\(\int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx\) [281]

Optimal result

Integrand size = 21, antiderivative size = 70 \[ \int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx=-a^2 x-\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 \tan (c+d x)}{d}+\frac {a b \sec (c+d x) \tan (c+d x)}{d}+\frac {b^2 \tan ^3(c+d x)}{3 d} \]

[Out]

-a^2*x-a*b*arctanh(sin(d*x+c))/d+a^2*tan(d*x+c)/d+a*b*sec(d*x+c)*tan(d*x+c)/d+1/3*b^2*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3971, 3554, 8, 2691, 3855, 2687, 30} \[ \int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {a^2 \tan (c+d x)}{d}-a^2 x-\frac {a b \text {arctanh}(\sin (c+d x))}{d}+\frac {a b \tan (c+d x) \sec (c+d x)}{d}+\frac {b^2 \tan ^3(c+d x)}{3 d} \]

[In]

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

[Out]

-(a^2*x) - (a*b*ArcTanh[Sin[c + d*x]])/d + (a^2*Tan[c + d*x])/d + (a*b*Sec[c + d*x]*Tan[c + d*x])/d + (b^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 2687

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 2691

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 3554

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 3855

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

Rule 3971

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

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {-3 a^2 \arctan (\tan (c+d x))-3 a b \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 a^2+3 a b \sec (c+d x)+b^2 \tan ^2(c+d x)\right )}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.30

method	result	size
parts	\(\frac {a^{2} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {b^{2} \tan \left (d x +c \right )^{3}}{3 d}+\frac {2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(91\)
derivativedivides	\(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {b^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(92\)
default	\(\frac {a^{2} \left (\tan \left (d x +c \right )-d x -c \right )+2 a b \left (\frac {\sin \left (d x +c \right )^{3}}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {b^{2} \sin \left (d x +c \right )^{3}}{3 \cos \left (d x +c \right )^{3}}}{d}\)	\(92\)
risch	\(-a^{2} x -\frac {2 i \left (3 a b \,{\mathrm e}^{5 i \left (d x +c \right )}-3 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+3 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-6 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-3 a b \,{\mathrm e}^{i \left (d x +c \right )}-3 a^{2}+b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\)	\(143\)

[In]

int((a+b*sec(d*x+c))^2*tan(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.64 \[ \int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {6 \, a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, a b \cos \left (d x + c\right ) + {\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

Sympy [F]

\[ \int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \tan ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.17 \[ \int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx=\frac {2 \, b^{2} \tan \left (d x + c\right )^{3} - 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 3 \, a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 158 vs. \(2 (68) = 136\).

Time = 0.57 (sec) , antiderivative size = 158, normalized size of antiderivative = 2.26 \[ \int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {3 \, {\left (d x + c\right )} a^{2} + 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

[In]

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

[Out]

-1/3*(3*(d*x + c)*a^2 + 3*a*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*a*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) +
2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 3*a*b*tan(1/2*d*x + 1/2*c)^5 - 6*a^2*tan(1/2*d*x + 1/2*c)^3 + 4*b^2*tan(1/2*
d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) + 3*a*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d

Mupad [B] (verification not implemented)

Time = 14.66 (sec) , antiderivative size = 227, normalized size of antiderivative = 3.24 \[ \int (a+b \sec (c+d x))^2 \tan ^2(c+d x) \, dx=-\frac {\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{12}-\frac {b^2\,\sin \left (c+d\,x\right )}{4}-\frac {a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}-\frac {a^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,a^2\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}-\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,a\,b\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]

[In]

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

[Out]

-((b^2*sin(3*c + 3*d*x))/12 - (b^2*sin(c + d*x))/4 - (a^2*sin(3*c + 3*d*x))/4 - (a^2*sin(c + d*x))/4 + (3*a^2*
cos(c + d*x)*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))
*cos(3*c + 3*d*x))/2 - (a*b*sin(2*c + 2*d*x))/2 + (3*a*b*cos(c + d*x)*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)
/2)))/2 + (a*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(3*c + 3*d*x))/2)/(d*((3*cos(c + d*x))/4 + cos(
3*c + 3*d*x)/4))