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

Optimal. Leaf size=172 \[ \frac{a^2 (4 A+3 C) \tan (c+d x)}{3 d}+\frac{a^2 (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 (4 A+3 C) \tan (c+d x) \sec (c+d x)}{12 d}+\frac{(10 A+3 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{30 d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^2}{5 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^3}{10 a d} \]

[Out]

(a^2*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(4*d) + (a^2*(4*A + 3*C)*Tan[c + d*x])/(3*d) + (a^2*(4*A + 3*C)*Sec[c
+ d*x]*Tan[c + d*x])/(12*d) + ((10*A + 3*C)*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(30*d) + (C*Sec[c + d*x]^2*(a
+ a*Sec[c + d*x])^2*Tan[c + d*x])/(5*d) + (C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(10*a*d)

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Rubi [A]  time = 0.385351, antiderivative size = 172, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4089, 4010, 4001, 3788, 3767, 8, 4046, 3770} \[ \frac{a^2 (4 A+3 C) \tan (c+d x)}{3 d}+\frac{a^2 (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 (4 A+3 C) \tan (c+d x) \sec (c+d x)}{12 d}+\frac{(10 A+3 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{30 d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^2}{5 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^3}{10 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(4*d) + (a^2*(4*A + 3*C)*Tan[c + d*x])/(3*d) + (a^2*(4*A + 3*C)*Sec[c
+ d*x]*Tan[c + d*x])/(12*d) + ((10*A + 3*C)*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(30*d) + (C*Sec[c + d*x]^2*(a
+ a*Sec[c + d*x])^2*Tan[c + d*x])/(5*d) + (C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(10*a*d)

Rule 4089

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

Rule 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4001

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

Rule 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{\int \sec ^2(c+d x) (a+a \sec (c+d x))^2 (a (5 A+2 C)+2 a C \sec (c+d x)) \, dx}{5 a}\\ &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^2 \left (6 a^2 C+2 a^2 (10 A+3 C) \sec (c+d x)\right ) \, dx}{20 a^2}\\ &=\frac{(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}+\frac{1}{6} (4 A+3 C) \int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx\\ &=\frac{(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}+\frac{1}{6} (4 A+3 C) \int \sec (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac{1}{3} \left (a^2 (4 A+3 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (4 A+3 C) \sec (c+d x) \tan (c+d x)}{12 d}+\frac{(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}+\frac{1}{4} \left (a^2 (4 A+3 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^2 (4 A+3 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac{a^2 (4 A+3 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 (4 A+3 C) \tan (c+d x)}{3 d}+\frac{a^2 (4 A+3 C) \sec (c+d x) \tan (c+d x)}{12 d}+\frac{(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}\\ \end{align*}

Mathematica [A]  time = 1.78983, size = 321, normalized size = 1.87 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (240 (4 A+3 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-120 (3 A+C) \sin (2 c+d x)+120 A \sin (c+2 d x)+120 A \sin (3 c+2 d x)+440 A \sin (2 c+3 d x)-60 A \sin (4 c+3 d x)+60 A \sin (3 c+4 d x)+60 A \sin (5 c+4 d x)+100 A \sin (4 c+5 d x)+40 (16 A+15 C) \sin (d x)+210 C \sin (c+2 d x)+210 C \sin (3 c+2 d x)+360 C \sin (2 c+3 d x)+45 C \sin (3 c+4 d x)+45 C \sin (5 c+4 d x)+72 C \sin (4 c+5 d x))\right )}{1920 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(1 + Cos[c + d*x])^2*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^4*Sec[c + d*x]^5*(240*(4*A + 3*C)*Cos[c + d
*x]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(40*(16*A
+ 15*C)*Sin[d*x] - 120*(3*A + C)*Sin[2*c + d*x] + 120*A*Sin[c + 2*d*x] + 210*C*Sin[c + 2*d*x] + 120*A*Sin[3*c
+ 2*d*x] + 210*C*Sin[3*c + 2*d*x] + 440*A*Sin[2*c + 3*d*x] + 360*C*Sin[2*c + 3*d*x] - 60*A*Sin[4*c + 3*d*x] +
60*A*Sin[3*c + 4*d*x] + 45*C*Sin[3*c + 4*d*x] + 60*A*Sin[5*c + 4*d*x] + 45*C*Sin[5*c + 4*d*x] + 100*A*Sin[4*c
+ 5*d*x] + 72*C*Sin[4*c + 5*d*x])))/(1920*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.119, size = 210, normalized size = 1.2 \begin{align*}{\frac{5\,{a}^{2}A\tan \left ( dx+c \right ) }{3\,d}}+{\frac{6\,{a}^{2}C\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,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))^2*(A+C*sec(d*x+c)^2),x)

[Out]

5/3/d*a^2*A*tan(d*x+c)+6/5/d*a^2*C*tan(d*x+c)+3/5/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2+1/d*a^2*A*sec(d*x+c)*tan(d*x
+c)+1/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a^2*C*tan(d*x+c)*sec(d*x+c)^3+3/4/d*a^2*C*sec(d*x+c)*tan(d*x+c)+
3/4/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+1/3/d*a^2*A*tan(d*x+c)*sec(d*x+c)^2+1/5/d*a^2*C*tan(d*x+c)*sec(d*x+c)^4

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Maxima [A]  time = 0.944061, size = 294, normalized size = 1.71 \begin{align*} \frac{40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 8 \,{\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^{2} + 40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, C a^{2}{\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^{2}{\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 )} + 120 \, A a^{2} \tan \left (d x + c\right )}{120 \, 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))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/120*(40*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))
*C*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 15*C*a^2*(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*(2*sin(d*x + c
)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*A*a^2*tan(d*x + c))/d

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Fricas [A]  time = 0.520133, size = 409, normalized size = 2.38 \begin{align*} \frac{15 \,{\left (4 \, A + 3 \, C\right )} a^{2} \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^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (4 \,{\left (25 \, A + 18 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \,{\left (4 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \,{\left (5 \, A + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, C a^{2} \cos \left (d x + c\right ) + 12 \, C a^{2}\right )} \sin \left (d x + c\right )}{120 \, 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)^2*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(4*A + 3*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*A + 3*C)*a^2*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(4*(25*A + 18*C)*a^2*cos(d*x + c)^4 + 15*(4*A + 3*C)*a^2*cos(d*x + c)^3 + 4*(5*A + 9*C)*a^2*co
s(d*x + c)^2 + 30*C*a^2*cos(d*x + c) + 12*C*a^2)*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^{2} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 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)**2*(a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

a**2*(Integral(A*sec(c + d*x)**2, x) + Integral(2*A*sec(c + d*x)**3, x) + Integral(A*sec(c + d*x)**4, x) + Int
egral(C*sec(c + d*x)**4, x) + Integral(2*C*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))

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Giac [A]  time = 1.26917, size = 332, normalized size = 1.93 \begin{align*} \frac{15 \,{\left (4 \, A a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (4 \, A a^{2} + 3 \, C a^{2}\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^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 45 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 280 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 210 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 560 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 432 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 520 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 270 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 180 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 195 \, C a^{2} \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}}}{60 \, 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))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/60*(15*(4*A*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(60*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 45*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 280*A*a^2*tan(1/2*d*x +
1/2*c)^7 - 210*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 560*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 432*C*a^2*tan(1/2*d*x + 1/2*
c)^5 - 520*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 270*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 180*A*a^2*tan(1/2*d*x + 1/2*c) +
195*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d