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

Optimal. Leaf size=35 \[ \frac {(a C+b B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \tan (c+d x)}{d}+b C x \]

[Out]

b*C*x+(B*b+C*a)*arctanh(sin(d*x+c))/d+a*B*tan(d*x+c)/d

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Rubi [A]  time = 0.17, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3029, 2968, 3021, 2735, 3770} \[ \frac {(a C+b B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \tan (c+d x)}{d}+b C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

b*C*x + ((b*B + a*C)*ArcTanh[Sin[c + d*x]])/d + (a*B*Tan[c + d*x])/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 2968

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 3021

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 3029

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

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

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Mathematica [A]  time = 0.01, size = 43, normalized size = 1.23 \[ \frac {a B \tan (c+d x)}{d}+\frac {a C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b B \tanh ^{-1}(\sin (c+d x))}{d}+b C x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 85, normalized size = 2.43 \[ \frac {2 \, C b d x \cos \left (d x + c\right ) + {\left (C a + B b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (C a + B b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*C*b*d*x*cos(d*x + c) + (C*a + B*b)*cos(d*x + c)*log(sin(d*x + c) + 1) - (C*a + B*b)*cos(d*x + c)*log(-s
in(d*x + c) + 1) + 2*B*a*sin(d*x + c))/(d*cos(d*x + c))

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giac [B]  time = 0.26, size = 84, normalized size = 2.40 \[ \frac {{\left (d x + c\right )} C b + {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.23, size = 65, normalized size = 1.86 \[ b C x +\frac {a B \tan \left (d x +c \right )}{d}+\frac {B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {C b c}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.32, size = 73, normalized size = 2.09 \[ \frac {2 \, {\left (d x + c\right )} C b + C a {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + B b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a \tan \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.83, size = 114, normalized size = 3.26 \[ \frac {2\,C\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}-\frac {B\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}-\frac {C\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*C*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (C*a*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
2i)/d - (B*b*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d + (B*a*sin(c + d*x))/(d*cos(c + d*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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