3.277 \(\int \sec ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=114 \[ \frac {(a B+A b) \tan ^3(c+d x)}{3 d}+\frac {(a B+A b) \tan (c+d x)}{d}+\frac {(4 a A+3 b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 a A+3 b B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[Out]

1/8*(4*A*a+3*B*b)*arctanh(sin(d*x+c))/d+(A*b+B*a)*tan(d*x+c)/d+1/8*(4*A*a+3*B*b)*sec(d*x+c)*tan(d*x+c)/d+1/4*b
*B*sec(d*x+c)^3*tan(d*x+c)/d+1/3*(A*b+B*a)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.15, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3997, 3787, 3768, 3770, 3767} \[ \frac {(a B+A b) \tan ^3(c+d x)}{3 d}+\frac {(a B+A b) \tan (c+d x)}{d}+\frac {(4 a A+3 b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 a A+3 b B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b B \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

((4*a*A + 3*b*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((A*b + a*B)*Tan[c + d*x])/d + ((4*a*A + 3*b*B)*Sec[c + d*x]*T
an[c + d*x])/(8*d) + (b*B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((A*b + a*B)*Tan[c + d*x]^3)/(3*d)

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

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 ^3(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx &=\frac {b B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \sec ^3(c+d x) (4 a A+3 b B+4 (A b+a B) \sec (c+d x)) \, dx\\ &=\frac {b B \sec ^3(c+d x) \tan (c+d x)}{4 d}+(A b+a B) \int \sec ^4(c+d x) \, dx+\frac {1}{4} (4 a A+3 b B) \int \sec ^3(c+d x) \, dx\\ &=\frac {(4 a A+3 b B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} (4 a A+3 b B) \int \sec (c+d x) \, dx-\frac {(A b+a B) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d}\\ &=\frac {(4 a A+3 b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {(4 a A+3 b B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {(A b+a B) \tan ^3(c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 85, normalized size = 0.75 \[ \frac {3 (4 a A+3 b B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x) \left (8 (a B+A b) (\cos (2 (c+d x))+2) \sec (c+d x)+12 a A+6 b B \sec ^2(c+d x)+9 b B\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x]),x]

[Out]

(3*(4*a*A + 3*b*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(12*a*A + 9*b*B + 8*(A*b + a*B)*(2 + Cos[2*(c + d*x)])
*Sec[c + d*x] + 6*b*B*Sec[c + d*x]^2)*Tan[c + d*x])/(24*d)

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fricas [A]  time = 0.97, size = 136, normalized size = 1.19 \[ \frac {3 \, {\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A a + 3 \, B b\right )} \cos \left (d x + c\right )^{2} + 6 \, B b + 8 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(3*(4*A*a + 3*B*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A*a + 3*B*b)*cos(d*x + c)^4*log(-sin(d*x +
c) + 1) + 2*(16*(B*a + A*b)*cos(d*x + c)^3 + 3*(4*A*a + 3*B*b)*cos(d*x + c)^2 + 6*B*b + 8*(B*a + A*b)*cos(d*x
+ c))*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [B]  time = 0.30, size = 304, normalized size = 2.67 \[ \frac {3 \, {\left (4 \, A a + 3 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, A a + 3 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(4*A*a + 3*B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A*a + 3*B*b)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) + 2*(12*A*a*tan(1/2*d*x + 1/2*c)^7 - 24*B*a*tan(1/2*d*x + 1/2*c)^7 - 24*A*b*tan(1/2*d*x + 1/2*c)^7 + 15*B*
b*tan(1/2*d*x + 1/2*c)^7 - 12*A*a*tan(1/2*d*x + 1/2*c)^5 + 40*B*a*tan(1/2*d*x + 1/2*c)^5 + 40*A*b*tan(1/2*d*x
+ 1/2*c)^5 + 9*B*b*tan(1/2*d*x + 1/2*c)^5 - 12*A*a*tan(1/2*d*x + 1/2*c)^3 - 40*B*a*tan(1/2*d*x + 1/2*c)^3 - 40
*A*b*tan(1/2*d*x + 1/2*c)^3 + 9*B*b*tan(1/2*d*x + 1/2*c)^3 + 12*A*a*tan(1/2*d*x + 1/2*c) + 24*B*a*tan(1/2*d*x
+ 1/2*c) + 24*A*b*tan(1/2*d*x + 1/2*c) + 15*B*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 1.25, size = 171, normalized size = 1.50 \[ \frac {a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a B \tan \left (d x +c \right )}{3 d}+\frac {a B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 A b \tan \left (d x +c \right )}{3 d}+\frac {A b \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {b B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 b B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

1/2*a*A*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a*A*ln(sec(d*x+c)+tan(d*x+c))+2/3/d*a*B*tan(d*x+c)+1/3/d*a*B*tan(d*x+c)*
sec(d*x+c)^2+2/3*A*b*tan(d*x+c)/d+1/3*A*b*sec(d*x+c)^2*tan(d*x+c)/d+1/4*b*B*sec(d*x+c)^3*tan(d*x+c)/d+3/8*b*B*
sec(d*x+c)*tan(d*x+c)/d+3/8/d*B*b*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.88, size = 163, normalized size = 1.43 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b - 3 \, B b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b - 3*B*b*(2*(3*sin(d*
x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x +
c) - 1)) - 12*A*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d

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mupad [B]  time = 5.73, size = 194, normalized size = 1.70 \[ \frac {\left (A\,a-2\,A\,b-2\,B\,a+\frac {5\,B\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {10\,A\,b}{3}-A\,a+\frac {10\,B\,a}{3}+\frac {3\,B\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,B\,b}{4}-\frac {10\,A\,b}{3}-\frac {10\,B\,a}{3}-A\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a+2\,A\,b+2\,B\,a+\frac {5\,B\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A\,a+\frac {3\,B\,b}{4}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B/cos(c + d*x))*(a + b/cos(c + d*x)))/cos(c + d*x)^3,x)

[Out]

(tan(c/2 + (d*x)/2)*(A*a + 2*A*b + 2*B*a + (5*B*b)/4) + tan(c/2 + (d*x)/2)^7*(A*a - 2*A*b - 2*B*a + (5*B*b)/4)
- tan(c/2 + (d*x)/2)^3*(A*a + (10*A*b)/3 + (10*B*a)/3 - (3*B*b)/4) + tan(c/2 + (d*x)/2)^5*((10*A*b)/3 - A*a +
(10*B*a)/3 + (3*B*b)/4))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c
/2 + (d*x)/2)^8 + 1)) + (atanh(tan(c/2 + (d*x)/2))*(A*a + (3*B*b)/4))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))*sec(c + d*x)**3, x)

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