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

Optimal. Leaf size=116 \[ \frac{2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{3 d}+\frac{\left (2 a^2 A+2 a b B+A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b (2 a B+3 A b) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{B \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.179885, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {4002, 3997, 3787, 3770, 3767, 8} \[ \frac{2 \left (a^2 B+3 a A b+b^2 B\right ) \tan (c+d x)}{3 d}+\frac{\left (2 a^2 A+2 a b B+A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b (2 a B+3 A b) \tan (c+d x) \sec (c+d x)}{6 d}+\frac{B \tan (c+d x) (a+b \sec (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4002

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 3770

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

Rubi steps

\begin{align*} \int \sec (c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac{B (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{3} \int \sec (c+d x) (a+b \sec (c+d x)) (3 a A+2 b B+(3 A b+2 a B) \sec (c+d x)) \, dx\\ &=\frac{b (3 A b+2 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{B (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{6} \int \sec (c+d x) \left (3 \left (2 a^2 A+A b^2+2 a b B\right )+4 \left (3 a A b+a^2 B+b^2 B\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b (3 A b+2 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{B (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac{1}{2} \left (2 a^2 A+A b^2+2 a b B\right ) \int \sec (c+d x) \, dx+\frac{1}{3} \left (2 \left (3 a A b+a^2 B+b^2 B\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{\left (2 a^2 A+A b^2+2 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{b (3 A b+2 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{B (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}-\frac{\left (2 \left (3 a A b+a^2 B+b^2 B\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{\left (2 a^2 A+A b^2+2 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{2 \left (3 a A b+a^2 B+b^2 B\right ) \tan (c+d x)}{3 d}+\frac{b (3 A b+2 a B) \sec (c+d x) \tan (c+d x)}{6 d}+\frac{B (a+b \sec (c+d x))^2 \tan (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.465689, size = 92, normalized size = 0.79 \[ \frac{3 \left (2 a^2 A+2 a b B+A b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (2 \left (3 a^2 B+6 a A b+b^2 B \tan ^2(c+d x)+3 b^2 B\right )+3 b (2 a B+A b) \sec (c+d x)\right )}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 174, normalized size = 1.5 \begin{align*}{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{B{a}^{2}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{Aab\tan \left ( dx+c \right ) }{d}}+{\frac{Bab\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{Bab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,B{b}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{B{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{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))^2*(A+B*sec(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.00315, size = 223, normalized size = 1.92 \begin{align*} \frac{4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{2} - 6 \, B a 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 )} - 3 \, A b^{2}{\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^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, B a^{2} \tan \left (d x + c\right ) + 24 \, A 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))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*b^2 - 6*B*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x +
c) + 1) + log(sin(d*x + c) - 1)) - 3*A*b^2*(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^2*log(sec(d*x + c) + tan(d*x + c)) + 12*B*a^2*tan(d*x + c) + 24*A*a*b*tan(d*x + c
))/d

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Fricas [A]  time = 0.504269, size = 371, normalized size = 3.2 \begin{align*} \frac{3 \,{\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (2 \, B b^{2} + 2 \,{\left (3 \, B a^{2} + 6 \, A a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (2 \, B a b + A b^{2}\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))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(3*(2*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*A*a^2 + 2*B*a*b + A*b^2)*cos(d
*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*B*b^2 + 2*(3*B*a^2 + 6*A*a*b + 2*B*b^2)*cos(d*x + c)^2 + 3*(2*B*a*b +
A*b^2)*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 )}\right )^{2} \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))**2*(A+B*sec(d*x+c)),x)

[Out]

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

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Giac [B]  time = 1.20154, size = 397, normalized size = 3.42 \begin{align*} \frac{3 \,{\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\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^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, B 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} + 6 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B 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 ) + 6 \, B 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}}}{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))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/6*(3*(2*A*a^2 + 2*B*a*b + A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a^2 + 2*B*a*b + A*b^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*A*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*B*a*b*tan
(1/2*d*x + 1/2*c)^5 - 3*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*B*b^2*tan(1/2*d*x + 1/2*c)^5 - 12*B*a^2*tan(1/2*d*x +
1/2*c)^3 - 24*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 4*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*B*a^2*tan(1/2*d*x + 1/2*c) +
12*A*a*b*tan(1/2*d*x + 1/2*c) + 6*B*a*b*tan(1/2*d*x + 1/2*c) + 3*A*b^2*tan(1/2*d*x + 1/2*c) + 6*B*b^2*tan(1/2*
d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d