3.59 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=102 \[ \frac{2 a^2 (2 A+3 B) \sin (c+d x)}{3 d}+\frac{a^2 (2 A+3 B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{1}{2} a^2 x (2 A+3 B)+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

[Out]

(a^2*(2*A + 3*B)*x)/2 + (2*a^2*(2*A + 3*B)*Sin[c + d*x])/(3*d) + (a^2*(2*A + 3*B)*Cos[c + d*x]*Sin[c + d*x])/(
6*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.153287, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {4013, 3788, 2637, 4045, 8} \[ \frac{2 a^2 (2 A+3 B) \sin (c+d x)}{3 d}+\frac{a^2 (2 A+3 B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{1}{2} a^2 x (2 A+3 B)+\frac{A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(2*A + 3*B)*x)/2 + (2*a^2*(2*A + 3*B)*Sin[c + d*x])/(3*d) + (a^2*(2*A + 3*B)*Cos[c + d*x]*Sin[c + d*x])/(
6*d) + (A*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d)

Rule 4013

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

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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 8

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

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} (2 A+3 B) \int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \, dx\\ &=\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} (2 A+3 B) \int \cos ^2(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac{1}{3} \left (2 a^2 (2 A+3 B)\right ) \int \cos (c+d x) \, dx\\ &=\frac{2 a^2 (2 A+3 B) \sin (c+d x)}{3 d}+\frac{a^2 (2 A+3 B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{2} \left (a^2 (2 A+3 B)\right ) \int 1 \, dx\\ &=\frac{1}{2} a^2 (2 A+3 B) x+\frac{2 a^2 (2 A+3 B) \sin (c+d x)}{3 d}+\frac{a^2 (2 A+3 B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.171502, size = 61, normalized size = 0.6 \[ \frac{a^2 (3 (7 A+8 B) \sin (c+d x)+3 (2 A+B) \sin (2 (c+d x))+A \sin (3 (c+d x))+12 A d x+18 B d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A*d*x + 18*B*d*x + 3*(7*A + 8*B)*Sin[c + d*x] + 3*(2*A + B)*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)]))/(
12*d)

________________________________________________________________________________________

Maple [A]  time = 0.078, size = 116, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}A \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,{a}^{2}A \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +B{a}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +{a}^{2}A\sin \left ( dx+c \right ) +2\,B{a}^{2}\sin \left ( dx+c \right ) +B{a}^{2} \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+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.01302, size = 149, normalized size = 1.46 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 6 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 12 \,{\left (d x + c\right )} B a^{2} - 12 \, A a^{2} \sin \left (d x + c\right ) - 24 \, B a^{2} \sin \left (d x + c\right )}{12 \, 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))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 - 3*(2*d*x + 2*c +
sin(2*d*x + 2*c))*B*a^2 - 12*(d*x + c)*B*a^2 - 12*A*a^2*sin(d*x + c) - 24*B*a^2*sin(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 0.462819, size = 165, normalized size = 1.62 \begin{align*} \frac{3 \,{\left (2 \, A + 3 \, B\right )} a^{2} d x +{\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 2 \,{\left (5 \, A + 6 \, B\right )} a^{2}\right )} \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+a*sec(d*x+c))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*(2*A + 3*B)*a^2*d*x + (2*A*a^2*cos(d*x + c)^2 + 3*(2*A + B)*a^2*cos(d*x + c) + 2*(5*A + 6*B)*a^2)*sin(d
*x + c))/d

________________________________________________________________________________________

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))**2*(A+B*sec(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.40303, size = 192, normalized size = 1.88 \begin{align*} \frac{3 \,{\left (2 \, A a^{2} + 3 \, B a^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 16 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, B a^{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}}}{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))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(3*(2*A*a^2 + 3*B*a^2)*(d*x + c) + 2*(6*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 16
*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 18*A*a^2*tan(1/2*d*x + 1/2*c) + 15*B*a^2*tan
(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d