3.243 \(\int \cos (a+b x) \sec ^3(c+b x) \, dx\)

Optimal. Leaf size=38 \[ \frac {\cos (a-c) \tan (b x+c)}{b}-\frac {\sin (a-c) \sec ^2(b x+c)}{2 b} \]

[Out]

-1/2*sec(b*x+c)^2*sin(a-c)/b+cos(a-c)*tan(b*x+c)/b

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Rubi [A]  time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4583, 2606, 30, 3767, 8} \[ \frac {\cos (a-c) \tan (b x+c)}{b}-\frac {\sin (a-c) \sec ^2(b x+c)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Sec[c + b*x]^3,x]

[Out]

-(Sec[c + b*x]^2*Sin[a - c])/(2*b) + (Cos[a - c]*Tan[c + b*x])/b

Rule 8

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 4583

Int[Cos[v_]*Sec[w_]^(n_.), x_Symbol] :> -Dist[Sin[v - w], Int[Tan[w]*Sec[w]^(n - 1), x], x] + Dist[Cos[v - w],
Int[Sec[w]^(n - 1), x], x] /; GtQ[n, 0] && FreeQ[v - w, x] && NeQ[w, v]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 35, normalized size = 0.92 \[ -\frac {\sec (c) \sec ^2(b x+c) (\sin (a)-\cos (a-c) \sin (2 b x+c))}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Sec[c + b*x]^3,x]

[Out]

-1/2*(Sec[c]*Sec[c + b*x]^2*(Sin[a] - Cos[a - c]*Sin[c + 2*b*x]))/b

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fricas [A]  time = 0.51, size = 40, normalized size = 1.05 \[ \frac {2 \, \cos \left (b x + c\right ) \cos \left (-a + c\right ) \sin \left (b x + c\right ) + \sin \left (-a + c\right )}{2 \, b \cos \left (b x + c\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 5.75, size = 315, normalized size = 8.29 \[ -\frac {\tan \left (\frac {1}{2} \, a\right )^{6} \tan \left (\frac {1}{2} \, c\right )^{6} + 3 \, \tan \left (\frac {1}{2} \, a\right )^{6} \tan \left (\frac {1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac {1}{2} \, a\right )^{4} \tan \left (\frac {1}{2} \, c\right )^{6} + 3 \, \tan \left (\frac {1}{2} \, a\right )^{6} \tan \left (\frac {1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac {1}{2} \, a\right )^{4} \tan \left (\frac {1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{6} + \tan \left (\frac {1}{2} \, a\right )^{6} + 9 \, \tan \left (\frac {1}{2} \, a\right )^{4} \tan \left (\frac {1}{2} \, c\right )^{2} + 9 \, \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{4} + \tan \left (\frac {1}{2} \, c\right )^{6} + 3 \, \tan \left (\frac {1}{2} \, a\right )^{4} + 9 \, \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 3 \, \tan \left (\frac {1}{2} \, c\right )^{4} + 3 \, \tan \left (\frac {1}{2} \, a\right )^{2} + 3 \, \tan \left (\frac {1}{2} \, c\right )^{2} + 1}{4 \, {\left (2 \, \tan \left (b x + a\right ) \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - 2 \, \tan \left (b x + a\right ) \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right )^{2} + 2 \, \tan \left (b x + a\right ) \tan \left (\frac {1}{2} \, a\right ) - \tan \left (\frac {1}{2} \, a\right )^{2} - 2 \, \tan \left (b x + a\right ) \tan \left (\frac {1}{2} \, c\right ) + 4 \, \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right ) - \tan \left (\frac {1}{2} \, c\right )^{2} + 1\right )}^{2} {\left (\tan \left (\frac {1}{2} \, a\right )^{2} \tan \left (\frac {1}{2} \, c\right ) - \tan \left (\frac {1}{2} \, a\right ) \tan \left (\frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, a\right ) - \tan \left (\frac {1}{2} \, c\right )\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(tan(1/2*a)^6*tan(1/2*c)^6 + 3*tan(1/2*a)^6*tan(1/2*c)^4 + 3*tan(1/2*a)^4*tan(1/2*c)^6 + 3*tan(1/2*a)^6*t
an(1/2*c)^2 + 9*tan(1/2*a)^4*tan(1/2*c)^4 + 3*tan(1/2*a)^2*tan(1/2*c)^6 + tan(1/2*a)^6 + 9*tan(1/2*a)^4*tan(1/
2*c)^2 + 9*tan(1/2*a)^2*tan(1/2*c)^4 + tan(1/2*c)^6 + 3*tan(1/2*a)^4 + 9*tan(1/2*a)^2*tan(1/2*c)^2 + 3*tan(1/2
*c)^4 + 3*tan(1/2*a)^2 + 3*tan(1/2*c)^2 + 1)/((2*tan(b*x + a)*tan(1/2*a)^2*tan(1/2*c) - 2*tan(b*x + a)*tan(1/2
*a)*tan(1/2*c)^2 + tan(1/2*a)^2*tan(1/2*c)^2 + 2*tan(b*x + a)*tan(1/2*a) - tan(1/2*a)^2 - 2*tan(b*x + a)*tan(1
/2*c) + 4*tan(1/2*a)*tan(1/2*c) - tan(1/2*c)^2 + 1)^2*(tan(1/2*a)^2*tan(1/2*c) - tan(1/2*a)*tan(1/2*c)^2 + tan
(1/2*a) - tan(1/2*c))*b)

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maple [A]  time = 3.67, size = 56, normalized size = 1.47 \[ \frac {1}{2 b \left (\cos \relax (a ) \sin \relax (c )-\sin \relax (a ) \cos \relax (c )\right ) \left (-\tan \left (b x +a \right ) \cos \relax (a ) \sin \relax (c )+\tan \left (b x +a \right ) \sin \relax (a ) \cos \relax (c )+\cos \relax (a ) \cos \relax (c )+\sin \relax (a ) \sin \relax (c )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b/(cos(a)*sin(c)-sin(a)*cos(c))/(-tan(b*x+a)*cos(a)*sin(c)+tan(b*x+a)*sin(a)*cos(c)+cos(a)*cos(c)+sin(a)*s
in(c))^2

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maxima [B]  time = 0.33, size = 382, normalized size = 10.05 \[ -\frac {{\left (2 \, \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) + \sin \left (2 \, a\right ) + \sin \left (2 \, c\right )\right )} \cos \left (4 \, b x + a + 5 \, c\right ) + 2 \, {\left (2 \, \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) + \sin \left (2 \, a\right ) + \sin \left (2 \, c\right )\right )} \cos \left (2 \, b x + a + 3 \, c\right ) + {\left (\sin \left (2 \, a\right ) + \sin \left (2 \, c\right )\right )} \cos \left (a + c\right ) - {\left (2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) + \cos \left (2 \, a\right ) + \cos \left (2 \, c\right )\right )} \sin \left (4 \, b x + a + 5 \, c\right ) + 2 \, \cos \left (a + c\right ) \sin \left (2 \, b x + 2 \, a + 2 \, c\right ) - 2 \, {\left (2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) + \cos \left (2 \, a\right ) + \cos \left (2 \, c\right )\right )} \sin \left (2 \, b x + a + 3 \, c\right ) - {\left (\cos \left (2 \, a\right ) + \cos \left (2 \, c\right )\right )} \sin \left (a + c\right ) - 2 \, \cos \left (2 \, b x + 2 \, a + 2 \, c\right ) \sin \left (a + c\right )}{b \cos \left (4 \, b x + a + 5 \, c\right )^{2} + 4 \, b \cos \left (2 \, b x + a + 3 \, c\right )^{2} + 4 \, b \cos \left (2 \, b x + a + 3 \, c\right ) \cos \left (a + c\right ) + b \cos \left (a + c\right )^{2} + b \sin \left (4 \, b x + a + 5 \, c\right )^{2} + 4 \, b \sin \left (2 \, b x + a + 3 \, c\right )^{2} + 4 \, b \sin \left (2 \, b x + a + 3 \, c\right ) \sin \left (a + c\right ) + b \sin \left (a + c\right )^{2} + 2 \, {\left (2 \, b \cos \left (2 \, b x + a + 3 \, c\right ) + b \cos \left (a + c\right )\right )} \cos \left (4 \, b x + a + 5 \, c\right ) + 2 \, {\left (2 \, b \sin \left (2 \, b x + a + 3 \, c\right ) + b \sin \left (a + c\right )\right )} \sin \left (4 \, b x + a + 5 \, c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.03 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)/cos(c + b*x)^3,x)

[Out]

\text{Hanged}

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed

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