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

Optimal. Leaf size=132 \[ \frac{a^2 (12 A+7 C) \tan (c+d x)}{6 d}+\frac{a^2 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (12 A+7 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}-\frac{C \tan (c+d x) (a \sec (c+d x)+a)^2}{12 d} \]

[Out]

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

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Rubi [A]  time = 0.212292, antiderivative size = 132, 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 = {4083, 4001, 3788, 3767, 8, 4046, 3770} \[ \frac{a^2 (12 A+7 C) \tan (c+d x)}{6 d}+\frac{a^2 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (12 A+7 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}-\frac{C \tan (c+d x) (a \sec (c+d x)+a)^2}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4083

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(
m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)),
Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C*Csc[e + f*x], x], x], x] /; FreeQ
[{a, b, e, f, A, C, m}, x] &&  !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 (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C (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 (a (4 A+3 C)-a C \sec (c+d x)) \, dx}{4 a}\\ &=-\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac{1}{12} (12 A+7 C) \int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx\\ &=-\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac{1}{12} (12 A+7 C) \int \sec (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac{1}{6} \left (a^2 (12 A+7 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (12 A+7 C) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}+\frac{1}{8} \left (a^2 (12 A+7 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^2 (12 A+7 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac{a^2 (12 A+7 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (12 A+7 C) \tan (c+d x)}{6 d}+\frac{a^2 (12 A+7 C) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d}\\ \end{align*}

Mathematica [B]  time = 1.40833, size = 291, normalized size = 2.2 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^4(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (24 (12 A+7 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-48 (3 A+2 C) \sin (c)+12 A \sin (2 c+d x)+144 A \sin (c+2 d x)-48 A \sin (3 c+2 d x)+12 A \sin (2 c+3 d x)+12 A \sin (4 c+3 d x)+48 A \sin (3 c+4 d x)+3 (4 A+15 C) \sin (d x)+45 C \sin (2 c+d x)+128 C \sin (c+2 d x)+21 C \sin (2 c+3 d x)+21 C \sin (4 c+3 d x)+32 C \sin (3 c+4 d x))\right )}{384 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(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]^4*(24*(12*A + 7*C)*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]*(-48*(3*A
+ 2*C)*Sin[c] + 3*(4*A + 15*C)*Sin[d*x] + 12*A*Sin[2*c + d*x] + 45*C*Sin[2*c + d*x] + 144*A*Sin[c + 2*d*x] +
128*C*Sin[c + 2*d*x] - 48*A*Sin[3*c + 2*d*x] + 12*A*Sin[2*c + 3*d*x] + 21*C*Sin[2*c + 3*d*x] + 12*A*Sin[4*c +
3*d*x] + 21*C*Sin[4*c + 3*d*x] + 48*A*Sin[3*c + 4*d*x] + 32*C*Sin[3*c + 4*d*x])))/(384*d*(A + 2*C + A*Cos[2*(c
+ d*x)]))

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Maple [A]  time = 0.049, size = 166, normalized size = 1.3 \begin{align*}{\frac{3\,{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{7\,{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{7\,{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+2\,{\frac{{a}^{2}A\tan \left ( dx+c \right ) }{d}}+{\frac{4\,{a}^{2}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{2}C\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)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.943957, size = 306, normalized size = 2.32 \begin{align*} \frac{32 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 3 \, 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 )} - 12 \, 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 \, C 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} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, 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)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 3*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)) - 12*A*a^2*(2*sin(d*x + c)/(s
in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*C*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*log(sec(d*x + c) + tan(d*x + c)) + 96*A*a^2*
tan(d*x + c))/d

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Fricas [A]  time = 0.51219, size = 356, normalized size = 2.7 \begin{align*} \frac{3 \,{\left (12 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (12 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 16 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C 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)*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(3*(12*A + 7*C)*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(12*A + 7*C)*a^2*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(16*(3*A + 2*C)*a^2*cos(d*x + c)^3 + 3*(4*A + 7*C)*a^2*cos(d*x + c)^2 + 16*C*a^2*cos(d*x + c) +
6*C*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{\left (c + d x \right )}\, dx + \int 2 A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 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)*(a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.24423, size = 286, normalized size = 2.17 \begin{align*} \frac{3 \,{\left (12 \, A a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (12 \, A a^{2} + 7 \, 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 (36 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 21 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 132 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 77 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 156 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 83 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 75 \, 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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(12*A*a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*A*a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(36*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 21*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 132*A*a^2*tan(1/2*d*x +
1/2*c)^5 - 77*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 156*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 83*C*a^2*tan(1/2*d*x + 1/2*c)
^3 - 60*A*a^2*tan(1/2*d*x + 1/2*c) - 75*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d