3.1.98 \(\int \cos ^3(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [98]

Optimal. Leaf size=110 \[ a^2 (A+2 C) x+\frac {a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a^2 (A+C) \sin (c+d x)}{d}+\frac {A \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A \cos (c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{3 d} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4172, 4102, 4081, 3855} \begin {gather*} \frac {a^2 (A+C) \sin (c+d x)}{d}+\frac {A \sin (c+d x) \cos (c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{3 d}+a^2 x (A+2 C)+\frac {a^2 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3855

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

Rule 4081

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(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) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rule 4102

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[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2
- b^2, 0] && GtQ[m, 1/2] && LtQ[n, -1]

Rule 4172

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/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps

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

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Mathematica [A]
time = 0.22, size = 109, normalized size = 0.99 \begin {gather*} \frac {a^2 \left (12 A d x+24 C d x-12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 (7 A+4 C) \sin (c+d x)+6 A \sin (2 (c+d x))+A \sin (3 (c+d x))\right )}{12 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A*d*x + 24*C*d*x - 12*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*C*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + 3*(7*A + 4*C)*Sin[c + d*x] + 6*A*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)]))/(12*d)

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Maple [A]
time = 0.57, size = 108, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {a^{2} A \sin \left (d x +c \right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} C \left (d x +c \right )+\frac {a^{2} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} C \sin \left (d x +c \right )}{d}\)	\(108\)
default	\(\frac {a^{2} A \sin \left (d x +c \right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} C \left (d x +c \right )+\frac {a^{2} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{2} C \sin \left (d x +c \right )}{d}\)	\(108\)
risch	\(a^{2} A x +2 a^{2} x C -\frac {7 i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{2 d}+\frac {7 i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {a^{2} A \sin \left (3 d x +3 c \right )}{12 d}+\frac {a^{2} A \sin \left (2 d x +2 c \right )}{2 d}\)	\(170\)
norman	\(\frac {\left (-a^{2} A -2 a^{2} C \right ) x +\left (-3 a^{2} A -6 a^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (a^{2} A +2 a^{2} C \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 a^{2} A +6 a^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 a^{2} \left (A -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {2 a^{2} \left (A +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (3 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} \left (19 A +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 \left (A +C \right ) a^{2} \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 \left (A +C \right ) a^{2} \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {a^{2} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(307\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 114, normalized size = 1.04 \begin {gather*} -\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^{2} - 12 \, {\left (d x + c\right )} C a^{2} - 3 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A a^{2} \sin \left (d x + c\right ) - 6 \, C a^{2} \sin \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 3.15, size = 95, normalized size = 0.86 \begin {gather*} \frac {6 \, {\left (A + 2 \, C\right )} a^{2} d x + 3 \, C a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, C a^{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^{2} \cos \left (d x + c\right ) + {\left (5 \, A + 3 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*(A + 2*C)*a^2*d*x + 3*C*a^2*log(sin(d*x + c) + 1) - 3*C*a^2*log(-sin(d*x + c) + 1) + 2*(A*a^2*cos(d*x +
c)^2 + 3*A*a^2*cos(d*x + c) + (5*A + 3*C)*a^2)*sin(d*x + c))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int A \cos ^{3}{\left (c + d x \right )}\, dx + \int 2 A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{3}{\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(A*cos(c + d*x)**3, x) + Integral(2*A*cos(c + d*x)**3*sec(c + d*x), x) + Integral(A*cos(c + d*x)
**3*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(2*C*cos(c + d*x)**3*sec(c
+ d*x)**3, x) + Integral(C*cos(c + d*x)**3*sec(c + d*x)**4, x))

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Giac [A]
time = 0.49, size = 179, normalized size = 1.63 \begin {gather*} \frac {3 \, C a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, C a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (A a^{2} + 2 \, C a^{2}\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} + 8 \, 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} + 9 \, 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 )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*C*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(A*a^2 + 2*C*
a^2)*(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 + 8*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 + 9*A*a^2*tan(1/2*d*x + 1/2*c) + 3*C*a^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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Mupad [B]
time = 2.68, size = 159, normalized size = 1.45 \begin {gather*} \frac {7\,A\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^2\,\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^2\,\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 {2\,C\,a^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 (2\,c+2\,d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \end {gather*}

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 + a/cos(c + d*x))^2,x)

[Out]

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

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