3.463 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=58 \[ \frac{\left (a^2+b^2\right ) \sin (c+d x)}{d}-\frac{a^2 \sin ^3(c+d x)}{3 d}+\frac{a b \sin (c+d x) \cos (c+d x)}{d}+a b x \]

[Out]

a*b*x + ((a^2 + b^2)*Sin[c + d*x])/d + (a*b*Cos[c + d*x]*Sin[c + d*x])/d - (a^2*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.089153, antiderivative size = 58, 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 = {3788, 2635, 8, 4044, 3013} \[ \frac{\left (a^2+b^2\right ) \sin (c+d x)}{d}-\frac{a^2 \sin ^3(c+d x)}{3 d}+\frac{a b \sin (c+d x) \cos (c+d x)}{d}+a b x \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2,x]

[Out]

a*b*x + ((a^2 + b^2)*Sin[c + d*x])/d + (a*b*Cos[c + d*x]*Sin[c + d*x])/d - (a^2*Sin[c + d*x]^3)/(3*d)

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), 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 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

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

Mathematica [A]  time = 0.168356, size = 59, normalized size = 1.02 \[ \frac{3 \left (3 a^2+4 b^2\right ) \sin (c+d x)+a (a \sin (3 (c+d x))+12 b (c+d x)+6 b \sin (2 (c+d x)))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2,x]

[Out]

(3*(3*a^2 + 4*b^2)*Sin[c + d*x] + a*(12*b*(c + d*x) + 6*b*Sin[2*(c + d*x)] + a*Sin[3*(c + d*x)]))/(12*d)

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Maple [A]  time = 0.051, size = 63, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) }{3}}+2\,ab \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{b}^{2}\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.0886, size = 81, normalized size = 1.4 \begin{align*} -\frac{2 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 6 \, b^{2} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/6*(2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^2 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*a*b - 6*b^2*sin(d*x + c))/d

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Fricas [A]  time = 1.6223, size = 124, normalized size = 2.14 \begin{align*} \frac{3 \, a b d x +{\left (a^{2} \cos \left (d x + c\right )^{2} + 3 \, a b \cos \left (d x + c\right ) + 2 \, a^{2} + 3 \, b^{2}\right )} \sin \left (d x + c\right )}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/3*(3*a*b*d*x + (a^2*cos(d*x + c)^2 + 3*a*b*cos(d*x + c) + 2*a^2 + 3*b^2)*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)**3*(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.24984, size = 207, normalized size = 3.57 \begin{align*} \frac{3 \,{\left (d x + c\right )} a b + \frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, b^{2} \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}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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