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

Optimal. Leaf size=147 \[ \frac{2 a^2 (2 A+3 C) \tan (c+d x)}{3 d}+\frac{a^2 (7 A+12 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (11 A+12 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

[Out]

(a^2*(7*A + 12*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a^2*(2*A + 3*C)*Tan[c + d*x])/(3*d) + (a^2*(11*A + 12*C)*S
ec[c + d*x]*Tan[c + d*x])/(24*d) + (A*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) + (A*(a + a*
Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.449169, antiderivative size = 147, 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 = {3044, 2975, 2968, 3021, 2748, 3767, 8, 3770} \[ \frac{2 a^2 (2 A+3 C) \tan (c+d x)}{3 d}+\frac{a^2 (7 A+12 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (11 A+12 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{A \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{6 d}+\frac{A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*A + 12*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (2*a^2*(2*A + 3*C)*Tan[c + d*x])/(3*d) + (a^2*(11*A + 12*C)*S
ec[c + d*x]*Tan[c + d*x])/(24*d) + (A*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) + (A*(a + a*
Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 3044

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/(b*d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^
m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n +
2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b
*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0
])

Rule 2975

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

Rule 2968

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

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+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx &=\frac{A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x))^2 (2 a A+a (A+4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx}{4 a}\\ &=\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int (a+a \cos (c+d x)) \left (a^2 (11 A+12 C)+a^2 (5 A+12 C) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \left (a^3 (11 A+12 C)+\left (a^3 (5 A+12 C)+a^3 (11 A+12 C)\right ) \cos (c+d x)+a^3 (5 A+12 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx}{12 a}\\ &=\frac{a^2 (11 A+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{\int \left (16 a^3 (2 A+3 C)+3 a^3 (7 A+12 C) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{24 a}\\ &=\frac{a^2 (11 A+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{3} \left (2 a^2 (2 A+3 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (a^2 (7 A+12 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{a^2 (7 A+12 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (11 A+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{\left (2 a^2 (2 A+3 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{a^2 (7 A+12 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{2 a^2 (2 A+3 C) \tan (c+d x)}{3 d}+\frac{a^2 (11 A+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac{A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 1.13972, size = 262, normalized size = 1.78 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (24 (7 A+12 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-48 (2 A+3 C) \sin (c)+45 A \sin (2 c+d x)+128 A \sin (c+2 d x)+21 A \sin (2 c+3 d x)+21 A \sin (4 c+3 d x)+32 A \sin (3 c+4 d x)+3 (15 A+4 C) \sin (d x)+12 C \sin (2 c+d x)+144 C \sin (c+2 d x)-48 C \sin (3 c+2 d x)+12 C \sin (2 c+3 d x)+12 C \sin (4 c+3 d x)+48 C \sin (3 c+4 d x))\right )}{768 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*Sec[c + d*x]^4*(24*(7*A + 12*C)*Cos[c + d*x]^4*(Log[Cos[(c + d*x
)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-48*(2*A + 3*C)*Sin[c] + 3*(15*
A + 4*C)*Sin[d*x] + 45*A*Sin[2*c + d*x] + 12*C*Sin[2*c + d*x] + 128*A*Sin[c + 2*d*x] + 144*C*Sin[c + 2*d*x] -
48*C*Sin[3*c + 2*d*x] + 21*A*Sin[2*c + 3*d*x] + 12*C*Sin[2*c + 3*d*x] + 21*A*Sin[4*c + 3*d*x] + 12*C*Sin[4*c +
3*d*x] + 32*A*Sin[3*c + 4*d*x] + 48*C*Sin[3*c + 4*d*x])))/(768*d)

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Maple [A]  time = 0.076, size = 166, normalized size = 1.1 \begin{align*}{\frac{7\,A{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{7\,A{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{4\,A{a}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+2\,{\frac{{a}^{2}C\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x)

[Out]

7/8/d*A*a^2*sec(d*x+c)*tan(d*x+c)+7/8/d*A*a^2*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+
4/3/d*A*a^2*tan(d*x+c)+2/3/d*A*a^2*tan(d*x+c)*sec(d*x+c)^2+2/d*a^2*C*tan(d*x+c)+1/4/d*A*a^2*tan(d*x+c)*sec(d*x
+c)^3+1/2/d*a^2*C*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 1.19544, size = 316, normalized size = 2.15 \begin{align*} \frac{32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} - 3 \, A a^{2}{\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 \, A 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 )} - 12 \, C 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 )} + 24 \, C a^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, C a^{2} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 - 3*A*a^2*(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*A*a^2*(2*sin(d*x + c)/(s
in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*C*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)) + 24*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) -
1)) + 96*C*a^2*tan(d*x + c))/d

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Fricas [A]  time = 1.46305, size = 356, normalized size = 2.42 \begin{align*} \frac{3 \,{\left (7 \, A + 12 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (7 \, A + 12 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (2 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (7 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 16 \, A a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\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+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(3*(7*A + 12*C)*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(7*A + 12*C)*a^2*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(16*(2*A + 3*C)*a^2*cos(d*x + c)^3 + 3*(7*A + 4*C)*a^2*cos(d*x + c)^2 + 16*A*a^2*cos(d*x + c) +
6*A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

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

[Out]

Timed out

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Giac [A]  time = 1.28254, size = 286, normalized size = 1.95 \begin{align*} \frac{3 \,{\left (7 \, A a^{2} + 12 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (7 \, A a^{2} + 12 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (21 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 36 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 77 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 132 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 83 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 156 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 75 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 60 \, C a^{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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(7*A*a^2 + 12*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(7*A*a^2 + 12*C*a^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(21*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 36*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 77*A*a^2*tan(1/2*d*x +
1/2*c)^5 - 132*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 156*C*a^2*tan(1/2*d*x + 1/2*c)
^3 - 75*A*a^2*tan(1/2*d*x + 1/2*c) - 60*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d