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

Optimal. Leaf size=140 \[ \frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {b (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {b (5 A+4 C) \tan (c+d x)}{5 d}+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4077, 4047, 3767, 4046, 3768, 3770} \[ \frac {a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {b (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {b (5 A+4 C) \tan (c+d x)}{5 d}+\frac {b C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (b*(5*A + 4*C)*Tan[c + d*x])/(5*d) + (a*(4*A + 3*C)*Sec[c + d*x]
*Tan[c + d*x])/(8*d) + (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) + (b*
(5*A + 4*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 4046

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> -Simp[(C*Cot[
e + f*x]*(b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x]
/; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4077

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 2)), x] + Dist[1/(
n + 2), Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 2) + b*(C*(n + 1) + A*(n + 2))*Csc[e + f*x] + a*C*(n + 2)*Csc[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !LtQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.92, size = 96, normalized size = 0.69 \[ \frac {\tan (c+d x) \left (15 a (4 A+3 C) \sec (c+d x)+30 a C \sec ^3(c+d x)+8 b \left (5 (A+2 C) \tan ^2(c+d x)+15 (A+C)+3 C \tan ^4(c+d x)\right )\right )+15 a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{120 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.46, size = 147, normalized size = 1.05 \[ \frac {15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} b \cos \left (d x + c\right )^{2} + 30 \, C a \cos \left (d x + c\right ) + 24 \, C b\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(4*A + 3*C)*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*A + 3*C)*a*cos(d*x + c)^5*log(-sin(d*x +
c) + 1) + 2*(16*(5*A + 4*C)*b*cos(d*x + c)^4 + 15*(4*A + 3*C)*a*cos(d*x + c)^3 + 8*(5*A + 4*C)*b*cos(d*x + c)^
2 + 30*C*a*cos(d*x + c) + 24*C*b)*sin(d*x + c))/(d*cos(d*x + c)^5)

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giac [B]  time = 0.25, size = 334, normalized size = 2.39 \[ \frac {15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 400 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 464 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 75 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) + 2*(60*A*a*tan(1/2*d*x + 1/2*c)^9 + 75*C*a*tan(1/2*d*x + 1/2*c)^9 - 120*A*b*tan(1/2*d*x + 1/2*c)^9 - 1
20*C*b*tan(1/2*d*x + 1/2*c)^9 - 120*A*a*tan(1/2*d*x + 1/2*c)^7 - 30*C*a*tan(1/2*d*x + 1/2*c)^7 + 320*A*b*tan(1
/2*d*x + 1/2*c)^7 + 160*C*b*tan(1/2*d*x + 1/2*c)^7 - 400*A*b*tan(1/2*d*x + 1/2*c)^5 - 464*C*b*tan(1/2*d*x + 1/
2*c)^5 + 120*A*a*tan(1/2*d*x + 1/2*c)^3 + 30*C*a*tan(1/2*d*x + 1/2*c)^3 + 320*A*b*tan(1/2*d*x + 1/2*c)^3 + 160
*C*b*tan(1/2*d*x + 1/2*c)^3 - 60*A*a*tan(1/2*d*x + 1/2*c) - 75*C*a*tan(1/2*d*x + 1/2*c) - 120*A*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 = 1.38, size = 192, normalized size = 1.37 \[ \frac {a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a C \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 a C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 A b \tan \left (d x +c \right )}{3 d}+\frac {A b \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {8 b C \tan \left (d x +c \right )}{15 d}+\frac {b C \left (\sec ^{4}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{5 d}+\frac {4 b C \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sec(d*x+c))*(A+C*sec(d*x+c)^2),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))+1/4*a*C*sec(d*x+c)^3*tan(d*x+c)/d+3/8*a*C*
sec(d*x+c)*tan(d*x+c)/d+3/8/d*a*C*ln(sec(d*x+c)+tan(d*x+c))+2/3*A*b*tan(d*x+c)/d+1/3*A*b*sec(d*x+c)^2*tan(d*x+
c)/d+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 = 0.36, size = 175, normalized size = 1.25 \[ \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b + 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)) - 60*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 = 7.68, size = 232, normalized size = 1.66 \[ \frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{4\,d}-\frac {\left (2\,A\,b-A\,a-\frac {5\,C\,a}{4}+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,a-\frac {16\,A\,b}{3}+\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,b}{3}+\frac {116\,C\,b}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-2\,A\,a-\frac {16\,A\,b}{3}-\frac {C\,a}{2}-\frac {8\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a+2\,A\,b+\frac {5\,C\,a}{4}+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*atanh(tan(c/2 + (d*x)/2))*(4*A + 3*C))/(4*d) - (tan(c/2 + (d*x)/2)*(A*a + 2*A*b + (5*C*a)/4 + 2*C*b) + tan(
c/2 + (d*x)/2)^5*((20*A*b)/3 + (116*C*b)/15) - tan(c/2 + (d*x)/2)^9*(A*a - 2*A*b + (5*C*a)/4 - 2*C*b) - tan(c/
2 + (d*x)/2)^3*(2*A*a + (16*A*b)/3 + (C*a)/2 + (8*C*b)/3) + tan(c/2 + (d*x)/2)^7*(2*A*a - (16*A*b)/3 + (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 \[ \int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))*sec(c + d*x)**3, x)

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