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

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Rubi [A]  time = 0.11, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 8

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

Rule 30

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

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 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 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 3770

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

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]

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*}

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Mathematica [B]  time = 6.33, 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]

-1/4*(x*Cos[c + d*x]^2*Sec[c/2 + (d*x)/2]^4*(a + a*Sec[c + d*x])^2) + (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*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
*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] + S
in[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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fricas [A]  time = 0.73, size = 111, normalized size = 1.54 \[ -\frac {6 \, a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, a^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} \cos \left (d x + c\right ) + a^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

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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giac [A]  time = 2.33, size = 99, normalized size = 1.38 \[ -\frac {3 \, {\left (d x + c\right )} a^{2} + 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, {\left (a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

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

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maple [A]  time = 0.49, size = 112, normalized size = 1.56 \[ -a^{2} x +\frac {a^{2} \tan \left (d x +c \right )}{d}-\frac {a^{2} c}{d}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {a^{2} \sin \left (d x +c \right )}{d}-\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}} \]

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+a^2*sin(d*x+c)/d-1/d*a^2*ln(sec(d*x+c)+tan
(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 = 0.82, size = 83, normalized size = 1.15 \[ \frac {2 \, a^{2} \tan \left (d x + c\right )^{3} - 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 3 \, a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \]

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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mupad [B]  time = 1.21, size = 101, normalized size = 1.40 \[ \frac {\frac {4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-a^2\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((4*a^2*tan(c/2 + (d*x)/2)^3)/3 - 4*a^2*tan(c/2 + (d*x)/2))/(d*(3*tan(c/2 + (d*x)/2)^2 - 3*tan(c/2 + (d*x)/2)^
4 + tan(c/2 + (d*x)/2)^6 - 1)) - (2*a^2*atanh(tan(c/2 + (d*x)/2)))/d - a^2*x

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int 2 \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{2}{\left (c + d x \right )}\, dx\right ) \]

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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