3.765 \(\int \sec ^3(c+d x) (a+b \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=145 \[ \frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d}+\frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {3 (a C+b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(a C+b B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 (a C+b B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

[Out]

3/8*(B*b+C*a)*arctanh(sin(d*x+c))/d+1/5*(5*B*a+4*C*b)*tan(d*x+c)/d+3/8*(B*b+C*a)*sec(d*x+c)*tan(d*x+c)/d+1/4*(
B*b+C*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*b*C*sec(d*x+c)^4*tan(d*x+c)/d+1/15*(5*B*a+4*C*b)*tan(d*x+c)^3/d

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Rubi [A]  time = 0.20, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4072, 3997, 3787, 3767, 3768, 3770} \[ \frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d}+\frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {3 (a C+b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(a C+b B) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 (a C+b B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(3*(b*B + a*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((5*a*B + 4*b*C)*Tan[c + d*x])/(5*d) + (3*(b*B + a*C)*Sec[c + d*
x]*Tan[c + d*x])/(8*d) + ((b*B + a*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (b*C*Sec[c + d*x]^4*Tan[c + d*x])/(
5*d) + ((5*a*B + 4*b*C)*Tan[c + d*x]^3)/(15*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]

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^4(c+d x) (a+b \sec (c+d x)) (B+C \sec (c+d x)) \, dx\\ &=\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^4(c+d x) (5 a B+4 b C+5 (b B+a C) \sec (c+d x)) \, dx\\ &=\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+(b B+a C) \int \sec ^5(c+d x) \, dx+\frac {1}{5} (5 a B+4 b C) \int \sec ^4(c+d x) \, dx\\ &=\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (3 (b B+a C)) \int \sec ^3(c+d x) \, dx-\frac {(5 a B+4 b C) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {3 (b B+a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d}+\frac {1}{8} (3 (b B+a C)) \int \sec (c+d x) \, dx\\ &=\frac {3 (b B+a C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(5 a B+4 b C) \tan (c+d x)}{5 d}+\frac {3 (b B+a C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {b C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {(5 a B+4 b C) \tan ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 1.07, size = 106, normalized size = 0.73 \[ \frac {45 (a C+b B) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 (a B+2 b C) \tan ^2(c+d x)+15 (a B+b C)+3 b C \tan ^4(c+d x)\right )+30 (a C+b B) \sec ^3(c+d x)+45 (a C+b B) \sec (c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])*(B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(45*(b*B + a*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(45*(b*B + a*C)*Sec[c + d*x] + 30*(b*B + a*C)*Sec[c + d*x
]^3 + 8*(15*(a*B + b*C) + 5*(a*B + 2*b*C)*Tan[c + d*x]^2 + 3*b*C*Tan[c + d*x]^4)))/(120*d)

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fricas [A]  time = 0.98, size = 151, normalized size = 1.04 \[ \frac {45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, B a + 4 \, C b\right )} \cos \left (d x + c\right )^{4} + 45 \, {\left (C a + B b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, B a + 4 \, C b\right )} \cos \left (d x + c\right )^{2} + 24 \, C b + 30 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(45*(C*a + B*b)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*(C*a + B*b)*cos(d*x + c)^5*log(-sin(d*x + c) +
1) + 2*(16*(5*B*a + 4*C*b)*cos(d*x + c)^4 + 45*(C*a + B*b)*cos(d*x + c)^3 + 8*(5*B*a + 4*C*b)*cos(d*x + c)^2
+ 24*C*b + 30*(C*a + B*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [B]  time = 0.27, size = 330, normalized size = 2.28 \[ \frac {45 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, {\left (C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 320 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 320 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(45*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*(C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
2*(120*B*a*tan(1/2*d*x + 1/2*c)^9 - 75*C*a*tan(1/2*d*x + 1/2*c)^9 - 75*B*b*tan(1/2*d*x + 1/2*c)^9 + 120*C*b*t
an(1/2*d*x + 1/2*c)^9 - 320*B*a*tan(1/2*d*x + 1/2*c)^7 + 30*C*a*tan(1/2*d*x + 1/2*c)^7 + 30*B*b*tan(1/2*d*x +
1/2*c)^7 - 160*C*b*tan(1/2*d*x + 1/2*c)^7 + 400*B*a*tan(1/2*d*x + 1/2*c)^5 + 464*C*b*tan(1/2*d*x + 1/2*c)^5 -
320*B*a*tan(1/2*d*x + 1/2*c)^3 - 30*C*a*tan(1/2*d*x + 1/2*c)^3 - 30*B*b*tan(1/2*d*x + 1/2*c)^3 - 160*C*b*tan(1
/2*d*x + 1/2*c)^3 + 120*B*a*tan(1/2*d*x + 1/2*c) + 75*C*a*tan(1/2*d*x + 1/2*c) + 75*B*b*tan(1/2*d*x + 1/2*c) +
120*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

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maple [A]  time = 1.26, size = 213, normalized size = 1.47 \[ \frac {2 a B \tan \left (d x +c \right )}{3 d}+\frac {a B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a C \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {b B \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 b B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {8 b C \tan \left (d x +c \right )}{15 d}+\frac {b C \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 b C \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

2/3/d*a*B*tan(d*x+c)+1/3/d*a*B*tan(d*x+c)*sec(d*x+c)^2+1/4*a*C*sec(d*x+c)^3*tan(d*x+c)/d+3/8*a*C*sec(d*x+c)*ta
n(d*x+c)/d+3/8/d*a*C*ln(sec(d*x+c)+tan(d*x+c))+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))+8/15*b*C*tan(d*x+c)/d+1/5*b*C*sec(d*x+c)^4*tan(d*x+c)/d+4/15*b*C*sec(d*
x+c)^2*tan(d*x+c)/d

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maxima [A]  time = 0.35, size = 200, normalized size = 1.38 \[ \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, B b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*
C*b - 15*C*a*(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)) - 15*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)))/d

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mupad [B]  time = 7.95, size = 234, normalized size = 1.61 \[ \frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,B\,b}{4}+\frac {3\,C\,a}{4}\right )}{d}-\frac {\left (2\,B\,a-\frac {5\,B\,b}{4}-\frac {5\,C\,a}{4}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,b}{2}-\frac {16\,B\,a}{3}+\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,B\,a}{3}+\frac {116\,C\,b}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,B\,a}{3}-\frac {B\,b}{2}-\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a+\frac {5\,B\,b}{4}+\frac {5\,C\,a}{4}+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((B/cos(c + d*x) + C/cos(c + d*x)^2)*(a + b/cos(c + d*x)))/cos(c + d*x)^3,x)

[Out]

(atanh(tan(c/2 + (d*x)/2))*((3*B*b)/4 + (3*C*a)/4))/d - (tan(c/2 + (d*x)/2)*(2*B*a + (5*B*b)/4 + (5*C*a)/4 + 2
*C*b) + tan(c/2 + (d*x)/2)^5*((20*B*a)/3 + (116*C*b)/15) + tan(c/2 + (d*x)/2)^9*(2*B*a - (5*B*b)/4 - (5*C*a)/4
+ 2*C*b) - tan(c/2 + (d*x)/2)^3*((16*B*a)/3 + (B*b)/2 + (C*a)/2 + (8*C*b)/3) - tan(c/2 + (d*x)/2)^7*((16*B*a)
/3 - (B*b)/2 - (C*a)/2 + (8*C*b)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)
/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((B + C*sec(c + d*x))*(a + b*sec(c + d*x))*sec(c + d*x)**4, x)

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