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

Optimal. Leaf size=60 \[ \frac{a^2 A \tan (c+d x)}{d}+\frac{a (a B+2 A b) \tanh ^{-1}(\sin (c+d x))}{d}+b x (2 a B+A b)+\frac{b^2 B \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.168526, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {2988, 3023, 2735, 3770} \[ \frac{a^2 A \tan (c+d x)}{d}+\frac{a (a B+2 A b) \tanh ^{-1}(\sin (c+d x))}{d}+b x (2 a B+A b)+\frac{b^2 B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2988

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/
(f*d^2*(n + 1)*(c^2 - d^2)), x] - Dist[1/(d^2*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*Simp[d*(n
+ 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n +
1))) + 2*a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x
]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && LtQ[n, -1]

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

Mathematica [A]  time = 0.473277, size = 109, normalized size = 1.82 \[ \frac{a^2 A \tan (c+d x)+b (c+d x) (2 a B+A b)-a (a B+2 A b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+a (a B+2 A b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+b^2 B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.071, size = 104, normalized size = 1.7 \begin{align*} A{b}^{2}x+2\,Babx+{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}+2\,{\frac{Aab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{2}c}{d}}+{\frac{{b}^{2}B\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{Babc}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.969159, size = 139, normalized size = 2.32 \begin{align*} \frac{4 \,{\left (d x + c\right )} B a b + 2 \,{\left (d x + c\right )} A b^{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 )} + 2 \, A a b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B b^{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+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.40506, size = 294, normalized size = 4.9 \begin{align*} \frac{2 \,{\left (2 \, B a b + A b^{2}\right )} d x \cos \left (d x + c\right ) +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (B b^{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+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

1/2*(2*(2*B*a*b + A*b^2)*d*x*cos(d*x + c) + (B*a^2 + 2*A*a*b)*cos(d*x + c)*log(sin(d*x + c) + 1) - (B*a^2 + 2*
A*a*b)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(B*b^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+b*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 = 1.51963, size = 205, normalized size = 3.42 \begin{align*} \frac{{\left (2 \, B a b + A b^{2}\right )}{\left (d x + c\right )} +{\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (B a^{2} + 2 \, A a b\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 b^{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 b^{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+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^2,x, algorithm="giac")

[Out]

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