3.1.26 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\) [26]

Optimal. Leaf size=63 \[ \frac {5 a^3 x}{2}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3876, 2717, 2715, 8, 2713} \begin {gather*} -\frac {a^3 \sin ^3(c+d x)}{3 d}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {5 a^3 x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*x)/2 + (4*a^3*Sin[c + d*x])/d + (3*a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (a^3*Sin[c + d*x]^3)/(3*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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3876

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx &=\int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=a^3 x+a^3 \int \cos ^3(c+d x) \, dx+\left (3 a^3\right ) \int \cos (c+d x) \, dx+\left (3 a^3\right ) \int \cos ^2(c+d x) \, dx\\ &=a^3 x+\frac {3 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {1}{2} \left (3 a^3\right ) \int 1 \, dx-\frac {a^3 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {5 a^3 x}{2}+\frac {4 a^3 \sin (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac {a^3 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 44, normalized size = 0.70 \begin {gather*} \frac {a^3 (30 c+30 d x+45 \sin (c+d x)+9 \sin (2 (c+d x))+\sin (3 (c+d x)))}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(30*c + 30*d*x + 45*Sin[c + d*x] + 9*Sin[2*(c + d*x)] + Sin[3*(c + d*x)]))/(12*d)

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Maple [A]
time = 0.10, size = 74, normalized size = 1.17

method	result	size
risch	\(\frac {5 a^{3} x}{2}+\frac {15 a^{3} \sin \left (d x +c \right )}{4 d}+\frac {a^{3} \sin \left (3 d x +3 c \right )}{12 d}+\frac {3 a^{3} \sin \left (2 d x +2 c \right )}{4 d}\)	\(56\)
derivativedivides	\(\frac {\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} \sin \left (d x +c \right )+a^{3} \left (d x +c \right )}{d}\)	\(74\)
default	\(\frac {\frac {a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} \sin \left (d x +c \right )+a^{3} \left (d x +c \right )}{d}\)	\(74\)
norman	\(\frac {\frac {5 a^{3} x}{2}+\frac {11 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {26 a^{3} \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {32 a^{3} \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {10 a^{3} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {5 a^{3} \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {5 a^{3} x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}-5 a^{3} x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-5 a^{3} x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {5 a^{3} x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{3} x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/3*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+3*a^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+3*a^3*sin(d*x+c)+a^3*
(d*x+c))

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Maxima [A]
time = 0.29, size = 71, normalized size = 1.13 \begin {gather*} -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{3} - 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 12 \, {\left (d x + c\right )} a^{3} - 36 \, a^{3} \sin \left (d x + c\right )}{12 \, 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))^3,x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^3 - 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3 - 12*(d*x + c)*a^3 - 3
6*a^3*sin(d*x + c))/d

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Fricas [A]
time = 3.42, size = 50, normalized size = 0.79 \begin {gather*} \frac {15 \, a^{3} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{2} + 9 \, a^{3} \cos \left (d x + c\right ) + 22 \, a^{3}\right )} \sin \left (d x + c\right )}{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))^3,x, algorithm="fricas")

[Out]

1/6*(15*a^3*d*x + (2*a^3*cos(d*x + c)^2 + 9*a^3*cos(d*x + c) + 22*a^3)*sin(d*x + c))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int 3 \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \cos ^{3}{\left (c + d x \right )}\, dx\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))**3,x)

[Out]

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

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Giac [A]
time = 0.46, size = 80, normalized size = 1.27 \begin {gather*} \frac {15 \, {\left (d x + c\right )} a^{3} + \frac {2 \, {\left (15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, a^{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 )}^{3}}}{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))^3,x, algorithm="giac")

[Out]

1/6*(15*(d*x + c)*a^3 + 2*(15*a^3*tan(1/2*d*x + 1/2*c)^5 + 40*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*a^3*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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Mupad [B]
time = 0.67, size = 63, normalized size = 1.00 \begin {gather*} \frac {5\,a^3\,x}{2}+\frac {11\,a^3\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {3\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \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))^3,x)

[Out]

(5*a^3*x)/2 + (11*a^3*sin(c + d*x))/(3*d) + (a^3*cos(c + d*x)^2*sin(c + d*x))/(3*d) + (3*a^3*cos(c + d*x)*sin(
c + d*x))/(2*d)

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