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

Optimal. Leaf size=104 \[ \frac{a^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 d}+\frac{b^2 \left (12 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+4 a^3 b x+\frac{3 a b^3 \tan (c+d x)}{d}+\frac{b^2 \sin (c+d x) (a+b \sec (c+d x))^2}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.21183, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {3842, 4076, 4047, 8, 4045, 3770} \[ \frac{a^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{2 d}+\frac{b^2 \left (12 a^2+b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+4 a^3 b x+\frac{3 a b^3 \tan (c+d x)}{d}+\frac{b^2 \sin (c+d x) (a+b \sec (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3842

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(b^2*
Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*(m + n - 1)), x] + Dist[1/(d*(m + n - 1)), In
t[(a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a*b^2*d*n + b*(b^2*d*(m + n - 2) +
3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2*d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] &&  !Integ
erQ[m])

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

Mathematica [B]  time = 0.520243, size = 280, normalized size = 2.69 \[ \frac{\sec ^2(c+d x) \left (\left (a^4+2 b^4\right ) \sin (c+d x)-12 a^2 b^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 a^2 b^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+b \cos (2 (c+d x)) \left (-b \left (12 a^2+b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+b \left (12 a^2+b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 a^3 (c+d x)\right )+8 a^3 b c+8 a^3 b d x+a^4 \sin (3 (c+d x))+8 a b^3 \sin (2 (c+d x))-b^4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+b^4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.048, size = 114, normalized size = 1.1 \begin{align*}{\frac{{a}^{4}\sin \left ( dx+c \right ) }{d}}+4\,{a}^{3}bx+4\,{\frac{{a}^{3}bc}{d}}+6\,{\frac{{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{a{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.34408, size = 155, normalized size = 1.49 \begin{align*} \frac{16 \,{\left (d x + c\right )} a^{3} b - b^{4}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{2} b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, a^{4} \sin \left (d x + c\right ) + 16 \, a b^{3} \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)*(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/4*(16*(d*x + c)*a^3*b - b^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) + 12*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*a^4*sin(d*x + c) + 16*a*b^3*tan(d*x + c
))/d

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Fricas [A]  time = 1.71756, size = 324, normalized size = 3.12 \begin{align*} \frac{16 \, a^{3} b d x \cos \left (d x + c\right )^{2} +{\left (12 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (12 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, a^{4} \cos \left (d x + c\right )^{2} + 8 \, a b^{3} \cos \left (d x + c\right ) + b^{4}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(16*a^3*b*d*x*cos(d*x + c)^2 + (12*a^2*b^2 + b^4)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (12*a^2*b^2 + b^4
)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*a^4*cos(d*x + c)^2 + 8*a*b^3*cos(d*x + c) + b^4)*sin(d*x + c))/
(d*cos(d*x + c)^2)

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

[Out]

Timed out

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Giac [A]  time = 1.28211, size = 242, normalized size = 2.33 \begin{align*} \frac{8 \,{\left (d x + c\right )} a^{3} b + \frac{4 \, a^{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 (12 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (12 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (8 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b^{4} \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)*(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

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