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

Optimal. Leaf size=155 \[ \frac{2 b^2 \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (a^2-2 b^2\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{b^2 \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.407008, antiderivative size = 155, normalized size of antiderivative = 1., 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 = {2802, 3055, 3001, 3770, 2659, 205} \[ \frac{2 b^2 \left (3 a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (a^2-2 b^2\right ) \tan (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac{b^2 \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac{2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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 205

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

Rubi steps

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

Mathematica [A]  time = 0.891426, size = 163, normalized size = 1.05 \[ \frac{-\frac{2 b^2 \left (2 b^2-3 a^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{3/2}}-\frac{a b^3 \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}+a \tan (c+d x)+2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b^2*(-3*a^2 + 2*b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(3/2) + 2*b*Log[C
os[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - (a*b^3*Sin[c + d*x])/((a
- b)*(a + b)*(a + b*Cos[c + d*x])) + a*Tan[c + d*x])/(a^3*d)

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Maple [A]  time = 0.115, size = 271, normalized size = 1.8 \begin{align*} 2\,{\frac{b\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }{d{a}^{3}}}-{\frac{1}{{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{1}{{a}^{2}d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-2\,{\frac{b\ln \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }{d{a}^{3}}}-2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) }{{a}^{2}d \left ({a}^{2}-{b}^{2} \right ) \left ( \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 \right ) }}+6\,{\frac{{b}^{2}}{da \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }-4\,{\frac{{b}^{4}}{d{a}^{3} \left ( a-b \right ) \left ( a+b \right ) \sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( 1/2\,dx+c/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*b/a^3*ln(tan(1/2*d*x+1/2*c)-1)-1/d/a^2/(tan(1/2*d*x+1/2*c)-1)-1/d/a^2/(tan(1/2*d*x+1/2*c)+1)-2/d*b/a^3*ln(
tan(1/2*d*x+1/2*c)+1)-2/d*b^3/a^2/(a^2-b^2)*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)+6/d*b^2/a/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-4/d*b^4/a^
3/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))

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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+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.83338, size = 1689, normalized size = 10.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(((3*a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + (3*a^3*b^2 - 2*a*b^4)*cos(d*x + c))*sqrt(-a^2 + b^2)*log((2*a*b*c
os(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 + 2*sqrt(-a^2 + b^2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^
2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) + 2*((a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + (a^5*b -
2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) - 2*((a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x + c)^2 + (a^5
*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(a^6 - 2*a^4*b^2 + a^2*b^4 + (a^5*b - 3*a^3*b
^3 + 2*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b - 2*a^5*b^3 + a^3*b^5)*d*cos(d*x + c)^2 + (a^8 - 2*a^6*b^2 +
a^4*b^4)*d*cos(d*x + c)), (((3*a^2*b^3 - 2*b^5)*cos(d*x + c)^2 + (3*a^3*b^2 - 2*a*b^4)*cos(d*x + c))*sqrt(a^2
- b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c))) - ((a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x +
c)^2 + (a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + ((a^4*b^2 - 2*a^2*b^4 + b^6)*cos(d*x
+ c)^2 + (a^5*b - 2*a^3*b^3 + a*b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) + (a^6 - 2*a^4*b^2 + a^2*b^4 + (a^5*
b - 3*a^3*b^3 + 2*a*b^5)*cos(d*x + c))*sin(d*x + c))/((a^7*b - 2*a^5*b^3 + a^3*b^5)*d*cos(d*x + c)^2 + (a^8 -
2*a^6*b^2 + a^4*b^4)*d*cos(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 + b \cos{\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+b*cos(d*x+c))**2,x)

[Out]

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

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Giac [B]  time = 1.27567, size = 448, normalized size = 2.89 \begin{align*} -\frac{2 \,{\left (\frac{{\left (3 \, a^{2} b^{2} - 2 \, b^{4}\right )}{\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 )}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 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 )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}{\left (a^{4} - a^{2} b^{2}\right )}} + \frac{b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac{b \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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