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

Optimal. Leaf size=82 \[ \frac{a^2 (2 A+3 B) \sin (c+d x)}{2 d}+\frac{1}{2} a^2 x (4 A+3 B)+\frac{a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.192693, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2976, 2968, 3023, 2735, 3770} \[ \frac{a^2 (2 A+3 B) \sin (c+d x)}{2 d}+\frac{1}{2} a^2 x (4 A+3 B)+\frac{a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac{B \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.171159, size = 96, normalized size = 1.17 \[ \frac{a^2 \left (4 (A+2 B) \sin (c+d x)-4 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 A d x+B \sin (2 (c+d x))+6 B d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.081, size = 108, normalized size = 1.3 \begin{align*}{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{d}}+{\frac{B{a}^{2}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}Bx}{2}}+{\frac{3\,B{a}^{2}c}{2\,d}}+2\,{a}^{2}Ax+2\,{\frac{A{a}^{2}c}{d}}+2\,{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}A\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+cos(d*x+c)*a)^2*(A+B*cos(d*x+c))*sec(d*x+c),x)

[Out]

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

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Maxima [A]  time = 0.970566, size = 127, normalized size = 1.55 \begin{align*} \frac{8 \,{\left (d x + c\right )} A a^{2} +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 4 \,{\left (d x + c\right )} B a^{2} + 4 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 4 \, A a^{2} \sin \left (d x + c\right ) + 8 \, B a^{2} \sin \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))*sec(d*x+c),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.44351, size = 194, normalized size = 2.37 \begin{align*} \frac{{\left (4 \, A + 3 \, B\right )} a^{2} d x + A a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - A a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (B a^{2} \cos \left (d x + c\right ) + 2 \,{\left (A + 2 \, B\right )} a^{2}\right )} \sin \left (d x + c\right )}{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))*sec(d*x+c),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int A \sec{\left (c + d x \right )}\, dx + \int 2 A \cos{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int B \cos{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx\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))*sec(d*x+c),x)

[Out]

a**2*(Integral(A*sec(c + d*x), x) + Integral(2*A*cos(c + d*x)*sec(c + d*x), x) + Integral(A*cos(c + d*x)**2*se
c(c + d*x), x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(2*B*cos(c + d*x)**2*sec(c + d*x), x) + In
tegral(B*cos(c + d*x)**3*sec(c + d*x), x))

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Giac [A]  time = 1.22461, size = 196, normalized size = 2.39 \begin{align*} \frac{2 \, A a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, A a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) +{\left (4 \, A a^{2} + 3 \, B a^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, B 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))*sec(d*x+c),x, algorithm="giac")

[Out]

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