3.299 \(\int (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x) \, dx\)

Optimal. Leaf size=97 \[ \frac{(3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(3 A+4 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{A \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d} \]

[Out]

((3*A + 4*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (B*Tan[c + d*x])/d + ((3*A + 4*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d)
+ (A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (B*Tan[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.103897, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {3021, 2748, 3767, 3768, 3770} \[ \frac{(3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{(3 A+4 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{A \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{B \tan ^3(c+d x)}{3 d}+\frac{B \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

((3*A + 4*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (B*Tan[c + d*x])/d + ((3*A + 4*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d)
+ (A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (B*Tan[c + d*x]^3)/(3*d)

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 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 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 ^5(c+d x) \, dx &=\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{4} \int (4 B+(3 A+4 C) \cos (c+d x)) \sec ^4(c+d x) \, dx\\ &=\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 d}+B \int \sec ^4(c+d x) \, dx+\frac{1}{4} (3 A+4 C) \int \sec ^3(c+d x) \, dx\\ &=\frac{(3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{1}{8} (3 A+4 C) \int \sec (c+d x) \, dx-\frac{B \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac{(3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{B \tan (c+d x)}{d}+\frac{(3 A+4 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{B \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.256505, size = 71, normalized size = 0.73 \[ \frac{\tan (c+d x) \left (3 (3 A+4 C) \sec (c+d x)+6 A \sec ^3(c+d x)+8 B \left (\tan ^2(c+d x)+3\right )\right )+3 (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^5,x]

[Out]

(3*(3*A + 4*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*(3*A + 4*C)*Sec[c + d*x] + 6*A*Sec[c + d*x]^3 + 8*B*(3
+ Tan[c + d*x]^2)))/(24*d)

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Maple [A]  time = 0.04, size = 130, normalized size = 1.3 \begin{align*}{\frac{A \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,B\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

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)^5,x)

[Out]

1/4*A*sec(d*x+c)^3*tan(d*x+c)/d+3/8*A*sec(d*x+c)*tan(d*x+c)/d+3/8/d*A*ln(sec(d*x+c)+tan(d*x+c))+2/3*B*tan(d*x+
c)/d+1/3/d*B*tan(d*x+c)*sec(d*x+c)^2+1/2*C*sec(d*x+c)*tan(d*x+c)/d+1/2/d*C*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 1.00276, size = 188, normalized size = 1.94 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B - 3 \, A{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C{\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 )}}{48 \, d} \end{align*}

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)^5,x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B - 3*A*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*
sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*C*(2*sin(d*x + c)/(sin(d*x + c)^
2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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Fricas [A]  time = 2.00681, size = 306, normalized size = 3.15 \begin{align*} \frac{3 \,{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \, B \cos \left (d x + c\right )^{3} + 3 \,{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, B \cos \left (d x + c\right ) + 6 \, A\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

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)^5,x, algorithm="fricas")

[Out]

1/48*(3*(3*A + 4*C)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*A + 4*C)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)
+ 2*(16*B*cos(d*x + c)^3 + 3*(3*A + 4*C)*cos(d*x + c)^2 + 8*B*cos(d*x + c) + 6*A)*sin(d*x + c))/(d*cos(d*x +
c)^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.32384, size = 311, normalized size = 3.21 \begin{align*} \frac{3 \,{\left (3 \, A + 4 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (3 \, A + 4 \, C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, B \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} + 15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, 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 )}^{4}}}{24 \, d} \end{align*}

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)^5,x, algorithm="giac")

[Out]

1/24*(3*(3*A + 4*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A + 4*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*
(15*A*tan(1/2*d*x + 1/2*c)^7 - 24*B*tan(1/2*d*x + 1/2*c)^7 + 12*C*tan(1/2*d*x + 1/2*c)^7 + 9*A*tan(1/2*d*x + 1
/2*c)^5 + 40*B*tan(1/2*d*x + 1/2*c)^5 - 12*C*tan(1/2*d*x + 1/2*c)^5 + 9*A*tan(1/2*d*x + 1/2*c)^3 - 40*B*tan(1/
2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2*c)^3 + 15*A*tan(1/2*d*x + 1/2*c) + 24*B*tan(1/2*d*x + 1/2*c) + 12*C*
tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d