3.783 \(\int (a+b \cos (c+d x))^2 (B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=80 \[ \frac{\left (a^2 B+4 a b C+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 B \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a (a C+2 b B) \tan (c+d x)}{d}+b^2 C x \]

[Out]

b^2*C*x + ((a^2*B + 2*b^2*B + 4*a*b*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*b*B + a*C)*Tan[c + d*x])/d + (a^2*
B*Sec[c + d*x]*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.279569, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3029, 2988, 3021, 2735, 3770} \[ \frac{\left (a^2 B+4 a b C+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 B \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a (a C+2 b B) \tan (c+d x)}{d}+b^2 C x \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

b^2*C*x + ((a^2*B + 2*b^2*B + 4*a*b*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (a*(2*b*B + a*C)*Tan[c + d*x])/d + (a^2*
B*Sec[c + d*x]*Tan[c + d*x])/(2*d)

Rule 3029

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

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 3021

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

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

Mathematica [A]  time = 0.257165, size = 67, normalized size = 0.84 \[ \frac{\left (a^2 B+4 a b C+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+a \tan (c+d x) (a B \sec (c+d x)+2 a C+4 b B)+2 b^2 C d x}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(2*b^2*C*d*x + (a^2*B + 2*b^2*B + 4*a*b*C)*ArcTanh[Sin[c + d*x]] + a*(4*b*B + 2*a*C + a*B*Sec[c + d*x])*Tan[c
+ d*x])/(2*d)

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Maple [A]  time = 0.051, size = 133, normalized size = 1.7 \begin{align*}{b}^{2}Cx+{\frac{C{b}^{2}c}{d}}+{\frac{{b}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{abB\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}B\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}B\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)

[Out]

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

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Maxima [A]  time = 1.14292, size = 189, normalized size = 2.36 \begin{align*} \frac{4 \,{\left (d x + c\right )} C b^{2} - B 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 )} + 4 \, C 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}{\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} \tan \left (d x + c\right ) + 8 \, B a b \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.50077, size = 335, normalized size = 4.19 \begin{align*} \frac{4 \, C b^{2} d x \cos \left (d x + c\right )^{2} +{\left (B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (B a^{2} + 2 \,{\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )\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+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/4*(4*C*b^2*d*x*cos(d*x + c)^2 + (B*a^2 + 4*C*a*b + 2*B*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (B*a^2 +
4*C*a*b + 2*B*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(B*a^2 + 2*(C*a^2 + 2*B*a*b)*cos(d*x + c))*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+b*cos(d*x+c))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.47228, size = 257, normalized size = 3.21 \begin{align*} \frac{2 \,{\left (d x + c\right )} C b^{2} +{\left (B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B a b \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 ) + 2 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, B a b \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+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

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