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

Optimal. Leaf size=73 \[ \frac {a^2 (A-B) \sin (c+d x)}{d}+\frac {a^2 (A+2 B) \tanh ^{-1}(\sin (c+d x))}{d}+a^2 x (2 A+B)+\frac {B \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{d} \]

[Out]

a^2*(2*A+B)*x+a^2*(A+2*B)*arctanh(sin(d*x+c))/d+a^2*(A-B)*sin(d*x+c)/d+B*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.13, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {4018, 3996, 3770} \[ \frac {a^2 (A-B) \sin (c+d x)}{d}+\frac {a^2 (A+2 B) \tanh ^{-1}(\sin (c+d x))}{d}+a^2 x (2 A+B)+\frac {B \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*(2*A + B)*x + (a^2*(A + 2*B)*ArcTanh[Sin[c + d*x]])/d + (a^2*(A - B)*Sin[c + d*x])/d + (B*(a^2 + a^2*Sec[c
+ d*x])*Sin[c + d*x])/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]

Rubi steps

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

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Mathematica [B]  time = 1.66, size = 258, normalized size = 3.53 \[ \frac {a^2 \cos ^3(c+d x) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^2 (A+B \sec (c+d x)) \left (-\frac {(A+2 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {(A+2 B) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+x (2 A+B)+\frac {A \sin (c) \cos (d x)}{d}+\frac {A \cos (c) \sin (d x)}{d}+\frac {B \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {B \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{4 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 108, normalized size = 1.48 \[ \frac {2 \, {\left (2 \, A + B\right )} a^{2} d x \cos \left (d x + c\right ) + {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{2} \cos \left (d x + c\right ) + B a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.96, size = 157, normalized size = 2.15 \[ \frac {{\left (2 \, A a^{2} + B a^{2}\right )} {\left (d x + c\right )} + {\left (A a^{2} + 2 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{2} + 2 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*A*a^2 + B*a^2)*(d*x + c) + (A*a^2 + 2*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (A*a^2 + 2*B*a^2)*log(ab
s(tan(1/2*d*x + 1/2*c) - 1)) + 2*(A*a^2*tan(1/2*d*x + 1/2*c)^3 - B*a^2*tan(1/2*d*x + 1/2*c)^3 - A*a^2*tan(1/2*
d*x + 1/2*c) - B*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1))/d

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maple [A]  time = 0.93, size = 107, normalized size = 1.47 \[ 2 a^{2} A x +a^{2} B x +\frac {a^{2} A \sin \left (d x +c \right )}{d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 A \,a^{2} c}{d}+\frac {a^{2} B \tan \left (d x +c \right )}{d}+\frac {2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B \,a^{2} c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 105, normalized size = 1.44 \[ \frac {4 \, {\left (d x + c\right )} A a^{2} + 2 \, {\left (d x + c\right )} B a^{2} + A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} \sin \left (d x + c\right ) + 2 \, B a^{2} \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.01, size = 161, normalized size = 2.21 \[ \frac {A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {4\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,B\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*cos(c + d*x), x) + Integral(2*A*cos(c + d*x)*sec(c + d*x), x) + Integral(A*cos(c + d*x)*sec(c
+ d*x)**2, x) + Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(2*B*cos(c + d*x)*sec(c + d*x)**2, x) + In
tegral(B*cos(c + d*x)*sec(c + d*x)**3, x))

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