3.466 \(\int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx\)

Optimal. Leaf size=189 \[ -\frac{\left (-52 a^2 b^2+3 a^4-16 b^4\right ) \tan (c+d x)}{30 b d}+\frac{a \left (4 a^2+9 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}-\frac{a \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}-\frac{a \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d} \]

[Out]

(a*(4*a^2 + 9*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) - ((3*a^4 - 52*a^2*b^2 - 16*b^4)*Tan[c + d*x])/(30*b*d) - (a*(
6*a^2 - 71*b^2)*Sec[c + d*x]*Tan[c + d*x])/(120*d) - ((3*a^2 - 16*b^2)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(6
0*b*d) - (a*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(20*b*d) + ((a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*b*d)

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Rubi [A]  time = 0.312282, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3840, 4002, 3997, 3787, 3770, 3767, 8} \[ -\frac{\left (-52 a^2 b^2+3 a^4-16 b^4\right ) \tan (c+d x)}{30 b d}+\frac{a \left (4 a^2+9 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\left (3 a^2-16 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}-\frac{a \left (6 a^2-71 b^2\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac{\tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}-\frac{a \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^3,x]

[Out]

(a*(4*a^2 + 9*b^2)*ArcTanh[Sin[c + d*x]])/(8*d) - ((3*a^4 - 52*a^2*b^2 - 16*b^4)*Tan[c + d*x])/(30*b*d) - (a*(
6*a^2 - 71*b^2)*Sec[c + d*x]*Tan[c + d*x])/(120*d) - ((3*a^2 - 16*b^2)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(6
0*b*d) - (a*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(20*b*d) + ((a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*b*d)

Rule 3840

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(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*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4002

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[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 3997

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> -Simp[(b*B*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*(n + 1)), x] + Dist[1/(n + 1), Int[(d*C
sc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f
, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 3787

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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]

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^3 \, dx &=\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (4 b-a \sec (c+d x)) (a+b \sec (c+d x))^3 \, dx}{5 b}\\ &=-\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (13 a b-\left (3 a^2-16 b^2\right ) \sec (c+d x)\right ) \, dx}{20 b}\\ &=-\frac{\left (3 a^2-16 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (33 a^2+32 b^2\right )-a \left (6 a^2-71 b^2\right ) \sec (c+d x)\right ) \, dx}{60 b}\\ &=-\frac{a \left (6 a^2-71 b^2\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac{\left (3 a^2-16 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\int \sec (c+d x) \left (15 a b \left (4 a^2+9 b^2\right )-4 \left (3 a^4-52 a^2 b^2-16 b^4\right ) \sec (c+d x)\right ) \, dx}{120 b}\\ &=-\frac{a \left (6 a^2-71 b^2\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac{\left (3 a^2-16 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{1}{8} \left (a \left (4 a^2+9 b^2\right )\right ) \int \sec (c+d x) \, dx-\frac{\left (3 a^4-52 a^2 b^2-16 b^4\right ) \int \sec ^2(c+d x) \, dx}{30 b}\\ &=\frac{a \left (4 a^2+9 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{a \left (6 a^2-71 b^2\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac{\left (3 a^2-16 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac{\left (3 a^4-52 a^2 b^2-16 b^4\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b d}\\ &=\frac{a \left (4 a^2+9 b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{\left (3 a^4-52 a^2 b^2-16 b^4\right ) \tan (c+d x)}{30 b d}-\frac{a \left (6 a^2-71 b^2\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac{\left (3 a^2-16 b^2\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac{a (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac{(a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.856563, size = 120, normalized size = 0.63 \[ \frac{15 a \left (4 a^2+9 b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 b \left (5 \left (3 a^2+2 b^2\right ) \tan ^2(c+d x)+15 \left (3 a^2+b^2\right )+3 b^2 \tan ^4(c+d x)\right )+15 a \left (4 a^2+9 b^2\right ) \sec (c+d x)+90 a b^2 \sec ^3(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sec[c + d*x])^3,x]

[Out]

(15*a*(4*a^2 + 9*b^2)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*a*(4*a^2 + 9*b^2)*Sec[c + d*x] + 90*a*b^2*Sec[c
+ d*x]^3 + 8*b*(15*(3*a^2 + b^2) + 5*(3*a^2 + 2*b^2)*Tan[c + d*x]^2 + 3*b^2*Tan[c + d*x]^4)))/(120*d)

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Maple [A]  time = 0.033, size = 206, normalized size = 1.1 \begin{align*}{\frac{{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+2\,{\frac{{a}^{2}b\tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,a{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,a{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,a{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,{b}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sec(d*x+c))^3,x)

[Out]

1/2*a^3*sec(d*x+c)*tan(d*x+c)/d+1/2/d*a^3*ln(sec(d*x+c)+tan(d*x+c))+2/d*a^2*b*tan(d*x+c)+1/d*a^2*b*tan(d*x+c)*
sec(d*x+c)^2+3/4/d*a*b^2*tan(d*x+c)*sec(d*x+c)^3+9/8*a*b^2*sec(d*x+c)*tan(d*x+c)/d+9/8/d*a*b^2*ln(sec(d*x+c)+t
an(d*x+c))+8/15/d*b^3*tan(d*x+c)+1/5/d*b^3*tan(d*x+c)*sec(d*x+c)^4+4/15/d*b^3*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 1.18682, size = 244, normalized size = 1.29 \begin{align*} \frac{240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} b + 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 )} b^{3} - 45 \, a b^{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^{3}{\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+b*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*a^2*b + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*b^3 - 45*a*b^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^3*(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 = 1.75416, size = 421, normalized size = 2.23 \begin{align*} \frac{15 \,{\left (4 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (4 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (15 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{4} + 90 \, a b^{2} \cos \left (d x + c\right ) + 15 \,{\left (4 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, b^{3} + 8 \,{\left (15 \, a^{2} b + 4 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\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+b*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/240*(15*(4*a^3 + 9*a*b^2)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*a^3 + 9*a*b^2)*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 2*(16*(15*a^2*b + 4*b^3)*cos(d*x + c)^4 + 90*a*b^2*cos(d*x + c) + 15*(4*a^3 + 9*a*b^2)*cos(d
*x + c)^3 + 24*b^3 + 8*(15*a^2*b + 4*b^3)*cos(d*x + c)^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*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{3} \sec ^{3}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))**3,x)

[Out]

Integral((a + b*sec(c + d*x))**3*sec(c + d*x)**3, x)

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Giac [B]  time = 1.33344, size = 495, normalized size = 2.62 \begin{align*} \frac{15 \,{\left (4 \, a^{3} + 9 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (4 \, a^{3} + 9 \, a b^{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^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 360 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 225 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 120 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 120 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 960 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 90 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 160 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 1200 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 464 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 120 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 960 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 90 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 160 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 60 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 360 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 225 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 120 \, b^{3} \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+b*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/120*(15*(4*a^3 + 9*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*a^3 + 9*a*b^2)*log(abs(tan(1/2*d*x + 1/
2*c) - 1)) + 2*(60*a^3*tan(1/2*d*x + 1/2*c)^9 - 360*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 225*a*b^2*tan(1/2*d*x + 1/2
*c)^9 - 120*b^3*tan(1/2*d*x + 1/2*c)^9 - 120*a^3*tan(1/2*d*x + 1/2*c)^7 + 960*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 9
0*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 160*b^3*tan(1/2*d*x + 1/2*c)^7 - 1200*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 464*b^3*
tan(1/2*d*x + 1/2*c)^5 + 120*a^3*tan(1/2*d*x + 1/2*c)^3 + 960*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 90*a*b^2*tan(1/2*
d*x + 1/2*c)^3 + 160*b^3*tan(1/2*d*x + 1/2*c)^3 - 60*a^3*tan(1/2*d*x + 1/2*c) - 360*a^2*b*tan(1/2*d*x + 1/2*c)
- 225*a*b^2*tan(1/2*d*x + 1/2*c) - 120*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d