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

Optimal. Leaf size=74 \[ -\frac{a^2 (A-B) \sin (c+d x)}{d}+\frac{a^2 (2 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+a^2 x (A+2 B)+\frac{A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.209642, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {2975, 2968, 3023, 2735, 3770} \[ -\frac{a^2 (A-B) \sin (c+d x)}{d}+\frac{a^2 (2 A+B) \tanh ^{-1}(\sin (c+d x))}{d}+a^2 x (A+2 B)+\frac{A \tan (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.311407, size = 143, normalized size = 1.93 \[ \frac{a^2 \left (A \tan (c+d x)-2 A \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 A \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+A c+A d x+B \sin (c+d x)-B \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+B \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+2 B c+2 B d x\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.092, size = 107, normalized size = 1.5 \begin{align*}{a}^{2}Ax+2\,{a}^{2}Bx+{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{a}^{2}c}{d}}+{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{B{a}^{2}c}{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)^2,x)

[Out]

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

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Maxima [A]  time = 1.00087, size = 142, normalized size = 1.92 \begin{align*} \frac{2 \,{\left (d x + c\right )} A a^{2} + 4 \,{\left (d x + c\right )} B a^{2} + 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 )} + 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 )} + 2 \, B a^{2} \sin \left (d x + c\right ) + 2 \, A a^{2} \tan \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.42364, size = 278, normalized size = 3.76 \begin{align*} \frac{2 \,{\left (A + 2 \, B\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 ) + 2 \,{\left (B a^{2} \cos \left (d x + c\right ) + 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))*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

1/2*(2*(A + 2*B)*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) + 2*(B*a^2*cos(d*x + c) + 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))*sec(d*x+c)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.24293, size = 209, normalized size = 2.82 \begin{align*} \frac{{\left (A a^{2} + 2 \, B a^{2}\right )}{\left (d x + c\right )} +{\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 ) -{\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 (A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1}}{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)^2,x, algorithm="giac")

[Out]

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