3.478 \(\int \sec (c+d x) (a+b \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=146 \[ \frac{a b \left (19 a^2+16 b^2\right ) \tan (c+d x)}{6 d}+\frac{\left (24 a^2 b^2+8 a^4+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \left (26 a^2+9 b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{b \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{7 a b \tan (c+d x) (a+b \sec (c+d x))^2}{12 d} \]

[Out]

((8*a^4 + 24*a^2*b^2 + 3*b^4)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*b*(19*a^2 + 16*b^2)*Tan[c + d*x])/(6*d) + (b^2
*(26*a^2 + 9*b^2)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + (7*a*b*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + (b*
(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.242574, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {3830, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac{a b \left (19 a^2+16 b^2\right ) \tan (c+d x)}{6 d}+\frac{\left (24 a^2 b^2+8 a^4+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \left (26 a^2+9 b^2\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{b \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}+\frac{7 a b \tan (c+d x) (a+b \sec (c+d x))^2}{12 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + b*Sec[c + d*x])^4,x]

[Out]

((8*a^4 + 24*a^2*b^2 + 3*b^4)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*b*(19*a^2 + 16*b^2)*Tan[c + d*x])/(6*d) + (b^2
*(26*a^2 + 9*b^2)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + (7*a*b*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) + (b*
(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d)

Rule 3830

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

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 (c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac{b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (4 a^2+3 b^2+7 a b \sec (c+d x)\right ) \, dx\\ &=\frac{7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{12} \int \sec (c+d x) (a+b \sec (c+d x)) \left (a \left (12 a^2+23 b^2\right )+b \left (26 a^2+9 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b^2 \left (26 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{24} \int \sec (c+d x) \left (3 \left (8 a^4+24 a^2 b^2+3 b^4\right )+4 a b \left (19 a^2+16 b^2\right ) \sec (c+d x)\right ) \, dx\\ &=\frac{b^2 \left (26 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{6} \left (a b \left (19 a^2+16 b^2\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (8 a^4+24 a^2 b^2+3 b^4\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (8 a^4+24 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{b^2 \left (26 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{\left (a b \left (19 a^2+16 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac{\left (8 a^4+24 a^2 b^2+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a b \left (19 a^2+16 b^2\right ) \tan (c+d x)}{6 d}+\frac{b^2 \left (26 a^2+9 b^2\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{7 a b (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac{b (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.515846, size = 101, normalized size = 0.69 \[ \frac{3 \left (24 a^2 b^2+8 a^4+3 b^4\right ) \tanh ^{-1}(\sin (c+d x))+b \tan (c+d x) \left (32 a \left (3 \left (a^2+b^2\right )+b^2 \tan ^2(c+d x)\right )+9 b \left (8 a^2+b^2\right ) \sec (c+d x)+6 b^3 \sec ^3(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])^4,x]

[Out]

(3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*ArcTanh[Sin[c + d*x]] + b*Tan[c + d*x]*(9*b*(8*a^2 + b^2)*Sec[c + d*x] + 6*b^3
*Sec[c + d*x]^3 + 32*a*(3*(a^2 + b^2) + b^2*Tan[c + d*x]^2)))/(24*d)

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Maple [A]  time = 0.034, size = 188, normalized size = 1.3 \begin{align*}{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{{a}^{3}b\tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+3\,{\frac{{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{8\,a{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,a{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{b}^{4}\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)*(a+b*sec(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 1.19612, size = 243, normalized size = 1.66 \begin{align*} \frac{64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a b^{3} - 3 \, b^{4}{\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 )} - 72 \, a^{2} b^{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^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 192 \, a^{3} b \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+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

1/48*(64*(tan(d*x + c)^3 + 3*tan(d*x + c))*a*b^3 - 3*b^4*(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)) - 72*a^2*b^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)) + 48*a^4*log(sec(d*x + c) + tan(d*x + c))
+ 192*a^3*b*tan(d*x + c))/d

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Fricas [A]  time = 1.72338, size = 394, normalized size = 2.7 \begin{align*} \frac{3 \,{\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (32 \, a b^{3} \cos \left (d x + c\right ) + 6 \, b^{4} + 32 \,{\left (3 \, a^{3} b + 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} + 9 \,{\left (8 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{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+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/48*(3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*cos
(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(32*a*b^3*cos(d*x + c) + 6*b^4 + 32*(3*a^3*b + 2*a*b^3)*cos(d*x + c)^3
+ 9*(8*a^2*b^2 + b^4)*cos(d*x + c)^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*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{4} \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sec(d*x+c))**4,x)

[Out]

Integral((a + b*sec(c + d*x))**4*sec(c + d*x), x)

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Giac [B]  time = 1.3343, size = 486, normalized size = 3.33 \begin{align*} \frac{3 \,{\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (8 \, a^{4} + 24 \, a^{2} b^{2} + 3 \, b^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (96 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 96 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 15 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 288 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 160 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 9 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 288 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 160 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 9 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 96 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 72 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 96 \, a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 15 \, b^{4} \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+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/24*(3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(8*a^4 + 24*a^2*b^2 + 3*b^4)*log(a
bs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(96*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 96*a*
b^3*tan(1/2*d*x + 1/2*c)^7 - 15*b^4*tan(1/2*d*x + 1/2*c)^7 - 288*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 72*a^2*b^2*tan
(1/2*d*x + 1/2*c)^5 - 160*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 9*b^4*tan(1/2*d*x + 1/2*c)^5 + 288*a^3*b*tan(1/2*d*x
+ 1/2*c)^3 + 72*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 160*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 9*b^4*tan(1/2*d*x + 1/2*c)
^3 - 96*a^3*b*tan(1/2*d*x + 1/2*c) - 72*a^2*b^2*tan(1/2*d*x + 1/2*c) - 96*a*b^3*tan(1/2*d*x + 1/2*c) - 15*b^4*
tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d