3.1.58 \(\int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [58]

Optimal. Leaf size=72 \[ -\frac {2 x}{a^2}+\frac {10 \sin (c+d x)}{3 a^2 d}-\frac {2 \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

-2*x/a^2+10/3*sin(d*x+c)/a^2/d-2*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*sin(d*x+c)/d/(a+a*sec(d*x+c))^2

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Rubi [A]
time = 0.09, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3902, 4105, 3872, 2717, 8} \begin {gather*} \frac {10 \sin (c+d x)}{3 a^2 d}-\frac {2 \sin (c+d x)}{a^2 d (\sec (c+d x)+1)}-\frac {2 x}{a^2}-\frac {\sin (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/(a + a*Sec[c + d*x])^2,x]

[Out]

(-2*x)/a^2 + (10*Sin[c + d*x])/(3*a^2*d) - (2*Sin[c + d*x])/(a^2*d*(1 + Sec[c + d*x])) - Sin[c + d*x]/(3*d*(a
+ a*Sec[c + d*x])^2)

Rule 8

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3902

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[
e + f*x])*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(2*m + 1))), x] + Dist[1/(a^2*(2*m + 1)), Int[(a + b*C
sc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a*(2*m + n + 1) - b*(m + n + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b,
d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m])

Rule 4105

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(
2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*
(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[
A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac {\sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\cos (c+d x) (-4 a+2 a \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \cos (c+d x) \left (-10 a^2+6 a^2 \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {2 \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {2 \int 1 \, dx}{a^2}+\frac {10 \int \cos (c+d x) \, dx}{3 a^2}\\ &=-\frac {2 x}{a^2}+\frac {10 \sin (c+d x)}{3 a^2 d}-\frac {2 \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(151\) vs. \(2(72)=144\).
time = 0.51, size = 151, normalized size = 2.10 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-36 d x \cos \left (\frac {d x}{2}\right )-36 d x \cos \left (c+\frac {d x}{2}\right )-12 d x \cos \left (c+\frac {3 d x}{2}\right )-12 d x \cos \left (2 c+\frac {3 d x}{2}\right )+66 \sin \left (\frac {d x}{2}\right )-30 \sin \left (c+\frac {d x}{2}\right )+41 \sin \left (c+\frac {3 d x}{2}\right )+9 \sin \left (2 c+\frac {3 d x}{2}\right )+3 \sin \left (2 c+\frac {5 d x}{2}\right )+3 \sin \left (3 c+\frac {5 d x}{2}\right )\right )}{48 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]/(a + a*Sec[c + d*x])^2,x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(-36*d*x*Cos[(d*x)/2] - 36*d*x*Cos[c + (d*x)/2] - 12*d*x*Cos[c + (3*d*x)/2] - 12*
d*x*Cos[2*c + (3*d*x)/2] + 66*Sin[(d*x)/2] - 30*Sin[c + (d*x)/2] + 41*Sin[c + (3*d*x)/2] + 9*Sin[2*c + (3*d*x)
/2] + 3*Sin[2*c + (5*d*x)/2] + 3*Sin[3*c + (5*d*x)/2]))/(48*a^2*d)

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Maple [A]
time = 0.07, size = 72, normalized size = 1.00

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(72\)
default	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(72\)
risch	\(-\frac {2 x}{a^{2}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {2 i \left (9 \,{\mathrm e}^{2 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}+8\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(90\)
norman	\(\frac {-\frac {2 x}{a}+\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}-\frac {2 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(a+a*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/2/d/a^2*(-1/3*tan(1/2*d*x+1/2*c)^3+5*tan(1/2*d*x+1/2*c)+4*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^2)-8*arct
an(tan(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.50, size = 118, normalized size = 1.64 \begin {gather*} \frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)))
/d

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Fricas [A]
time = 2.81, size = 90, normalized size = 1.25 \begin {gather*} -\frac {6 \, d x \cos \left (d x + c\right )^{2} + 12 \, d x \cos \left (d x + c\right ) + 6 \, d x - {\left (3 \, \cos \left (d x + c\right )^{2} + 14 \, \cos \left (d x + c\right ) + 10\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(6*d*x*cos(d*x + c)^2 + 12*d*x*cos(d*x + c) + 6*d*x - (3*cos(d*x + c)^2 + 14*cos(d*x + c) + 10)*sin(d*x +
c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\cos {\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sec(d*x+c))**2,x)

[Out]

Integral(cos(c + d*x)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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Giac [A]
time = 0.46, size = 79, normalized size = 1.10 \begin {gather*} -\frac {\frac {12 \, {\left (d x + c\right )}}{a^{2}} - \frac {12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(12*(d*x + c)/a^2 - 12*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) + (a^4*tan(1/2*d*x + 1/2*c
)^3 - 15*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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Mupad [B]
time = 0.70, size = 91, normalized size = 1.26 \begin {gather*} -\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (c+d\,x\right )}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)/(a + a/cos(c + d*x))^2,x)

[Out]

-(sin(c/2 + (d*x)/2) - 16*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) - 12*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)
+ 12*cos(c/2 + (d*x)/2)^3*(c + d*x))/(6*a^2*d*cos(c/2 + (d*x)/2)^3)

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