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

Optimal. Leaf size=108 \[ -\frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{2 d}+\frac{1}{2} a^2 x \left (a^2+12 b^2\right )+\frac{3 a^3 b \sin (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{4 a b^3 \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(a^2*(a^2 + 12*b^2)*x)/2 + (4*a*b^3*ArcTanh[Sin[c + d*x]])/d + (3*a^3*b*Sin[c + d*x])/d + (a^2*Cos[c + d*x]*(a
+ b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - (b^2*(a^2 - 2*b^2)*Tan[c + d*x])/(2*d)

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Rubi [A]  time = 0.217404, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3841, 4076, 4047, 8, 4045, 3770} \[ -\frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{2 d}+\frac{1}{2} a^2 x \left (a^2+12 b^2\right )+\frac{3 a^3 b \sin (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d}+\frac{4 a b^3 \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a^2 + 12*b^2)*x)/2 + (4*a*b^3*ArcTanh[Sin[c + d*x]])/d + (3*a^3*b*Sin[c + d*x])/d + (a^2*Cos[c + d*x]*(a
+ b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - (b^2*(a^2 - 2*b^2)*Tan[c + d*x])/(2*d)

Rule 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4076

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

Rule 4047

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac{a^2 \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+b \sec (c+d x)) \left (6 a^2 b+a \left (a^2+6 b^2\right ) \sec (c+d x)-b \left (a^2-2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) \left (6 a^3 b+a^2 \left (a^2+12 b^2\right ) \sec (c+d x)+8 a b^3 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) \left (6 a^3 b+8 a b^3 \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (a^2 \left (a^2+12 b^2\right )\right ) \int 1 \, dx\\ &=\frac{1}{2} a^2 \left (a^2+12 b^2\right ) x+\frac{3 a^3 b \sin (c+d x)}{d}+\frac{a^2 \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{2 d}+\left (4 a b^3\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^2 \left (a^2+12 b^2\right ) x+\frac{4 a b^3 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{3 a^3 b \sin (c+d x)}{d}+\frac{a^2 \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{b^2 \left (a^2-2 b^2\right ) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.670839, size = 119, normalized size = 1.1 \[ \frac{2 a \left (a \left (a^2+12 b^2\right ) (c+d x)-8 b^3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+8 b^3 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+16 a^3 b \sin (c+d x)+a^4 \sin (2 (c+d x))+4 b^4 \tan (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(a*(a^2 + 12*b^2)*(c + d*x) - 8*b^3*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*b^3*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) + 16*a^3*b*Sin[c + d*x] + a^4*Sin[2*(c + d*x)] + 4*b^4*Tan[c + d*x])/(4*d)

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Maple [A]  time = 0.05, size = 109, normalized size = 1. \begin{align*}{\frac{{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}x}{2}}+{\frac{{a}^{4}c}{2\,d}}+4\,{\frac{{a}^{3}b\sin \left ( dx+c \right ) }{d}}+6\,{a}^{2}{b}^{2}x+6\,{\frac{{a}^{2}{b}^{2}c}{d}}+4\,{\frac{a{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{4}\tan \left ( dx+c \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*sec(d*x+c))^4,x)

[Out]

1/2*a^4*cos(d*x+c)*sin(d*x+c)/d+1/2*a^4*x+1/2/d*a^4*c+4*a^3*b*sin(d*x+c)/d+6*a^2*b^2*x+6/d*a^2*b^2*c+4/d*a*b^3
*ln(sec(d*x+c)+tan(d*x+c))+1/d*b^4*tan(d*x+c)

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Maxima [A]  time = 1.4042, size = 122, normalized size = 1.13 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} + 24 \,{\left (d x + c\right )} a^{2} b^{2} + 8 \, a b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 16 \, a^{3} b \sin \left (d x + c\right ) + 4 \, b^{4} \tan \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*a^4 + 24*(d*x + c)*a^2*b^2 + 8*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*
x + c) - 1)) + 16*a^3*b*sin(d*x + c) + 4*b^4*tan(d*x + c))/d

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Fricas [A]  time = 1.7507, size = 294, normalized size = 2.72 \begin{align*} \frac{4 \, a b^{3} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, a b^{3} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (a^{4} + 12 \, a^{2} b^{2}\right )} d x \cos \left (d x + c\right ) +{\left (a^{4} \cos \left (d x + c\right )^{2} + 8 \, a^{3} b \cos \left (d x + c\right ) + 2 \, b^{4}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*a*b^3*cos(d*x + c)*log(sin(d*x + c) + 1) - 4*a*b^3*cos(d*x + c)*log(-sin(d*x + c) + 1) + (a^4 + 12*a^2*
b^2)*d*x*cos(d*x + c) + (a^4*cos(d*x + c)^2 + 8*a^3*b*cos(d*x + c) + 2*b^4)*sin(d*x + c))/(d*cos(d*x + c))

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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)**2*(a+b*sec(d*x+c))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.32051, size = 230, normalized size = 2.13 \begin{align*} \frac{8 \, a b^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, a b^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} +{\left (a^{4} + 12 \, a^{2} b^{2}\right )}{\left (d x + c\right )} - \frac{2 \,{\left (a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, a^{3} b \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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(8*a*b^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*a*b^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 4*b^4*tan(1/2*d
*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + (a^4 + 12*a^2*b^2)*(d*x + c) - 2*(a^4*tan(1/2*d*x + 1/2*c)^3 - 8*a^
3*b*tan(1/2*d*x + 1/2*c)^3 - a^4*tan(1/2*d*x + 1/2*c) - 8*a^3*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2
+ 1)^2)/d