3.423 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=134 \[ \frac {a^2 (2 A+3 B+2 C) \sin (c+d x)}{2 d}+\frac {(2 A+3 B) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{6 d}+\frac {1}{2} a^2 x (2 A+3 B+4 C)+\frac {a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[Out]

1/2*a^2*(2*A+3*B+4*C)*x+a^2*C*arctanh(sin(d*x+c))/d+1/2*a^2*(2*A+3*B+2*C)*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+a
*sec(d*x+c))^2*sin(d*x+c)/d+1/6*(2*A+3*B)*cos(d*x+c)*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.29, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {4086, 4017, 3996, 3770} \[ \frac {a^2 (2 A+3 B+2 C) \sin (c+d x)}{2 d}+\frac {(2 A+3 B) \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{6 d}+\frac {1}{2} a^2 x (2 A+3 B+4 C)+\frac {a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

Rule 3770

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

Rule 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4017

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

Rule 4086

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

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

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Mathematica [A]  time = 0.30, size = 121, normalized size = 0.90 \[ \frac {a^2 \left (3 (7 A+8 B+4 C) \sin (c+d x)+3 (2 A+B) \sin (2 (c+d x))+A \sin (3 (c+d x))+12 A d x+18 B d x-12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+24 C d x\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(12*A*d*x + 18*B*d*x + 24*C*d*x - 12*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*C*Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]] + 3*(7*A + 8*B + 4*C)*Sin[c + d*x] + 3*(2*A + B)*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)])
)/(12*d)

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fricas [A]  time = 0.52, size = 108, normalized size = 0.81 \[ \frac {3 \, {\left (2 \, A + 3 \, B + 4 \, C\right )} a^{2} d x + 3 \, C a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (5 \, A + 6 \, B + 3 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(3*(2*A + 3*B + 4*C)*a^2*d*x + 3*C*a^2*log(sin(d*x + c) + 1) - 3*C*a^2*log(-sin(d*x + c) + 1) + (2*A*a^2*c
os(d*x + c)^2 + 3*(2*A + B)*a^2*cos(d*x + c) + 2*(5*A + 6*B + 3*C)*a^2)*sin(d*x + c))/d

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giac [A]  time = 0.29, size = 235, normalized size = 1.75 \[ \frac {6 \, C a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, C a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (2 \, A a^{2} + 3 \, B a^{2} + 4 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(6*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(2*A*a^2 + 3*
B*a^2 + 4*C*a^2)*(d*x + c) + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(
1/2*d*x + 1/2*c)^5 + 16*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*tan(1/2*d*x
+ 1/2*c)^3 + 18*A*a^2*tan(1/2*d*x + 1/2*c) + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 6*C*a^2*tan(1/2*d*x + 1/2*c))/(ta
n(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 1.33, size = 181, normalized size = 1.35 \[ \frac {a^{2} A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {5 a^{2} A \sin \left (d x +c \right )}{3 d}+\frac {B \,a^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a^{2} B x}{2}+\frac {3 B \,a^{2} c}{2 d}+\frac {a^{2} C \sin \left (d x +c \right )}{d}+\frac {a^{2} A \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+a^{2} A x +\frac {A \,a^{2} c}{d}+\frac {2 B \,a^{2} \sin \left (d x +c \right )}{d}+2 a^{2} C x +\frac {2 C \,a^{2} c}{d}+\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/3*a^2*A*cos(d*x+c)^2*sin(d*x+c)/d+5/3/d*a^2*A*sin(d*x+c)+1/2/d*B*a^2*cos(d*x+c)*sin(d*x+c)+3/2*a^2*B*x+3/2/d
*B*a^2*c+1/d*a^2*C*sin(d*x+c)+a^2*A*cos(d*x+c)*sin(d*x+c)/d+a^2*A*x+1/d*A*a^2*c+2/d*B*a^2*sin(d*x+c)+2*a^2*C*x
+2/d*C*a^2*c+1/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.41, size = 160, normalized size = 1.19 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 12 \, {\left (d x + c\right )} B a^{2} - 24 \, {\left (d x + c\right )} C a^{2} - 6 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{2} \sin \left (d x + c\right ) - 24 \, B a^{2} \sin \left (d x + c\right ) - 12 \, C a^{2} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.36, size = 223, normalized size = 1.66 \[ \frac {7\,A\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(a + a/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(7*A*a^2*sin(c + d*x))/(4*d) + (2*B*a^2*sin(c + d*x))/d + (C*a^2*sin(c + d*x))/d + (2*A*a^2*atan(sin(c/2 + (d*
x)/2)/cos(c/2 + (d*x)/2)))/d + (3*B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (4*C*a^2*atan(sin(c/2
+ (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^2*sin(2*c
+ 2*d*x))/(2*d) + (A*a^2*sin(3*c + 3*d*x))/(12*d) + (B*a^2*sin(2*c + 2*d*x))/(4*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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