3.3.20 \(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\) [220]

Optimal. Leaf size=61 \[ \frac {(a A+2 b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3047, 3100, 2827, 3852, 8, 3855} \begin {gather*} \frac {(a B+A b) \tan (c+d x)}{d}+\frac {(a A+2 b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a A \tan (c+d x) \sec (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

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

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3047

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

Rule 3100

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

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

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

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Mathematica [A]
time = 0.04, size = 75, normalized size = 1.23 \begin {gather*} \frac {a A \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {b B \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A b \tan (c+d x)}{d}+\frac {a B \tan (c+d x)}{d}+\frac {a A \sec (c+d x) \tan (c+d x)}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.18, size = 75, normalized size = 1.23

method	result	size
derivativedivides	\(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a B \tan \left (d x +c \right )+A b \tan \left (d x +c \right )+B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(75\)
default	\(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a B \tan \left (d x +c \right )+A b \tan \left (d x +c \right )+B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(75\)
risch	\(-\frac {i \left (A a \,{\mathrm e}^{3 i \left (d x +c \right )}-2 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-a A \,{\mathrm e}^{i \left (d x +c \right )}-2 A b -2 a B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{d}-\frac {a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{d}\)	\(160\)
norman	\(\frac {\frac {\left (a A -2 A b -2 a B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (a A +2 A b +2 a B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 a A -2 A b -2 a B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {\left (3 a A +2 A b +2 a B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {\left (a A +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (a A +2 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))*(A+B*cos(d*x+c))*sec(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 95, normalized size = 1.56 \begin {gather*} -\frac {A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, B b {\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 \tan \left (d x + c\right ) - 4 \, A b \tan \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))*(A+B*cos(d*x+c))*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 96, normalized size = 1.57 \begin {gather*} \frac {{\left (A a + 2 \, B b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A a + 2 \, B b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a + 2 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 ) \sec ^{3}{\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))*(A+B*cos(d*x+c))*sec(d*x+c)**3,x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (57) = 114\).
time = 0.46, size = 151, normalized size = 2.48 \begin {gather*} \frac {{\left (A a + 2 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a + 2 \, B 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 \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((A*a + 2*B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (A*a + 2*B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(
A*a*tan(1/2*d*x + 1/2*c)^3 - 2*B*a*tan(1/2*d*x + 1/2*c)^3 - 2*A*b*tan(1/2*d*x + 1/2*c)^3 + A*a*tan(1/2*d*x + 1
/2*c) + 2*B*a*tan(1/2*d*x + 1/2*c) + 2*A*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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Mupad [B]
time = 1.27, size = 104, normalized size = 1.70 \begin {gather*} \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a+2\,A\,b+2\,B\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,b-A\,a+2\,B\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A\,a+2\,B\,b\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)))/cos(c + d*x)^3,x)

[Out]

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

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