3.473 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=123 \[ \frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{4 d}+\frac {3 a \left (a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} a x \left (a^2+4 b^2\right )-\frac {3 a^2 b \sin ^3(c+d x)}{4 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d} \]

[Out]

3/8*a*(a^2+4*b^2)*x+1/4*b*(11*a^2+4*b^2)*sin(d*x+c)/d+3/8*a*(a^2+4*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/4*a^2*cos(d*
x+c)^3*(a+b*sec(d*x+c))*sin(d*x+c)/d-3/4*a^2*b*sin(d*x+c)^3/d

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Rubi [A]  time = 0.18, antiderivative size = 123, 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 = {3841, 4047, 2635, 8, 4044, 3013} \[ \frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{4 d}+\frac {3 a \left (a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3}{8} a x \left (a^2+4 b^2\right )-\frac {3 a^2 b \sin ^3(c+d x)}{4 d}+\frac {a^2 \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3,x]

[Out]

(3*a*(a^2 + 4*b^2)*x)/8 + (b*(11*a^2 + 4*b^2)*Sin[c + d*x])/(4*d) + (3*a*(a^2 + 4*b^2)*Cos[c + d*x]*Sin[c + d*
x])/(8*d) + (a^2*Cos[c + d*x]^3*(a + b*Sec[c + d*x])*Sin[c + d*x])/(4*d) - (3*a^2*b*Sin[c + d*x]^3)/(4*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 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

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]

Rule 4047

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

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) \left (9 a^2 b+3 a \left (a^2+4 b^2\right ) \sec (c+d x)+2 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos ^3(c+d x) \left (9 a^2 b+2 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{4} \left (3 a \left (a^2+4 b^2\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {3 a \left (a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}+\frac {1}{4} \int \cos (c+d x) \left (2 b \left (a^2+2 b^2\right )+9 a^2 b \cos ^2(c+d x)\right ) \, dx+\frac {1}{8} \left (3 a \left (a^2+4 b^2\right )\right ) \int 1 \, dx\\ &=\frac {3}{8} a \left (a^2+4 b^2\right ) x+\frac {3 a \left (a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}-\frac {\operatorname {Subst}\left (\int \left (9 a^2 b+2 b \left (a^2+2 b^2\right )-9 a^2 b x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac {3}{8} a \left (a^2+4 b^2\right ) x+\frac {b \left (11 a^2+4 b^2\right ) \sin (c+d x)}{4 d}+\frac {3 a \left (a^2+4 b^2\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 \cos ^3(c+d x) (a+b \sec (c+d x)) \sin (c+d x)}{4 d}-\frac {3 a^2 b \sin ^3(c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 100, normalized size = 0.81 \[ \frac {8 b \left (9 a^2+4 b^2\right ) \sin (c+d x)+a \left (8 \left (a^2+3 b^2\right ) \sin (2 (c+d x))+a^2 \sin (4 (c+d x))+12 a^2 c+12 a^2 d x+8 a b \sin (3 (c+d x))+48 b^2 c+48 b^2 d x\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3,x]

[Out]

(8*b*(9*a^2 + 4*b^2)*Sin[c + d*x] + a*(12*a^2*c + 48*b^2*c + 12*a^2*d*x + 48*b^2*d*x + 8*(a^2 + 3*b^2)*Sin[2*(
c + d*x)] + 8*a*b*Sin[3*(c + d*x)] + a^2*Sin[4*(c + d*x)]))/(32*d)

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fricas [A]  time = 0.44, size = 84, normalized size = 0.68 \[ \frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} d x + {\left (2 \, a^{3} \cos \left (d x + c\right )^{3} + 8 \, a^{2} b \cos \left (d x + c\right )^{2} + 16 \, a^{2} b + 8 \, b^{3} + 3 \, {\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/8*(3*(a^3 + 4*a*b^2)*d*x + (2*a^3*cos(d*x + c)^3 + 8*a^2*b*cos(d*x + c)^2 + 16*a^2*b + 8*b^3 + 3*(a^3 + 4*a*
b^2)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.25, size = 297, normalized size = 2.41 \[ \frac {3 \, {\left (a^{3} + 4 \, a b^{2}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/8*(3*(a^3 + 4*a*b^2)*(d*x + c) - 2*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*a*b^
2*tan(1/2*d*x + 1/2*c)^7 - 8*b^3*tan(1/2*d*x + 1/2*c)^7 - 3*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*a^2*b*tan(1/2*d*x
+ 1/2*c)^5 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 24*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^3 -
40*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*b^3*tan(1/2*d*x + 1/2*c)^3 - 5*a^3*tan(
1/2*d*x + 1/2*c) - 24*a^2*b*tan(1/2*d*x + 1/2*c) - 12*a*b^2*tan(1/2*d*x + 1/2*c) - 8*b^3*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 = 1.00, size = 102, normalized size = 0.83 \[ \frac {a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 b^{2} a \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{3} \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sec(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.37, size = 95, normalized size = 0.77 \[ \frac {{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b^{2} + 32 \, b^{3} \sin \left (d x + c\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/32*((12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^3 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2*b
+ 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b^2 + 32*b^3*sin(d*x + c))/d

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mupad [B]  time = 3.83, size = 250, normalized size = 2.03 \[ \frac {\left (-\frac {5\,a^3}{4}+6\,a^2\,b-3\,a\,b^2+2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,a^3}{4}+10\,a^2\,b-3\,a\,b^2+6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {3\,a^3}{4}+10\,a^2\,b+3\,a\,b^2+6\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,a^3}{4}+6\,a^2\,b+3\,a\,b^2+2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,a\,\mathrm {atan}\left (\frac {3\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2+4\,b^2\right )}{4\,\left (\frac {3\,a^3}{4}+3\,a\,b^2\right )}\right )\,\left (a^2+4\,b^2\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + b/cos(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)^3*(3*a*b^2 + 10*a^2*b - (3*a^3)/4 + 6*b^3) - tan(c/2 + (d*x)/2)^7*(3*a*b^2 - 6*a^2*b + (5*
a^3)/4 - 2*b^3) + tan(c/2 + (d*x)/2)^5*(10*a^2*b - 3*a*b^2 + (3*a^3)/4 + 6*b^3) + tan(c/2 + (d*x)/2)*(3*a*b^2
+ 6*a^2*b + (5*a^3)/4 + 2*b^3))/(d*(4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 +
tan(c/2 + (d*x)/2)^8 + 1)) + (3*a*atan((3*a*tan(c/2 + (d*x)/2)*(a^2 + 4*b^2))/(4*(3*a*b^2 + (3*a^3)/4)))*(a^2
+ 4*b^2))/(4*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))**3,x)

[Out]

Timed out

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