3.50 \(\int \frac{\cos ^3(c+d x)}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=94 \[ -\frac{4 \sin ^3(c+d x)}{3 a d}+\frac{4 \sin (c+d x)}{a d}-\frac{3 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{\sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac{3 x}{2 a} \]

[Out]

(-3*x)/(2*a) + (4*Sin[c + d*x])/(a*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - (Cos[c + d*x]^2*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 = 0.0903989, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3819, 3787, 2633, 2635, 8} \[ -\frac{4 \sin ^3(c+d x)}{3 a d}+\frac{4 \sin (c+d x)}{a d}-\frac{3 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{\sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac{3 x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + a*Sec[c + d*x]),x]

[Out]

(-3*x)/(2*a) + (4*Sin[c + d*x])/(a*d) - (3*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - (Cos[c + d*x]^2*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 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 \frac{\cos ^3(c+d x)}{a+a \sec (c+d x)} \, dx &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac{\int \cos ^3(c+d x) (-4 a+3 a \sec (c+d x)) \, dx}{a^2}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac{3 \int \cos ^2(c+d x) \, dx}{a}+\frac{4 \int \cos ^3(c+d x) \, dx}{a}\\ &=-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{\cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac{3 \int 1 \, dx}{2 a}-\frac{4 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a d}\\ &=-\frac{3 x}{2 a}+\frac{4 \sin (c+d x)}{a d}-\frac{3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{\cos ^2(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 = 0.310296, size = 143, normalized size = 1.52 \[ \frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \left (21 \sin \left (c+\frac{d x}{2}\right )+18 \sin \left (c+\frac{3 d x}{2}\right )+18 \sin \left (2 c+\frac{3 d x}{2}\right )-2 \sin \left (2 c+\frac{5 d x}{2}\right )-2 \sin \left (3 c+\frac{5 d x}{2}\right )+\sin \left (3 c+\frac{7 d x}{2}\right )+\sin \left (4 c+\frac{7 d x}{2}\right )-36 d x \cos \left (c+\frac{d x}{2}\right )+69 \sin \left (\frac{d x}{2}\right )-36 d x \cos \left (\frac{d x}{2}\right )\right )}{48 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + a*Sec[c + d*x]),x]

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]*(-36*d*x*Cos[(d*x)/2] - 36*d*x*Cos[c + (d*x)/2] + 69*Sin[(d*x)/2] + 21*Sin[c + (d*x
)/2] + 18*Sin[c + (3*d*x)/2] + 18*Sin[2*c + (3*d*x)/2] - 2*Sin[2*c + (5*d*x)/2] - 2*Sin[3*c + (5*d*x)/2] + Sin
[3*c + (7*d*x)/2] + Sin[4*c + (7*d*x)/2]))/(48*a*d)

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Maple [A]  time = 0.057, size = 136, normalized size = 1.5 \begin{align*}{\frac{1}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+5\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}+{\frac{16}{3\,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 ) ^{-3}}+3\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/d*tan(1/2*d*x+1/2*c)+5/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5+16/3/a/d/(1+tan(1/2*d*x+1/2*c)^
2)^3*tan(1/2*d*x+1/2*c)^3+3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)-3/a/d*arctan(tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.74326, size = 238, normalized size = 2.53 \begin{align*} \frac{\frac{\frac{9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac{3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac{9 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac{3 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x
+ c) + 1)^5)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(d*
x + c)^6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 3*sin(d*x + c)/(a*(cos(d*x + c)
+ 1)))/d

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Fricas [A]  time = 1.66955, size = 180, normalized size = 1.91 \begin{align*} -\frac{9 \, d x \cos \left (d x + c\right ) + 9 \, d x -{\left (2 \, \cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )^{2} + 7 \, \cos \left (d x + c\right ) + 16\right )} \sin \left (d x + c\right )}{6 \,{\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)^3/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(9*d*x*cos(d*x + c) + 9*d*x - (2*cos(d*x + c)^3 - cos(d*x + c)^2 + 7*cos(d*x + c) + 16)*sin(d*x + c))/(a*
d*cos(d*x + c) + a*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)**3/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.32232, size = 119, normalized size = 1.27 \begin{align*} -\frac{\frac{9 \,{\left (d x + c\right )}}{a} - \frac{6 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a} - \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 16 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, \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 )}^{3} a}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(9*(d*x + c)/a - 6*tan(1/2*d*x + 1/2*c)/a - 2*(15*tan(1/2*d*x + 1/2*c)^5 + 16*tan(1/2*d*x + 1/2*c)^3 + 9*
tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d