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

Optimal. Leaf size=88 \[ \frac {a^2 (3 A+2 B) \sin (c+d x)}{2 d}+\frac {1}{2} a^2 x (3 A+4 B)+\frac {A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac {a^2 B \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {4017, 3996, 3770} \[ \frac {a^2 (3 A+2 B) \sin (c+d x)}{2 d}+\frac {1}{2} a^2 x (3 A+4 B)+\frac {A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{2 d}+\frac {a^2 B \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 96, normalized size = 1.09 \[ \frac {a^2 \left (4 (2 A+B) \sin (c+d x)+A \sin (2 (c+d x))+6 A d x-4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 B \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+8 B d x\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(6*A*d*x + 8*B*d*x - 4*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*B*Log[Cos[(c + d*x)/2] + Sin[(c + d
*x)/2]] + 4*(2*A + B)*Sin[c + d*x] + A*Sin[2*(c + d*x)]))/(4*d)

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fricas [A]  time = 0.46, size = 79, normalized size = 0.90 \[ \frac {{\left (3 \, A + 4 \, B\right )} a^{2} d x + B a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - B a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{2} \cos \left (d x + c\right ) + 2 \, {\left (2 \, A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]

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+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.28, size = 145, normalized size = 1.65 \[ \frac {2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, B a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (3 \, A a^{2} + 4 \, B a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

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+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/2*(2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*B*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (3*A*a^2 + 4*B*
a^2)*(d*x + c) + 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 5*A*a^2*tan(1/2*d*x + 1/
2*c) + 2*B*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 = 0.82, size = 108, normalized size = 1.23 \[ \frac {a^{2} A \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a^{2} A x}{2}+\frac {3 A \,a^{2} c}{2 d}+\frac {B \,a^{2} \sin \left (d x +c \right )}{d}+\frac {2 a^{2} A \sin \left (d x +c \right )}{d}+2 a^{2} B x +\frac {2 B \,a^{2} c}{d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),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+1/d*B*a^2*sin(d*x+c)+2/d*a^2*A*sin(d*x+c)+2*a^2*B*
x+2/d*B*a^2*c+1/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.34, size = 101, normalized size = 1.15 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 4 \, {\left (d x + c\right )} A a^{2} + 8 \, {\left (d x + c\right )} B a^{2} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, A a^{2} \sin \left (d x + c\right ) + 4 \, B a^{2} \sin \left (d x + c\right )}{4 \, d} \]

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+B*sec(d*x+c)),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 + 8*(d*x + c)*B*a^2 + 2*B*a^2*(log(sin(d*x + c
) + 1) - log(sin(d*x + c) - 1)) + 8*A*a^2*sin(d*x + c) + 4*B*a^2*sin(d*x + c))/d

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mupad [B]  time = 2.05, size = 141, normalized size = 1.60 \[ \frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {3\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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+B*sec(d*x+c)),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(B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(2*B*cos(c + d*x)**2*sec(c + d
*x)**2, x) + Integral(B*cos(c + d*x)**2*sec(c + d*x)**3, x))

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