3.482 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=115 \[ \frac{a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+2 a b x \left (a^2+2 b^2\right )+\frac{4 a^3 b \sin (c+d x) \cos (c+d x)}{3 d}+\frac{a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{b^4 \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

2*a*b*(a^2 + 2*b^2)*x + (b^4*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*a^2 + 17*b^2)*Sin[c + d*x])/(3*d) + (4*a^3*b*C
os[c + d*x]*Sin[c + d*x])/(3*d) + (a^2*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.239785, antiderivative size = 115, 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, 4074, 4047, 8, 4045, 3770} \[ \frac{a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+2 a b x \left (a^2+2 b^2\right )+\frac{4 a^3 b \sin (c+d x) \cos (c+d x)}{3 d}+\frac{a^2 \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{b^4 \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^4,x]

[Out]

2*a*b*(a^2 + 2*b^2)*x + (b^4*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*a^2 + 17*b^2)*Sin[c + d*x])/(3*d) + (4*a^3*b*C
os[c + d*x]*Sin[c + d*x])/(3*d) + (a^2*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*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 4074

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[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, 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 ^3(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac{1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (8 a^2 b+a \left (2 a^2+9 b^2\right ) \sec (c+d x)+3 b^3 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}-\frac{1}{6} \int \cos (c+d x) \left (-2 a^2 \left (2 a^2+17 b^2\right )-12 a b \left (a^2+2 b^2\right ) \sec (c+d x)-6 b^4 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}-\frac{1}{6} \int \cos (c+d x) \left (-2 a^2 \left (2 a^2+17 b^2\right )-6 b^4 \sec ^2(c+d x)\right ) \, dx+\left (2 a b \left (a^2+2 b^2\right )\right ) \int 1 \, dx\\ &=2 a b \left (a^2+2 b^2\right ) x+\frac{a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+\frac{4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+b^4 \int \sec (c+d x) \, dx\\ &=2 a b \left (a^2+2 b^2\right ) x+\frac{b^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \left (2 a^2+17 b^2\right ) \sin (c+d x)}{3 d}+\frac{4 a^3 b \cos (c+d x) \sin (c+d x)}{3 d}+\frac{a^2 \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.157782, size = 128, normalized size = 1.11 \[ \frac{24 a b \left (a^2+2 b^2\right ) (c+d x)+9 a^2 \left (a^2+8 b^2\right ) \sin (c+d x)+12 a^3 b \sin (2 (c+d x))+a^4 \sin (3 (c+d x))-12 b^4 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 b^4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^4,x]

[Out]

(24*a*b*(a^2 + 2*b^2)*(c + d*x) - 12*b^4*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*b^4*Log[Cos[(c + d*x)/2
] + Sin[(c + d*x)/2]] + 9*a^2*(a^2 + 8*b^2)*Sin[c + d*x] + 12*a^3*b*Sin[2*(c + d*x)] + a^4*Sin[3*(c + d*x)])/(
12*d)

________________________________________________________________________________________

Maple [A]  time = 0.053, size = 131, normalized size = 1.1 \begin{align*}{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}}+{\frac{2\,{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+2\,{\frac{{a}^{3}b\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{d}}+2\,{a}^{3}bx+2\,{\frac{{a}^{3}bc}{d}}+6\,{\frac{{a}^{2}{b}^{2}\sin \left ( dx+c \right ) }{d}}+4\,a{b}^{3}x+4\,{\frac{a{b}^{3}c}{d}}+{\frac{{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \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))^4,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.3234, size = 138, normalized size = 1.2 \begin{align*} -\frac{2 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4} - 6 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} a^{3} b - 24 \,{\left (d x + c\right )} a b^{3} - 3 \, b^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, a^{2} 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))^4,x, algorithm="maxima")

[Out]

-1/6*(2*(sin(d*x + c)^3 - 3*sin(d*x + c))*a^4 - 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*a^3*b - 24*(d*x + c)*a*b^3
- 3*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 36*a^2*b^2*sin(d*x + c))/d

________________________________________________________________________________________

Fricas [A]  time = 1.7455, size = 239, normalized size = 2.08 \begin{align*} \frac{3 \, b^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, b^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 12 \,{\left (a^{3} b + 2 \, a b^{3}\right )} d x + 2 \,{\left (a^{4} \cos \left (d x + c\right )^{2} + 6 \, a^{3} b \cos \left (d x + c\right ) + 2 \, a^{4} + 18 \, a^{2} b^{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+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/6*(3*b^4*log(sin(d*x + c) + 1) - 3*b^4*log(-sin(d*x + c) + 1) + 12*(a^3*b + 2*a*b^3)*d*x + 2*(a^4*cos(d*x +
c)^2 + 6*a^3*b*cos(d*x + c) + 2*a^4 + 18*a^2*b^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+b*sec(d*x+c))**4,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.33233, size = 286, normalized size = 2.49 \begin{align*} \frac{3 \, b^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, b^{4} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 6 \,{\left (a^{3} b + 2 \, a b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, a^{2} 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))^4,x, algorithm="giac")

[Out]

1/3*(3*b^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*b^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(a^3*b + 2*a*b^3)
*(d*x + c) + 2*(3*a^4*tan(1/2*d*x + 1/2*c)^5 - 6*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 18*a^2*b^2*tan(1/2*d*x + 1/2*c
)^5 + 2*a^4*tan(1/2*d*x + 1/2*c)^3 + 36*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 3*a^4*tan(1/2*d*x + 1/2*c) + 6*a^3*b*
tan(1/2*d*x + 1/2*c) + 18*a^2*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d