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

Optimal. Leaf size=70 \[ \frac {a^2 \tan (c+d x)}{d}-a^2 x-\frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a b \tan (c+d x) \sec (c+d x)}{d}+\frac {b^2 \tan ^3(c+d x)}{3 d} \]

[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

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Rubi [A]  time = 0.11, antiderivative size = 70, 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} \[ \frac {a^2 \tan (c+d x)}{d}-a^2 x-\frac {a b \tanh ^{-1}(\sin (c+d x))}{d}+\frac {a b \tan (c+d x) \sec (c+d x)}{d}+\frac {b^2 \tan ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

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

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Mathematica [B]  time = 1.20, size = 201, normalized size = 2.87 \[ \frac {\sec ^3(c+d x) \left (2 \sin (c+d x) \left (\left (3 a^2-b^2\right ) \cos (2 (c+d x))+3 a^2+6 a b \cos (c+d x)+b^2\right )-9 a \cos (c+d x) \left (a (c+d x)-b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )-3 a \cos (3 (c+d x)) \left (a (c+d x)-b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(-9*a*Cos[c + d*x]*(a*(c + d*x) - b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + b*Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]]) - 3*a*Cos[3*(c + d*x)]*(a*(c + d*x) - b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]
+ b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + 2*(3*a^2 + b^2 + 6*a*b*Cos[c + d*x] + (3*a^2 - b^2)*Cos[2*(c +
d*x)])*Sin[c + d*x]))/(12*d)

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fricas [A]  time = 0.63, size = 115, normalized size = 1.64 \[ -\frac {6 \, a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, a b \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a b \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, a b \cos \left (d x + c\right ) + {\left (3 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [B]  time = 1.67, size = 158, normalized size = 2.26 \[ -\frac {3 \, {\left (d x + c\right )} a^{2} + 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, a b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.39, size = 109, normalized size = 1.56 \[ -a^{2} x +\frac {a^{2} \tan \left (d x +c \right )}{d}-\frac {a^{2} c}{d}+\frac {a b \left (\sin ^{3}\left (d x +c \right )\right )}{d \cos \left (d x +c \right )^{2}}+\frac {a b \sin \left (d x +c \right )}{d}-\frac {a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.66, size = 82, normalized size = 1.17 \[ \frac {2 \, b^{2} \tan \left (d x + c\right )^{3} - 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 3 \, a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.67, size = 227, normalized size = 3.24 \[ -\frac {\frac {b^2\,\sin \left (3\,c+3\,d\,x\right )}{12}-\frac {b^2\,\sin \left (c+d\,x\right )}{4}-\frac {a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}-\frac {a^2\,\sin \left (c+d\,x\right )}{4}+\frac {3\,a^2\,\cos \left (c+d\,x\right )\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}-\frac {a\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,a\,b\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (3\,c+3\,d\,x\right )}{2}}{d\,\left (\frac {3\,\cos \left (c+d\,x\right )}{4}+\frac {\cos \left (3\,c+3\,d\,x\right )}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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