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

Optimal. Leaf size=117 \[ \frac{a (3 A+2 C) \tan (c+d x)}{3 d}+\frac{a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{a C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.170044, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {4077, 4047, 3768, 3770, 4046, 3767, 8} \[ \frac{a (3 A+2 C) \tan (c+d x)}{3 d}+\frac{a (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{a C \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{a C \tan (c+d x) \sec ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*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) + (a*(3*A + 2*C)*Tan[c + d*x])/(3*d) + (a*(4*A + 3*C)*Sec[c + d*x]
*Tan[c + d*x])/(8*d) + (a*C*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + (a*C*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

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]

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 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 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.428452, size = 75, normalized size = 0.64 \[ \frac{a \left (3 (4 A+3 C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 (4 A+3 C) \sec (c+d x)+24 (A+C)+8 C \tan ^2(c+d x)+6 C \sec ^3(c+d x)\right )\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*(4*A + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24*(A + C) + 3*(4*A + 3*C)*Sec[c + d*x] + 6*C*Sec[c +
d*x]^3 + 8*C*Tan[c + d*x]^2)))/(24*d)

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Maple [A]  time = 0.043, size = 149, normalized size = 1.3 \begin{align*}{\frac{Aa\tan \left ( dx+c \right ) }{d}}+{\frac{2\,aC\tan \left ( dx+c \right ) }{3\,d}}+{\frac{aC \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\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{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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*A*a*tan(d*x+c)+2/3*a*C*tan(d*x+c)/d+1/3*a*C*sec(d*x+c)^2*tan(d*x+c)/d+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))+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))

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Maxima [A]  time = 0.941002, size = 205, normalized size = 1.75 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, 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 )} - 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 \, A a \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a - 3*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)) - 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)) + 48*A*a*tan(d*x + c))/d

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Fricas [A]  time = 0.509292, size = 335, normalized size = 2.86 \begin{align*} \frac{3 \,{\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (3 \, A + 2 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, C a \cos \left (d x + c\right ) + 6 \, C a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\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)**2*(a+a*sec(d*x+c))*(A+C*sec(d*x+c)**2),x)

[Out]

a*(Integral(A*sec(c + d*x)**2, x) + Integral(A*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**4, x) + Integral
(C*sec(c + d*x)**5, x))

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Giac [A]  time = 1.25697, size = 254, normalized size = 2.17 \begin{align*} \frac{3 \,{\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 ) - 3 \,{\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 (12 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 9 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 60 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 49 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 84 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 31 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 36 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 39 \, 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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) - 2*(12*A*a*tan(1/2*d*x + 1/2*c)^7 + 9*C*a*tan(1/2*d*x + 1/2*c)^7 - 60*A*a*tan(1/2*d*x + 1/2*c)^5 - 49*C*a
*tan(1/2*d*x + 1/2*c)^5 + 84*A*a*tan(1/2*d*x + 1/2*c)^3 + 31*C*a*tan(1/2*d*x + 1/2*c)^3 - 36*A*a*tan(1/2*d*x +
 1/2*c) - 39*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d