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

Optimal. Leaf size=87 \[ -\frac{4 a^4 \sin ^3(c+d x)}{3 d}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{a^4 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{27 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{35 a^4 x}{8} \]

[Out]

(35*a^4*x)/8 + (8*a^4*Sin[c + d*x])/d + (27*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4*Cos[c + d*x]^3*Sin[c +
d*x])/(4*d) - (4*a^4*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.098184, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3791, 2637, 2635, 8, 2633} \[ -\frac{4 a^4 \sin ^3(c+d x)}{3 d}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{a^4 \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{27 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{35 a^4 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*a^4*x)/8 + (8*a^4*Sin[c + d*x])/d + (27*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^4*Cos[c + d*x]^3*Sin[c +
d*x])/(4*d) - (4*a^4*Sin[c + d*x]^3)/(3*d)

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, 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))^4 \, dx &=\int \left (a^4+4 a^4 \cos (c+d x)+6 a^4 \cos ^2(c+d x)+4 a^4 \cos ^3(c+d x)+a^4 \cos ^4(c+d x)\right ) \, dx\\ &=a^4 x+a^4 \int \cos ^4(c+d x) \, dx+\left (4 a^4\right ) \int \cos (c+d x) \, dx+\left (4 a^4\right ) \int \cos ^3(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^2(c+d x) \, dx\\ &=a^4 x+\frac{4 a^4 \sin (c+d x)}{d}+\frac{3 a^4 \cos (c+d x) \sin (c+d x)}{d}+\frac{a^4 \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{1}{4} \left (3 a^4\right ) \int \cos ^2(c+d x) \, dx+\left (3 a^4\right ) \int 1 \, dx-\frac{\left (4 a^4\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=4 a^4 x+\frac{8 a^4 \sin (c+d x)}{d}+\frac{27 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^4 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{4 a^4 \sin ^3(c+d x)}{3 d}+\frac{1}{8} \left (3 a^4\right ) \int 1 \, dx\\ &=\frac{35 a^4 x}{8}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{27 a^4 \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^4 \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{4 a^4 \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.100717, size = 56, normalized size = 0.64 \[ \frac{a^4 (672 \sin (c+d x)+168 \sin (2 (c+d x))+32 \sin (3 (c+d x))+3 \sin (4 (c+d x))+420 c+420 d x)}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(420*c + 420*d*x + 672*Sin[c + d*x] + 168*Sin[2*(c + d*x)] + 32*Sin[3*(c + d*x)] + 3*Sin[4*(c + d*x)]))/(
96*d)

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Maple [A]  time = 0.074, size = 111, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({a}^{4} \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{4\,{a}^{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) }{3}}+6\,{a}^{4} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +4\,{a}^{4}\sin \left ( dx+c \right ) +{a}^{4} \left ( dx+c \right ) \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))^4,x)

[Out]

1/d*(a^4*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+4/3*a^4*(cos(d*x+c)^2+2)*sin(d*x+c)+6*a^
4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*a^4*sin(d*x+c)+a^4*(d*x+c))

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Maxima [A]  time = 1.09911, size = 140, normalized size = 1.61 \begin{align*} -\frac{128 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} - 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^{4} - 144 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 96 \,{\left (d x + c\right )} a^{4} - 384 \, a^{4} \sin \left (d x + c\right )}{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))^4,x, algorithm="maxima")

[Out]

-1/96*(128*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 - 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a
^4 - 144*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^4 - 96*(d*x + c)*a^4 - 384*a^4*sin(d*x + c))/d

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Fricas [A]  time = 1.71224, size = 157, normalized size = 1.8 \begin{align*} \frac{105 \, a^{4} d x +{\left (6 \, a^{4} \cos \left (d x + c\right )^{3} + 32 \, a^{4} \cos \left (d x + c\right )^{2} + 81 \, a^{4} \cos \left (d x + c\right ) + 160 \, a^{4}\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))^4,x, algorithm="fricas")

[Out]

1/24*(105*a^4*d*x + (6*a^4*cos(d*x + c)^3 + 32*a^4*cos(d*x + c)^2 + 81*a^4*cos(d*x + c) + 160*a^4)*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))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.41584, size = 130, normalized size = 1.49 \begin{align*} \frac{105 \,{\left (d x + c\right )} a^{4} + \frac{2 \,{\left (105 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 385 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 511 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 279 \, a^{4} \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))^4,x, algorithm="giac")

[Out]

1/24*(105*(d*x + c)*a^4 + 2*(105*a^4*tan(1/2*d*x + 1/2*c)^7 + 385*a^4*tan(1/2*d*x + 1/2*c)^5 + 511*a^4*tan(1/2
*d*x + 1/2*c)^3 + 279*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d