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

Optimal. Leaf size=131 \[ \frac {a \csc (c+d x)}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\csc (c+d x) \left (3 a b-\left (2 a^2+b^2\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )^2}-\frac {6 a^2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}} \]

[Out]

-6*a^2*b*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/d+a*csc(d*x+c)/(a^2-b^2)/
d/(b+a*cos(d*x+c))+(3*a*b-(2*a^2+b^2)*cos(d*x+c))*csc(d*x+c)/(a^2-b^2)^2/d

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Rubi [A]  time = 0.33, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {4397, 2694, 2866, 12, 2659, 208} \[ \frac {a \csc (c+d x)}{d \left (a^2-b^2\right ) (a \cos (c+d x)+b)}+\frac {\csc (c+d x) \left (3 a b-\left (2 a^2+b^2\right ) \cos (c+d x)\right )}{d \left (a^2-b^2\right )^2}-\frac {6 a^2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a^2*b*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/((a - b)^(5/2)*(a + b)^(5/2)*d) + (a*Csc[c + d*
x])/((a^2 - b^2)*d*(b + a*Cos[c + d*x])) + ((3*a*b - (2*a^2 + b^2)*Cos[c + d*x])*Csc[c + d*x])/((a^2 - b^2)^2*
d)

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 2694

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1))/(f*g*(a^2 - b^2)*(m + 1)), x] + Dist[1/((a^2 - b^2)*(m +
1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(a*(m + 1) - b*(m + p + 2)*Sin[e + f*x]), x], x] /; F
reeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*p]

Rule 2866

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c - a*d - (a*c -
b*d)*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p
+ 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] && IntegerQ[2*m]

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx &=\int \frac {\csc ^2(c+d x)}{(b+a \cos (c+d x))^2} \, dx\\ &=\frac {a \csc (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\int \frac {(-b+2 a \cos (c+d x)) \csc ^2(c+d x)}{b+a \cos (c+d x)} \, dx}{a^2-b^2}\\ &=\frac {a \csc (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\left (3 a b-\left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{\left (a^2-b^2\right )^2 d}+\frac {\int -\frac {3 a^2 b}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {a \csc (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\left (3 a b-\left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{\left (a^2-b^2\right )^2 d}-\frac {\left (3 a^2 b\right ) \int \frac {1}{b+a \cos (c+d x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=\frac {a \csc (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\left (3 a b-\left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{\left (a^2-b^2\right )^2 d}-\frac {\left (6 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{a+b+(-a+b) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{\left (a^2-b^2\right )^2 d}\\ &=-\frac {6 a^2 b \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}+\frac {a \csc (c+d x)}{\left (a^2-b^2\right ) d (b+a \cos (c+d x))}+\frac {\left (3 a b-\left (2 a^2+b^2\right ) \cos (c+d x)\right ) \csc (c+d x)}{\left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 121, normalized size = 0.92 \[ \frac {\frac {\frac {2 a^3 \sin (c+d x)}{(a+b)^2 (a \cos (c+d x)+b)}+\tan \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2}+\frac {12 a^2 b \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((12*a^2*b*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - Cot[(c + d*x)/2]/(a + b)^
2 + ((2*a^3*Sin[c + d*x])/((a + b)^2*(b + a*Cos[c + d*x])) + Tan[(c + d*x)/2])/(a - b)^2)/(2*d)

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fricas [A]  time = 0.71, size = 516, normalized size = 3.94 \[ \left [\frac {2 \, a^{5} + 2 \, a^{3} b^{2} - 4 \, a b^{4} + 3 \, {\left (a^{3} b \cos \left (d x + c\right ) + a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}, \frac {a^{5} + a^{3} b^{2} - 2 \, a b^{4} - 3 \, {\left (a^{3} b \cos \left (d x + c\right ) + a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )}{{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*a^5 + 2*a^3*b^2 - 4*a*b^4 + 3*(a^3*b*cos(d*x + c) + a^2*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))*sin(d*x + c) - 2*(2*a^5 - a^3*b^2 - a*b^4)*cos(d*x + c)^2 + 2*(a^4*b - 2
*a^2*b^3 + b^5)*cos(d*x + c))/(((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*x + c) + (a^6*b - 3*a^4*b^3 + 3*
a^2*b^5 - b^7)*d)*sin(d*x + c)), (a^5 + a^3*b^2 - 2*a*b^4 - 3*(a^3*b*cos(d*x + c) + a^2*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)))*sin(d*x + c) - (2*a^5 - a^3*b^2 - a*
b^4)*cos(d*x + c)^2 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))/(((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*cos(d*
x + c) + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d)*sin(d*x + c))]

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giac [B]  time = 0.56, size = 284, normalized size = 2.17 \[ \frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a^{2} b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} - 2 \, a b + b^{2}} - \frac {5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} + a^{2} b + a b^{2} - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(12*(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)))*a^2*b/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + tan(1/2*d*x + 1/2*c)/(a^2 - 2*a*b
+ b^2) - (5*a^3*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 3*a*b^2*tan(1/2*d*x + 1/2*c)^2 - b^3
*tan(1/2*d*x + 1/2*c)^2 - a^3 + a^2*b + a*b^2 - b^3)/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^3 - b*ta
n(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))))/d

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maple [A]  time = 0.22, size = 155, normalized size = 1.18 \[ \frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a^{2}-4 a b +2 b^{2}}+\frac {2 a^{2} \left (-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b}-\frac {3 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{2 \left (a +b \right )^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(1/2/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)+2*a^2/(a-b)^2/(a+b)^2*(-a*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2
*a-b*tan(1/2*d*x+1/2*c)^2-a-b)-3*b/((a+b)*(a-b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a+b)*(a-b))^(1/2)))-
1/2/(a+b)^2/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(sec(d*x+c)^2/(a*sin(d*x+c)+b*tan(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 = 1.08, size = 215, normalized size = 1.64 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2\,d\,{\left (a-b\right )}^2}+\frac {\frac {a^2-2\,a\,b+b^2}{a+b}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (5\,a^3-3\,a^2\,b+3\,a\,b^2-b^3\right )}{{\left (a+b\right )}^2}}{d\,\left (\left (2\,a^3-6\,a^2\,b+6\,a\,b^2-2\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,a^3+2\,a^2\,b+2\,a\,b^2-2\,b^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}-\frac {6\,a^2\,b\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{3/2}}\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*(a*sin(c + d*x) + b*tan(c + d*x))^2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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