3.790 \(\int \cos ^4(c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=145 \[ \frac{a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (3 a^2 b B+a^3 C+6 a b^2 C+2 b^3 B\right )+\frac{a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{b^3 C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

((3*a^2*b*B + 2*b^3*B + a^3*C + 6*a*b^2*C)*x)/2 + (b^3*C*ArcTanh[Sin[c + d*x]])/d + (a*(2*a^2*B + 8*b^2*B + 9*
a*b*C)*Sin[c + d*x])/(3*d) + (a^2*(5*b*B + 3*a*C)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (a*B*Cos[c + d*x]^2*(a +
b*Sec[c + d*x])^2*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.4148, antiderivative size = 145, 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, 4074, 4047, 8, 4045, 3770} \[ \frac{a \left (2 a^2 B+9 a b C+8 b^2 B\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (3 a^2 b B+a^3 C+6 a b^2 C+2 b^3 B\right )+\frac{a^2 (3 a C+5 b B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{a B \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac{b^3 C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

((3*a^2*b*B + 2*b^3*B + a^3*C + 6*a*b^2*C)*x)/2 + (b^3*C*ArcTanh[Sin[c + d*x]])/d + (a*(2*a^2*B + 8*b^2*B + 9*
a*b*C)*Sin[c + d*x])/(3*d) + (a^2*(5*b*B + 3*a*C)*Cos[c + d*x]*Sin[c + d*x])/(6*d) + (a*B*Cos[c + d*x]^2*(a +
b*Sec[c + d*x])^2*Sin[c + d*x])/(3*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 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 ^4(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 ^3(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{a B \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 (-a (5 b B+3 a C)-\left (2 a^2 B+3 b^2 B+6 a b C\right ) \sec (c+d x)-3 b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 (5 b B+3 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a B \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 \left (2 a^2 B+8 b^2 B+9 a b C\right )+3 \left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) \sec (c+d x)+6 b^3 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 (5 b B+3 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a B \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 \left (2 a^2 B+8 b^2 B+9 a b C\right )+6 b^3 C \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) \int 1 \, dx\\ &=\frac{1}{2} \left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) x+\frac{a \left (2 a^2 B+8 b^2 B+9 a b C\right ) \sin (c+d x)}{3 d}+\frac{a^2 (5 b B+3 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a B \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\left (b^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} \left (3 a^2 b B+2 b^3 B+a^3 C+6 a b^2 C\right ) x+\frac{b^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a \left (2 a^2 B+8 b^2 B+9 a b C\right ) \sin (c+d x)}{3 d}+\frac{a^2 (5 b B+3 a C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac{a B \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.368836, size = 159, normalized size = 1.1 \[ \frac{6 (c+d x) \left (3 a^2 b B+a^3 C+6 a b^2 C+2 b^3 B\right )+9 a \left (a^2 B+4 a b C+4 b^2 B\right ) \sin (c+d x)+3 a^2 (a C+3 b B) \sin (2 (c+d x))+a^3 B \sin (3 (c+d x))-12 b^3 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+12 b^3 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(6*(3*a^2*b*B + 2*b^3*B + a^3*C + 6*a*b^2*C)*(c + d*x) - 12*b^3*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 1
2*b^3*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*a*(a^2*B + 4*b^2*B + 4*a*b*C)*Sin[c + d*x] + 3*a^2*(3*b*B
+ a*C)*Sin[2*(c + d*x)] + a^3*B*Sin[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.064, size = 207, normalized size = 1.4 \begin{align*}{\frac{B \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{3}}{3\,d}}+{\frac{2\,B{a}^{3}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}Cx}{2}}+{\frac{{a}^{3}Cc}{2\,d}}+{\frac{3\,B{a}^{2}b\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{3\,B{a}^{2}bx}{2}}+{\frac{3\,B{a}^{2}bc}{2\,d}}+3\,{\frac{{a}^{2}bC\sin \left ( dx+c \right ) }{d}}+3\,{\frac{Ba{b}^{2}\sin \left ( dx+c \right ) }{d}}+3\,Ca{b}^{2}x+3\,{\frac{Ca{b}^{2}c}{d}}+B{b}^{3}x+{\frac{B{b}^{3}c}{d}}+{\frac{C{b}^{3}\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)^4*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.973428, size = 205, normalized size = 1.41 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 9 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} b - 36 \,{\left (d x + c\right )} C a b^{2} - 12 \,{\left (d x + c\right )} B b^{3} - 6 \, C b^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{2} b \sin \left (d x + c\right ) - 36 \, B a b^{2} \sin \left (d x + c\right )}{12 \, 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))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 - 9*(2*d*x + 2*c +
sin(2*d*x + 2*c))*B*a^2*b - 36*(d*x + c)*C*a*b^2 - 12*(d*x + c)*B*b^3 - 6*C*b^3*(log(sin(d*x + c) + 1) - log(
sin(d*x + c) - 1)) - 36*C*a^2*b*sin(d*x + c) - 36*B*a*b^2*sin(d*x + c))/d

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Fricas [A]  time = 0.544111, size = 317, normalized size = 2.19 \begin{align*} \frac{3 \, C b^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C b^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )} d x +{\left (2 \, B a^{3} \cos \left (d x + c\right )^{2} + 4 \, B a^{3} + 18 \, C a^{2} b + 18 \, B a b^{2} + 3 \,{\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\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)^4*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.27011, size = 424, normalized size = 2.92 \begin{align*} \frac{6 \, C b^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, C b^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + 3 \,{\left (C a^{3} + 3 \, B a^{2} b + 6 \, C a b^{2} + 2 \, B b^{3}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 36 \, B a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, B a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, C a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 18 \, B a 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}}}{6 \, 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))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(6*C*b^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*C*b^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(C*a^3 + 3*B*
a^2*b + 6*C*a*b^2 + 2*B*b^3)*(d*x + c) + 2*(6*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^3*tan(1/2*d*x + 1/2*c)^5 -
9*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 4*B
*a^3*tan(1/2*d*x + 1/2*c)^3 + 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^3*
tan(1/2*d*x + 1/2*c) + 3*C*a^3*tan(1/2*d*x + 1/2*c) + 9*B*a^2*b*tan(1/2*d*x + 1/2*c) + 18*C*a^2*b*tan(1/2*d*x
+ 1/2*c) + 18*B*a*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d