3.1.59 \(\int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [59]

Optimal. Leaf size=110 \[ \frac {7 x}{2 a^2}-\frac {16 \sin (c+d x)}{3 a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]

[Out]

7/2*x/a^2-16/3*sin(d*x+c)/a^2/d+7/2*cos(d*x+c)*sin(d*x+c)/a^2/d-8/3*cos(d*x+c)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))
-1/3*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2

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Rubi [A]
time = 0.12, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3902, 4105, 3872, 2715, 8, 2717} \begin {gather*} -\frac {16 \sin (c+d x)}{3 a^2 d}+\frac {7 \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {8 \sin (c+d x) \cos (c+d x)}{3 a^2 d (\sec (c+d x)+1)}+\frac {7 x}{2 a^2}-\frac {\sin (c+d x) \cos (c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*x)/(2*a^2) - (16*Sin[c + d*x])/(3*a^2*d) + (7*Cos[c + d*x]*Sin[c + d*x])/(2*a^2*d) - (8*Cos[c + d*x]*Sin[c
+ d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - (Cos[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 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 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 ^2(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac {\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\cos ^2(c+d x) (-5 a+3 a \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \cos ^2(c+d x) \left (-21 a^2+16 a^2 \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {16 \int \cos (c+d x) \, dx}{3 a^2}+\frac {7 \int \cos ^2(c+d x) \, dx}{a^2}\\ &=-\frac {16 \sin (c+d x)}{3 a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}+\frac {7 \int 1 \, dx}{2 a^2}\\ &=\frac {7 x}{2 a^2}-\frac {16 \sin (c+d x)}{3 a^2 d}+\frac {7 \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {8 \cos (c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos (c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 177, normalized size = 1.61 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (252 d x \cos \left (\frac {d x}{2}\right )+252 d x \cos \left (c+\frac {d x}{2}\right )+84 d x \cos \left (c+\frac {3 d x}{2}\right )+84 d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 \sin \left (\frac {d x}{2}\right )+147 \sin \left (c+\frac {d x}{2}\right )-239 \sin \left (c+\frac {3 d x}{2}\right )-63 \sin \left (2 c+\frac {3 d x}{2}\right )-15 \sin \left (2 c+\frac {5 d x}{2}\right )-15 \sin \left (3 c+\frac {5 d x}{2}\right )+3 \sin \left (3 c+\frac {7 d x}{2}\right )+3 \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{192 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(252*d*x*Cos[(d*x)/2] + 252*d*x*Cos[c + (d*x)/2] + 84*d*x*Cos[c + (3*d*x)/2] + 84
*d*x*Cos[2*c + (3*d*x)/2] - 381*Sin[(d*x)/2] + 147*Sin[c + (d*x)/2] - 239*Sin[c + (3*d*x)/2] - 63*Sin[2*c + (3
*d*x)/2] - 15*Sin[2*c + (5*d*x)/2] - 15*Sin[3*c + (5*d*x)/2] + 3*Sin[3*c + (7*d*x)/2] + 3*Sin[4*c + (7*d*x)/2]
))/(192*a^2*d)

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Maple [A]
time = 0.08, size = 88, normalized size = 0.80

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+14 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(88\)
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+14 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(88\)
risch	\(\frac {7 x}{2 a^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{2} d}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{a^{2} d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{a^{2} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{2} d}-\frac {2 i \left (12 \,{\mathrm e}^{2 i \left (d x +c \right )}+21 \,{\mathrm e}^{i \left (d x +c \right )}+11\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}\)	\(126\)
norman	\(\frac {\frac {7 x}{2 a}-\frac {13 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {71 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {19 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}+\frac {7 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {7 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(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)^3-7*tan(1/2*d*x+1/2*c)+8*(-5/4*tan(1/2*d*x+1/2*c)^3-3/4*tan(1/2*d*x+1/2*c))/
(1+tan(1/2*d*x+1/2*c)^2)^2+14*arctan(tan(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.51, size = 164, normalized size = 1.49 \begin {gather*} -\frac {\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \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 {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(
d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]
time = 2.48, size = 99, normalized size = 0.90 \begin {gather*} \frac {21 \, d x \cos \left (d x + c\right )^{2} + 42 \, d x \cos \left (d x + c\right ) + 21 \, d x + {\left (3 \, \cos \left (d x + c\right )^{3} - 6 \, \cos \left (d x + c\right )^{2} - 43 \, \cos \left (d x + c\right ) - 32\right )} \sin \left (d x + c\right )}{6 \, {\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)^2/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(21*d*x*cos(d*x + c)^2 + 42*d*x*cos(d*x + c) + 21*d*x + (3*cos(d*x + c)^3 - 6*cos(d*x + c)^2 - 43*cos(d*x
+ c) - 32)*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 ^{2}{\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)**2/(a+a*sec(d*x+c))**2,x)

[Out]

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

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Giac [A]
time = 0.44, size = 95, normalized size = 0.86 \begin {gather*} \frac {\frac {21 \, {\left (d x + c\right )}}{a^{2}} - \frac {6 \, {\left (5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, \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 )}^{2} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 21 \, 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)^2/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/6*(21*(d*x + c)/a^2 - 6*(5*tan(1/2*d*x + 1/2*c)^3 + 3*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*
a^2) + (a^4*tan(1/2*d*x + 1/2*c)^3 - 21*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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Mupad [B]
time = 0.73, size = 113, normalized size = 1.03 \begin {gather*} \frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-22\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-30\,{\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 )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+21\,{\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)^2/(a + a/cos(c + d*x))^2,x)

[Out]

(sin(c/2 + (d*x)/2) - 22*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) - 30*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)
+ 12*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) + 21*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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