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

Optimal. Leaf size=116 \[ \frac{\left (2 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x)}{3 d}+\frac{\left (a^2 B+2 a A b+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac{a (a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

((2*a*A*b + a^2*B + 2*b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + ((2*a^2*A + 3*A*b^2 + 6*a*b*B)*Tan[c + d*x])/(3*d)
+ (a*(2*A*b + a*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^2*A*Sec[c + d*x]^2*Tan[c + d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.270161, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2988, 3021, 2748, 3767, 8, 3770} \[ \frac{\left (2 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x)}{3 d}+\frac{\left (a^2 B+2 a A b+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^2 A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac{a (a B+2 A b) \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*A*b + a^2*B + 2*b^2*B)*ArcTanh[Sin[c + d*x]])/(2*d) + ((2*a^2*A + 3*A*b^2 + 6*a*b*B)*Tan[c + d*x])/(3*d)
+ (a*(2*A*b + a*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^2*A*Sec[c + d*x]^2*Tan[c + d*x])/(3*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 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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 ^4(c+d x) \, dx &=\frac{a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{3} \int \left (-3 a (2 A b+a B)-\left (2 a^2 A+3 A b^2+6 a b B\right ) \cos (c+d x)-3 b^2 B \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{6} \int \left (-2 \left (2 a^2 A+3 A b^2+6 a b B\right )-3 \left (2 a A b+a^2 B+2 b^2 B\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac{a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{1}{3} \left (-2 a^2 A-3 A b^2-6 a b B\right ) \int \sec ^2(c+d x) \, dx-\frac{1}{2} \left (-2 a A b-a^2 B-2 b^2 B\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (2 a A b+a^2 B+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}-\frac{\left (2 a^2 A+3 A b^2+6 a b B\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{\left (2 a A b+a^2 B+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\left (2 a^2 A+3 A b^2+6 a b B\right ) \tan (c+d x)}{3 d}+\frac{a (2 A b+a B) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^2 A \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.450794, size = 92, normalized size = 0.79 \[ \frac{3 \left (a^2 B+2 a A b+2 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (2 \left (a^2 A \tan ^2(c+d x)+3 a^2 A+6 a b B+3 A b^2\right )+3 a (a B+2 A b) \sec (c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(2*a*A*b + a^2*B + 2*b^2*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*a*(2*A*b + a*B)*Sec[c + d*x] + 2*(3*a^2
*A + 3*A*b^2 + 6*a*b*B + a^2*A*Tan[c + d*x]^2)))/(6*d)

________________________________________________________________________________________

Maple [A]  time = 0.083, size = 174, normalized size = 1.5 \begin{align*}{\frac{2\,{a}^{2}A\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}A \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{Aab\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{Aab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Bab\tan \left ( dx+c \right ) }{d}}+{\frac{A{b}^{2}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{2}B\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+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^4,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.05812, size = 232, normalized size = 2. \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} - 3 \, 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 )} - 6 \, A a b{\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 )} + 6 \, 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 )} + 24 \, B a b \tan \left (d x + c\right ) + 12 \, A b^{2} \tan \left (d x + c\right )}{12 \, 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)^4,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 - 3*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)) - 6*A*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log
(sin(d*x + c) - 1)) + 6*B*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*B*a*b*tan(d*x + c) + 12*A*b
^2*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.45872, size = 371, normalized size = 3.2 \begin{align*} \frac{3 \,{\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (B a^{2} + 2 \, A a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, A a^{2} + 2 \,{\left (2 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \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)^4,x, algorithm="fricas")

[Out]

1/12*(3*(B*a^2 + 2*A*a*b + 2*B*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(B*a^2 + 2*A*a*b + 2*B*b^2)*cos(d
*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*A*a^2 + 2*(2*A*a^2 + 6*B*a*b + 3*A*b^2)*cos(d*x + c)^2 + 3*(B*a^2 + 2*
A*a*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)

________________________________________________________________________________________

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)**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.49652, size = 397, normalized size = 3.42 \begin{align*} \frac{3 \,{\left (B a^{2} + 2 \, A 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 ) - 3 \,{\left (B a^{2} + 2 \, A 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 (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, 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 ) + 6 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A b^{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 )}^{3}}}{6 \, 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)^4,x, algorithm="giac")

[Out]

1/6*(3*(B*a^2 + 2*A*a*b + 2*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(B*a^2 + 2*A*a*b + 2*B*b^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*A*a*b*tan(
1/2*d*x + 1/2*c)^5 + 12*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*a^2*tan(1/2*d*x +
1/2*c)^3 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c) +
3*B*a^2*tan(1/2*d*x + 1/2*c) + 6*A*a*b*tan(1/2*d*x + 1/2*c) + 12*B*a*b*tan(1/2*d*x + 1/2*c) + 6*A*b^2*tan(1/2*
d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d