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

Optimal. Leaf size=124 \[ \frac {1}{2} a x \left (a^2 A+6 a b B+6 A b^2\right )+\frac {a^2 (a B+2 A b) \sin (c+d x)}{d}-\frac {b^2 (a A-2 b B) \tan (c+d x)}{2 d}+\frac {b^2 (3 a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4025, 4076, 4047, 8, 4045, 3770} \[ \frac {1}{2} a x \left (a^2 A+6 a b B+6 A b^2\right )+\frac {a^2 (a B+2 A b) \sin (c+d x)}{d}-\frac {b^2 (a A-2 b B) \tan (c+d x)}{2 d}+\frac {b^2 (3 a B+A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3770

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

Rule 4025

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[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4076

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_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.71, size = 217, normalized size = 1.75 \[ \frac {a^3 A \sin (2 (c+d x))+2 a (c+d x) \left (a^2 A+6 a b B+6 A b^2\right )+4 a^2 (a B+3 A b) \sin (c+d x)-4 b^2 (3 a B+A b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b^2 (3 a B+A b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 b^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 b^3 B \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(a^2*A + 6*A*b^2 + 6*a*b*B)*(c + d*x) - 4*b^2*(A*b + 3*a*B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*
b^2*(A*b + 3*a*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (4*b^3*B*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]) + (4*b^3*B*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*a^2*(3*A*b + a*B)*Sin[c
+ d*x] + a^3*A*Sin[2*(c + d*x)])/(4*d)

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fricas [A]  time = 0.47, size = 152, normalized size = 1.23 \[ \frac {{\left (A a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (A a^{3} \cos \left (d x + c\right )^{2} + 2 \, B b^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.35, size = 234, normalized size = 1.89 \[ -\frac {\frac {4 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (A a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*B*b^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - (A*a^3 + 6*B*a^2*b + 6*A*a*b^2)*(d*x + c) -
2*(3*B*a*b^2 + A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 2*(3*B*a*b^2 + A*b^3)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(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 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 -
A*a^3*tan(1/2*d*x + 1/2*c) - 2*B*a^3*tan(1/2*d*x + 1/2*c) - 6*A*a^2*b*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 = 0.95, size = 168, normalized size = 1.35 \[ \frac {A \,a^{3} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {a^{3} A x}{2}+\frac {A \,a^{3} c}{2 d}+\frac {a^{3} B \sin \left (d x +c \right )}{d}+\frac {3 A \,a^{2} b \sin \left (d x +c \right )}{d}+3 B x \,a^{2} b +\frac {3 B \,a^{2} b c}{d}+3 A x a \,b^{2}+\frac {3 A a \,b^{2} c}{d}+\frac {3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{3} B \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.81, size = 144, normalized size = 1.16 \[ \frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 12 \, {\left (d x + c\right )} B a^{2} b + 12 \, {\left (d x + c\right )} A a b^{2} + 6 \, B a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, A a^{2} b \sin \left (d x + c\right ) + 4 \, B b^{3} \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 3.33, size = 236, normalized size = 1.90 \[ \frac {A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}+6\,A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,B\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-B\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{8}+B\,b^3\,\sin \left (c+d\,x\right )+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}}{d\,\cos \left (c+d\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - A*b^3*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i
+ 6*A*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 6*B*a^2*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
)) - B*a*b^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*6i)/d + ((A*a^3*sin(3*c + 3*d*x))/8 + (B*a^3*sin
(2*c + 2*d*x))/2 + (A*a^3*sin(c + d*x))/8 + B*b^3*sin(c + d*x) + (3*A*a^2*b*sin(2*c + 2*d*x))/2)/(d*cos(c + d*
x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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