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

Optimal. Leaf size=138 \[ \frac{a^2 (8 A+7 B) \tan (c+d x)}{6 d}+\frac{a^2 (8 A+7 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (8 A+7 B) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(4 A-B) \tan (c+d x) (a \sec (c+d x)+a)^2}{12 d}+\frac{B \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d} \]

[Out]

(a^2*(8*A + 7*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(8*A + 7*B)*Tan[c + d*x])/(6*d) + (a^2*(8*A + 7*B)*Sec[c
+ d*x]*Tan[c + d*x])/(24*d) + ((4*A - B)*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + (B*(a + a*Sec[c + d*x])
^3*Tan[c + d*x])/(4*a*d)

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Rubi [A]  time = 0.228472, antiderivative size = 138, 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 = {4010, 4001, 3788, 3767, 8, 4046, 3770} \[ \frac{a^2 (8 A+7 B) \tan (c+d x)}{6 d}+\frac{a^2 (8 A+7 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (8 A+7 B) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(4 A-B) \tan (c+d x) (a \sec (c+d x)+a)^2}{12 d}+\frac{B \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(8*A + 7*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(8*A + 7*B)*Tan[c + d*x])/(6*d) + (a^2*(8*A + 7*B)*Sec[c
+ d*x]*Tan[c + d*x])/(24*d) + ((4*A - B)*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + (B*(a + a*Sec[c + d*x])
^3*Tan[c + d*x])/(4*a*d)

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 (A+B \sec (c+d x)) \, dx &=\frac{B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^2 (3 a B+a (4 A-B) \sec (c+d x)) \, dx}{4 a}\\ &=\frac{(4 A-B) (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac{1}{12} (8 A+7 B) \int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx\\ &=\frac{(4 A-B) (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac{1}{12} (8 A+7 B) \int \sec (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac{1}{6} \left (a^2 (8 A+7 B)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (8 A+7 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 A-B) (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac{1}{8} \left (a^2 (8 A+7 B)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^2 (8 A+7 B)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac{a^2 (8 A+7 B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (8 A+7 B) \tan (c+d x)}{6 d}+\frac{a^2 (8 A+7 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(4 A-B) (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{B (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}\\ \end{align*}

Mathematica [A]  time = 1.17248, size = 262, normalized size = 1.9 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (24 (8 A+7 B) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-24 (5 A+4 B) \sin (c)+3 (8 A+15 B) \sin (d x)+24 A \sin (2 c+d x)+136 A \sin (c+2 d x)-24 A \sin (3 c+2 d x)+24 A \sin (2 c+3 d x)+24 A \sin (4 c+3 d x)+40 A \sin (3 c+4 d x)+45 B \sin (2 c+d x)+128 B \sin (c+2 d x)+21 B \sin (2 c+3 d x)+21 B \sin (4 c+3 d x)+32 B \sin (3 c+4 d x))\right )}{768 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*Sec[c + d*x]^4*(24*(8*A + 7*B)*Cos[c + d*x]^4*(Log[Cos[(c + d*x)
/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-24*(5*A + 4*B)*Sin[c] + 3*(8*A
+ 15*B)*Sin[d*x] + 24*A*Sin[2*c + d*x] + 45*B*Sin[2*c + d*x] + 136*A*Sin[c + 2*d*x] + 128*B*Sin[c + 2*d*x] - 2
4*A*Sin[3*c + 2*d*x] + 24*A*Sin[2*c + 3*d*x] + 21*B*Sin[2*c + 3*d*x] + 24*A*Sin[4*c + 3*d*x] + 21*B*Sin[4*c +
3*d*x] + 40*A*Sin[3*c + 4*d*x] + 32*B*Sin[3*c + 4*d*x])))/(768*d)

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Maple [A]  time = 0.043, size = 187, normalized size = 1.4 \begin{align*}{\frac{5\,{a}^{2}A\tan \left ( dx+c \right ) }{3\,d}}+{\frac{7\,B{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{7\,B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,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{4\,B{a}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,B{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{B{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,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+B*sec(d*x+c)),x)

[Out]

5/3/d*a^2*A*tan(d*x+c)+7/8/d*B*a^2*sec(d*x+c)*tan(d*x+c)+7/8/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))+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))+4/3/d*B*a^2*tan(d*x+c)+2/3/d*B*a^2*tan(d*x+c)*sec(d*x+c)^
2+1/3/d*a^2*A*tan(d*x+c)*sec(d*x+c)^2+1/4/d*B*a^2*tan(d*x+c)*sec(d*x+c)^3

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Maxima [A]  time = 1.00926, size = 311, normalized size = 2.25 \begin{align*} \frac{16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 3 \, B 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 )} - 24 \, 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 )} - 12 \, B 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 )} + 48 \, A a^{2} \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))^2*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 32*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 - 3*B*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)) - 24*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
)) - 12*B*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)) + 48*A*a^2
*tan(d*x + c))/d

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Fricas [A]  time = 0.494235, size = 362, normalized size = 2.62 \begin{align*} \frac{3 \,{\left (8 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (5 \, A + 4 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \,{\left (A + 2 \, B\right )} a^{2} \cos \left (d x + c\right ) + 6 \, B a^{2}\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))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(3*(8*A + 7*B)*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(8*A + 7*B)*a^2*cos(d*x + c)^4*log(-sin(d*x +
c) + 1) + 2*(8*(5*A + 4*B)*a^2*cos(d*x + c)^3 + 3*(8*A + 7*B)*a^2*cos(d*x + c)^2 + 8*(A + 2*B)*a^2*cos(d*x +
c) + 6*B*a^2)*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^{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 B \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \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))**2*(A+B*sec(d*x+c)),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(B*sec(c + d*x)**3, x) + Integral(2*B*sec(c + d*x)**4, x) + Integral(B*sec(c + d*x)**5, x))

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Giac [A]  time = 1.36399, size = 286, normalized size = 2.07 \begin{align*} \frac{3 \,{\left (8 \, A a^{2} + 7 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (8 \, A a^{2} + 7 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (24 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 21 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 88 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 77 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 136 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 83 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 75 \, B 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 )}^{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))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(8*A*a^2 + 7*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(8*A*a^2 + 7*B*a^2)*log(abs(tan(1/2*d*x + 1
/2*c) - 1)) - 2*(24*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 21*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 88*A*a^2*tan(1/2*d*x + 1/
2*c)^5 - 77*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 136*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 83*B*a^2*tan(1/2*d*x + 1/2*c)^3
- 72*A*a^2*tan(1/2*d*x + 1/2*c) - 75*B*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d