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

Optimal. Leaf size=182 \[ \frac{b \left (a^2 (4 A+6 C)+A b^2\right ) \sin (c+d x)}{2 d}+\frac{a \left (a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a x \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right )+\frac{A b \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{b^3 C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

(a*(12*b^2*(A + 2*C) + a^2*(3*A + 4*C))*x)/8 + (b^3*C*ArcTanh[Sin[c + d*x]])/d + (b*(A*b^2 + a^2*(4*A + 6*C))*
Sin[c + d*x])/(2*d) + (a*(2*A*b^2 + a^2*(3*A + 4*C))*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (A*b*Cos[c + d*x]^2*(a
+ b*Sec[c + d*x])^2*Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*d)

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Rubi [A]  time = 0.558659, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4095, 4094, 4074, 4047, 8, 4045, 3770} \[ \frac{b \left (a^2 (4 A+6 C)+A b^2\right ) \sin (c+d x)}{2 d}+\frac{a \left (a^2 (3 A+4 C)+2 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} a x \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right )+\frac{A b \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{4 d}+\frac{A \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{b^3 C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(12*b^2*(A + 2*C) + a^2*(3*A + 4*C))*x)/8 + (b^3*C*ArcTanh[Sin[c + d*x]])/d + (b*(A*b^2 + a^2*(4*A + 6*C))*
Sin[c + d*x])/(2*d) + (a*(2*A*b^2 + a^2*(3*A + 4*C))*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (A*b*Cos[c + d*x]^2*(a
+ b*Sec[c + d*x])^2*Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(4*d)

Rule 4095

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*(C*n + A*(n + 1))*Csc[e +
f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4094

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_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && 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 (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x))^2 \left (3 A b+a (3 A+4 C) \sec (c+d x)+4 b C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac{A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac{1}{12} \int \cos ^2(c+d x) (a+b \sec (c+d x)) \left (3 \left (2 A b^2+a^2 (3 A+4 C)\right )+3 a b (5 A+8 C) \sec (c+d x)+12 b^2 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac{A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac{1}{24} \int \cos (c+d x) \left (-12 b \left (A b^2+a^2 (4 A+6 C)\right )-3 a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \sec (c+d x)-24 b^3 C \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac{A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}-\frac{1}{24} \int \cos (c+d x) \left (-12 b \left (A b^2+a^2 (4 A+6 C)\right )-24 b^3 C \sec ^2(c+d x)\right ) \, dx+\frac{1}{8} \left (a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right )\right ) \int 1 \, dx\\ &=\frac{1}{8} a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac{b \left (A b^2+a^2 (4 A+6 C)\right ) \sin (c+d x)}{2 d}+\frac{a \left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac{A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}+\left (b^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{8} a \left (12 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) x+\frac{b^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac{b \left (A b^2+a^2 (4 A+6 C)\right ) \sin (c+d x)}{2 d}+\frac{a \left (2 A b^2+a^2 (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{A b \cos ^2(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{4 d}+\frac{A \cos ^3(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.55403, size = 177, normalized size = 0.97 \[ \frac{4 a (c+d x) \left (a^2 (3 A+4 C)+12 b^2 (A+2 C)\right )+8 a \left (a^2 (A+C)+3 A b^2\right ) \sin (2 (c+d x))+8 b \left (3 a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x)+8 a^2 A b \sin (3 (c+d x))+a^3 A \sin (4 (c+d x))-32 b^3 C \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+32 b^3 C \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(12*b^2*(A + 2*C) + a^2*(3*A + 4*C))*(c + d*x) - 32*b^3*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*b
^3*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*b*(4*A*b^2 + 3*a^2*(3*A + 4*C))*Sin[c + d*x] + 8*a*(3*A*b^2
+ a^2*(A + C))*Sin[2*(c + d*x)] + 8*a^2*A*b*Sin[3*(c + d*x)] + a^3*A*Sin[4*(c + d*x)])/(32*d)

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Maple [A]  time = 0.073, size = 252, normalized size = 1.4 \begin{align*}{\frac{A{a}^{3}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,A{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{3}Ax}{8}}+{\frac{3\,A{a}^{3}c}{8\,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{A \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ){a}^{2}b}{d}}+2\,{\frac{A{a}^{2}b\sin \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}bC\sin \left ( dx+c \right ) }{d}}+{\frac{3\,Aa{b}^{2}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{3\,Aa{b}^{2}x}{2}}+{\frac{3\,Aa{b}^{2}c}{2\,d}}+3\,Ca{b}^{2}x+3\,{\frac{Ca{b}^{2}c}{d}}+{\frac{A{b}^{3}\sin \left ( dx+c \right ) }{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*(A+C*sec(d*x+c)^2),x)

[Out]

1/4/d*A*a^3*sin(d*x+c)*cos(d*x+c)^3+3/8/d*A*a^3*sin(d*x+c)*cos(d*x+c)+3/8*a^3*A*x+3/8/d*A*a^3*c+1/2/d*a^3*C*si
n(d*x+c)*cos(d*x+c)+1/2*a^3*C*x+1/2/d*C*a^3*c+1/d*A*cos(d*x+c)^2*sin(d*x+c)*a^2*b+2/d*A*a^2*b*sin(d*x+c)+3/d*a
^2*b*C*sin(d*x+c)+3/2/d*A*a*b^2*sin(d*x+c)*cos(d*x+c)+3/2*A*a*b^2*x+3/2/d*A*a*b^2*c+3*C*a*b^2*x+3/d*C*a*b^2*c+
1/d*A*b^3*sin(d*x+c)+1/d*C*b^3*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.993957, size = 235, normalized size = 1.29 \begin{align*} \frac{{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 8 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} b + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} + 96 \,{\left (d x + c\right )} C a b^{2} + 16 \, 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 )} + 96 \, C a^{2} b \sin \left (d x + c\right ) + 32 \, A b^{3} \sin \left (d x + c\right )}{32 \, 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*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/32*((12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3 + 8*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3
- 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2*b + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a*b^2 + 96*(d*x + c)*C
*a*b^2 + 16*C*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 96*C*a^2*b*sin(d*x + c) + 32*A*b^3*sin(d*x
+ c))/d

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Fricas [A]  time = 0.554237, size = 354, normalized size = 1.95 \begin{align*} \frac{4 \, C b^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, C b^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \,{\left (A + 2 \, C\right )} a b^{2}\right )} d x +{\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \, A a^{2} b \cos \left (d x + c\right )^{2} + 8 \,{\left (2 \, A + 3 \, C\right )} a^{2} b + 8 \, A b^{3} +{\left ({\left (3 \, A + 4 \, C\right )} a^{3} + 12 \, A a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, 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*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.26949, size = 679, normalized size = 3.73 \begin{align*} \text{result too large to display} \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*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/8*(8*C*b^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*C*b^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (3*A*a^3 + 4*C*
a^3 + 12*A*a*b^2 + 24*C*a*b^2)*(d*x + c) - 2*(5*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 4*C*a^3*tan(1/2*d*x + 1/2*c)^7
- 24*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 -
8*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 4*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*A*a^2*b*
tan(1/2*d*x + 1/2*c)^5 - 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 24*A*b^3*tan(
1/2*d*x + 1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 4*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*A*a^2*b*tan(1/2*d*x
+ 1/2*c)^3 - 72*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*A*b^3*tan(1/2*d*x + 1/
2*c)^3 - 5*A*a^3*tan(1/2*d*x + 1/2*c) - 4*C*a^3*tan(1/2*d*x + 1/2*c) - 24*A*a^2*b*tan(1/2*d*x + 1/2*c) - 24*C*
a^2*b*tan(1/2*d*x + 1/2*c) - 12*A*a*b^2*tan(1/2*d*x + 1/2*c) - 8*A*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/
2*c)^2 + 1)^4)/d