*c+1/2/d*a^4*C*cos(d*x+c)*sin(d*x+c)+13/2*a^4*C*x+13/2/d*a^4*C*c+6/d*A*a^4*ln(sec(d*x+c)+tan(
d*x+c))+4/d*a^4*C*sin(d*x+c)+20/3/d*A*a^4*tan(d*x+c)+2/d*A*a^4*sec(d*x+c)*tan(d*x+c)+4/d*a^4*C*ln(sec(d*x+c)+t
an(d*x+c))+1/3/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+1/d*a^4*C*tan(d*x+c)

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maxima [A]  time = 0.34, size = 211, normalized size = 1.07 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 12 \, {\left (d x + c\right )} A a^{4} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} + 72 \, {\left (d x + c\right )} C a^{4} - 12 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C a^{4} \sin \left (d x + c\right ) + 72 \, A a^{4} \tan \left (d x + c\right ) + 12 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^4*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^4 + 12*(d*x + c)*A*a^4 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^
4 + 72*(d*x + c)*C*a^4 - 12*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x +
c) - 1)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*C*a^4*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) + 48*C*a^4*sin(d*x + c) + 72*A*a^4*tan(d*x + c) + 12*C*a^4*tan(d*x + c))/d

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mupad [B]  time = 1.12, size = 252, normalized size = 1.27 \[ \frac {4\,C\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {12\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {8\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {20\,A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {2\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^4\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

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))^4)/cos(c + d*x)^4,x)

[Out]

(4*C*a^4*sin(c + d*x))/d + (2*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (12*A*a^4*atanh(sin(c/2 +
(d*x)/2)/cos(c/2 + (d*x)/2)))/d + (13*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (8*C*a^4*atanh(s
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (20*A*a^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (2*A*a^4*sin(c + d*x))
/(d*cos(c + d*x)^2) + (A*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (C*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (C*a
^4*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))**4*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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