3.1.69 \(\int \frac {\cos ^2(c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [69]

Optimal. Leaf size=147 \[ \frac {13 x}{2 a^3}-\frac {152 \sin (c+d x)}{15 a^3 d}+\frac {13 \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

13/2*x/a^3-152/15*sin(d*x+c)/a^3/d+13/2*cos(d*x+c)*sin(d*x+c)/a^3/d-1/5*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c
))^3-11/15*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2-76/15*cos(d*x+c)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]
time = 0.19, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {152 \sin (c+d x)}{15 a^3 d}+\frac {13 \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {76 \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {13 x}{2 a^3}-\frac {11 \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {\sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a + a*Sec[c + d*x])^3,x]

[Out]

(13*x)/(2*a^3) - (152*Sin[c + d*x])/(15*a^3*d) + (13*Cos[c + d*x]*Sin[c + d*x])/(2*a^3*d) - (Cos[c + d*x]*Sin[
c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - (11*Cos[c + d*x]*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (76*
Cos[c + d*x]*Sin[c + d*x])/(15*d*(a^3 + a^3*Sec[c + d*x]))

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))^3} \, dx &=-\frac {\cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\cos ^2(c+d x) (-7 a+4 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\cos ^2(c+d x) \left (-43 a^2+33 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {\cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \cos ^2(c+d x) \left (-195 a^3+152 a^3 \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {\cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {152 \int \cos (c+d x) \, dx}{15 a^3}+\frac {13 \int \cos ^2(c+d x) \, dx}{a^3}\\ &=-\frac {152 \sin (c+d x)}{15 a^3 d}+\frac {13 \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {13 \int 1 \, dx}{2 a^3}\\ &=\frac {13 x}{2 a^3}-\frac {152 \sin (c+d x)}{15 a^3 d}+\frac {13 \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {\cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {11 \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {76 \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.58, size = 181, normalized size = 1.23 \begin {gather*} \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (-3 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+46 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-508 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+15 \cos ^5\left (\frac {1}{2} (c+d x)\right ) (26 d x-12 \sin (c+d x)+\sin (2 (c+d x)))-3 \cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )+46 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{15 a^3 d (1+\sec (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a + a*Sec[c + d*x])^3,x]

[Out]

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^3*(-3*Sec[c/2]*Sin[(d*x)/2] + 46*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] - 5
08*Cos[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] + 15*Cos[(c + d*x)/2]^5*(26*d*x - 12*Sin[c + d*x] + Sin[2*(c + d*x
)]) - 3*Cos[(c + d*x)/2]*Tan[c/2] + 46*Cos[(c + d*x)/2]^3*Tan[c/2]))/(15*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]
time = 0.09, size = 101, normalized size = 0.69

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-28 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+52 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(101\)
default	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {-28 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+52 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(101\)
risch	\(\frac {13 x}{2 a^{3}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{3} d}+\frac {3 i {\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{3} d}-\frac {2 i \left (150 \,{\mathrm e}^{4 i \left (d x +c \right )}+525 \,{\mathrm e}^{3 i \left (d x +c \right )}+745 \,{\mathrm e}^{2 i \left (d x +c \right )}+485 \,{\mathrm e}^{i \left (d x +c \right )}+127\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(148\)
norman	\(\frac {\frac {13 x}{2 a}-\frac {51 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {131 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 a d}-\frac {97 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 a d}+\frac {17 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{30 a d}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a d}+\frac {13 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {13 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} a^{2}}\)	\(154\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/4/d/a^3*(-1/5*tan(1/2*d*x+1/2*c)^5+8/3*tan(1/2*d*x+1/2*c)^3-31*tan(1/2*d*x+1/2*c)+16*(-7/4*tan(1/2*d*x+1/2*c
)^3-5/4*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^2+52*arctan(tan(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.52, size = 184, normalized size = 1.25 \begin {gather*} -\frac {\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{60 \, 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))^3,x, algorithm="maxima")

[Out]

-1/60*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x + c
)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) - 4
0*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 780*arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/a^3)/d

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Fricas [A]
time = 2.64, size = 135, normalized size = 0.92 \begin {gather*} \frac {195 \, d x \cos \left (d x + c\right )^{3} + 585 \, d x \cos \left (d x + c\right )^{2} + 585 \, d x \cos \left (d x + c\right ) + 195 \, d x + {\left (15 \, \cos \left (d x + c\right )^{4} - 45 \, \cos \left (d x + c\right )^{3} - 479 \, \cos \left (d x + c\right )^{2} - 717 \, \cos \left (d x + c\right ) - 304\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} 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))^3,x, algorithm="fricas")

[Out]

1/30*(195*d*x*cos(d*x + c)^3 + 585*d*x*cos(d*x + c)^2 + 585*d*x*cos(d*x + c) + 195*d*x + (15*cos(d*x + c)^4 -
45*cos(d*x + c)^3 - 479*cos(d*x + c)^2 - 717*cos(d*x + c) - 304)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d
*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*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 ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \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))**3,x)

[Out]

Integral(cos(c + d*x)**2/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [A]
time = 0.51, size = 113, normalized size = 0.77 \begin {gather*} \frac {\frac {390 \, {\left (d x + c\right )}}{a^{3}} - \frac {60 \, {\left (7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, \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^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, 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))^3,x, algorithm="giac")

[Out]

1/60*(390*(d*x + c)/a^3 - 60*(7*tan(1/2*d*x + 1/2*c)^3 + 5*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)
^2*a^3) - (3*a^12*tan(1/2*d*x + 1/2*c)^5 - 40*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*a^12*tan(1/2*d*x + 1/2*c))/a^1
5)/d

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Mupad [B]
time = 0.76, size = 137, normalized size = 0.93 \begin {gather*} -\frac {3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-46\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+508\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+420\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-120\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-390\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (c+d\,x\right )}{60\,a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \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))^3,x)

[Out]

-(3*sin(c/2 + (d*x)/2) - 46*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) + 508*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)
/2) + 420*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) - 120*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2) - 390*cos(c/2
+ (d*x)/2)^5*(c + d*x))/(60*a^3*d*cos(c/2 + (d*x)/2)^5)

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