3.945 \(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=137 \[ \frac{\tan (c+d x) (2 a B+2 A b+3 b C)}{3 d}+\frac{(3 a A+4 a C+4 b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) (3 a A+4 a C+4 b B)}{8 d}+\frac{(a B+A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac{a A \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[Out]

((3*a*A + 4*b*B + 4*a*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((2*A*b + 2*a*B + 3*b*C)*Tan[c + d*x])/(3*d) + ((3*a*A
+ 4*b*B + 4*a*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + ((A*b + a*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (a*A*Se
c[c + d*x]^3*Tan[c + d*x])/(4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.239113, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.18, Rules used = {3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{\tan (c+d x) (2 a B+2 A b+3 b C)}{3 d}+\frac{(3 a A+4 a C+4 b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\tan (c+d x) \sec (c+d x) (3 a A+4 a C+4 b B)}{8 d}+\frac{(a B+A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac{a A \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

((3*a*A + 4*b*B + 4*a*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((2*A*b + 2*a*B + 3*b*C)*Tan[c + d*x])/(3*d) + ((3*a*A
+ 4*b*B + 4*a*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + ((A*b + a*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (a*A*Se
c[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{a A \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{4} \int \left (-4 (A b+a B)-(3 a A+4 b B+4 a C) \cos (c+d x)-4 b C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{(A b+a B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{a A \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{12} \int (-3 (3 a A+4 b B+4 a C)-4 (2 A b+2 a B+3 b C) \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac{(A b+a B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{a A \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{4} (-3 a A-4 b B-4 a C) \int \sec ^3(c+d x) \, dx-\frac{1}{3} (-2 A b-2 a B-3 b C) \int \sec ^2(c+d x) \, dx\\ &=\frac{(3 a A+4 b B+4 a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(A b+a B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{a A \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{8} (-3 a A-4 b B-4 a C) \int \sec (c+d x) \, dx-\frac{(2 A b+2 a B+3 b C) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{(3 a A+4 b B+4 a C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(2 A b+2 a B+3 b C) \tan (c+d x)}{3 d}+\frac{(3 a A+4 b B+4 a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{(A b+a B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{a A \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.653965, size = 100, normalized size = 0.73 \[ \frac{3 (3 a A+4 a C+4 b B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \sec (c+d x) (3 a A+4 a C+4 b B)+8 (a B+A b) \tan ^2(c+d x)+24 (a B+A b+b C)+6 a A \sec ^3(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(3*(3*a*A + 4*b*B + 4*a*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24*(A*b + a*B + b*C) + 3*(3*a*A + 4*b*B + 4*a
*C)*Sec[c + d*x] + 6*a*A*Sec[c + d*x]^3 + 8*(A*b + a*B)*Tan[c + d*x]^2))/(24*d)

________________________________________________________________________________________

Maple [A]  time = 0.058, size = 223, normalized size = 1.6 \begin{align*}{\frac{2\,Ab\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Ab \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{bB\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{bB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{Cb\tan \left ( dx+c \right ) }{d}}+{\frac{aA \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,aA\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,Ba\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Ba \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{aC\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{aC\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))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

2/3*A*b*tan(d*x+c)/d+1/3*A*b*sec(d*x+c)^2*tan(d*x+c)/d+1/2/d*b*B*sec(d*x+c)*tan(d*x+c)+1/2/d*b*B*ln(sec(d*x+c)
+tan(d*x+c))+1/d*C*b*tan(d*x+c)+1/4*a*A*sec(d*x+c)^3*tan(d*x+c)/d+3/8*a*A*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a*A*ln
(sec(d*x+c)+tan(d*x+c))+2/3*a*B*tan(d*x+c)/d+1/3*a*B*sec(d*x+c)^2*tan(d*x+c)/d+1/2/d*a*C*tan(d*x+c)*sec(d*x+c)
+1/2/d*a*C*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.02229, size = 294, normalized size = 2.15 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b - 3 \, A a{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a{\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 )} - 12 \, B 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 )} + 48 \, C b \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b - 3*A*a*(2*(3*sin(d*
x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x +
c) - 1)) - 12*C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*
B*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*C*b*tan(d*x + c
))/d

________________________________________________________________________________________

Fricas [A]  time = 1.81406, size = 400, normalized size = 2.92 \begin{align*} \frac{3 \,{\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (2 \, B a +{\left (2 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} \cos \left (d x + c\right )^{2} + 6 \, A a + 8 \,{\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(3*((3*A + 4*C)*a + 4*B*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*((3*A + 4*C)*a + 4*B*b)*cos(d*x + c)^
4*log(-sin(d*x + c) + 1) + 2*(8*(2*B*a + (2*A + 3*C)*b)*cos(d*x + c)^3 + 3*((3*A + 4*C)*a + 4*B*b)*cos(d*x + c
)^2 + 6*A*a + 8*(B*a + A*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

________________________________________________________________________________________

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))*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**5,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.26706, size = 578, normalized size = 4.22 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(3*(3*A*a + 4*C*a + 4*B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a + 4*C*a + 4*B*b)*log(abs(tan(1/2
*d*x + 1/2*c) - 1)) + 2*(15*A*a*tan(1/2*d*x + 1/2*c)^7 - 24*B*a*tan(1/2*d*x + 1/2*c)^7 + 12*C*a*tan(1/2*d*x +
1/2*c)^7 - 24*A*b*tan(1/2*d*x + 1/2*c)^7 + 12*B*b*tan(1/2*d*x + 1/2*c)^7 - 24*C*b*tan(1/2*d*x + 1/2*c)^7 + 9*A
*a*tan(1/2*d*x + 1/2*c)^5 + 40*B*a*tan(1/2*d*x + 1/2*c)^5 - 12*C*a*tan(1/2*d*x + 1/2*c)^5 + 40*A*b*tan(1/2*d*x
+ 1/2*c)^5 - 12*B*b*tan(1/2*d*x + 1/2*c)^5 + 72*C*b*tan(1/2*d*x + 1/2*c)^5 + 9*A*a*tan(1/2*d*x + 1/2*c)^3 - 4
0*B*a*tan(1/2*d*x + 1/2*c)^3 - 12*C*a*tan(1/2*d*x + 1/2*c)^3 - 40*A*b*tan(1/2*d*x + 1/2*c)^3 - 12*B*b*tan(1/2*
d*x + 1/2*c)^3 - 72*C*b*tan(1/2*d*x + 1/2*c)^3 + 15*A*a*tan(1/2*d*x + 1/2*c) + 24*B*a*tan(1/2*d*x + 1/2*c) + 1
2*C*a*tan(1/2*d*x + 1/2*c) + 24*A*b*tan(1/2*d*x + 1/2*c) + 12*B*b*tan(1/2*d*x + 1/2*c) + 24*C*b*tan(1/2*d*x +
1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d