3.1.50 \(\int \frac {\cos ^3(c+d x)}{a+a \sec (c+d x)} \, dx\) [50]

Optimal. Leaf size=94 \[ -\frac {3 x}{2 a}+\frac {4 \sin (c+d x)}{a d}-\frac {3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {4 \sin ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3904, 3872, 2713, 2715, 8} \begin {gather*} -\frac {4 \sin ^3(c+d x)}{3 a d}+\frac {4 \sin (c+d x)}{a d}-\frac {3 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {\sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac {3 x}{2 a} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/(2*a) + (4*Sin[c + d*x])/(a*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - (Cos[c + d*x]^2*Sin[c + d*x])/
(d*(a + a*Sec[c + d*x])) - (4*Sin[c + d*x]^3)/(3*a*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 3904

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

Rubi steps

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

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Mathematica [A]
time = 0.33, size = 143, normalized size = 1.52 \begin {gather*} \frac {\sec \left (\frac {c}{2}\right ) \sec \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 )+69 \sin \left (\frac {d x}{2}\right )+21 \sin \left (c+\frac {d x}{2}\right )+18 \sin \left (c+\frac {3 d x}{2}\right )+18 \sin \left (2 c+\frac {3 d x}{2}\right )-2 \sin \left (2 c+\frac {5 d x}{2}\right )-2 \sin \left (3 c+\frac {5 d x}{2}\right )+\sin \left (3 c+\frac {7 d x}{2}\right )+\sin \left (4 c+\frac {7 d x}{2}\right )\right )}{48 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]*(-36*d*x*Cos[(d*x)/2] - 36*d*x*Cos[c + (d*x)/2] + 69*Sin[(d*x)/2] + 21*Sin[c + (d*x
)/2] + 18*Sin[c + (3*d*x)/2] + 18*Sin[2*c + (3*d*x)/2] - 2*Sin[2*c + (5*d*x)/2] - 2*Sin[3*c + (5*d*x)/2] + Sin
[3*c + (7*d*x)/2] + Sin[4*c + (7*d*x)/2]))/(48*a*d)

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Maple [A]
time = 0.07, size = 85, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 \left (-\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(85\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {8 \left (-\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8}\right )}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-3 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(85\)
risch	\(-\frac {3 x}{2 a}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )}}{8 a d}+\frac {2 i}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {\sin \left (3 d x +3 c \right )}{12 a d}-\frac {\sin \left (2 d x +2 c \right )}{4 a d}\)	\(100\)
norman	\(\frac {\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 x}{2 a}+\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {25 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {8 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {9 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {9 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {3 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 176, normalized size = 1.87 \begin {gather*} \frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \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}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{3 \, 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)),x, algorithm="maxima")

[Out]

1/3*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(d*
x + c)^6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 3*sin(d*x + c)/(a*(cos(d*x + c)
+ 1)))/d

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Fricas [A]
time = 3.10, size = 70, normalized size = 0.74 \begin {gather*} -\frac {9 \, d x \cos \left (d x + c\right ) + 9 \, d x - {\left (2 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 7 \, \cos \left (d x + c\right ) + 16\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a 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)),x, algorithm="fricas")

[Out]

-1/6*(9*d*x*cos(d*x + c) + 9*d*x - (2*cos(d*x + c)^3 - cos(d*x + c)^2 + 7*cos(d*x + c) + 16)*sin(d*x + c))/(a*
d*cos(d*x + c) + a*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)),x)

[Out]

Timed out

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Giac [A]
time = 0.44, size = 88, normalized size = 0.94 \begin {gather*} -\frac {\frac {9 \, {\left (d x + c\right )}}{a} - \frac {6 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, \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}}{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)),x, algorithm="giac")

[Out]

-1/6*(9*(d*x + c)/a - 6*tan(1/2*d*x + 1/2*c)/a - 2*(15*tan(1/2*d*x + 1/2*c)^5 + 16*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))/d

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Mupad [B]
time = 0.91, size = 70, normalized size = 0.74 \begin {gather*} \frac {\frac {15\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {3\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )}{4}-\frac {\sin \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )}{12}+\frac {\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )}{24}}{a\,d\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}-\frac {3\,x}{2\,a} \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)),x)

[Out]

((15*sin(c/2 + (d*x)/2))/8 + (3*sin((3*c)/2 + (3*d*x)/2))/4 - sin((5*c)/2 + (5*d*x)/2)/12 + sin((7*c)/2 + (7*d
*x)/2)/24)/(a*d*cos(c/2 + (d*x)/2)) - (3*x)/(2*a)

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