3.49 \(\int (a+a \sec (c+d x))^3 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=98 \[ \frac{a^3 \tan ^3(c+d x)}{d}+\frac{a^3 \tan (c+d x)}{d}-\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{11 a^3 \tan (c+d x) \sec (c+d x)}{8 d}-a^3 x \]

[Out]

-(a^3*x) - (13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*Tan[c + d*x])/d + (11*a^3*Sec[c + d*x]*Tan[c + d*x])/(8
*d) + (a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^3*Tan[c + d*x]^3)/d

________________________________________________________________________________________

Rubi [A]  time = 0.158117, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {3886, 3473, 8, 2611, 3770, 2607, 30, 3768} \[ \frac{a^3 \tan ^3(c+d x)}{d}+\frac{a^3 \tan (c+d x)}{d}-\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac{11 a^3 \tan (c+d x) \sec (c+d x)}{8 d}-a^3 x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

-(a^3*x) - (13*a^3*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*Tan[c + d*x])/d + (11*a^3*Sec[c + d*x]*Tan[c + d*x])/(8
*d) + (a^3*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^3*Tan[c + d*x]^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]

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]

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^3 \tan ^2(c+d x) \, dx &=\int \left (a^3 \tan ^2(c+d x)+3 a^3 \sec (c+d x) \tan ^2(c+d x)+3 a^3 \sec ^2(c+d x) \tan ^2(c+d x)+a^3 \sec ^3(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^3 \int \tan ^2(c+d x) \, dx+a^3 \int \sec ^3(c+d x) \tan ^2(c+d x) \, dx+\left (3 a^3\right ) \int \sec (c+d x) \tan ^2(c+d x) \, dx+\left (3 a^3\right ) \int \sec ^2(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac{a^3 \tan (c+d x)}{d}+\frac{3 a^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac{a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac{1}{4} a^3 \int \sec ^3(c+d x) \, dx-a^3 \int 1 \, dx-\frac{1}{2} \left (3 a^3\right ) \int \sec (c+d x) \, dx+\frac{\left (3 a^3\right ) \operatorname{Subst}\left (\int x^2 \, dx,x,\tan (c+d x)\right )}{d}\\ &=-a^3 x-\frac{3 a^3 \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{a^3 \tan (c+d x)}{d}+\frac{11 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a^3 \tan ^3(c+d x)}{d}-\frac{1}{8} a^3 \int \sec (c+d x) \, dx\\ &=-a^3 x-\frac{13 a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 \tan (c+d x)}{d}+\frac{11 a^3 \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^3 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac{a^3 \tan ^3(c+d x)}{d}\\ \end{align*}

Mathematica [B]  time = 0.834652, size = 230, normalized size = 2.35 \[ -\frac{a^3 \sec ^4(c+d x) \left (-38 \sin (c+d x)-32 \sin (2 (c+d x))-22 \sin (3 (c+d x))-39 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+39 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+4 \cos (2 (c+d x)) \left (-13 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+13 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 d x\right )+\cos (4 (c+d x)) \left (-13 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+13 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+8 d x\right )+24 d x\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3*Tan[c + d*x]^2,x]

[Out]

-(a^3*Sec[c + d*x]^4*(24*d*x - 39*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 39*Log[Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2]] + 4*Cos[2*(c + d*x)]*(8*d*x - 13*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 13*Log[Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2]]) + Cos[4*(c + d*x)]*(8*d*x - 13*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 13*Log[Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2]]) - 38*Sin[c + d*x] - 32*Sin[2*(c + d*x)] - 22*Sin[3*(c + d*x)]))/(64*d)

________________________________________________________________________________________

Maple [A]  time = 0.041, size = 137, normalized size = 1.4 \begin{align*} -{a}^{3}x+{\frac{{a}^{3}\tan \left ( dx+c \right ) }{d}}-{\frac{{a}^{3}c}{d}}+{\frac{13\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{13\,{a}^{3}\sin \left ( dx+c \right ) }{8\,d}}-{\frac{13\,{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3*tan(d*x+c)^2,x)

[Out]

-a^3*x+a^3*tan(d*x+c)/d-1/d*a^3*c+13/8/d*a^3*sin(d*x+c)^3/cos(d*x+c)^2+13/8/d*a^3*sin(d*x+c)-13/8/d*a^3*ln(sec
(d*x+c)+tan(d*x+c))+1/d*a^3*sin(d*x+c)^3/cos(d*x+c)^3+1/4/d*a^3*sin(d*x+c)^3/cos(d*x+c)^4

________________________________________________________________________________________

Maxima [A]  time = 1.89841, size = 198, normalized size = 2.02 \begin{align*} \frac{16 \, a^{3} \tan \left (d x + c\right )^{3} - 16 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} + a^{3}{\left (\frac{2 \,{\left (\sin \left (d x + c\right )^{3} + \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \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 )} - 12 \, 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 )}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^2,x, algorithm="maxima")

[Out]

1/16*(16*a^3*tan(d*x + c)^3 - 16*(d*x + c - tan(d*x + c))*a^3 + a^3*(2*(sin(d*x + c)^3 + sin(d*x + c))/(sin(d*
x + c)^4 - 2*sin(d*x + c)^2 + 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*a^3*(2*sin(d*x + c)/(si
n(d*x + c)^2 - 1) + log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)))/d

________________________________________________________________________________________

Fricas [A]  time = 1.18766, size = 293, normalized size = 2.99 \begin{align*} -\frac{16 \, a^{3} d x \cos \left (d x + c\right )^{4} + 13 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 13 \, a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (11 \, a^{3} \cos \left (d x + c\right )^{2} + 8 \, a^{3} \cos \left (d x + c\right ) + 2 \, a^{3}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^2,x, algorithm="fricas")

[Out]

-1/16*(16*a^3*d*x*cos(d*x + c)^4 + 13*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 13*a^3*cos(d*x + c)^4*log(-si
n(d*x + c) + 1) - 2*(11*a^3*cos(d*x + c)^2 + 8*a^3*cos(d*x + c) + 2*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} \left (\int 3 \tan ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 3 \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )} \sec ^{3}{\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))**3*tan(d*x+c)**2,x)

[Out]

a**3*(Integral(3*tan(c + d*x)**2*sec(c + d*x), x) + Integral(3*tan(c + d*x)**2*sec(c + d*x)**2, x) + Integral(
tan(c + d*x)**2*sec(c + d*x)**3, x) + Integral(tan(c + d*x)**2, x))

________________________________________________________________________________________

Giac [A]  time = 1.66335, size = 178, normalized size = 1.82 \begin{align*} -\frac{8 \,{\left (d x + c\right )} a^{3} + 13 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 13 \, a^{3} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (5 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 13 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, a^{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 )}^{4}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3*tan(d*x+c)^2,x, algorithm="giac")

[Out]

-1/8*(8*(d*x + c)*a^3 + 13*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 13*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1))
- 2*(5*a^3*tan(1/2*d*x + 1/2*c)^7 - 13*a^3*tan(1/2*d*x + 1/2*c)^5 + 3*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*a^3*tan(
1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d