3.10 \(\int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=122 \[ \frac{3 a^2 \tan ^3(c+d x)}{5 d}+\frac{9 a^2 \tan (c+d x)}{5 d}+\frac{3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac{a^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac{3 a^2 \tan (c+d x) \sec (c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.0946708, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {3788, 3768, 3770, 4046, 3767} \[ \frac{3 a^2 \tan ^3(c+d x)}{5 d}+\frac{9 a^2 \tan (c+d x)}{5 d}+\frac{3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^2 \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac{a^2 \tan (c+d x) \sec ^3(c+d x)}{2 d}+\frac{3 a^2 \tan (c+d x) \sec (c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^2,x]

[Out]

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

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

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

Rubi steps

\begin{align*} \int \sec ^4(c+d x) (a+a \sec (c+d x))^2 \, dx &=\left (2 a^2\right ) \int \sec ^5(c+d x) \, dx+\int \sec ^4(c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac{a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{2} \left (3 a^2\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{5} \left (9 a^2\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac{a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac{a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{1}{4} \left (3 a^2\right ) \int \sec (c+d x) \, dx-\frac{\left (9 a^2\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{3 a^2 \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{9 a^2 \tan (c+d x)}{5 d}+\frac{3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac{a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac{a^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac{3 a^2 \tan ^3(c+d x)}{5 d}\\ \end{align*}

Mathematica [B]  time = 1.51316, size = 487, normalized size = 3.99 \[ -\frac{a^2 \sec (c) \sec ^5(c+d x) \left (80 \sin (2 c+d x)-140 \sin (c+2 d x)-140 \sin (3 c+2 d x)-240 \sin (2 c+3 d x)-30 \sin (3 c+4 d x)-30 \sin (5 c+4 d x)-48 \sin (4 c+5 d x)+75 \cos (2 c+3 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+75 \cos (4 c+3 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+15 \cos (4 c+5 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+15 \cos (6 c+5 d x) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+150 \cos (d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+150 \cos (2 c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-75 \cos (2 c+3 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-75 \cos (4 c+3 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-15 \cos (4 c+5 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-15 \cos (6 c+5 d x) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-400 \sin (d x)\right )}{640 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sec[c + d*x])^2,x]

[Out]

-(a^2*Sec[c]*Sec[c + d*x]^5*(75*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 75*Cos[4*c + 3*d*x
]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 15*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 15
*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 150*Cos[d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x
)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 150*Cos[2*c + d*x]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2
]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - 75*Cos[2*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]
- 75*Cos[4*c + 3*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 15*Cos[4*c + 5*d*x]*Log[Cos[(c + d*x)/2] + Si
n[(c + d*x)/2]] - 15*Cos[6*c + 5*d*x]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 400*Sin[d*x] + 80*Sin[2*c + d
*x] - 140*Sin[c + 2*d*x] - 140*Sin[3*c + 2*d*x] - 240*Sin[2*c + 3*d*x] - 30*Sin[3*c + 4*d*x] - 30*Sin[5*c + 4*
d*x] - 48*Sin[4*c + 5*d*x]))/(640*d)

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Maple [A]  time = 0.036, size = 124, normalized size = 1. \begin{align*}{\frac{6\,{a}^{2}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{2}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{2\,d}}+{\frac{3\,{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{4}\tan \left ( dx+c \right ) }{5\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(a+a*sec(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 0.992602, size = 180, normalized size = 1.48 \begin{align*} \frac{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 )} a^{2} + 40 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} - 15 \, 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 )}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.72649, size = 320, normalized size = 2.62 \begin{align*} \frac{15 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, a^{2} \cos \left (d x + c\right )^{4} + 15 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} \cos \left (d x + c\right )^{2} + 10 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )} \sin \left (d x + c\right )}{40 \, 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)^4*(a+a*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

1/40*(15*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*a^2*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(24*a^2*c
os(d*x + c)^4 + 15*a^2*cos(d*x + c)^3 + 12*a^2*cos(d*x + c)^2 + 10*a^2*cos(d*x + c) + 4*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 \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 \sec ^{5}{\left (c + d x \right )}\, dx + \int \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)**4*(a+a*sec(d*x+c))**2,x)

[Out]

a**2*(Integral(sec(c + d*x)**4, x) + Integral(2*sec(c + d*x)**5, x) + Integral(sec(c + d*x)**6, x))

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Giac [A]  time = 1.38516, size = 186, normalized size = 1.52 \begin{align*} \frac{15 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 70 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 144 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 90 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 65 \, 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}}}{20 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(a+a*sec(d*x+c))^2,x, algorithm="giac")

[Out]

1/20*(15*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*a^2*tan(1/
2*d*x + 1/2*c)^9 - 70*a^2*tan(1/2*d*x + 1/2*c)^7 + 144*a^2*tan(1/2*d*x + 1/2*c)^5 - 90*a^2*tan(1/2*d*x + 1/2*c
)^3 + 65*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d