3.309 \(\int \frac {\tan ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=85 \[ \frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {x}{a^2}+\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3894, 4061, 12, 3783, 2659, 208} \[ \frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a-b} \sqrt {a+b}}-\frac {x}{a^2}+\frac {\tan (c+d x)}{a d (a+b \sec (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 3894

Int[cot[(c_.) + (d_.)*(x_)]^2*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(-1 + Csc[c + d*x]
^2)*(a + b*Csc[c + d*x])^n, x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 - b^2, 0]

Rule 4061

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(
(A*b^2 + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(
a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*b*(A + C)*(m + 1)*Csc[e + f*x] +
(A*b^2 + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && Int
egerQ[2*m] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.26, size = 80, normalized size = 0.94 \[ -\frac {\frac {2 b \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {a \sin (c+d x)}{a \cos (c+d x)+b}+c+d x}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.53, size = 383, normalized size = 4.51 \[ \left [-\frac {2 \, {\left (a^{3} - a b^{2}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} d x - {\left (a b \cos \left (d x + c\right ) + b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \, {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b - a^{2} b^{3}\right )} d\right )}}, -\frac {{\left (a^{3} - a b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (a^{2} b - b^{3}\right )} d x - {\left (a b \cos \left (d x + c\right ) + b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (a^{3} - a b^{2}\right )} \sin \left (d x + c\right )}{{\left (a^{5} - a^{3} b^{2}\right )} d \cos \left (d x + c\right ) + {\left (a^{4} b - a^{2} b^{3}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(2*(a^3 - a*b^2)*d*x*cos(d*x + c) + 2*(a^2*b - b^3)*d*x - (a*b*cos(d*x + c) + b^2)*sqrt(a^2 - b^2)*log((
2*a*b*cos(d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^
2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 2*(a^3 - a*b^2)*sin(d*x + c))/((a^5 - a^3*b^2)*d*c
os(d*x + c) + (a^4*b - a^2*b^3)*d), -((a^3 - a*b^2)*d*x*cos(d*x + c) + (a^2*b - b^3)*d*x - (a*b*cos(d*x + c) +
b^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (a^3 - a*b^
2)*sin(d*x + c))/((a^5 - a^3*b^2)*d*cos(d*x + c) + (a^4*b - a^2*b^3)*d)]

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giac [A]  time = 0.66, size = 144, normalized size = 1.69 \[ \frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b}{\sqrt {-a^{2} + b^{2}} a^{2}} - \frac {d x + c}{a^{2}} - \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} a}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c
))/sqrt(-a^2 + b^2)))*b/(sqrt(-a^2 + b^2)*a^2) - (d*x + c)/a^2 - 2*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*
c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)*a))/d

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maple [A]  time = 0.43, size = 120, normalized size = 1.41 \[ -\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {2 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/a*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)+2/d*b/a^2/((a-b)*(a+b))^(1/2)*ar
ctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-2/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 2.14, size = 551, normalized size = 6.48 \[ -\frac {2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^2\,d}-\frac {b^2\,\left (a\,\sin \left (c+d\,x\right )+2\,\mathrm {atanh}\left (\frac {2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}-a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+2\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-3\,a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (b\,\left (a^2-b^2\right )+a\,b^2-a^2\,b-a^3+b^3\right )}\right )\,\sqrt {a^2-b^2}\right )-a^3\,\sin \left (c+d\,x\right )+2\,a\,b\,\cos \left (c+d\,x\right )\,\mathrm {atanh}\left (\frac {2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left (a^2-b^2\right )}^{3/2}-a^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+2\,b^5\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}-3\,a^2\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^3\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}+a^4\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a\,b^2-a^3\right )\,\left (b\,\left (a^2-b^2\right )+a\,b^2-a^2\,b-a^3+b^3\right )}\right )\,\sqrt {a^2-b^2}}{a^2\,d\,\left (a^2-b^2\right )\,\left (b+a\,\cos \left (c+d\,x\right )\right )} \]

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]

- (2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^2*d) - (b^2*(a*sin(c + d*x) + 2*atanh((2*b^3*sin(c/2 + (d
*x)/2)*(a^2 - b^2)^(3/2) - a^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 2*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/
2) - 3*a^2*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + a^3*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + a^4*b*sin
(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(b*(a^2 - b^2) + a*b^2 - a^2*b - a^3 + b^
3)))*(a^2 - b^2)^(1/2)) - a^3*sin(c + d*x) + 2*a*b*cos(c + d*x)*atanh((2*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3
/2) - a^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 2*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 3*a^2*b^3*sin(c/
2 + (d*x)/2)*(a^2 - b^2)^(1/2) + a^3*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + a^4*b*sin(c/2 + (d*x)/2)*(a^2
- b^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(b*(a^2 - b^2) + a*b^2 - a^2*b - a^3 + b^3)))*(a^2 - b^2)^(1/2
))/(a^2*d*(a^2 - b^2)*(b + a*cos(c + d*x)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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