3.298 \(\int (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=78 \[ \frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

1/2*B*arctanh(sin(d*x+c))/d+1/3*(2*A+3*C)*tan(d*x+c)/d+1/2*B*sec(d*x+c)*tan(d*x+c)/d+1/3*A*sec(d*x+c)^2*tan(d*
x+c)/d

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Rubi [A]  time = 0.10, antiderivative size = 78, 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 = {3021, 2748, 3768, 3770, 3767, 8} \[ \frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {A \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {B \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(B*ArcTanh[Sin[c + d*x]])/(2*d) + ((2*A + 3*C)*Tan[c + d*x])/(3*d) + (B*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (A*
Sec[c + d*x]^2*Tan[c + d*x])/(3*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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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 ^4(c+d x) \, dx &=\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (3 B+(2 A+3 C) \cos (c+d x)) \sec ^3(c+d x) \, dx\\ &=\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d}+B \int \sec ^3(c+d x) \, dx+\frac {1}{3} (2 A+3 C) \int \sec ^2(c+d x) \, dx\\ &=\frac {B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} B \int \sec (c+d x) \, dx-\frac {(2 A+3 C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {(2 A+3 C) \tan (c+d x)}{3 d}+\frac {B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {A \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 51, normalized size = 0.65 \[ \frac {\tan (c+d x) \left (2 A \tan ^2(c+d x)+6 (A+C)+3 B \sec (c+d x)\right )+3 B \tanh ^{-1}(\sin (c+d x))}{6 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(3*B*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*(A + C) + 3*B*Sec[c + d*x] + 2*A*Tan[c + d*x]^2))/(6*d)

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fricas [A]  time = 0.50, size = 94, normalized size = 1.21 \[ \frac {3 \, B \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, B \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 2 \, A\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

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)^4,x, algorithm="fricas")

[Out]

1/12*(3*B*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*B*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(2*A + 3*C)*
cos(d*x + c)^2 + 3*B*cos(d*x + c) + 2*A)*sin(d*x + c))/(d*cos(d*x + c)^3)

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giac [B]  time = 0.46, size = 162, normalized size = 2.08 \[ \frac {3 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, B \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

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)^4,x, algorithm="giac")

[Out]

1/6*(3*B*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*B*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*tan(1/2*d*x + 1/
2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5 - 4*A*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d
*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) + 3*B*tan(1/2*d*x + 1/2*c) + 6*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x +
1/2*c)^2 - 1)^3)/d

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maple [A]  time = 0.30, size = 83, normalized size = 1.06 \[ \frac {2 A \tan \left (d x +c \right )}{3 d}+\frac {A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {C \tan \left (d x +c \right )}{d} \]

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)^4,x)

[Out]

2/3*A*tan(d*x+c)/d+1/3*A*sec(d*x+c)^2*tan(d*x+c)/d+1/2*B*sec(d*x+c)*tan(d*x+c)/d+1/2/d*B*ln(sec(d*x+c)+tan(d*x
+c))+C*tan(d*x+c)/d

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maxima [A]  time = 0.52, size = 79, normalized size = 1.01 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A - 3 \, B {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C \tan \left (d x + c\right )}{12 \, d} \]

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)^4,x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*A - 3*B*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1)
+ log(sin(d*x + c) - 1)) + 12*C*tan(d*x + c))/d

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mupad [B]  time = 3.07, size = 123, normalized size = 1.58 \[ \frac {B\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {\left (2\,A-B+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {4\,A}{3}-4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+B+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

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)^4,x)

[Out]

(B*atanh(tan(c/2 + (d*x)/2)))/d - (tan(c/2 + (d*x)/2)*(2*A + B + 2*C) - tan(c/2 + (d*x)/2)^3*((4*A)/3 + 4*C) +
tan(c/2 + (d*x)/2)^5*(2*A - B + 2*C))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2
)^6 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

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)**4,x)

[Out]

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sec(c + d*x)**4, x)

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