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

Optimal. Leaf size=117 \[ \frac {a^3 (A+3 B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {(A-2 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+\frac {1}{2} a^3 x (7 A+6 B)+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

[Out]

1/2*a^3*(7*A+6*B)*x+a^3*(A+3*B)*arctanh(sin(d*x+c))/d+5/2*a^3*A*sin(d*x+c)/d+1/2*a*A*cos(d*x+c)*(a+a*sec(d*x+c
))^2*sin(d*x+c)/d-1/2*(A-2*B)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4017, 4018, 3996, 3770} \[ \frac {a^3 (A+3 B) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {(A-2 B) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+\frac {1}{2} a^3 x (7 A+6 B)+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {a A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(7*A + 6*B)*x)/2 + (a^3*(A + 3*B)*ArcTanh[Sin[c + d*x]])/d + (5*a^3*A*Sin[c + d*x])/(2*d) + (a*A*Cos[c +
d*x]*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - ((A - 2*B)*(a^3 + a^3*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]

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 ^2(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {a A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {1}{2} \int \cos (c+d x) (a+a \sec (c+d x))^2 (2 a (2 A+B)-a (A-2 B) \sec (c+d x)) \, dx\\ &=\frac {a A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 B) \left (a^3+a^3 \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)) \left (5 a^2 A+2 a^2 (A+3 B) \sec (c+d x)\right ) \, dx\\ &=\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac {1}{2} \int \left (-a^3 (7 A+6 B)-2 a^3 (A+3 B) \sec (c+d x)\right ) \, dx\\ &=\frac {1}{2} a^3 (7 A+6 B) x+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\left (a^3 (A+3 B)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} a^3 (7 A+6 B) x+\frac {a^3 (A+3 B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 A \sin (c+d x)}{2 d}+\frac {a A \cos (c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac {(A-2 B) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 5.09, size = 302, normalized size = 2.58 \[ \frac {a^3 \cos ^4(c+d x) \sec ^6\left (\frac {1}{2} (c+d x)\right ) (\sec (c+d x)+1)^3 (A+B \sec (c+d x)) \left (\frac {4 (3 A+B) \sin (c) \cos (d x)}{d}+\frac {4 (3 A+B) \cos (c) \sin (d x)}{d}-\frac {4 (A+3 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {4 (A+3 B) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{d}+2 x (7 A+6 B)+\frac {A \sin (2 c) \cos (2 d x)}{d}+\frac {A \cos (2 c) \sin (2 d x)}{d}+\frac {4 B \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {4 B \sin \left (\frac {d x}{2}\right )}{d \left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{32 (A \cos (c+d x)+B)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[c + d*x]^4*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(A + B*Sec[c + d*x])*(2*(7*A + 6*B)*x - (4*(A + 3*
B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (4*(A + 3*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4
*(3*A + B)*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2*c])/d + (4*(3*A + B)*Cos[c]*Sin[d*x])/d + (A*Cos[2*c]*Sin[
2*d*x])/d + (4*B*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (4*B*Sin[(d*x
)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(32*(B + A*Cos[c + d*x]))

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 127, normalized size = 1.09 \[ \frac {{\left (7 \, A + 6 \, B\right )} a^{3} d x \cos \left (d x + c\right ) + {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 2 \, B a^{3}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((7*A + 6*B)*a^3*d*x*cos(d*x + c) + (A + 3*B)*a^3*cos(d*x + c)*log(sin(d*x + c) + 1) - (A + 3*B)*a^3*cos(d
*x + c)*log(-sin(d*x + c) + 1) + (A*a^3*cos(d*x + c)^2 + 2*(3*A + B)*a^3*cos(d*x + c) + 2*B*a^3)*sin(d*x + c))
/(d*cos(d*x + c))

________________________________________________________________________________________

giac [A]  time = 0.34, size = 192, normalized size = 1.64 \[ -\frac {\frac {4 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (7 \, A a^{3} + 6 \, B a^{3}\right )} {\left (d x + c\right )} - 2 \, {\left (A a^{3} + 3 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (A a^{3} + 3 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*B*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (7*A*a^3 + 6*B*a^3)*(d*x + c) - 2*(A*a^3 + 3
*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(A*a^3 + 3*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*A*a
^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 7*A*a^3*tan(1/2*d*x + 1/2*c) + 2*B*a^3*tan(1/2*d*
x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d

________________________________________________________________________________________

maple [A]  time = 0.88, size = 145, normalized size = 1.24 \[ \frac {A \,a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {7 a^{3} A x}{2}+\frac {7 A \,a^{3} c}{2 d}+\frac {a^{3} B \sin \left (d x +c \right )}{d}+\frac {3 a^{3} A \sin \left (d x +c \right )}{d}+3 a^{3} B x +\frac {3 a^{3} B c}{d}+\frac {3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{3} B \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*A*a^3*cos(d*x+c)*sin(d*x+c)+7/2*a^3*A*x+7/2/d*A*a^3*c+a^3*B*sin(d*x+c)/d+3*a^3*A*sin(d*x+c)/d+3*a^3*B*x+
3/d*a^3*B*c+3/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*a^3*B*tan(d*x+c)

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 140, normalized size = 1.20 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 12 \, {\left (d x + c\right )} A a^{3} + 12 \, {\left (d x + c\right )} B a^{3} + 2 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^3*(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^3 + 12*(d*x + c)*A*a^3 + 12*(d*x + c)*B*a^3 + 2*A*a^3*(log(sin(d*x +
c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x
+ c) + 4*B*a^3*sin(d*x + c) + 4*B*a^3*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 2.08, size = 197, normalized size = 1.68 \[ \frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {7\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*A*a^3*sin(c + d*x))/d + (B*a^3*sin(c + d*x))/d + (7*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d +
(2*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (6*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (B*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (A*a
^3*cos(c + d*x)*sin(c + d*x))/(2*d)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________