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

Optimal. Leaf size=152 \[ \frac{a (5 A+5 B+4 C) \tan ^3(c+d x)}{15 d}+\frac{a (5 A+5 B+4 C) \tan (c+d x)}{5 d}+\frac{a (4 A+3 (B+C)) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (4 A+3 (B+C)) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a (B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{a C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

[Out]

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

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Rubi [A]  time = 0.213046, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4076, 4047, 3767, 4046, 3768, 3770} \[ \frac{a (5 A+5 B+4 C) \tan ^3(c+d x)}{15 d}+\frac{a (5 A+5 B+4 C) \tan (c+d x)}{5 d}+\frac{a (4 A+3 (B+C)) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (4 A+3 (B+C)) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a (B+C) \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{a C \tan (c+d x) \sec ^4(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4076

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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*a*(n + 2) + b*(C*(n + 1)
+ A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*(n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C,
n}, x] &&  !LtQ[n, -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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 1.00636, size = 101, normalized size = 0.66 \[ \frac{a \left (15 (4 A+3 (B+C)) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 (A+B+2 C) \tan ^2(c+d x)+15 (A+B+C)+3 C \tan ^4(c+d x)\right )+15 (4 A+3 (B+C)) \sec (c+d x)+30 (B+C) \sec ^3(c+d x)\right )\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.056, size = 287, normalized size = 1.9 \begin{align*}{\frac{Aa\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{Aa\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,Ba\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Ba\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{aC \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,aC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,aC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{2\,Aa\tan \left ( dx+c \right ) }{3\,d}}+{\frac{Aa\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{Ba\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,Ba\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,Ba\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,aC\tan \left ( dx+c \right ) }{15\,d}}+{\frac{aC \left ( \sec \left ( dx+c \right ) \right ) ^{4}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{4\,aC \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.953166, size = 359, normalized size = 2.36 \begin{align*} \frac{80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 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 a - 15 \, B 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 \, 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sec(d*x+c))*(A+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))*A*a + 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*a - 15*B*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*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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Fricas [A]  time = 0.531814, size = 437, normalized size = 2.88 \begin{align*} \frac{15 \,{\left (4 \, A + 3 \, B + 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 \, B + 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 + 5 \, B + 4 \, C\right )} a \cos \left (d x + c\right )^{4} + 15 \,{\left (4 \, A + 3 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 5 \, B + 4 \, C\right )} a \cos \left (d x + c\right )^{2} + 30 \,{\left (B + C\right )} a \cos \left (d x + c\right ) + 24 \, C a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(Integral(A*sec(c + d*x)**3, x) + Integral(A*sec(c + d*x)**4, x) + Integral(B*sec(c + d*x)**4, x) + Integral
(B*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))

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Giac [B]  time = 1.30164, size = 404, normalized size = 2.66 \begin{align*} \frac{15 \,{\left (4 \, A a + 3 \, B 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 \, B 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} + 45 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 45 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 200 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 290 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 130 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 400 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 400 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 464 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 440 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 350 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 190 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 180 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 195 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 195 \, C 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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*A*a + 3*B*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a + 3*B*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 + 45*B*a*tan(1/2*d*x + 1/2*c)^9 + 45*C*a*tan(1/2*d*x
+ 1/2*c)^9 - 200*A*a*tan(1/2*d*x + 1/2*c)^7 - 290*B*a*tan(1/2*d*x + 1/2*c)^7 - 130*C*a*tan(1/2*d*x + 1/2*c)^7
+ 400*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*B*a*tan(1/2*d*x + 1/2*c)^5 + 464*C*a*tan(1/2*d*x + 1/2*c)^5 - 440*A*a*
tan(1/2*d*x + 1/2*c)^3 - 350*B*a*tan(1/2*d*x + 1/2*c)^3 - 190*C*a*tan(1/2*d*x + 1/2*c)^3 + 180*A*a*tan(1/2*d*x
+ 1/2*c) + 195*B*a*tan(1/2*d*x + 1/2*c) + 195*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d