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

Optimal. Leaf size=121 \[ -\frac{a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}+\frac{a^2 (2 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{2} a^2 x (2 A+4 B+3 C)-\frac{(2 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.383819, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {3043, 2976, 2968, 3023, 2735, 3770} \[ -\frac{a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}+\frac{a^2 (2 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{2} a^2 x (2 A+4 B+3 C)-\frac{(2 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+\frac{A \tan (c+d x) (a \cos (c+d x)+a)^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3043

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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 -
B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(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, B, 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 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 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 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 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 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 (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x))^2 (a (2 A+B)-a (2 A-C) \cos (c+d x)) \sec (c+d x) \, dx}{a}\\ &=-\frac{(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\frac{\int (a+a \cos (c+d x)) \left (2 a^2 (2 A+B)-a^2 (2 A-2 B-3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac{(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\frac{\int \left (2 a^3 (2 A+B)+\left (2 a^3 (2 A+B)-a^3 (2 A-2 B-3 C)\right ) \cos (c+d x)-a^3 (2 A-2 B-3 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac{a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}-\frac{(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\frac{\int \left (2 a^3 (2 A+B)+a^3 (2 A+4 B+3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=\frac{1}{2} a^2 (2 A+4 B+3 C) x-\frac{a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}-\frac{(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\left (a^2 (2 A+B)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^2 (2 A+4 B+3 C) x+\frac{a^2 (2 A+B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}-\frac{(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.579331, size = 174, normalized size = 1.44 \[ \frac{a^2 \left (4 A \tan (c+d x)-8 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+8 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 A c+4 A d x+4 (B+2 C) \sin (c+d x)-4 B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 B c+8 B d x+C \sin (2 (c+d x))+6 c C+6 C d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(4*A*c + 8*B*c + 6*c*C + 4*A*d*x + 8*B*d*x + 6*C*d*x - 8*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 4*B
*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 4*B*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]] + 4*(B + 2*C)*Sin[c + d*x] + C*Sin[2*(c + d*x)] + 4*A*Tan[c + d*x]))/(4*d)

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Maple [A]  time = 0.072, size = 160, normalized size = 1.3 \begin{align*}{\frac{A{a}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{3\,{a}^{2}Cx}{2}}+{\frac{3\,{a}^{2}Cc}{2\,d}}+2\,{\frac{A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{a}^{2}Bx+2\,{\frac{B{a}^{2}c}{d}}+2\,{\frac{{a}^{2}C\sin \left ( dx+c \right ) }{d}}+A{a}^{2}x+{\frac{A{a}^{2}c}{d}}+{\frac{{a}^{2}B\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}C\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.03136, size = 204, normalized size = 1.69 \begin{align*} \frac{4 \,{\left (d x + c\right )} A a^{2} + 8 \,{\left (d x + c\right )} B a^{2} +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 4 \,{\left (d x + c\right )} C a^{2} + 4 \, A a^{2}{\left (\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 \, B a^{2} \sin \left (d x + c\right ) + 8 \, C a^{2} \sin \left (d x + c\right ) + 4 \, A a^{2} \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.03978, size = 331, normalized size = 2.74 \begin{align*} \frac{{\left (2 \, A + 4 \, B + 3 \, C\right )} a^{2} d x \cos \left (d x + c\right ) +{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (C a^{2} \cos \left (d x + c\right )^{2} + 2 \,{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.19574, size = 267, normalized size = 2.21 \begin{align*} -\frac{\frac{4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} -{\left (2 \, A a^{2} + 4 \, B a^{2} + 3 \, C a^{2}\right )}{\left (d x + c\right )} - 2 \,{\left (2 \, A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 2 \,{\left (2 \, A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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