3.297 \(\int (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=51 \[ \frac {(A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}+\frac {B \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3021, 2748, 3767, 8, 3770} \[ \frac {(A+2 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}+\frac {B \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A + 2*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (B*Tan[c + d*x])/d + (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 2748

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 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 3767

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

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Mathematica [A]  time = 0.02, size = 59, normalized size = 1.16 \[ \frac {A \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {A \tan (c+d x) \sec (c+d x)}{2 d}+\frac {B \tan (c+d x)}{d}+\frac {C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 82, normalized size = 1.61 \[ \frac {{\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A + 2 \, C\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B \cos \left (d x + c\right ) + A\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.50, size = 113, normalized size = 2.22 \[ \frac {{\left (A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A + 2 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (A + 2*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(A*tan(1/
2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + A*tan(1/2*d*x + 1/2*c) + 2*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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maple [A]  time = 0.23, size = 70, normalized size = 1.37 \[ \frac {A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {B \tan \left (d x +c \right )}{d}+\frac {C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 82, normalized size = 1.61 \[ -\frac {A {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 2 \, C {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, B \tan \left (d x + c\right )}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(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*C*(log(sin(d
*x + c) + 1) - log(sin(d*x + c) - 1)) - 4*B*tan(d*x + c))/d

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mupad [B]  time = 1.68, size = 85, normalized size = 1.67 \[ \frac {\left (A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A+2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+2\,C\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(A + 2*B) + tan(c/2 + (d*x)/2)^3*(A - 2*B))/(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 + 2*C))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**3, x)

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