3.17 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=57 \[ -\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}+a^2 x \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 57, 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 = {3788, 2635, 8, 4044, 3013} \[ -\frac {a^2 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{d}+a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2635

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] && Integer
Q[2*n]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 3788

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

Rule 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 41, normalized size = 0.72 \[ \frac {a^2 (21 \sin (c+d x)+6 \sin (2 (c+d x))+\sin (3 (c+d x))+12 d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*d*x + 21*Sin[c + d*x] + 6*Sin[2*(c + d*x)] + Sin[3*(c + d*x)]))/(12*d)

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fricas [A]  time = 0.55, size = 49, normalized size = 0.86 \[ \frac {3 \, a^{2} d x + {\left (a^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} \cos \left (d x + c\right ) + 5 \, a^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \]

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

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giac [A]  time = 1.80, size = 80, normalized size = 1.40 \[ \frac {3 \, {\left (d x + c\right )} a^{2} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 8 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

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]

1/3*(3*(d*x + c)*a^2 + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 8*a^2*tan(1/2*d*x + 1/2*c)^3 + 9*a^2*tan(1/2*d*x + 1/
2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.90, size = 64, normalized size = 1.12 \[ \frac {\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{2} \sin \left (d x +c \right )}{d} \]

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)

[Out]

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

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maxima [A]  time = 0.73, size = 61, normalized size = 1.07 \[ -\frac {2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 6 \, a^{2} \sin \left (d x + c\right )}{6 \, d} \]

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*(2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2 - 6*a^2*sin(d*x + c))/d

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mupad [B]  time = 0.66, size = 61, normalized size = 1.07 \[ a^2\,x+\frac {5\,a^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d}+\frac {a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{d} \]

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]

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

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

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]

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

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