3.51 \(\int \frac{\cos ^4(c+d x)}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=118 \[ \frac{4 \sin ^3(c+d x)}{3 a d}-\frac{4 \sin (c+d x)}{a d}+\frac{5 \sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac{15 \sin (c+d x) \cos (c+d x)}{8 a d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac{15 x}{8 a} \]

[Out]

(15*x)/(8*a) - (4*Sin[c + d*x])/(a*d) + (15*Cos[c + d*x]*Sin[c + d*x])/(8*a*d) + (5*Cos[c + d*x]^3*Sin[c + d*x
])/(4*a*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + (4*Sin[c + d*x]^3)/(3*a*d)

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Rubi [A]  time = 0.101267, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3819, 3787, 2635, 8, 2633} \[ \frac{4 \sin ^3(c+d x)}{3 a d}-\frac{4 \sin (c+d x)}{a d}+\frac{5 \sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac{15 \sin (c+d x) \cos (c+d x)}{8 a d}-\frac{\sin (c+d x) \cos ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac{15 x}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*x)/(8*a) - (4*Sin[c + d*x])/(a*d) + (15*Cos[c + d*x]*Sin[c + d*x])/(8*a*d) + (5*Cos[c + d*x]^3*Sin[c + d*x
])/(4*a*d) - (Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) + (4*Sin[c + d*x]^3)/(3*a*d)

Rule 3819

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(f*(a + b*Csc[e + f*x])), x] - Dist[1/a^2, Int[(d*Csc[e + f*x])^n*(a*(n - 1) - b*n*Csc[
e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

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 \frac{\cos ^4(c+d x)}{a+a \sec (c+d x)} \, dx &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac{\int \cos ^4(c+d x) (-5 a+4 a \sec (c+d x)) \, dx}{a^2}\\ &=-\frac{\cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac{4 \int \cos ^3(c+d x) \, dx}{a}+\frac{5 \int \cos ^4(c+d x) \, dx}{a}\\ &=\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{15 \int \cos ^2(c+d x) \, dx}{4 a}+\frac{4 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a d}\\ &=-\frac{4 \sin (c+d x)}{a d}+\frac{15 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{4 \sin ^3(c+d x)}{3 a d}+\frac{15 \int 1 \, dx}{8 a}\\ &=\frac{15 x}{8 a}-\frac{4 \sin (c+d x)}{a d}+\frac{15 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac{5 \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{\cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{4 \sin ^3(c+d x)}{3 a d}\\ \end{align*}

Mathematica [A]  time = 0.309493, size = 173, normalized size = 1.47 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (-168 \sin \left (c+\frac{d x}{2}\right )-120 \sin \left (c+\frac{3 d x}{2}\right )-120 \sin \left (2 c+\frac{3 d x}{2}\right )+40 \sin \left (2 c+\frac{5 d x}{2}\right )+40 \sin \left (3 c+\frac{5 d x}{2}\right )-5 \sin \left (3 c+\frac{7 d x}{2}\right )-5 \sin \left (4 c+\frac{7 d x}{2}\right )+3 \sin \left (4 c+\frac{9 d x}{2}\right )+3 \sin \left (5 c+\frac{9 d x}{2}\right )+360 d x \cos \left (c+\frac{d x}{2}\right )-552 \sin \left (\frac{d x}{2}\right )+360 d x \cos \left (\frac{d x}{2}\right )\right )}{384 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]*(360*d*x*Cos[(d*x)/2] + 360*d*x*Cos[c + (d*x)/2] - 552*Sin[(d*x)/2] - 168*Sin[c + (
d*x)/2] - 120*Sin[c + (3*d*x)/2] - 120*Sin[2*c + (3*d*x)/2] + 40*Sin[2*c + (5*d*x)/2] + 40*Sin[3*c + (5*d*x)/2
] - 5*Sin[3*c + (7*d*x)/2] - 5*Sin[4*c + (7*d*x)/2] + 3*Sin[4*c + (9*d*x)/2] + 3*Sin[5*c + (9*d*x)/2]))/(384*a
*d)

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Maple [A]  time = 0.054, size = 171, normalized size = 1.5 \begin{align*} -{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{25}{4\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-{\frac{115}{12\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-{\frac{109}{12\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}-{\frac{7}{4\,da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) ^{-4}}+{\frac{15}{4\,da}\arctan \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \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)),x)

[Out]

-1/a/d*tan(1/2*d*x+1/2*c)-25/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7-115/12/a/d/(1+tan(1/2*d*x+1
/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5-109/12/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3-7/4/a/d/(1+tan(1/2*
d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)+15/4/a/d*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.69611, size = 293, normalized size = 2.48 \begin{align*} -\frac{\frac{\frac{21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{109 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{115 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{75 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a + \frac{4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}} - \frac{45 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{12 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{12 \, 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]

-1/12*((21*sin(d*x + c)/(cos(d*x + c) + 1) + 109*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 115*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5 + 75*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a*
sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8) - 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 12*sin(d*x + c)/(a*(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.67651, size = 213, normalized size = 1.81 \begin{align*} \frac{45 \, d x \cos \left (d x + c\right ) + 45 \, d x +{\left (6 \, \cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{3} + 13 \, \cos \left (d x + c\right )^{2} - 19 \, \cos \left (d x + c\right ) - 64\right )} \sin \left (d x + c\right )}{24 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \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*(45*d*x*cos(d*x + c) + 45*d*x + (6*cos(d*x + c)^4 - 2*cos(d*x + c)^3 + 13*cos(d*x + c)^2 - 19*cos(d*x + c
) - 64)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\cos ^{4}{\left (c + d x \right )}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \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]

Integral(cos(c + d*x)**4/(sec(c + d*x) + 1), x)/a

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Giac [A]  time = 1.30793, size = 136, normalized size = 1.15 \begin{align*} \frac{\frac{45 \,{\left (d x + c\right )}}{a} - \frac{24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{2 \,{\left (75 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 115 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 109 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, \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} a}}{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*(45*(d*x + c)/a - 24*tan(1/2*d*x + 1/2*c)/a - 2*(75*tan(1/2*d*x + 1/2*c)^7 + 115*tan(1/2*d*x + 1/2*c)^5 +
109*tan(1/2*d*x + 1/2*c)^3 + 21*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d