3.789 \(\int \cos ^3(c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=124 \[ \frac{1}{2} a x \left (a^2 B+6 a b C+6 b^2 B\right )+\frac{a^2 (a C+2 b B) \sin (c+d x)}{d}-\frac{b^2 (a B-2 b C) \tan (c+d x)}{2 d}+\frac{b^2 (3 a C+b B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d} \]

[Out]

(a*(a^2*B + 6*b^2*B + 6*a*b*C)*x)/2 + (b^2*(b*B + 3*a*C)*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*b*B + a*C)*Sin[c +
d*x])/d + (a*B*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - (b^2*(a*B - 2*b*C)*Tan[c + d*x])/(2*
d)

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Rubi [A]  time = 0.404287, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4025, 4076, 4047, 8, 4045, 3770} \[ \frac{1}{2} a x \left (a^2 B+6 a b C+6 b^2 B\right )+\frac{a^2 (a C+2 b B) \sin (c+d x)}{d}-\frac{b^2 (a B-2 b C) \tan (c+d x)}{2 d}+\frac{b^2 (3 a C+b B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a B \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(a^2*B + 6*b^2*B + 6*a*b*C)*x)/2 + (b^2*(b*B + 3*a*C)*ArcTanh[Sin[c + d*x]])/d + (a^2*(2*b*B + a*C)*Sin[c +
d*x])/d + (a*B*Cos[c + d*x]*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) - (b^2*(a*B - 2*b*C)*Tan[c + d*x])/(2*
d)

Rule 4072

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

Rule 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && 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 ^3(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^2(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \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 (-2 a (2 b B+a C)-\left (a^2 B+2 b^2 B+4 a b C\right ) \sec (c+d x)+b (a B-2 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{b^2 (a B-2 b C) \tan (c+d x)}{2 d}-\frac{1}{2} \int \cos (c+d x) \left (-2 a^2 (2 b B+a C)-a \left (a^2 B+6 b^2 B+6 a b C\right ) \sec (c+d x)-2 b^2 (b B+3 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{b^2 (a B-2 b C) \tan (c+d x)}{2 d}-\frac{1}{2} \int \cos (c+d x) \left (-2 a^2 (2 b B+a C)-2 b^2 (b B+3 a C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (a \left (a^2 B+6 b^2 B+6 a b C\right )\right ) \int 1 \, dx\\ &=\frac{1}{2} a \left (a^2 B+6 b^2 B+6 a b C\right ) x+\frac{a^2 (2 b B+a C) \sin (c+d x)}{d}+\frac{a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{b^2 (a B-2 b C) \tan (c+d x)}{2 d}+\left (b^2 (b B+3 a C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a \left (a^2 B+6 b^2 B+6 a b C\right ) x+\frac{b^2 (b B+3 a C) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 (2 b B+a C) \sin (c+d x)}{d}+\frac{a B \cos (c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}-\frac{b^2 (a B-2 b C) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.672071, size = 217, normalized size = 1.75 \[ \frac{2 a (c+d x) \left (a^2 B+6 a b C+6 b^2 B\right )+4 a^2 (a C+3 b B) \sin (c+d x)+a^3 B \sin (2 (c+d x))-4 b^2 (3 a C+b B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+4 b^2 (3 a C+b B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{4 b^3 C \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b^3 C \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(a^2*B + 6*b^2*B + 6*a*b*C)*(c + d*x) - 4*b^2*(b*B + 3*a*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4*
b^2*(b*B + 3*a*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (4*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]) + (4*b^3*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*a^2*(3*b*B + a*C)*Sin[c
+ d*x] + a^3*B*Sin[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.059, size = 168, normalized size = 1.4 \begin{align*}{\frac{B{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}Bx}{2}}+{\frac{B{a}^{3}c}{2\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) }{d}}+3\,{\frac{B{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,{a}^{2}bCx+3\,{\frac{C{a}^{2}bc}{d}}+3\,Ba{b}^{2}x+3\,{\frac{Ba{b}^{2}c}{d}}+3\,{\frac{Ca{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{C{b}^{3}\tan \left ( dx+c \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))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.997251, size = 194, normalized size = 1.56 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 12 \,{\left (d x + c\right )} C a^{2} b + 12 \,{\left (d x + c\right )} B a b^{2} + 6 \, C a b^{2}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{3} \sin \left (d x + c\right ) + 12 \, B a^{2} b \sin \left (d x + c\right ) + 4 \, C 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)^3*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 + 12*(d*x + c)*C*a^2*b + 12*(d*x + c)*B*a*b^2 + 6*C*a*b^2*(log(sin
(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*B*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*C*a^3*si
n(d*x + c) + 12*B*a^2*b*sin(d*x + c) + 4*C*b^3*tan(d*x + c))/d

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Fricas [A]  time = 0.550354, size = 369, normalized size = 2.98 \begin{align*} \frac{{\left (B a^{3} + 6 \, C a^{2} b + 6 \, B a b^{2}\right )} d x \cos \left (d x + c\right ) +{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (B a^{3} \cos \left (d x + c\right )^{2} + 2 \, C b^{3} + 2 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\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)^3*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.22009, size = 316, normalized size = 2.55 \begin{align*} -\frac{\frac{4 \, C b^{3} \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 (B a^{3} + 6 \, C a^{2} b + 6 \, B a b^{2}\right )}{\left (d x + c\right )} - 2 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 2 \,{\left (3 \, C a b^{2} + B b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, B a^{2} 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)^3*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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