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

Optimal. Leaf size=112 \[ \frac {\left (a^2 (2 A+3 C)+2 A b^2\right ) \sin (c+d x)}{3 d}+\frac {a A b \sin (c+d x) \cos (c+d x)}{3 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+a b x (A+2 C)+\frac {b^2 C \tanh ^{-1}(\sin (c+d x))}{d} \]

[Out]

a*b*(A+2*C)*x+b^2*C*arctanh(sin(d*x+c))/d+1/3*(2*A*b^2+a^2*(2*A+3*C))*sin(d*x+c)/d+1/3*a*A*b*cos(d*x+c)*sin(d*
x+c)/d+1/3*A*cos(d*x+c)^2*(a+b*sec(d*x+c))^2*sin(d*x+c)/d

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Rubi [A]  time = 0.29, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4095, 4074, 4047, 8, 4045, 3770} \[ \frac {\left (a^2 (2 A+3 C)+2 A b^2\right ) \sin (c+d x)}{3 d}+\frac {a A b \sin (c+d x) \cos (c+d x)}{3 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^2}{3 d}+a b x (A+2 C)+\frac {b^2 C \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.25, size = 144, normalized size = 1.29 \[ \frac {3 \left (a^2 (3 A+4 C)+4 A b^2\right ) \sin (c+d x)+a^2 A \sin (3 (c+d x))+6 a A b \sin (2 (c+d x))+12 a A b c+12 a A b d x+24 a b c C+24 a b C d x-12 b^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 b^2 C \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*a*A*b*c + 24*a*b*c*C + 12*a*A*b*d*x + 24*a*b*C*d*x - 12*b^2*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 1
2*b^2*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 3*(4*A*b^2 + a^2*(3*A + 4*C))*Sin[c + d*x] + 6*a*A*b*Sin[2*
(c + d*x)] + a^2*A*Sin[3*(c + d*x)])/(12*d)

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fricas [A]  time = 0.50, size = 99, normalized size = 0.88 \[ \frac {6 \, {\left (A + 2 \, C\right )} a b d x + 3 \, C b^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C b^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{2} \cos \left (d x + c\right )^{2} + 3 \, A a b \cos \left (d x + c\right ) + {\left (2 \, A + 3 \, C\right )} a^{2} + 3 \, A b^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.26, size = 256, normalized size = 2.29 \[ \frac {3 \, C b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, C b^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (A a b + 2 \, C a b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.96, size = 137, normalized size = 1.22 \[ \frac {a^{2} A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {2 a^{2} A \sin \left (d x +c \right )}{3 d}+\frac {a^{2} C \sin \left (d x +c \right )}{d}+\frac {a A b \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d}+a A b x +\frac {A a b c}{d}+2 a b C x +\frac {2 C a b c}{d}+\frac {A \,b^{2} \sin \left (d x +c \right )}{d}+\frac {b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.35, size = 112, normalized size = 1.00 \[ -\frac {2 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 12 \, {\left (d x + c\right )} C a b - 3 \, C b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, C a^{2} \sin \left (d x + c\right ) - 6 \, A b^{2} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.81, size = 170, normalized size = 1.52 \[ \frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,C\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {2\,A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^2,x)

[Out]

(3*A*a^2*sin(c + d*x))/(4*d) + (A*b^2*sin(c + d*x))/d + (C*a^2*sin(c + d*x))/d + (2*C*b^2*atanh(sin(c/2 + (d*x
)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^2*sin(3*c + 3*d*x))/(12*d) + (2*A*a*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*
x)/2)))/d + (4*C*a*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a*b*sin(2*c + 2*d*x))/(2*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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