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]

1/2*(2*A*a^2+A*b^2+2*B*a*b)*arctanh(sin(d*x+c))/d+2/3*(3*A*a*b+B*a^2+B*b^2)*tan(d*x+c)/d+1/6*b*(3*A*b+2*B*a)*s
ec(d*x+c)*tan(d*x+c)/d+1/3*B*(a+b*sec(d*x+c))^2*tan(d*x+c)/d

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Rubi [A]  time = 0.18, antiderivative size = 116, normalized size of antiderivative = 1.00, 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 3770

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

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 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 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]

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*}

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Mathematica [A]  time = 0.49, 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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fricas [A]  time = 0.45, size = 150, normalized size = 1.29 \[ \frac {3 \, {\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, B b^{2} + 2 \, {\left (3 \, B a^{2} + 6 \, A a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

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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giac [B]  time = 0.43, size = 294, normalized size = 2.53 \[ \frac {3 \, {\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

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

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maple [A]  time = 1.18, size = 174, normalized size = 1.50 \[ \frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} B \tan \left (d x +c \right )}{d}+\frac {2 a A b \tan \left (d x +c \right )}{d}+\frac {B a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {A \,b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 b^{2} B \tan \left (d x +c \right )}{3 d}+\frac {b^{2} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d} \]

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

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maxima [A]  time = 1.66, size = 165, normalized size = 1.42 \[ \frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{2} - 6 \, B a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, A b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, B a^{2} \tan \left (d x + c\right ) + 24 \, A a b \tan \left (d x + c\right )}{12 \, d} \]

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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mupad [B]  time = 5.44, size = 227, normalized size = 1.96 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a^2+B\,a\,b+\frac {A\,b^2}{2}\right )}{4\,A\,a^2+4\,B\,a\,b+2\,A\,b^2}\right )\,\left (2\,A\,a^2+2\,B\,a\,b+A\,b^2\right )}{d}-\frac {\left (2\,B\,a^2-A\,b^2+2\,B\,b^2+4\,A\,a\,b-2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,B\,a^2-8\,A\,a\,b-\frac {4\,B\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,b^2+2\,B\,a^2+2\,B\,b^2+4\,A\,a\,b+2\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x))^2)/cos(c + d*x),x)

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*(A*a^2 + (A*b^2)/2 + B*a*b))/(4*A*a^2 + 2*A*b^2 + 4*B*a*b))*(2*A*a^2 + A*b^2 + 2*
B*a*b))/d - (tan(c/2 + (d*x)/2)*(A*b^2 + 2*B*a^2 + 2*B*b^2 + 4*A*a*b + 2*B*a*b) - tan(c/2 + (d*x)/2)^3*(4*B*a^
2 + (4*B*b^2)/3 + 8*A*a*b) + tan(c/2 + (d*x)/2)^5*(2*B*a^2 - A*b^2 + 2*B*b^2 + 4*A*a*b - 2*B*a*b))/(d*(3*tan(c
/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec {\left (c + d x \right )}\, dx \]

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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