3.46 \(\int \sec (c+d x) (a+a \sec (c+d x)) (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=56 \[ \frac {a (A+B) \tan (c+d x)}{d}+\frac {a (2 A+B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d} \]

[Out]

1/2*a*(2*A+B)*arctanh(sin(d*x+c))/d+a*(A+B)*tan(d*x+c)/d+1/2*a*B*sec(d*x+c)*tan(d*x+c)/d

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Rubi [A]  time = 0.07, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {3997, 3787, 3770, 3767, 8} \[ \frac {a (A+B) \tan (c+d x)}{d}+\frac {a (2 A+B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 75, normalized size = 1.34 \[ \frac {a A \tan (c+d x)}{d}+\frac {a A \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a B \tan (c+d x)}{d}+\frac {a B \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {a B \tan (c+d x) \sec (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 89, normalized size = 1.59 \[ \frac {{\left (2 \, A + B\right )} a \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + B\right )} a \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + B a\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

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)),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.28, size = 124, normalized size = 2.21 \[ \frac {{\left (2 \, A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]

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)),x, algorithm="giac")

[Out]

1/2*((2*A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (2*A*a + B*a)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(
2*A*a*tan(1/2*d*x + 1/2*c)^3 + B*a*tan(1/2*d*x + 1/2*c)^3 - 2*A*a*tan(1/2*d*x + 1/2*c) - 3*B*a*tan(1/2*d*x + 1
/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d

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maple [A]  time = 0.94, size = 86, normalized size = 1.54 \[ \frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a B \tan \left (d x +c \right )}{d}+\frac {a A \tan \left (d x +c \right )}{d}+\frac {a B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d} \]

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)),x)

[Out]

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

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maxima [A]  time = 0.32, size = 88, normalized size = 1.57 \[ -\frac {B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 4 \, A a \tan \left (d x + c\right ) - 4 \, B a \tan \left (d x + c\right )}{4 \, d} \]

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)),x, algorithm="maxima")

[Out]

-1/4*(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)) - 4*A*a*log(se
c(d*x + c) + tan(d*x + c)) - 4*A*a*tan(d*x + c) - 4*B*a*tan(d*x + c))/d

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mupad [B]  time = 2.73, size = 94, normalized size = 1.68 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a+3\,B\,a\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,a+B\,a\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (2\,A+B\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int A \sec {\left (c + d x \right )}\, dx + \int A \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

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)),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))

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