3.3.27 \(\int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^2(c+d x) \, dx\) [227]

Optimal. Leaf size=60 \[ b (A b+2 a B) x+\frac {a (2 A b+a B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b^2 B \sin (c+d x)}{d}+\frac {a^2 A \tan (c+d x)}{d} \]

[Out]

b*(A*b+2*B*a)*x+a*(2*A*b+B*a)*arctanh(sin(d*x+c))/d+b^2*B*sin(d*x+c)/d+a^2*A*tan(d*x+c)/d

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Rubi [A]
time = 0.11, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3067, 3102, 2814, 3855} \begin {gather*} \frac {a^2 A \tan (c+d x)}{d}+\frac {a (a B+2 A b) \tanh ^{-1}(\sin (c+d x))}{d}+b x (2 a B+A b)+\frac {b^2 B \sin (c+d x)}{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]^2,x]

[Out]

b*(A*b + 2*a*B)*x + (a*(2*A*b + a*B)*ArcTanh[Sin[c + d*x]])/d + (b^2*B*Sin[c + d*x])/d + (a^2*A*Tan[c + d*x])/
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 3067

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(
f*d^2*(n + 1)*(c^2 - d^2))), x] - Dist[1/(d^2*(n + 1)*(c^2 - d^2)), Int[(c + d*Sin[e + f*x])^(n + 1)*Simp[d*(n
+ 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n +
1))) + 2*a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x
]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && LtQ[n, -1]

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

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Mathematica [A]
time = 0.55, size = 109, normalized size = 1.82 \begin {gather*} \frac {b (A b+2 a B) (c+d x)-a (2 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 )+a (2 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 )+b^2 B \sin (c+d x)+a^2 A \tan (c+d x)}{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]^2,x]

[Out]

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

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Maple [A]
time = 0.19, size = 86, normalized size = 1.43

method	result	size
derivativedivides	\(\frac {a^{2} A \tan \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B a b \left (d x +c \right )+A \,b^{2} \left (d x +c \right )+B \,b^{2} \sin \left (d x +c \right )}{d}\)	\(86\)
default	\(\frac {a^{2} A \tan \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B a b \left (d x +c \right )+A \,b^{2} \left (d x +c \right )+B \,b^{2} \sin \left (d x +c \right )}{d}\)	\(86\)
risch	\(x A \,b^{2}+2 x B a b -\frac {i B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i B \,b^{2} {\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {2 i a^{2} A}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(160\)
norman	\(\frac {\left (-A \,b^{2}-2 B a b \right ) x +\left (-2 A \,b^{2}-4 B a b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,b^{2}+2 B a b \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 A \,b^{2}+4 B a b \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 \left (a^{2} A -B \,b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (a^{2} A +B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 \left (3 a^{2} A -B \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 \left (3 a^{2} A +B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a \left (2 A b +a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \left (2 A b +a B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(283\)

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)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 103, normalized size = 1.72 \begin {gather*} \frac {4 \, {\left (d x + c\right )} B a b + 2 \, {\left (d x + c\right )} A b^{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 b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B b^{2} \sin \left (d x + c\right ) + 2 \, A a^{2} \tan \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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.37, size = 117, normalized size = 1.95 \begin {gather*} \frac {2 \, {\left (2 \, B a b + A b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (B b^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \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)^2,x, algorithm="fricas")

[Out]

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

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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 ^{2}{\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)**2,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (60) = 120\).
time = 0.46, size = 152, normalized size = 2.53 \begin {gather*} \frac {{\left (2 \, B a b + A b^{2}\right )} {\left (d x + c\right )} + {\left (B a^{2} + 2 \, A a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (B a^{2} + 2 \, A a b\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 b^{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 b^{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} \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)^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.88, size = 169, normalized size = 2.82 \begin {gather*} \frac {A\,a^2\,\mathrm {tan}\left (c+d\,x\right )}{d}+\frac {2\,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 )}{d}+\frac {B\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d\,\cos \left (c+d\,x\right )}+\frac {4\,B\,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}-\frac {B\,a^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 )\,2{}\mathrm {i}}{d}-\frac {A\,a\,b\,\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 )\,4{}\mathrm {i}}{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)^2,x)

[Out]

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

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