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

Optimal. Leaf size=101 \[ \frac{\tan (c+d x) (3 a B+3 A b+2 b C)}{3 d}+\frac{(a (2 A+C)+b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(a C+b B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.13822, antiderivative size = 101, 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} \[ \frac{\tan (c+d x) (3 a B+3 A b+2 b C)}{3 d}+\frac{(a (2 A+C)+b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(a C+b B) \tan (c+d x) \sec (c+d x)}{2 d}+\frac{b C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*B + a*(2*A + C))*ArcTanh[Sin[c + d*x]])/(2*d) + ((3*A*b + 3*a*B + 2*b*C)*Tan[c + d*x])/(3*d) + ((b*B + a*C
)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (b*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+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{b 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+3 a B+2 b C) \sec (c+d x)+3 (b B+a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{b 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 (b B+a C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{3} (3 A b+3 a B+2 b C) \int \sec ^2(c+d x) \, dx\\ &=\frac{(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{b C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac{1}{2} (b B+a (2 A+C)) \int \sec (c+d x) \, dx-\frac{(3 A b+3 a B+2 b C) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{(b B+a (2 A+C)) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{(3 A b+3 a B+2 b C) \tan (c+d x)}{3 d}+\frac{(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac{b C \sec ^2(c+d x) \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.370176, size = 75, normalized size = 0.74 \[ \frac{3 (a (2 A+C)+b B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (a C+b B) \sec (c+d x)+6 a B+6 A b+2 b C \tan ^2(c+d x)+6 b C\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*B + a*(2*A + C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*A*b + 6*a*B + 6*b*C + 3*(b*B + a*C)*Sec[c + d*x
] + 2*b*C*Tan[c + d*x]^2))/(6*d)

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Maple [A]  time = 0.038, size = 160, normalized size = 1.6 \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{Ab\tan \left ( dx+c \right ) }{d}}+{\frac{Bb\tan \left ( dx+c \right ) \sec \left ( dx+c \right ) }{2\,d}}+{\frac{Bb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,Cb\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Cb \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+b*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*b*tan(d*x+c)+1/2/d*B*b*tan(d*x+c)*sec(d*x+c)+1/2/d*B*b*ln(sec(d*x+c)+tan(d*x+c))+2/3*b*C*tan(
d*x+c)/d+1/3*b*C*sec(d*x+c)^2*tan(d*x+c)/d

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Maxima [A]  time = 1.05017, size = 209, normalized size = 2.07 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b - 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 )} - 3 \, B b{\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 \, B a \tan \left (d x + c\right ) + 12 \, A b \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+b*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*b - 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)) - 3*B*b*(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*B*a*tan(d*x + c) + 12*A*b*tan(d*x + c))/d

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Fricas [A]  time = 0.597705, size = 331, normalized size = 3.28 \begin{align*} \frac{3 \,{\left ({\left (2 \, A + C\right )} a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left ({\left (2 \, A + C\right )} a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \,{\left (3 \, B a +{\left (3 \, A + 2 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 2 \, C b + 3 \,{\left (C a + B b\right )} \cos \left (d x + c\right )\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+b*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 + C)*a + B*b)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*((2*A + C)*a + B*b)*cos(d*x + c)^3*log(-s
in(d*x + c) + 1) + 2*(2*(3*B*a + (3*A + 2*C)*b)*cos(d*x + c)^2 + 2*C*b + 3*(C*a + B*b)*cos(d*x + c))*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*} \int \left (a + b \sec{\left (c + d x \right )}\right ) \left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.25643, size = 352, normalized size = 3.49 \begin{align*} \frac{3 \,{\left (2 \, A a + C a + B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, A a + C a + B b\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (6 \, 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} + 6 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b \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+b*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 + C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a + C*a + B*b)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) - 2*(6*B*a*tan(1/2*d*x + 1/2*c)^5 - 3*C*a*tan(1/2*d*x + 1/2*c)^5 + 6*A*b*tan(1/2*d*x + 1/2*c)^5 - 3
*B*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*tan(1/2*d*x + 1/2*c)^3 - 12*A*b*tan(1/2*d*
x + 1/2*c)^3 - 4*C*b*tan(1/2*d*x + 1/2*c)^3 + 6*B*a*tan(1/2*d*x + 1/2*c) + 3*C*a*tan(1/2*d*x + 1/2*c) + 6*A*b*
tan(1/2*d*x + 1/2*c) + 3*B*b*tan(1/2*d*x + 1/2*c) + 6*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3
)/d