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

Optimal. Leaf size=88 \[ \frac {a^2 (3 B+2 C) \tan (c+d x)}{2 d}+\frac {a^2 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+a^2 C x \]

[Out]

a^2*C*x+1/2*a^2*(3*B+4*C)*arctanh(sin(d*x+c))/d+1/2*a^2*(3*B+2*C)*tan(d*x+c)/d+1/2*B*(a^2+a^2*cos(d*x+c))*sec(
d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.30, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {a^2 (3 B+2 C) \tan (c+d x)}{2 d}+\frac {a^2 (3 B+4 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+a^2 C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 1.36, size = 277, normalized size = 3.15 \[ \frac {1}{16} a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (\frac {4 (2 B+C) \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 (2 B+C) \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {2 (3 B+4 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {2 (3 B+4 C) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {B}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {B}{d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+4 C x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(4*C*x - (2*(3*B + 4*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])
/d + (2*(3*B + 4*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + B/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2
) + (4*(2*B + C)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - B/(d*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^2) + (4*(2*B + C)*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin
[(c + d*x)/2]))))/16

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fricas [A]  time = 0.50, size = 119, normalized size = 1.35 \[ \frac {4 \, C a^{2} d x \cos \left (d x + c\right )^{2} + {\left (3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) + B a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.52, size = 154, normalized size = 1.75 \[ \frac {2 \, {\left (d x + c\right )} C a^{2} + {\left (3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (3 \, B a^{2} + 4 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.29, size = 113, normalized size = 1.28 \[ a^{2} C x +\frac {a^{2} C c}{d}+\frac {3 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{2} B \tan \left (d x +c \right )}{d}+\frac {a^{2} C \tan \left (d x +c \right )}{d}+\frac {a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.04, size = 142, normalized size = 1.61 \[ \frac {4 \, {\left (d x + c\right )} C a^{2} - B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, B a^{2} \tan \left (d x + c\right ) + 4 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(d*x + c)*C*a^2 - B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) + 2*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*
x + c) - 1)) + 8*B*a^2*tan(d*x + c) + 4*C*a^2*tan(d*x + c))/d

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mupad [B]  time = 1.12, size = 162, normalized size = 1.84 \[ \frac {3\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

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))^2)/cos(c + d*x)^4,x)

[Out]

(3*B*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (4*C*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*a^2*sin(c + d*x))/(d*cos(c + d*x)) + (B
*a^2*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (C*a^2*sin(c + d*x))/(d*cos(c + d*x))

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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))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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