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

Optimal. Leaf size=112 \[ -\frac{a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac{a^2 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d}+2 a^2 C x+\frac{A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.358719, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3044, 2975, 2968, 3023, 2735, 3770} \[ -\frac{a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac{a^2 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d}+2 a^2 C x+\frac{A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int (a+a \cos (c+d x))^2 (2 a A-a (A-2 C) \cos (c+d x)) \sec ^2(c+d x) \, dx}{2 a}\\ &=\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int (a+a \cos (c+d x)) \left (a^2 (3 A+2 C)-a^2 (3 A-2 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int \left (a^3 (3 A+2 C)+\left (-a^3 (3 A-2 C)+a^3 (3 A+2 C)\right ) \cos (c+d x)-a^3 (3 A-2 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=-\frac{a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{\int \left (a^3 (3 A+2 C)+4 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a}\\ &=2 a^2 C x-\frac{a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{1}{2} \left (a^2 (3 A+2 C)\right ) \int \sec (c+d x) \, dx\\ &=2 a^2 C x+\frac{a^2 (3 A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac{A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 2.1381, size = 293, normalized size = 2.62 \[ \frac{1}{16} a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{2 (3 A+2 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (3 A+2 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{8 A \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{8 A \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{A}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{A}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{4 C \sin (c) \cos (d x)}{d}+\frac{4 C \cos (c) \sin (d x)}{d}+8 C x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(8*C*x - (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 + (4*C*Cos[d*x]*Sin[c])/d + (4*C*Cos[c]*Sin[d*
x])/d + A/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (8*A*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c +
d*x)/2] - Sin[(c + d*x)/2])) - A/(d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (8*A*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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Maple [A]  time = 0.071, size = 114, normalized size = 1. \begin{align*}{\frac{3\,A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{{a}^{2}C\sin \left ( dx+c \right ) }{d}}+2\,{\frac{A{a}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{a}^{2}Cx+2\,{\frac{C{a}^{2}c}{d}}+{\frac{A{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.08465, size = 192, normalized size = 1.71 \begin{align*} \frac{8 \,{\left (d x + c\right )} C a^{2} - A 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 \, 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 \, 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 )} + 4 \, C a^{2} \sin \left (d x + c\right ) + 8 \, 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+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.50356, size = 320, normalized size = 2.86 \begin{align*} \frac{8 \, C a^{2} d x \cos \left (d x + c\right )^{2} +{\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, C a^{2} \cos \left (d x + c\right )^{2} + 4 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(8*C*a^2*d*x*cos(d*x + c)^2 + (3*A + 2*C)*a^2*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (3*A + 2*C)*a^2*cos(d
*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*a^2*cos(d*x + c)^2 + 4*A*a^2*cos(d*x + c) + A*a^2)*sin(d*x + c))/(d*
cos(d*x + c)^2)

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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+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.23067, size = 205, normalized size = 1.83 \begin{align*} \frac{4 \,{\left (d x + c\right )} C a^{2} + \frac{4 \, C 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 (3 \, A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (3 \, A a^{2} + 2 \, 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 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 5 \, A 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+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

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