3.250 \(\int (a+a \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=125 \[ \frac {5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac {a^3 (5 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(5 B+3 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+a^3 C x+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

a^3*C*x+1/2*a^3*(5*B+7*C)*arctanh(sin(d*x+c))/d+5/2*a^3*(B+C)*tan(d*x+c)/d+1/6*(5*B+3*C)*(a^3+a^3*cos(d*x+c))*
sec(d*x+c)*tan(d*x+c)/d+1/3*a*B*(a+a*cos(d*x+c))^2*sec(d*x+c)^2*tan(d*x+c)/d

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Rubi [A]  time = 0.42, antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3029, 2975, 2968, 3021, 2735, 3770} \[ \frac {5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac {a^3 (5 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(5 B+3 C) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+a^3 C x+\frac {a B \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

a^3*C*x + (a^3*(5*B + 7*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(B + C)*Tan[c + d*x])/(2*d) + ((5*B + 3*C)*(a
^3 + a^3*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (a*B*(a + a*Cos[c + d*x])^2*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 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 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 (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))^3 (B+C \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac {a B (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+a \cos (c+d x))^2 (a (5 B+3 C)+3 a C \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac {(5 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int (a+a \cos (c+d x)) \left (15 a^2 (B+C)+6 a^2 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {(5 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (15 a^3 (B+C)+\left (6 a^3 C+15 a^3 (B+C)\right ) \cos (c+d x)+6 a^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac {(5 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (3 a^3 (5 B+7 C)+6 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=a^3 C x+\frac {5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac {(5 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} \left (a^3 (5 B+7 C)\right ) \int \sec (c+d x) \, dx\\ &=a^3 C x+\frac {a^3 (5 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {5 a^3 (B+C) \tan (c+d x)}{2 d}+\frac {(5 B+3 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [B]  time = 6.38, size = 786, normalized size = 6.29 \[ \frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (11 B \sin \left (\frac {d x}{2}\right )+9 C \sin \left (\frac {d x}{2}\right )\right )}{24 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (11 B \sin \left (\frac {d x}{2}\right )+9 C \sin \left (\frac {d x}{2}\right )\right )}{24 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}+\frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (-8 B \sin \left (\frac {c}{2}\right )+10 B \cos \left (\frac {c}{2}\right )-3 C \sin \left (\frac {c}{2}\right )+3 C \cos \left (\frac {c}{2}\right )\right )}{96 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {\sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \left (-8 B \sin \left (\frac {c}{2}\right )-10 B \cos \left (\frac {c}{2}\right )-3 C \sin \left (\frac {c}{2}\right )-3 C \cos \left (\frac {c}{2}\right )\right )}{96 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^2}+\frac {(-5 B-7 C) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \log \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{16 d}+\frac {(5 B+7 C) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \log \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )}{16 d}+\frac {B \sin \left (\frac {d x}{2}\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{48 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}+\frac {d x}{2}\right )-\sin \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {B \sin \left (\frac {d x}{2}\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3}{48 d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}+\frac {d x}{2}\right )+\cos \left (\frac {c}{2}+\frac {d x}{2}\right )\right )^3}+\frac {1}{8} C x \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) (a \cos (c+d x)+a)^3 \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])^3*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(C*x*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6)/8 + ((-5*B - 7*C)*(a + a*Cos[c + d*x])^3*Log[Cos[c/2 + (d*x)
/2] - Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6)/(16*d) + ((5*B + 7*C)*(a + a*Cos[c + d*x])^3*Log[Cos[c/2 + (d*
x)/2] + Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^6)/(16*d) + (B*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*Sin[
(d*x)/2])/(48*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) + ((a + a*Cos[c + d*x])^3*S
ec[c/2 + (d*x)/2]^6*(10*B*Cos[c/2] + 3*C*Cos[c/2] - 8*B*Sin[c/2] - 3*C*Sin[c/2]))/(96*d*(Cos[c/2] - Sin[c/2])*
(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + ((a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(11*B*Sin[(d*x)/2]
+ 9*C*Sin[(d*x)/2]))/(24*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (B*(a + a*Cos[c
+ d*x])^3*Sec[c/2 + (d*x)/2]^6*Sin[(d*x)/2])/(48*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x
)/2])^3) + ((a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-10*B*Cos[c/2] - 3*C*Cos[c/2] - 8*B*Sin[c/2] - 3*C*Si
n[c/2]))/(96*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + ((a + a*Cos[c + d*x])^3*Se
c[c/2 + (d*x)/2]^6*(11*B*Sin[(d*x)/2] + 9*C*Sin[(d*x)/2]))/(24*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + S
in[c/2 + (d*x)/2]))

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fricas [A]  time = 0.48, size = 141, normalized size = 1.13 \[ \frac {12 \, C a^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (11 \, B + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, B a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/12*(12*C*a^3*d*x*cos(d*x + c)^3 + 3*(5*B + 7*C)*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(5*B + 7*C)*a^3
*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(11*B + 9*C)*a^3*cos(d*x + c)^2 + 3*(3*B + C)*a^3*cos(d*x + c) +
2*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [A]  time = 0.36, size = 189, normalized size = 1.51 \[ \frac {6 \, {\left (d x + c\right )} C a^{3} + 3 \, {\left (5 \, B a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, B a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*C*a^3 + 3*(5*B*a^3 + 7*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(5*B*a^3 + 7*C*a^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*B*a
^3*tan(1/2*d*x + 1/2*c)^3 - 36*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*B*a^3*tan(1/2*d*x + 1/2*c) + 21*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 = 0.36, size = 158, normalized size = 1.26 \[ a^{3} C x +\frac {C \,a^{3} c}{d}+\frac {5 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {7 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {11 a^{3} B \tan \left (d x +c \right )}{3 d}+\frac {3 C \,a^{3} \tan \left (d x +c \right )}{d}+\frac {3 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {C \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

a^3*C*x+1/d*C*a^3*c+5/2/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+7/2/d*C*a^3*ln(sec(d*x+c)+tan(d*x+c))+11/3/d*a^3*B*t
an(d*x+c)+3/d*C*a^3*tan(d*x+c)+3/2/d*a^3*B*sec(d*x+c)*tan(d*x+c)+1/2/d*C*a^3*sec(d*x+c)*tan(d*x+c)+1/3/d*a^3*B
*tan(d*x+c)*sec(d*x+c)^2

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maxima [A]  time = 0.33, size = 212, normalized size = 1.70 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 12 \, {\left (d x + c\right )} C a^{3} - 9 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{3} \tan \left (d x + c\right ) + 36 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))^3*(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^3 + 12*(d*x + c)*C*a^3 - 9*B*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)) - 3*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(si
n(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*C*a^3*
(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*B*a^3*tan(d*x + c) + 36*C*a^3*tan(d*x + c))/d

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mupad [B]  time = 1.20, size = 209, normalized size = 1.67 \[ \frac {5\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {11\,B\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]

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))^3)/cos(c + d*x)^5,x)

[Out]

(5*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (7*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (11*B*a^3*sin(c + d*x))/(3*d*cos(c + d*x)) +
(3*B*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (B*a^3*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (3*C*a^3*sin(c + d*
x))/(d*cos(c + d*x)) + (C*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2)

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

[Out]

Timed out

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