3.18 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=87 \[ -\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {7 a^2 x}{8} \]

[Out]

7/8*a^2*x+2*a^2*sin(d*x+c)/d+7/8*a^2*cos(d*x+c)*sin(d*x+c)/d+1/4*a^2*cos(d*x+c)^3*sin(d*x+c)/d-2/3*a^2*sin(d*x
+c)^3/d

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Rubi [A]  time = 0.08, antiderivative size = 87, 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, 2633, 4045, 2635, 8} \[ -\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {7 a^2 \sin (c+d x) \cos (c+d x)}{8 d}+\frac {7 a^2 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2,x]

[Out]

(7*a^2*x)/8 + (2*a^2*Sin[c + d*x])/d + (7*a^2*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*Cos[c + d*x]^3*Sin[c + d
*x])/(4*d) - (2*a^2*Sin[c + d*x]^3)/(3*d)

Rule 8

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

Rule 2633

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 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 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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \, dx &=\left (2 a^2\right ) \int \cos ^3(c+d x) \, dx+\int \cos ^4(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {1}{4} \left (7 a^2\right ) \int \cos ^2(c+d x) \, dx-\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}+\frac {1}{8} \left (7 a^2\right ) \int 1 \, dx\\ &=\frac {7 a^2 x}{8}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {7 a^2 \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 53, normalized size = 0.61 \[ \frac {a^2 (144 \sin (c+d x)+48 \sin (2 (c+d x))+16 \sin (3 (c+d x))+3 \sin (4 (c+d x))+84 d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2,x]

[Out]

(a^2*(84*d*x + 144*Sin[c + d*x] + 48*Sin[2*(c + d*x)] + 16*Sin[3*(c + d*x)] + 3*Sin[4*(c + d*x)]))/(96*d)

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fricas [A]  time = 0.72, size = 63, normalized size = 0.72 \[ \frac {21 \, a^{2} d x + {\left (6 \, a^{2} \cos \left (d x + c\right )^{3} + 16 \, a^{2} \cos \left (d x + c\right )^{2} + 21 \, a^{2} \cos \left (d x + c\right ) + 32 \, a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(21*a^2*d*x + (6*a^2*cos(d*x + c)^3 + 16*a^2*cos(d*x + c)^2 + 21*a^2*cos(d*x + c) + 32*a^2)*sin(d*x + c))
/d

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giac [A]  time = 3.92, size = 96, normalized size = 1.10 \[ \frac {21 \, {\left (d x + c\right )} a^{2} + \frac {2 \, {\left (21 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 77 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(21*(d*x + c)*a^2 + 2*(21*a^2*tan(1/2*d*x + 1/2*c)^7 + 77*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*a^2*tan(1/2*d*x
+ 1/2*c)^3 + 75*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 0.98, size = 90, normalized size = 1.03 \[ \frac {a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^2,x)

[Out]

1/d*(a^2*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+2/3*a^2*(2+cos(d*x+c)^2)*sin(d*x+c)+a^2*
(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 0.90, size = 83, normalized size = 0.95 \[ -\frac {64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/96*(64*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^
2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^2)/d

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mupad [B]  time = 4.19, size = 89, normalized size = 1.02 \[ \frac {7\,a^2\,x}{8}+\frac {\frac {7\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {77\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{12}+\frac {83\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{12}+\frac {25\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a/cos(c + d*x))^2,x)

[Out]

(7*a^2*x)/8 + ((83*a^2*tan(c/2 + (d*x)/2)^3)/12 + (77*a^2*tan(c/2 + (d*x)/2)^5)/12 + (7*a^2*tan(c/2 + (d*x)/2)
^7)/4 + (25*a^2*tan(c/2 + (d*x)/2))/4)/(d*(tan(c/2 + (d*x)/2)^2 + 1)^4)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \cos ^{4}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(2*cos(c + d*x)**4*sec(c + d*x), x) + Integral(cos(c + d*x)**4*sec(c + d*x)**2, x) + Integral(co
s(c + d*x)**4, x))

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