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

Optimal. Leaf size=171 \[ \frac{a^3 (2 A+6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(A-2 B-4 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+\frac{1}{2} a^3 x (7 A+6 B+2 C)+\frac{5 a^3 (A-C) \sin (c+d x)}{2 d}-\frac{(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 a d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d} \]

[Out]

(a^3*(7*A + 6*B + 2*C)*x)/2 + (a^3*(2*A + 6*B + 7*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(A - C)*Sin[c + d*x
])/(2*d) + (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) - ((A - C)*(a^2 + a^2*Sec[c + d*x])^2*Si
n[c + d*x])/(2*a*d) - ((A - 2*B - 4*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.425446, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {4086, 4018, 3996, 3770} \[ \frac{a^3 (2 A+6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(A-2 B-4 C) \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+\frac{1}{2} a^3 x (7 A+6 B+2 C)+\frac{5 a^3 (A-C) \sin (c+d x)}{2 d}-\frac{(A-C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 a d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^3}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(7*A + 6*B + 2*C)*x)/2 + (a^3*(2*A + 6*B + 7*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(A - C)*Sin[c + d*x
])/(2*d) + (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) - ((A - C)*(a^2 + a^2*Sec[c + d*x])^2*Si
n[c + d*x])/(2*a*d) - ((A - 2*B - 4*C)*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(2*d)

Rule 4086

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

Rule 4018

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

Rule 3996

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 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 ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^3 (a (3 A+2 B)-2 a (A-C) \sec (c+d x)) \, dx}{2 a}\\ &=\frac{A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^2 (4 A+2 B-C)-2 a^2 (A-2 B-4 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac{A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac{(A-2 B-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (10 a^3 (A-C)+2 a^3 (2 A+6 B+7 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac{5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac{(A-2 B-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}-\frac{\int \left (-2 a^4 (7 A+6 B+2 C)-2 a^4 (2 A+6 B+7 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac{1}{2} a^3 (7 A+6 B+2 C) x+\frac{5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac{(A-2 B-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}+\frac{1}{2} \left (a^3 (2 A+6 B+7 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^3 (7 A+6 B+2 C) x+\frac{a^3 (2 A+6 B+7 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a^3 (A-C) \sin (c+d x)}{2 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac{(A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 a d}-\frac{(A-2 B-4 C) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 5.89112, size = 406, normalized size = 2.37 \[ \frac{a^3 \cos ^5(c+d x) \sec ^6\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac{2 (2 A+6 B+7 C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{2 (2 A+6 B+7 C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{4 (3 A+B) \sin (c) \cos (d x)}{d}+\frac{4 (3 A+B) \cos (c) \sin (d x)}{d}+2 x (7 A+6 B+2 C)+\frac{A \sin (2 c) \cos (2 d x)}{d}+\frac{A \cos (2 c) \sin (2 d x)}{d}+\frac{4 (B+3 C) \sin \left (\frac{d x}{2}\right )}{d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (B+3 C) \sin \left (\frac{d x}{2}\right )}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{C}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{C}{d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{16 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Cos[c + d*x]^5*Sec[(c + d*x)/2]^6*(1 + Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(2*(7*A +
6*B + 2*C)*x - (2*(2*A + 6*B + 7*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/d + (2*(2*A + 6*B + 7*C)*Log[Cos
[(c + d*x)/2] + Sin[(c + d*x)/2]])/d + (4*(3*A + B)*Cos[d*x]*Sin[c])/d + (A*Cos[2*d*x]*Sin[2*c])/d + (4*(3*A +
B)*Cos[c]*Sin[d*x])/d + (A*Cos[2*c]*Sin[2*d*x])/d + C/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*(B + 3
*C)*Sin[(d*x)/2])/(d*(Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) - C/(d*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^2) + (4*(B + 3*C)*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]
))))/(16*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.108, size = 219, normalized size = 1.3 \begin{align*}{\frac{A{a}^{3}\sin \left ( dx+c \right ) \cos \left ( dx+c \right ) }{2\,d}}+{\frac{7\,{a}^{3}Ax}{2}}+{\frac{7\,A{a}^{3}c}{2\,d}}+{\frac{B{a}^{3}\sin \left ( dx+c \right ) }{d}}+{a}^{3}Cx+{\frac{C{a}^{3}c}{d}}+3\,{\frac{A{a}^{3}\sin \left ( dx+c \right ) }{d}}+3\,{a}^{3}Bx+3\,{\frac{B{a}^{3}c}{d}}+{\frac{7\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+3\,{\frac{B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{{a}^{3}C\tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d*A*a^3*sin(d*x+c)*cos(d*x+c)+7/2*a^3*A*x+7/2/d*A*a^3*c+a^3*B*sin(d*x+c)/d+a^3*C*x+1/d*C*a^3*c+3*a^3*A*sin
(d*x+c)/d+3*a^3*B*x+3/d*B*a^3*c+7/2/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+3/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+3*a^
3*C*tan(d*x+c)/d+1/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*B*a^3*tan(d*x+c)+1/2/d*a^3*C*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 0.964186, size = 320, normalized size = 1.87 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 12 \,{\left (d x + c\right )} A a^{3} + 12 \,{\left (d x + c\right )} B a^{3} + 4 \,{\left (d x + c\right )} C a^{3} - C a^{3}{\left (\frac{2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3}{\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^{3}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right ) + 12 \, C a^{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)^2*(a+a*sec(d*x+c))^3*(A+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))*A*a^3 + 12*(d*x + c)*A*a^3 + 12*(d*x + c)*B*a^3 + 4*(d*x + c)*C*a^3 - C*
a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 2*A*a^3*(log(sin(d
*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*C*a^3*(log
(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*A*a^3*sin(d*x + c) + 4*B*a^3*sin(d*x + c) + 4*B*a^3*tan(d*x +
c) + 12*C*a^3*tan(d*x + c))/d

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Fricas [A]  time = 0.553861, size = 410, normalized size = 2.4 \begin{align*} \frac{2 \,{\left (7 \, A + 6 \, B + 2 \, C\right )} a^{3} d x \cos \left (d x + c\right )^{2} +{\left (2 \, A + 6 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (2 \, A + 6 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (A a^{3} \cos \left (d x + c\right )^{3} + 2 \,{\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{2} + 2 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + C a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*(7*A + 6*B + 2*C)*a^3*d*x*cos(d*x + c)^2 + (2*A + 6*B + 7*C)*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) -
(2*A + 6*B + 7*C)*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^3*cos(d*x + c)^3 + 2*(3*A + B)*a^3*cos(d
*x + c)^2 + 2*(B + 3*C)*a^3*cos(d*x + c) + C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^2)

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

[Out]

Timed out

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Giac [A]  time = 1.24962, size = 378, normalized size = 2.21 \begin{align*} \frac{{\left (7 \, A a^{3} + 6 \, B a^{3} + 2 \, C a^{3}\right )}{\left (d x + c\right )} +{\left (2 \, A a^{3} + 6 \, B a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) -{\left (2 \, A a^{3} + 6 \, B a^{3} + 7 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (5 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 5 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 3 \, A a^{3} \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 )^{5} - 3 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 7 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7 \, C a^{3} \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 )^{4} - 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((7*A*a^3 + 6*B*a^3 + 2*C*a^3)*(d*x + c) + (2*A*a^3 + 6*B*a^3 + 7*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)
) - (2*A*a^3 + 6*B*a^3 + 7*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(5*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 5*C
*a^3*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 4*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*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 9*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 7*A*a^3*tan(1/2*d*x + 1/2
*c) + 4*B*a^3*tan(1/2*d*x + 1/2*c) + 7*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d