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

Optimal. Leaf size=86 \[ \frac {1}{2} x \left (2 a^2 B+4 a A b+b^2 B\right )+\frac {a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b (3 a B+2 A b) \sin (c+d x)}{2 d}+\frac {b B \sin (c+d x) (a+b \cos (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2990, 3023, 2735, 3770} \[ \frac {1}{2} x \left (2 a^2 B+4 a A b+b^2 B\right )+\frac {a^2 A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b (3 a B+2 A b) \sin (c+d x)}{2 d}+\frac {b B \sin (c+d x) (a+b \cos (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2990

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x]
)^(n + 1))/(d*f*(m + n + 1)), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x
])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c -
b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /; F
reeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m,
1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 3770

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

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 120, normalized size = 1.40 \[ \frac {2 (c+d x) \left (2 a^2 B+4 a A b+b^2 B\right )-4 a^2 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a^2 A \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+4 b (2 a B+A b) \sin (c+d x)+b^2 B \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 87, normalized size = 1.01 \[ \frac {A a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - A a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} d x + {\left (B b^{2} \cos \left (d x + c\right ) + 4 \, B a b + 2 \, A b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.45, size = 178, normalized size = 2.07 \[ \frac {2 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 2 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (2 \, B a^{2} + 4 \, A a b + B b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (4 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 2*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (2*B*a^2 + 4*A*
a*b + B*b^2)*(d*x + c) + 2*(4*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 2*A*b^2*tan(1/2*d*x + 1/2*c)^3 - B*b^2*tan(1/2*d*
x + 1/2*c)^3 + 4*B*a*b*tan(1/2*d*x + 1/2*c) + 2*A*b^2*tan(1/2*d*x + 1/2*c) + B*b^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 = 0.10, size = 120, normalized size = 1.40 \[ \frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+a^{2} B x +\frac {B \,a^{2} c}{d}+2 A a b x +\frac {2 A a b c}{d}+\frac {2 B a b \sin \left (d x +c \right )}{d}+\frac {A \,b^{2} \sin \left (d x +c \right )}{d}+\frac {b^{2} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {b^{2} B x}{2}+\frac {b^{2} B c}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.21, size = 92, normalized size = 1.07 \[ \frac {4 \, {\left (d x + c\right )} B a^{2} + 8 \, {\left (d x + c\right )} A a b + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 4 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 8 \, B a b \sin \left (d x + c\right ) + 4 \, A b^{2} \sin \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.69, size = 169, normalized size = 1.97 \[ \frac {A\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {2\,B\,a\,b\,\sin \left (c+d\,x\right )}{d}+\frac {4\,A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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