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

Optimal. Leaf size=111 \[ \frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {(3 B+5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {a^3 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} a^3 x (7 B+5 C)+\frac {a C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.38, antiderivative size = 111, 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, 2976, 2968, 3023, 2735, 3770} \[ \frac {5 a^3 (B+C) \sin (c+d x)}{2 d}+\frac {(3 B+5 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {a^3 B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} a^3 x (7 B+5 C)+\frac {a C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

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]^2,x]

[Out]

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

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Mathematica [A]  time = 0.26, size = 113, normalized size = 1.02 \[ \frac {a^3 \left (9 (4 B+5 C) \sin (c+d x)+3 (B+3 C) \sin (2 (c+d x))-12 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+42 B d x+C \sin (3 (c+d x))+30 C d x\right )}{12 d} \]

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]^2,x]

[Out]

(a^3*(42*B*d*x + 30*C*d*x - 12*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*B*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + 9*(4*B + 5*C)*Sin[c + d*x] + 3*(B + 3*C)*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)]))/(12*d)

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fricas [A]  time = 0.43, size = 102, normalized size = 0.92 \[ \frac {3 \, {\left (7 \, B + 5 \, C\right )} a^{3} d x + 3 \, B a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 2 \, {\left (9 \, B + 11 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{6 \, d} \]

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)^2,x, algorithm="fricas")

[Out]

1/6*(3*(7*B + 5*C)*a^3*d*x + 3*B*a^3*log(sin(d*x + c) + 1) - 3*B*a^3*log(-sin(d*x + c) + 1) + (2*C*a^3*cos(d*x
+ c)^2 + 3*(B + 3*C)*a^3*cos(d*x + c) + 2*(9*B + 11*C)*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.59, size = 180, normalized size = 1.62 \[ \frac {6 \, B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, B a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (7 \, B a^{3} + 5 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*B*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*B*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(7*B*a^3 + 5*
C*a^3)*(d*x + c) + 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 + 36*B*a^3*tan(1/2*d*x
+ 1/2*c)^3 + 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*B*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 = 0.28, size = 153, normalized size = 1.38 \[ \frac {C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) a^{3}}{3 d}+\frac {11 a^{3} C \sin \left (d x +c \right )}{3 d}+\frac {a^{3} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {7 a^{3} B x}{2}+\frac {7 a^{3} B c}{2 d}+\frac {3 C \,a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {5 a^{3} C x}{2}+\frac {5 C \,a^{3} c}{2 d}+\frac {3 a^{3} B \sin \left (d x +c \right )}{d}+\frac {a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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+1/2/d*a^3*B*cos(d*x+c)*sin(d*x+c)+7/2*a^3*B*x+7/2/
d*a^3*B*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*a^3*B*sin(d*x+c)/d+1/d*a^3*B*ln(sec(d*
x+c)+tan(d*x+c))

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maxima [A]  time = 0.47, size = 148, normalized size = 1.33 \[ \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 36 \, {\left (d x + c\right )} B a^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 12 \, {\left (d x + c\right )} C a^{3} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{3} \sin \left (d x + c\right ) + 36 \, C a^{3} \sin \left (d x + c\right )}{12 \, d} \]

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)^2,x, algorithm="maxima")

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 36*(d*x + c)*B*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 + 6*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d
*x + c) - 1)) + 36*B*a^3*sin(d*x + c) + 36*C*a^3*sin(d*x + c))/d

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mupad [B]  time = 1.31, size = 178, normalized size = 1.60 \[ \frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {15\,C\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {7\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

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)^2,x)

[Out]

(3*B*a^3*sin(c + d*x))/d + (15*C*a^3*sin(c + d*x))/(4*d) + (7*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (2*B*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 + (B*a^3*sin(2*c + 2*d*x))/(4*d) + (3*C*a^3*sin(2*c + 2*d*x))/(4*d) + (C*a^3*sin(3*c + 3*d*x))/(12
*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*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

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