3.1.44 \(\int \cos ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [44]

Optimal. Leaf size=54 \[ \frac {C x}{2}+\frac {B \sin (c+d x)}{d}+\frac {C \cos (c+d x) \sin (c+d x)}{2 d}-\frac {B \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4132, 2713, 12, 2715, 8} \begin {gather*} -\frac {B \sin ^3(c+d x)}{3 d}+\frac {B \sin (c+d x)}{d}+\frac {C \sin (c+d x) \cos (c+d x)}{2 d}+\frac {C x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x)/2 + (B*Sin[c + d*x])/d + (C*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (B*Sin[c + d*x]^3)/(3*d)

Rule 8

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=B \int \cos ^3(c+d x) \, dx+\int C \cos ^2(c+d x) \, dx\\ &=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 {C \cos (c+d x) \sin (c+d x)}{2 d}-\frac {B \sin ^3(c+d x)}{3 d}+\frac {1}{2} C \int 1 \, dx\\ &=\frac {C x}{2}+\frac {B \sin (c+d x)}{d}+\frac {C \cos (c+d x) \sin (c+d x)}{2 d}-\frac {B \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 57, normalized size = 1.06 \begin {gather*} \frac {C (c+d x)}{2 d}+\frac {B \sin (c+d x)}{d}-\frac {B \sin ^3(c+d x)}{3 d}+\frac {C \sin (2 (c+d x))}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(c + d*x))/(2*d) + (B*Sin[c + d*x])/d - (B*Sin[c + d*x]^3)/(3*d) + (C*Sin[2*(c + d*x)])/(4*d)

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Maple [A]
time = 0.46, size = 49, normalized size = 0.91

method	result	size
risch	\(\frac {C x}{2}+\frac {3 B \sin \left (d x +c \right )}{4 d}+\frac {B \sin \left (3 d x +3 c \right )}{12 d}+\frac {C \sin \left (2 d x +2 c \right )}{4 d}\)	\(48\)
derivativedivides	\(\frac {\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(49\)
default	\(\frac {\frac {B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(49\)
norman	\(\frac {C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (2 B -C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {C x}{2}-\frac {4 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {4 B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {3 C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {3 C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {C x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {\left (2 B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 46, normalized size = 0.85 \begin {gather*} -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.90, size = 42, normalized size = 0.78 \begin {gather*} \frac {3 \, C d x + {\left (2 \, B \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + 4 \, B\right )} \sin \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*C*d*x + (2*B*cos(d*x + c)^2 + 3*C*cos(d*x + c) + 4*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 (B + C \sec {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )} \sec {\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*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (48) = 96\).
time = 0.47, size = 98, normalized size = 1.81 \begin {gather*} \frac {3 \, {\left (d x + c\right )} C + \frac {2 \, {\left (6 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(d*x + c)*C + 2*(6*B*tan(1/2*d*x + 1/2*c)^5 - 3*C*tan(1/2*d*x + 1/2*c)^5 + 4*B*tan(1/2*d*x + 1/2*c)^3 +
6*B*tan(1/2*d*x + 1/2*c) + 3*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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Mupad [B]
time = 2.51, size = 55, normalized size = 1.02 \begin {gather*} \frac {C\,x}{2}+\frac {2\,B\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {B\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(C*x)/2 + (2*B*sin(c + d*x))/(3*d) + (C*cos(c + d*x)*sin(c + d*x))/(2*d) + (B*cos(c + d*x)^2*sin(c + d*x))/(3*
d)

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