3.1.62 \(\int \cos ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [62]

Optimal. Leaf size=88 \[ \frac {1}{8} (3 A+4 C) x+\frac {B \sin (c+d x)}{d}+\frac {(3 A+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {B \sin ^3(c+d x)}{3 d} \]

[Out]

1/8*(3*A+4*C)*x+B*sin(d*x+c)/d+1/8*(3*A+4*C)*cos(d*x+c)*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*sin(d*x+c)/d-1/3*B*sin
(d*x+c)^3/d

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Rubi [A]
time = 0.06, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 = {4132, 2713, 4130, 2715, 8} \begin {gather*} \frac {(3 A+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {A \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{8} x (3 A+4 C)-\frac {B \sin ^3(c+d x)}{3 d}+\frac {B \sin (c+d x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A + 4*C)*x)/8 + (B*Sin[c + d*x])/d + ((3*A + 4*C)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (A*Cos[c + d*x]^3*Sin
[c + d*x])/(4*d) - (B*Sin[c + d*x]^3)/(3*d)

Rule 8

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

Rule 2713

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

Rule 2715

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

Rule 4130

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e
+ f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 70, normalized size = 0.80 \begin {gather*} \frac {36 A c+48 c C+36 A d x+48 C d x+96 B \sin (c+d x)-32 B \sin ^3(c+d x)+24 (A+C) \sin (2 (c+d x))+3 A \sin (4 (c+d x))}{96 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(36*A*c + 48*c*C + 36*A*d*x + 48*C*d*x + 96*B*Sin[c + d*x] - 32*B*Sin[c + d*x]^3 + 24*(A + C)*Sin[2*(c + d*x)]
+ 3*A*Sin[4*(c + d*x)])/(96*d)

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Maple [A]
time = 0.69, size = 84, normalized size = 0.95

method	result	size
risch	\(\frac {3 A x}{8}+\frac {C x}{2}+\frac {3 B \sin \left (d x +c \right )}{4 d}+\frac {A \sin \left (4 d x +4 c \right )}{32 d}+\frac {B \sin \left (3 d x +3 c \right )}{12 d}+\frac {A \sin \left (2 d x +2 c \right )}{4 d}+\frac {C \sin \left (2 d x +2 c \right )}{4 d}\)	\(82\)
derivativedivides	\(\frac {A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(84\)
default	\(\frac {A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(84\)
norman	\(\frac {\left (-\frac {3 A}{8}-\frac {C}{2}\right ) x +\left (-\frac {9 A}{8}-\frac {3 C}{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {3 A}{4}-C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3 A}{4}+C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {3 A}{8}+\frac {C}{2}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {9 A}{8}+\frac {3 C}{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {2 \left (3 A -2 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 \left (3 A +2 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (3 A -4 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}-\frac {\left (5 A -8 B +4 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {\left (5 A +8 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/d*(A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/3*B*(2+cos(d*x+c)^2)*sin(d*x+c)+C*(1/2*c
os(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]
time = 0.28, size = 77, normalized size = 0.88 \begin {gather*} \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{96 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B + 2
4*(2*d*x + 2*c + sin(2*d*x + 2*c))*C)/d

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Fricas [A]
time = 2.89, size = 65, normalized size = 0.74 \begin {gather*} \frac {3 \, {\left (3 \, A + 4 \, C\right )} d x + {\left (6 \, A \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right ) + 16 \, B\right )} \sin \left (d x + c\right )}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*(3*A + 4*C)*d*x + (6*A*cos(d*x + c)^3 + 8*B*cos(d*x + c)^2 + 3*(3*A + 4*C)*cos(d*x + c) + 16*B)*sin(d*
x + c))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 200 vs. \(2 (80) = 160\).
time = 0.43, size = 200, normalized size = 2.27 \begin {gather*} \frac {3 \, {\left (d x + c\right )} {\left (3 \, A + 4 \, C\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(d*x + c)*(3*A + 4*C) - 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*ta
n(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

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Mupad [B]
time = 2.53, size = 81, normalized size = 0.92 \begin {gather*} \frac {3\,A\,x}{8}+\frac {C\,x}{2}+\frac {A\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,B\,\sin \left (c+d\,x\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(3*A*x)/8 + (C*x)/2 + (A*sin(2*c + 2*d*x))/(4*d) + (A*sin(4*c + 4*d*x))/(32*d) + (B*sin(3*c + 3*d*x))/(12*d) +
(C*sin(2*c + 2*d*x))/(4*d) + (3*B*sin(c + d*x))/(4*d)

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