3.1.60 \(\int \frac {\cos ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx\) [60]

Optimal. Leaf size=124 \[ -\frac {5 x}{a^2}+\frac {12 \sin (c+d x)}{a^2 d}-\frac {5 \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {10 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 \sin ^3(c+d x)}{a^2 d} \]

[Out]

-5*x/a^2+12*sin(d*x+c)/a^2/d-5*cos(d*x+c)*sin(d*x+c)/a^2/d-10/3*cos(d*x+c)^2*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1
/3*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^2-4*sin(d*x+c)^3/a^2/d

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Rubi [A]
time = 0.13, antiderivative size = 124, 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, 2713, 2715, 8} \begin {gather*} -\frac {4 \sin ^3(c+d x)}{a^2 d}+\frac {12 \sin (c+d x)}{a^2 d}-\frac {5 \sin (c+d x) \cos (c+d x)}{a^2 d}-\frac {10 \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {5 x}{a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*x)/a^2 + (12*Sin[c + d*x])/(a^2*d) - (5*Cos[c + d*x]*Sin[c + d*x])/(a^2*d) - (10*Cos[c + d*x]^2*Sin[c + d*
x])/(3*a^2*d*(1 + Sec[c + d*x])) - (Cos[c + d*x]^2*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) - (4*Sin[c + d*x
]^3)/(a^2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\cos ^3(c+d x) (-6 a+4 a \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {10 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \cos ^3(c+d x) \left (-36 a^2+30 a^2 \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {10 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {10 \int \cos ^2(c+d x) \, dx}{a^2}+\frac {12 \int \cos ^3(c+d x) \, dx}{a^2}\\ &=-\frac {5 \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {10 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {5 \int 1 \, dx}{a^2}-\frac {12 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac {5 x}{a^2}+\frac {12 \sin (c+d x)}{a^2 d}-\frac {5 \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {10 \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 \sin ^3(c+d x)}{a^2 d}\\ \end {align*}

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Mathematica [A]
time = 0.36, size = 199, normalized size = 1.60 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (-360 d x \cos \left (\frac {d x}{2}\right )-360 d x \cos \left (c+\frac {d x}{2}\right )-120 d x \cos \left (c+\frac {3 d x}{2}\right )-120 d x \cos \left (2 c+\frac {3 d x}{2}\right )+516 \sin \left (\frac {d x}{2}\right )-156 \sin \left (c+\frac {d x}{2}\right )+342 \sin \left (c+\frac {3 d x}{2}\right )+118 \sin \left (2 c+\frac {3 d x}{2}\right )+30 \sin \left (2 c+\frac {5 d x}{2}\right )+30 \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 )+\sin \left (4 c+\frac {9 d x}{2}\right )+\sin \left (5 c+\frac {9 d x}{2}\right )\right )}{192 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^3*(-360*d*x*Cos[(d*x)/2] - 360*d*x*Cos[c + (d*x)/2] - 120*d*x*Cos[c + (3*d*x)/2] -
120*d*x*Cos[2*c + (3*d*x)/2] + 516*Sin[(d*x)/2] - 156*Sin[c + (d*x)/2] + 342*Sin[c + (3*d*x)/2] + 118*Sin[2*c
+ (3*d*x)/2] + 30*Sin[2*c + (5*d*x)/2] + 30*Sin[3*c + (5*d*x)/2] - 3*Sin[3*c + (7*d*x)/2] - 3*Sin[4*c + (7*d*x
)/2] + Sin[4*c + (9*d*x)/2] + Sin[5*c + (9*d*x)/2]))/(192*a^2*d)

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Maple [A]
time = 0.08, size = 101, normalized size = 0.81

method	result	size
derivativedivides	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 \left (-\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-20 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(101\)
default	\(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 \left (-\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-\frac {10 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-20 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}\)	\(101\)
risch	\(-\frac {5 x}{a^{2}}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{4 a^{2} d}-\frac {15 i {\mathrm e}^{i \left (d x +c \right )}}{8 a^{2} d}+\frac {15 i {\mathrm e}^{-i \left (d x +c \right )}}{8 a^{2} d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{4 a^{2} d}+\frac {2 i \left (15 \,{\mathrm e}^{2 i \left (d x +c \right )}+27 \,{\mathrm e}^{i \left (d x +c \right )}+14\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {\sin \left (3 d x +3 c \right )}{12 a^{2} d}\)	\(143\)
norman	\(\frac {-\frac {5 x}{a}+\frac {21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {80 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {23 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {4 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 a d}-\frac {15 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {15 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {5 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a}\)	\(171\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(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+9*tan(1/2*d*x+1/2*c)-8*(-5/2*tan(1/2*d*x+1/2*c)^5-10/3*tan(1/2*d*x+1/2*c)
^3-3/2*tan(1/2*d*x+1/2*c))/(1+tan(1/2*d*x+1/2*c)^2)^3-20*arctan(tan(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.49, size = 207, normalized size = 1.67 \begin {gather*} \frac {\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \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 {60 \, \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)^3/(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*
x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 +
a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c)
+ 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2)/d

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Fricas [A]
time = 2.48, size = 108, normalized size = 0.87 \begin {gather*} -\frac {15 \, d x \cos \left (d x + c\right )^{2} + 30 \, d x \cos \left (d x + c\right ) + 15 \, d x - {\left (\cos \left (d x + c\right )^{4} - \cos \left (d x + c\right )^{3} + 6 \, \cos \left (d x + c\right )^{2} + 33 \, \cos \left (d x + c\right ) + 24\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)^3/(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(15*d*x*cos(d*x + c)^2 + 30*d*x*cos(d*x + c) + 15*d*x - (cos(d*x + c)^4 - cos(d*x + c)^3 + 6*cos(d*x + c)
^2 + 33*cos(d*x + c) + 24)*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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.46, size = 108, normalized size = 0.87 \begin {gather*} -\frac {\frac {30 \, {\left (d x + c\right )}}{a^{2}} - \frac {4 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, \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} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, 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)^3/(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(30*(d*x + c)/a^2 - 4*(15*tan(1/2*d*x + 1/2*c)^5 + 20*tan(1/2*d*x + 1/2*c)^3 + 9*tan(1/2*d*x + 1/2*c))/((
tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^2) + (a^4*tan(1/2*d*x + 1/2*c)^3 - 27*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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Mupad [B]
time = 0.78, size = 135, normalized size = 1.09 \begin {gather*} -\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-28\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-60\,{\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 )-16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+30\,{\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)^3/(a + a/cos(c + d*x))^2,x)

[Out]

-(sin(c/2 + (d*x)/2) - 28*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) - 60*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) - 16*cos(c/2 + (d*x)/2)^8*sin(c/2 + (d*x)/2) + 30*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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