3.1.68 \(\int \frac {\cos (c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [68]

Optimal. Leaf size=103 \[ -\frac {3 x}{a^3}+\frac {24 \sin (c+d x)}{5 a^3 d}-\frac {\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac {3 \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )} \]

[Out]

-3*x/a^3+24/5*sin(d*x+c)/a^3/d-1/5*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-3/5*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2-3*sin
(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]
time = 0.15, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {24 \sin (c+d x)}{5 a^3 d}-\frac {3 \sin (c+d x)}{d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {3 x}{a^3}-\frac {3 \sin (c+d x)}{5 a d (a \sec (c+d x)+a)^2}-\frac {\sin (c+d x)}{5 d (a \sec (c+d x)+a)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/a^3 + (24*Sin[c + d*x])/(5*a^3*d) - Sin[c + d*x]/(5*d*(a + a*Sec[c + d*x])^3) - (3*Sin[c + d*x])/(5*a*d
*(a + a*Sec[c + d*x])^2) - (3*Sin[c + d*x])/(d*(a^3 + a^3*Sec[c + d*x]))

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))^3} \, dx &=-\frac {\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {\int \frac {\cos (c+d x) (-6 a+3 a \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac {\int \frac {\cos (c+d x) \left (-27 a^2+18 a^2 \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac {3 \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {\int \cos (c+d x) \left (-72 a^3+45 a^3 \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac {3 \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {3 \int 1 \, dx}{a^3}+\frac {24 \int \cos (c+d x) \, dx}{5 a^3}\\ &=-\frac {3 x}{a^3}+\frac {24 \sin (c+d x)}{5 a^3 d}-\frac {\sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {3 \sin (c+d x)}{5 a d (a+a \sec (c+d x))^2}-\frac {3 \sin (c+d x)}{d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.57, size = 169, normalized size = 1.64 \begin {gather*} \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-12 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+96 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+20 \cos ^5\left (\frac {1}{2} (c+d x)\right ) (-3 d x+\sin (c+d x))+\cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )-12 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{5 a^3 d (1+\sec (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*Sec[c + d*x]^3*(Sec[c/2]*Sin[(d*x)/2] - 12*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 96*C
os[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] + 20*Cos[(c + d*x)/2]^5*(-3*d*x + Sin[c + d*x]) + Cos[(c + d*x)/2]*Tan
[c/2] - 12*Cos[(c + d*x)/2]^3*Tan[c/2]))/(5*a^3*d*(1 + Sec[c + d*x])^3)

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Maple [A]
time = 0.08, size = 85, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-24 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(85\)
default	\(\frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}-24 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\)	\(85\)
risch	\(-\frac {3 x}{a^{3}}-\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}+\frac {4 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+50 \,{\mathrm e}^{3 i \left (d x +c \right )}+70 \,{\mathrm e}^{2 i \left (d x +c \right )}+45 \,{\mathrm e}^{i \left (d x +c \right )}+12\right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(112\)
norman	\(\frac {-\frac {3 x}{a}+\frac {25 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {15 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 a d}-\frac {9 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 a d}+\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a d}-\frac {3 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(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)^5-2*tan(1/2*d*x+1/2*c)^3+17*tan(1/2*d*x+1/2*c)+8*tan(1/2*d*x+1/2*c)/(1+tan(1
/2*d*x+1/2*c)^2)-24*arctan(tan(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.49, size = 137, normalized size = 1.33 \begin {gather*} \frac {\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac {a^{3} \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 )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{20 \, 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))^3,x, algorithm="maxima")

[Out]

1/20*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c)/
(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 120*a
rctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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Fricas [A]
time = 2.03, size = 126, normalized size = 1.22 \begin {gather*} -\frac {15 \, d x \cos \left (d x + c\right )^{3} + 45 \, d x \cos \left (d x + c\right )^{2} + 45 \, d x \cos \left (d x + c\right ) + 15 \, d x - {\left (5 \, \cos \left (d x + c\right )^{3} + 39 \, \cos \left (d x + c\right )^{2} + 57 \, \cos \left (d x + c\right ) + 24\right )} \sin \left (d x + c\right )}{5 \, {\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)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/5*(15*d*x*cos(d*x + c)^3 + 45*d*x*cos(d*x + c)^2 + 45*d*x*cos(d*x + c) + 15*d*x - (5*cos(d*x + c)^3 + 39*co
s(d*x + c)^2 + 57*cos(d*x + c) + 24)*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*co
s(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 {\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)/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(cos(c + d*x)/(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 = 0.49, size = 96, normalized size = 0.93 \begin {gather*} -\frac {\frac {60 \, {\left (d x + c\right )}}{a^{3}} - \frac {40 \, \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^{3}} - \frac {a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 85 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{20 \, 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))^3,x, algorithm="giac")

[Out]

-1/20*(60*(d*x + c)/a^3 - 40*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^3) - (a^12*tan(1/2*d*x + 1/2
*c)^5 - 10*a^12*tan(1/2*d*x + 1/2*c)^3 + 85*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d

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Mupad [B]
time = 0.73, size = 113, normalized size = 1.10 \begin {gather*} \frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+96\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+40\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-60\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (c+d\,x\right )}{20\,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)/(a + a/cos(c + d*x))^3,x)

[Out]

(sin(c/2 + (d*x)/2) - 12*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) + 96*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)
+ 40*cos(c/2 + (d*x)/2)^6*sin(c/2 + (d*x)/2) - 60*cos(c/2 + (d*x)/2)^5*(c + d*x))/(20*a^3*d*cos(c/2 + (d*x)/2)
^5)

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