3.464 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=101 \[ \frac{\left (3 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (3 a^2+4 b^2\right )+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}+\frac{2 a b \sin (c+d x)}{d} \]

[Out]

((3*a^2 + 4*b^2)*x)/8 + (2*a*b*Sin[c + d*x])/d + ((3*a^2 + 4*b^2)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*Cos[
c + d*x]^3*Sin[c + d*x])/(4*d) - (2*a*b*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.0878394, antiderivative size = 101, 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, 2633, 4045, 2635, 8} \[ \frac{\left (3 a^2+4 b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (3 a^2+4 b^2\right )+\frac{a^2 \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac{2 a b \sin ^3(c+d x)}{3 d}+\frac{2 a b \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2,x]

[Out]

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

Mathematica [A]  time = 0.169573, size = 86, normalized size = 0.85 \[ \frac{24 \left (a^2+b^2\right ) \sin (2 (c+d x))+3 a^2 \sin (4 (c+d x))+36 a^2 c+36 a^2 d x-64 a b \sin ^3(c+d x)+192 a b \sin (c+d x)+48 b^2 c+48 b^2 d x}{96 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2,x]

[Out]

(36*a^2*c + 48*b^2*c + 36*a^2*d*x + 48*b^2*d*x + 192*a*b*Sin[c + d*x] - 64*a*b*Sin[c + d*x]^3 + 24*(a^2 + b^2)
*Sin[2*(c + d*x)] + 3*a^2*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.056, size = 89, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{2\,ab \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) }{3}}+{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\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+b*sec(d*x+c))^2,x)

[Out]

1/d*(a^2*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+2/3*a*b*(cos(d*x+c)^2+2)*sin(d*x+c)+b^2*
(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 1.0472, size = 111, normalized size = 1.1 \begin{align*} \frac{3 \,{\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^{2} - 64 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{2}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*a^2 - 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*a*b
+ 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*b^2)/d

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Fricas [A]  time = 1.64619, size = 184, normalized size = 1.82 \begin{align*} \frac{3 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} d x +{\left (6 \, a^{2} \cos \left (d x + c\right )^{3} + 16 \, a b \cos \left (d x + c\right )^{2} + 32 \, a b + 3 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/24*(3*(3*a^2 + 4*b^2)*d*x + (6*a^2*cos(d*x + c)^3 + 16*a*b*cos(d*x + c)^2 + 32*a*b + 3*(3*a^2 + 4*b^2)*cos(d
*x + c))*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)**4*(a+b*sec(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.3467, size = 302, normalized size = 2.99 \begin{align*} \frac{3 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )}{\left (d x + c\right )} - \frac{2 \,{\left (15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 80 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 48 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, 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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/24*(3*(3*a^2 + 4*b^2)*(d*x + c) - 2*(15*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*b^2*
tan(1/2*d*x + 1/2*c)^7 - 9*a^2*tan(1/2*d*x + 1/2*c)^5 - 80*a*b*tan(1/2*d*x + 1/2*c)^5 + 12*b^2*tan(1/2*d*x + 1
/2*c)^5 + 9*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*b^2*tan(1/2*d*x + 1/2*c)^3 - 15*a^
2*tan(1/2*d*x + 1/2*c) - 48*a*b*tan(1/2*d*x + 1/2*c) - 12*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 +
1)^4)/d