3.539 \(\int (a+b \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=187 \[ \frac{\left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac{\left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{a b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a b (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac{a A b \tan (c+d x) \sec ^3(c+d x)}{10 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[Out]

(a*b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]])/(4*d) + ((5*b^2*(2*A + 3*C) + 2*a^2*(4*A + 5*C))*Tan[c + d*x])/(15*d)
+ (a*b*(3*A + 4*C)*Sec[c + d*x]*Tan[c + d*x])/(4*d) + ((2*A*b^2 + a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x]
)/(15*d) + (a*A*b*Sec[c + d*x]^3*Tan[c + d*x])/(10*d) + (A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])
/(5*d)

________________________________________________________________________________________

Rubi [A]  time = 0.446714, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {3048, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac{\left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac{\left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac{a b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a b (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac{a A b \tan (c+d x) \sec ^3(c+d x)}{10 d}+\frac{A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(a*b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]])/(4*d) + ((5*b^2*(2*A + 3*C) + 2*a^2*(4*A + 5*C))*Tan[c + d*x])/(15*d)
+ (a*b*(3*A + 4*C)*Sec[c + d*x]*Tan[c + d*x])/(4*d) + ((2*A*b^2 + a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x]
)/(15*d) + (a*A*b*Sec[c + d*x]^3*Tan[c + d*x])/(10*d) + (A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])
/(5*d)

Rule 3048

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

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))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cos (c+d x)) \left (2 A b+a (4 A+5 C) \cos (c+d x)+b (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac{a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{20} \int \left (-4 \left (2 A b^2+a^2 (4 A+5 C)\right )-10 a b (3 A+4 C) \cos (c+d x)-4 b^2 (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac{\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac{1}{60} \int \left (-30 a b (3 A+4 C)-4 \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac{\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{2} (a b (3 A+4 C)) \int \sec ^3(c+d x) \, dx-\frac{1}{15} \left (-5 b^2 (2 A+3 C)-2 a^2 (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac{\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} (a b (3 A+4 C)) \int \sec (c+d x) \, dx-\frac{\left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac{a b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{\left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac{a b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac{\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac{a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac{A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.09355, size = 115, normalized size = 0.61 \[ \frac{\tan (c+d x) \left (20 \left (a^2 (2 A+C)+A b^2\right ) \tan ^2(c+d x)+60 \left (a^2+b^2\right ) (A+C)+12 a^2 A \tan ^4(c+d x)+15 a b (3 A+4 C) \sec (c+d x)+30 a A b \sec ^3(c+d x)\right )+15 a b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(15*a*b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(60*(a^2 + b^2)*(A + C) + 15*a*b*(3*A + 4*C)*Sec[c +
d*x] + 30*a*A*b*Sec[c + d*x]^3 + 20*(A*b^2 + a^2*(2*A + C))*Tan[c + d*x]^2 + 12*a^2*A*Tan[c + d*x]^4))/(60*d)

________________________________________________________________________________________

Maple [A]  time = 0.056, size = 257, normalized size = 1.4 \begin{align*}{\frac{2\,A{b}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{A{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{aAb \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,aAb\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,aAb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{abC\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{d}}+{\frac{abC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,A{a}^{2}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{2\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

2/3/d*A*b^2*tan(d*x+c)+1/3/d*A*b^2*tan(d*x+c)*sec(d*x+c)^2+1/d*b^2*C*tan(d*x+c)+1/2*a*A*b*sec(d*x+c)^3*tan(d*x
+c)/d+3/4*a*A*b*sec(d*x+c)*tan(d*x+c)/d+3/4/d*a*A*b*ln(sec(d*x+c)+tan(d*x+c))+1/d*a*b*C*tan(d*x+c)*sec(d*x+c)+
1/d*a*b*C*ln(sec(d*x+c)+tan(d*x+c))+8/15/d*A*a^2*tan(d*x+c)+1/5/d*A*a^2*tan(d*x+c)*sec(d*x+c)^4+4/15/d*A*a^2*t
an(d*x+c)*sec(d*x+c)^2+2/3/d*a^2*C*tan(d*x+c)+1/3/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2

________________________________________________________________________________________

Maxima [A]  time = 1.00044, size = 292, normalized size = 1.56 \begin{align*} \frac{8 \,{\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 15 \, A a b{\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 )} - 60 \, C 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 )} + 120 \, C b^{2} \tan \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/120*(8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))
*C*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^2 - 15*A*a*b*(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)) - 60*C*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)) + 120*C*b^2*tan(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.47705, size = 455, normalized size = 2.43 \begin{align*} \frac{15 \,{\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (15 \,{\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{3} + 4 \,{\left (2 \,{\left (4 \, A + 5 \, C\right )} a^{2} + 5 \,{\left (2 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, A a b \cos \left (d x + c\right ) + 12 \, A a^{2} + 4 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/120*(15*(3*A + 4*C)*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*A + 4*C)*a*b*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(15*(3*A + 4*C)*a*b*cos(d*x + c)^3 + 4*(2*(4*A + 5*C)*a^2 + 5*(2*A + 3*C)*b^2)*cos(d*x + c)^4
+ 30*A*a*b*cos(d*x + c) + 12*A*a^2 + 4*((4*A + 5*C)*a^2 + 5*A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x +
c)^5)

________________________________________________________________________________________

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+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.28288, size = 718, normalized size = 3.84 \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))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/60*(15*(3*A*a*b + 4*C*a*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*A*a*b + 4*C*a*b)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(60*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*A*a*b*tan(1/2*d*x +
1/2*c)^9 - 60*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*A*b^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*b^2*tan(1/2*d*x + 1/2*c)^9
- 80*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 160*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 30*A*a*b*tan(1/2*d*x + 1/2*c)^7 + 120*
C*a*b*tan(1/2*d*x + 1/2*c)^7 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 240*C*b^2*tan(1/2*d*x + 1/2*c)^7 + 232*A*a^2
*tan(1/2*d*x + 1/2*c)^5 + 200*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 360*C*b^2*tan(
1/2*d*x + 1/2*c)^5 - 80*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 160*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 30*A*a*b*tan(1/2*d*x
+ 1/2*c)^3 - 120*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 240*C*b^2*tan(1/2*d*x + 1/
2*c)^3 + 60*A*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*tan(1/2*d*x + 1/2*c) + 75*A*a*b*tan(1/2*d*x + 1/2*c) + 60*C*
a*b*tan(1/2*d*x + 1/2*c) + 60*A*b^2*tan(1/2*d*x + 1/2*c) + 60*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c
)^2 - 1)^5)/d