3.3.26 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [226]

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

[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 = 0.11, 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 = {3069, 3102, 2814, 3855} \begin {gather*} \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} \end {gather*}

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 2814

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 3069

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] /;
FreeQ[{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 3102

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 3855

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 = 0.26, size = 120, normalized size = 1.40 \begin {gather*} \frac {2 \left (4 a A b+2 a^2 B+b^2 B\right ) (c+d x)-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 (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 b (A b+2 a B) \sin (c+d x)+b^2 B \sin (2 (c+d x))}{4 d} \end {gather*}

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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Maple [A]
time = 0.17, size = 94, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+2 A a b \left (d x +c \right )+2 B a b \sin \left (d x +c \right )+A \,b^{2} \sin \left (d x +c \right )+B \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(94\)
default	\(\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+2 A a b \left (d x +c \right )+2 B a b \sin \left (d x +c \right )+A \,b^{2} \sin \left (d x +c \right )+B \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(94\)
risch	\(2 x A a b +a^{2} B x +\frac {b^{2} B x}{2}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B a b}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B a b}{d}+\frac {a^{2} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{2} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (2 d x +2 c \right ) B \,b^{2}}{4 d}\)	\(156\)
norman	\(\frac {\left (2 A a b +B \,a^{2}+\frac {1}{2} B \,b^{2}\right ) x +\left (2 A a b +B \,a^{2}+\frac {1}{2} B \,b^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 A a b +3 B \,a^{2}+\frac {3}{2} B \,b^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 A a b +3 B \,a^{2}+\frac {3}{2} B \,b^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (2 A b +4 a B -B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (2 A b +4 a B +B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 b \left (A b +2 a B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {a^{2} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(251\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 92, normalized size = 1.07 \begin {gather*} \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} \end {gather*}

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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Fricas [A]
time = 0.37, size = 87, normalized size = 1.01 \begin {gather*} \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} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (80) = 160\).
time = 0.45, size = 178, normalized size = 2.07 \begin {gather*} \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} \end {gather*}

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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Mupad [B]
time = 0.69, size = 169, normalized size = 1.97 \begin {gather*} \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} \end {gather*}

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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