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

Optimal. Leaf size=92 \[ \frac{a (3 A+3 B+2 C) \tan (c+d x)}{3 d}+\frac{a (2 A+B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (B+C) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.120209, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {4076, 4047, 3767, 8, 4046, 3770} \[ \frac{a (3 A+3 B+2 C) \tan (c+d x)}{3 d}+\frac{a (2 A+B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (B+C) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{a C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4076

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[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x
])^n)/(f*(n + 2)), x] + Dist[1/(n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + (B*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, 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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rule 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, 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 \sec (c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int \sec (c+d x) \left (3 a A+a (3 A+3 B+2 C) \sec (c+d x)+3 a (B+C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{3} \int \sec (c+d x) \left (3 a A+3 a (B+C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{3} (a (3 A+3 B+2 C)) \int \sec ^2(c+d x) \, dx\\ &=\frac{a (B+C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} (a (2 A+B+C)) \int \sec (c+d x) \, dx-\frac{(a (3 A+3 B+2 C)) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{a (2 A+B+C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a (3 A+3 B+2 C) \tan (c+d x)}{3 d}+\frac{a (B+C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a C \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 4.35116, size = 485, normalized size = 5.27 \[ \frac{a \cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4 (3 A+3 B+2 C) \sin \left (\frac{d x}{2}\right )}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 (3 A+3 B+2 C) \sin \left (\frac{d x}{2}\right )}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-6 (2 A+B+C) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 (2 A+B+C) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{(3 B+4 C) \cos \left (\frac{c}{2}\right )-(3 B+2 C) \sin \left (\frac{c}{2}\right )}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{(3 B+2 C) \sin \left (\frac{c}{2}\right )+(3 B+4 C) \cos \left (\frac{c}{2}\right )}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 C \sin \left (\frac{d x}{2}\right )}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 C \sin \left (\frac{d x}{2}\right )}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}\right )}{6 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.048, size = 160, normalized size = 1.7 \begin{align*}{\frac{Aa\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{Ba\tan \left ( dx+c \right ) }{d}}+{\frac{aC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{Aa\tan \left ( dx+c \right ) }{d}}+{\frac{Ba\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{Ba\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,aC\tan \left ( dx+c \right ) }{3\,d}}+{\frac{aC \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*A*a*ln(sec(d*x+c)+tan(d*x+c))+1/d*B*a*tan(d*x+c)+1/2*a*C*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a*C*ln(sec(d*x+c)+t
an(d*x+c))+1/d*A*a*tan(d*x+c)+1/2/d*B*a*sec(d*x+c)*tan(d*x+c)+1/2/d*B*a*ln(sec(d*x+c)+tan(d*x+c))+2/3*a*C*tan(
d*x+c)/d+1/3*a*C*sec(d*x+c)^2*tan(d*x+c)/d

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Maxima [A]  time = 0.947713, size = 209, normalized size = 2.27 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, B a{\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 )} - 3 \, C a{\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 )} + 12 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, A a \tan \left (d x + c\right ) + 12 \, B a \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a - 3*B*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) - 3*C*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d
*x + c) - 1)) + 12*A*a*log(sec(d*x + c) + tan(d*x + c)) + 12*A*a*tan(d*x + c) + 12*B*a*tan(d*x + c))/d

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Fricas [A]  time = 0.516922, size = 312, normalized size = 3.39 \begin{align*} \frac{3 \,{\left (2 \, A + B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, A + B + C\right )} a \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (3 \, A + 3 \, B + 2 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \,{\left (B + C\right )} a \cos \left (d x + c\right ) + 2 \, C a\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.24508, size = 277, normalized size = 3.01 \begin{align*} \frac{3 \,{\left (2 \, A a + B a + C a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, A a + B a + C a\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, C a \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(sec(d*x+c)*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(2*A*a + B*a + C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a + B*a + C*a)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) - 2*(6*A*a*tan(1/2*d*x + 1/2*c)^5 + 3*B*a*tan(1/2*d*x + 1/2*c)^5 + 3*C*a*tan(1/2*d*x + 1/2*c)^5 - 1
2*A*a*tan(1/2*d*x + 1/2*c)^3 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*tan(1/2*d*x + 1/2*c)^3 + 6*A*a*tan(1/2*d*
x + 1/2*c) + 9*B*a*tan(1/2*d*x + 1/2*c) + 9*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d