3.524 \(\int \sec ^5(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=131 \[ \frac{\left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (6 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{\left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{7 a b \sec ^5(c+d x)}{30 d}+\frac{b \sec ^5(c+d x) (a+b \tan (c+d x))}{6 d} \]

[Out]

((6*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(16*d) + (7*a*b*Sec[c + d*x]^5)/(30*d) + ((6*a^2 - b^2)*Sec[c + d*x]*Tan
[c + d*x])/(16*d) + ((6*a^2 - b^2)*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (b*Sec[c + d*x]^5*(a + b*Tan[c + d*x]
))/(6*d)

________________________________________________________________________________________

Rubi [A]  time = 0.115188, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3508, 3486, 3768, 3770} \[ \frac{\left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{\left (6 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac{\left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac{7 a b \sec ^5(c+d x)}{30 d}+\frac{b \sec ^5(c+d x) (a+b \tan (c+d x))}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

((6*a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(16*d) + (7*a*b*Sec[c + d*x]^5)/(30*d) + ((6*a^2 - b^2)*Sec[c + d*x]*Tan
[c + d*x])/(16*d) + ((6*a^2 - b^2)*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (b*Sec[c + d*x]^5*(a + b*Tan[c + d*x]
))/(6*d)

Rule 3508

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(d*Se
c[e + f*x])^m*(a + b*Tan[e + f*x]))/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(d*Sec[e + f*x])^m*(a^2*(m + 1) - b^
2 + a*b*(m + 2)*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && NeQ[a^2 + b^2, 0] && NeQ[m, -1]

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

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]

Rubi steps

\begin{align*} \int \sec ^5(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac{b \sec ^5(c+d x) (a+b \tan (c+d x))}{6 d}+\frac{1}{6} \int \sec ^5(c+d x) \left (6 a^2-b^2+7 a b \tan (c+d x)\right ) \, dx\\ &=\frac{7 a b \sec ^5(c+d x)}{30 d}+\frac{b \sec ^5(c+d x) (a+b \tan (c+d x))}{6 d}+\frac{1}{6} \left (6 a^2-b^2\right ) \int \sec ^5(c+d x) \, dx\\ &=\frac{7 a b \sec ^5(c+d x)}{30 d}+\frac{\left (6 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{b \sec ^5(c+d x) (a+b \tan (c+d x))}{6 d}+\frac{1}{8} \left (6 a^2-b^2\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{7 a b \sec ^5(c+d x)}{30 d}+\frac{\left (6 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{\left (6 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{b \sec ^5(c+d x) (a+b \tan (c+d x))}{6 d}+\frac{1}{16} \left (6 a^2-b^2\right ) \int \sec (c+d x) \, dx\\ &=\frac{\left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{7 a b \sec ^5(c+d x)}{30 d}+\frac{\left (6 a^2-b^2\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac{\left (6 a^2-b^2\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac{b \sec ^5(c+d x) (a+b \tan (c+d x))}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.504837, size = 104, normalized size = 0.79 \[ \frac{15 \left (6 a^2-b^2\right ) \tanh ^{-1}(\sin (c+d x))+10 \left (6 a^2-b^2\right ) \tan (c+d x) \sec ^3(c+d x)+15 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x)+8 b \sec ^5(c+d x) (12 a+5 b \tan (c+d x))}{240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + b*Tan[c + d*x])^2,x]

[Out]

(15*(6*a^2 - b^2)*ArcTanh[Sin[c + d*x]] + 15*(6*a^2 - b^2)*Sec[c + d*x]*Tan[c + d*x] + 10*(6*a^2 - b^2)*Sec[c
+ d*x]^3*Tan[c + d*x] + 8*b*Sec[c + d*x]^5*(12*a + 5*b*Tan[c + d*x]))/(240*d)

________________________________________________________________________________________

Maple [A]  time = 0.052, size = 189, normalized size = 1.4 \begin{align*}{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{16\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{\sin \left ( dx+c \right ){b}^{2}}{16\,d}}-{\frac{{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{2\,ab}{5\,d \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}+{\frac{{a}^{2} \left ( \sec \left ( dx+c \right ) \right ) ^{3}\tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,{a}^{2}\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)^5*(a+b*tan(d*x+c))^2,x)

[Out]

1/6/d*b^2*sin(d*x+c)^3/cos(d*x+c)^6+1/8/d*b^2*sin(d*x+c)^3/cos(d*x+c)^4+1/16/d*b^2*sin(d*x+c)^3/cos(d*x+c)^2+1
/16/d*sin(d*x+c)*b^2-1/16/d*b^2*ln(sec(d*x+c)+tan(d*x+c))+2/5/d*a*b/cos(d*x+c)^5+1/4*a^2*sec(d*x+c)^3*tan(d*x+
c)/d+3/8*a^2*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a^2*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

Maxima [A]  time = 1.41491, size = 243, normalized size = 1.85 \begin{align*} \frac{5 \, b^{2}{\left (\frac{2 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 8 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \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 )} - 30 \, 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 )} + \frac{192 \, a b}{\cos \left (d x + c\right )^{5}}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/480*(5*b^2*(2*(3*sin(d*x + c)^5 - 8*sin(d*x + c)^3 - 3*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*
sin(d*x + c)^2 - 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*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)) + 192*
a*b/cos(d*x + c)^5)/d

________________________________________________________________________________________

Fricas [A]  time = 1.93548, size = 343, normalized size = 2.62 \begin{align*} \frac{15 \,{\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 192 \, a b \cos \left (d x + c\right ) + 10 \,{\left (3 \,{\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (6 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, b^{2}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/480*(15*(6*a^2 - b^2)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(6*a^2 - b^2)*cos(d*x + c)^6*log(-sin(d*x +
c) + 1) + 192*a*b*cos(d*x + c) + 10*(3*(6*a^2 - b^2)*cos(d*x + c)^4 + 2*(6*a^2 - b^2)*cos(d*x + c)^2 + 8*b^2)*
sin(d*x + c))/(d*cos(d*x + c)^6)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{2} \sec ^{5}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+b*tan(d*x+c))**2,x)

[Out]

Integral((a + b*tan(c + d*x))**2*sec(c + d*x)**5, x)

________________________________________________________________________________________

Giac [B]  time = 1.55437, size = 463, normalized size = 3.53 \begin{align*} \frac{15 \,{\left (6 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (6 \, a^{2} - b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{2 \,{\left (150 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 15 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 480 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{10} - 210 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 235 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 480 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{8} + 60 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 390 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 960 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} + 60 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 390 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 960 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 210 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 235 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 96 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 150 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 15 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 96 \, a b\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

1/240*(15*(6*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(6*a^2 - b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1
)) + 2*(150*a^2*tan(1/2*d*x + 1/2*c)^11 + 15*b^2*tan(1/2*d*x + 1/2*c)^11 - 480*a*b*tan(1/2*d*x + 1/2*c)^10 - 2
10*a^2*tan(1/2*d*x + 1/2*c)^9 + 235*b^2*tan(1/2*d*x + 1/2*c)^9 + 480*a*b*tan(1/2*d*x + 1/2*c)^8 + 60*a^2*tan(1
/2*d*x + 1/2*c)^7 + 390*b^2*tan(1/2*d*x + 1/2*c)^7 - 960*a*b*tan(1/2*d*x + 1/2*c)^6 + 60*a^2*tan(1/2*d*x + 1/2
*c)^5 + 390*b^2*tan(1/2*d*x + 1/2*c)^5 + 960*a*b*tan(1/2*d*x + 1/2*c)^4 - 210*a^2*tan(1/2*d*x + 1/2*c)^3 + 235
*b^2*tan(1/2*d*x + 1/2*c)^3 - 96*a*b*tan(1/2*d*x + 1/2*c)^2 + 150*a^2*tan(1/2*d*x + 1/2*c) + 15*b^2*tan(1/2*d*
x + 1/2*c) + 96*a*b)/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d