3.31 \(\int (a+a \sec (c+d x))^2 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=72 \[ \frac{a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}-\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \tan (c+d x) \sec (c+d x)}{d}-a^2 x \]

[Out]

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

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Rubi [A]  time = 0.109061, antiderivative size = 72, 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 = {3886, 3473, 8, 2611, 3770, 2607, 30} \[ \frac{a^2 \tan ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x)}{d}-\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{a^2 \tan (c+d x) \sec (c+d x)}{d}-a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2611

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

Rule 3770

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

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [B]  time = 6.30994, size = 773, normalized size = 10.74 \[ -\frac{1}{4} x \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2+\frac{\sin \left (\frac{d x}{2}\right ) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{6 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{\sin \left (\frac{d x}{2}\right ) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{6 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{\left (7 \cos \left (\frac{c}{2}\right )-5 \sin \left (\frac{c}{2}\right )\right ) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{48 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}+\frac{\left (-5 \sin \left (\frac{c}{2}\right )-7 \cos \left (\frac{c}{2}\right )\right ) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{48 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^2}+\frac{\sin \left (\frac{d x}{2}\right ) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{24 d \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3}+\frac{\sin \left (\frac{d x}{2}\right ) \cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2}{24 d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )^3}+\frac{\cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \log \left (\cos \left (\frac{c}{2}+\frac{d x}{2}\right )-\sin \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{4 d}-\frac{\cos ^2(c+d x) \sec ^4\left (\frac{c}{2}+\frac{d x}{2}\right ) (a \sec (c+d x)+a)^2 \log \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/4 + (Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2] - S
in[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(4*d) - (Cos[c + d*x]^2*Log[Cos[c/2 + (d*x)/2]
+ Sin[c/2 + (d*x)/2]]*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2)/(4*d) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]
^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(24*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^
3) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(7*Cos[c/2] - 5*Sin[c/2]))/(48*d*(Cos[c/2] -
Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d
*x])^2*Sin[(d*x)/2])/(6*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (Cos[c + d*x]^2*S
ec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(24*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin
[c/2 + (d*x)/2])^3) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2*(-7*Cos[c/2] - 5*Sin[c/2]))/
(48*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2])^2) + (Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^
4*(a + a*Sec[c + d*x])^2*Sin[(d*x)/2])/(6*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]))

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Maple [A]  time = 0.038, size = 112, normalized size = 1.6 \begin{align*} -{a}^{2}x+{\frac{{a}^{2}\tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{2}c}{d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{d}}-{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a^2*x+a^2*tan(d*x+c)/d-1/d*a^2*c+1/d*a^2*sin(d*x+c)^3/cos(d*x+c)^2+1/d*a^2*sin(d*x+c)-1/d*a^2*ln(sec(d*x+c)+t
an(d*x+c))+1/3/d*a^2*sin(d*x+c)^3/cos(d*x+c)^3

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Maxima [A]  time = 1.8317, size = 112, normalized size = 1.56 \begin{align*} \frac{2 \, a^{2} \tan \left (d x + c\right )^{3} - 6 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 3 \, 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 )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a^2*tan(d*x + c)^3 - 6*(d*x + c - tan(d*x + c))*a^2 - 3*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)))/d

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Fricas [A]  time = 0.966053, size = 284, normalized size = 3.94 \begin{align*} -\frac{6 \, a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (2 \, a^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*a^2*d*x*cos(d*x + c)^3 + 3*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*a^2*cos(d*x + c)^3*log(-sin(d*
x + c) + 1) - 2*(2*a^2*cos(d*x + c)^2 + 3*a^2*cos(d*x + c) + a^2)*sin(d*x + c))/(d*cos(d*x + c)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(2*tan(c + d*x)**2*sec(c + d*x), x) + Integral(tan(c + d*x)**2*sec(c + d*x)**2, x) + Integral(ta
n(c + d*x)**2, x))

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Giac [A]  time = 1.82555, size = 134, normalized size = 1.86 \begin{align*} -\frac{3 \,{\left (d x + c\right )} a^{2} + 3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{4 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 3 \, 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 )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(d*x + c)*a^2 + 3*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a^2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) -
4*(a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d