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

Optimal. Leaf size=119 \[ \frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}-\frac {(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac {1}{2} a^2 x (3 A+2 C)+\frac {2 a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

[Out]

1/2*a^2*(3*A+2*C)*x+2*a^2*C*arctanh(sin(d*x+c))/d+1/2*a^2*(3*A-2*C)*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+a*sec(d*x
+c))^2*sin(d*x+c)/d-1/2*(A-2*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 = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4087, 4018, 3996, 3770} \[ \frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}-\frac {(A-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac {1}{2} a^2 x (3 A+2 C)+\frac {2 a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(3*A + 2*C)*x)/2 + (2*a^2*C*ArcTanh[Sin[c + d*x]])/d + (a^2*(3*A - 2*C)*Sin[c + d*x])/(2*d) + (A*Cos[c +
d*x]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - ((A - 2*C)*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(2*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 4018

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

Rule 4087

Int[((A_.) + 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] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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 ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^2 (2 a A-a (A-2 C) \sec (c+d x)) \, dx}{2 a}\\ &=\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (a^2 (3 A-2 C)+4 a^2 C \sec (c+d x)\right ) \, dx}{2 a}\\ &=\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac {\int \left (-a^3 (3 A+2 C)-4 a^3 C \sec (c+d x)\right ) \, dx}{2 a}\\ &=\frac {1}{2} a^2 (3 A+2 C) x+\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\left (2 a^2 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^2 (3 A+2 C) x+\frac {2 a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 (3 A-2 C) \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 C) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [B]  time = 1.25, size = 292, normalized size = 2.45 \[ -\frac {a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (4 \cos (d x) \left (3 A d x-4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 C d x\right )+4 \cos (2 c+d x) \left (3 A d x-4 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+2 C d x\right )+A \sin (2 c+d x)+8 A \sin (c+2 d x)+8 A \sin (3 c+2 d x)+A \sin (2 c+3 d x)+A \sin (4 c+3 d x)+A \sin (d x)+16 C \sin (d x)\right )}{16 d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\tan \left (\frac {1}{2} (c+d x)\right )-1\right ) \left (\tan \left (\frac {1}{2} (c+d x)\right )+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/16*(a^2*Sec[(c + d*x)/2]^2*(4*Cos[d*x]*(3*A*d*x + 2*C*d*x - 4*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] +
4*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 4*Cos[2*c + d*x]*(3*A*d*x + 2*C*d*x - 4*C*Log[Cos[(c + d*x)/2]
- Sin[(c + d*x)/2]] + 4*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + A*Sin[d*x] + 16*C*Sin[d*x] + A*Sin[2*c
+ d*x] + 8*A*Sin[c + 2*d*x] + 8*A*Sin[3*c + 2*d*x] + A*Sin[2*c + 3*d*x] + A*Sin[4*c + 3*d*x]))/(d*(Cos[c/2] -
Sin[c/2])*(Cos[c/2] + Sin[c/2])*(-1 + Tan[(c + d*x)/2])*(1 + Tan[(c + d*x)/2]))

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fricas [A]  time = 0.46, size = 116, normalized size = 0.97 \[ \frac {{\left (3 \, A + 2 \, C\right )} a^{2} d x \cos \left (d x + c\right ) + 2 \, C a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, C a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{2} \cos \left (d x + c\right )^{2} + 4 \, A a^{2} \cos \left (d x + c\right ) + 2 \, C a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.29, size = 143, normalized size = 1.20 \[ \frac {4 \, C a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, C a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + {\left (3 \, A a^{2} + 2 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 4*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 4*C*a^2*tan(1/2
*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + (3*A*a^2 + 2*C*a^2)*(d*x + c) + 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^3
+ 5*A*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d

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maple [A]  time = 0.90, size = 107, normalized size = 0.90 \[ \frac {a^{2} A \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a^{2} A x}{2}+\frac {3 A \,a^{2} c}{2 d}+a^{2} C x +\frac {C \,a^{2} c}{d}+\frac {2 a^{2} A \sin \left (d x +c \right )}{d}+\frac {2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} C \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 101, normalized size = 0.85 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 4 \, {\left (d x + c\right )} A a^{2} + 4 \, {\left (d x + c\right )} C a^{2} + 4 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \sin \left (d x + c\right ) + 4 \, C a^{2} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 4*(d*x + c)*A*a^2 + 4*(d*x + c)*C*a^2 + 4*C*a^2*(log(sin(d*x + c
) + 1) - log(sin(d*x + c) - 1)) + 8*A*a^2*sin(d*x + c) + 4*C*a^2*tan(d*x + c))/d

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mupad [B]  time = 2.64, size = 152, normalized size = 1.28 \[ \frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int A \cos ^{2}{\left (c + d x \right )}\, dx + \int 2 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*cos(c + d*x)**2, x) + Integral(2*A*cos(c + d*x)**2*sec(c + d*x), x) + Integral(A*cos(c + d*x)
**2*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(2*C*cos(c + d*x)**2*sec(c
+ d*x)**3, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**4, x))

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