3.40 \(\int \cos ^7(c+d x) (a+a \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=147 \[ -\frac{a^4 \sin ^7(c+d x)}{7 d}+\frac{9 a^4 \sin ^5(c+d x)}{5 d}-\frac{16 a^4 \sin ^3(c+d x)}{3 d}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{2 a^4 \sin (c+d x) \cos ^5(c+d x)}{3 d}+\frac{11 a^4 \sin (c+d x) \cos ^3(c+d x)}{6 d}+\frac{11 a^4 \sin (c+d x) \cos (c+d x)}{4 d}+\frac{11 a^4 x}{4} \]

[Out]

(11*a^4*x)/4 + (8*a^4*Sin[c + d*x])/d + (11*a^4*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (11*a^4*Cos[c + d*x]^3*Sin[
c + d*x])/(6*d) + (2*a^4*Cos[c + d*x]^5*Sin[c + d*x])/(3*d) - (16*a^4*Sin[c + d*x]^3)/(3*d) + (9*a^4*Sin[c + d
*x]^5)/(5*d) - (a^4*Sin[c + d*x]^7)/(7*d)

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Rubi [A]  time = 0.154827, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3791, 2633, 2635, 8} \[ -\frac{a^4 \sin ^7(c+d x)}{7 d}+\frac{9 a^4 \sin ^5(c+d x)}{5 d}-\frac{16 a^4 \sin ^3(c+d x)}{3 d}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{2 a^4 \sin (c+d x) \cos ^5(c+d x)}{3 d}+\frac{11 a^4 \sin (c+d x) \cos ^3(c+d x)}{6 d}+\frac{11 a^4 \sin (c+d x) \cos (c+d x)}{4 d}+\frac{11 a^4 x}{4} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^4,x]

[Out]

(11*a^4*x)/4 + (8*a^4*Sin[c + d*x])/d + (11*a^4*Cos[c + d*x]*Sin[c + d*x])/(4*d) + (11*a^4*Cos[c + d*x]^3*Sin[
c + d*x])/(6*d) + (2*a^4*Cos[c + d*x]^5*Sin[c + d*x])/(3*d) - (16*a^4*Sin[c + d*x]^3)/(3*d) + (9*a^4*Sin[c + d
*x]^5)/(5*d) - (a^4*Sin[c + d*x]^7)/(7*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 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 8

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

Rubi steps

\begin{align*} \int \cos ^7(c+d x) (a+a \sec (c+d x))^4 \, dx &=\int \left (a^4 \cos ^3(c+d x)+4 a^4 \cos ^4(c+d x)+6 a^4 \cos ^5(c+d x)+4 a^4 \cos ^6(c+d x)+a^4 \cos ^7(c+d x)\right ) \, dx\\ &=a^4 \int \cos ^3(c+d x) \, dx+a^4 \int \cos ^7(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^4(c+d x) \, dx+\left (4 a^4\right ) \int \cos ^6(c+d x) \, dx+\left (6 a^4\right ) \int \cos ^5(c+d x) \, dx\\ &=\frac{a^4 \cos ^3(c+d x) \sin (c+d x)}{d}+\frac{2 a^4 \cos ^5(c+d x) \sin (c+d x)}{3 d}+\left (3 a^4\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{3} \left (10 a^4\right ) \int \cos ^4(c+d x) \, dx-\frac{a^4 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{a^4 \operatorname{Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{d}-\frac{\left (6 a^4\right ) \operatorname{Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d}\\ &=\frac{8 a^4 \sin (c+d x)}{d}+\frac{3 a^4 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{11 a^4 \cos ^3(c+d x) \sin (c+d x)}{6 d}+\frac{2 a^4 \cos ^5(c+d x) \sin (c+d x)}{3 d}-\frac{16 a^4 \sin ^3(c+d x)}{3 d}+\frac{9 a^4 \sin ^5(c+d x)}{5 d}-\frac{a^4 \sin ^7(c+d x)}{7 d}+\frac{1}{2} \left (3 a^4\right ) \int 1 \, dx+\frac{1}{2} \left (5 a^4\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{3 a^4 x}{2}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{11 a^4 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{11 a^4 \cos ^3(c+d x) \sin (c+d x)}{6 d}+\frac{2 a^4 \cos ^5(c+d x) \sin (c+d x)}{3 d}-\frac{16 a^4 \sin ^3(c+d x)}{3 d}+\frac{9 a^4 \sin ^5(c+d x)}{5 d}-\frac{a^4 \sin ^7(c+d x)}{7 d}+\frac{1}{4} \left (5 a^4\right ) \int 1 \, dx\\ &=\frac{11 a^4 x}{4}+\frac{8 a^4 \sin (c+d x)}{d}+\frac{11 a^4 \cos (c+d x) \sin (c+d x)}{4 d}+\frac{11 a^4 \cos ^3(c+d x) \sin (c+d x)}{6 d}+\frac{2 a^4 \cos ^5(c+d x) \sin (c+d x)}{3 d}-\frac{16 a^4 \sin ^3(c+d x)}{3 d}+\frac{9 a^4 \sin ^5(c+d x)}{5 d}-\frac{a^4 \sin ^7(c+d x)}{7 d}\\ \end{align*}

Mathematica [A]  time = 0.26099, size = 83, normalized size = 0.56 \[ \frac{a^4 (33915 \sin (c+d x)+13020 \sin (2 (c+d x))+5495 \sin (3 (c+d x))+2100 \sin (4 (c+d x))+651 \sin (5 (c+d x))+140 \sin (6 (c+d x))+15 \sin (7 (c+d x))+18480 d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^4,x]

[Out]

(a^4*(18480*d*x + 33915*Sin[c + d*x] + 13020*Sin[2*(c + d*x)] + 5495*Sin[3*(c + d*x)] + 2100*Sin[4*(c + d*x)]
+ 651*Sin[5*(c + d*x)] + 140*Sin[6*(c + d*x)] + 15*Sin[7*(c + d*x)]))/(6720*d)

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Maple [A]  time = 0.128, size = 185, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{4}\sin \left ( dx+c \right ) }{7} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) }+4\,{a}^{4} \left ( 1/6\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) \sin \left ( dx+c \right ) +{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{6\,{a}^{4}\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+4\,{a}^{4} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{\frac{{a}^{4} \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)^7*(a+a*sec(d*x+c))^4,x)

[Out]

1/d*(1/7*a^4*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c)+4*a^4*(1/6*(cos(d*x+c)^5+5/4*cos
(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+6/5*a^4*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+
4*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)+1/3*a^4*(cos(d*x+c)^2+2)*sin(d*x+c))

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Maxima [A]  time = 1.16521, size = 252, normalized size = 1.71 \begin{align*} -\frac{48 \,{\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{4} - 672 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{4} + 35 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 560 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} - 210 \,{\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}}{1680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/1680*(48*(5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*a^4 - 672*(3*sin(d*x
+ c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^4 + 35*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4
*c) - 48*sin(2*d*x + 2*c))*a^4 + 560*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 - 210*(12*d*x + 12*c + sin(4*d*x +
4*c) + 8*sin(2*d*x + 2*c))*a^4)/d

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Fricas [A]  time = 1.73657, size = 267, normalized size = 1.82 \begin{align*} \frac{1155 \, a^{4} d x +{\left (60 \, a^{4} \cos \left (d x + c\right )^{6} + 280 \, a^{4} \cos \left (d x + c\right )^{5} + 576 \, a^{4} \cos \left (d x + c\right )^{4} + 770 \, a^{4} \cos \left (d x + c\right )^{3} + 908 \, a^{4} \cos \left (d x + c\right )^{2} + 1155 \, a^{4} \cos \left (d x + c\right ) + 1816 \, a^{4}\right )} \sin \left (d x + c\right )}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/420*(1155*a^4*d*x + (60*a^4*cos(d*x + c)^6 + 280*a^4*cos(d*x + c)^5 + 576*a^4*cos(d*x + c)^4 + 770*a^4*cos(d
*x + c)^3 + 908*a^4*cos(d*x + c)^2 + 1155*a^4*cos(d*x + c) + 1816*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)**7*(a+a*sec(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.2765, size = 194, normalized size = 1.32 \begin{align*} \frac{1155 \,{\left (d x + c\right )} a^{4} + \frac{2 \,{\left (1155 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 7700 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 21791 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 33792 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 31521 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 14700 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5565 \, 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 )}^{7}}}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/420*(1155*(d*x + c)*a^4 + 2*(1155*a^4*tan(1/2*d*x + 1/2*c)^13 + 7700*a^4*tan(1/2*d*x + 1/2*c)^11 + 21791*a^4
*tan(1/2*d*x + 1/2*c)^9 + 33792*a^4*tan(1/2*d*x + 1/2*c)^7 + 31521*a^4*tan(1/2*d*x + 1/2*c)^5 + 14700*a^4*tan(
1/2*d*x + 1/2*c)^3 + 5565*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d