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[(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)

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Rubi [A]  time = 0.334015, antiderivative size = 131, normalized size of antiderivative = 1., 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 4397

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

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 12

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

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 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]

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*}

Mathematica [A]  time = 0.762765, size = 121, normalized size = 0.92 \[ \frac{\frac{12 a^2 b \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{\frac{2 a^3 \sin (c+d x)}{(a+b)^2 (a \cos (c+d x)+b)}+\tan \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2}}{2 d} \]

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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Maple [A]  time = 0.188, size = 155, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{1}{2\,{a}^{2}-4\,ab+2\,{b}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+2\,{\frac{{a}^{2}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}} \left ( -{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) }{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b}}-3\,{\frac{b}{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \right ) }-{\frac{1}{2\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \right ) } \end{align*}

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-tan(1/2*d*x+1/2*c)^2*b-a-b)-3*b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 0.583734, size = 1131, normalized size = 8.63 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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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Giac [B]  time = 1.20543, size = 383, normalized size = 2.92 \begin{align*} \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} \end{align*}

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