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

Optimal. Leaf size=76 \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a x}{8} \]

[Out]

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

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Rubi [A]  time = 0.0550533, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3787, 2635, 8, 2633} \[ -\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}+\frac{a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{3 a \sin (c+d x) \cos (c+d x)}{8 d}+\frac{3 a x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3787

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

Rubi steps

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

Mathematica [A]  time = 0.0969441, size = 73, normalized size = 0.96 \[ \frac{3 a (c+d x)}{8 d}-\frac{a \sin ^3(c+d x)}{3 d}+\frac{a \sin (c+d x)}{d}+\frac{a \sin (2 (c+d x))}{4 d}+\frac{a \sin (4 (c+d x))}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*(c + d*x))/(8*d) + (a*Sin[c + d*x])/d - (a*Sin[c + d*x]^3)/(3*d) + (a*Sin[2*(c + d*x)])/(4*d) + (a*Sin[4*
(c + d*x)])/(32*d)

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Maple [A]  time = 0.103, size = 60, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{a \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.08461, size = 77, normalized size = 1.01 \begin{align*} -\frac{32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a - 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}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*a - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a)/d

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Fricas [A]  time = 1.72212, size = 136, normalized size = 1.79 \begin{align*} \frac{9 \, a d x +{\left (6 \, a \cos \left (d x + c\right )^{3} + 8 \, a \cos \left (d x + c\right )^{2} + 9 \, a \cos \left (d x + c\right ) + 16 \, a\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*a*d*x + (6*a*cos(d*x + c)^3 + 8*a*cos(d*x + c)^2 + 9*a*cos(d*x + c) + 16*a)*sin(d*x + c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.30253, size = 116, normalized size = 1.53 \begin{align*} \frac{9 \,{\left (d x + c\right )} a + \frac{2 \,{\left (9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 49 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 31 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 39 \, a \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*(d*x + c)*a + 2*(9*a*tan(1/2*d*x + 1/2*c)^7 + 49*a*tan(1/2*d*x + 1/2*c)^5 + 31*a*tan(1/2*d*x + 1/2*c)^
3 + 39*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d