3.607 \(\int \frac{(1-\cos ^2(c+d x)) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=118 \[ -\frac{2 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}-\frac{2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{2 \tan (c+d x)}{a^2 d}-\frac{\tan (c+d x)}{a d (a+b \cos (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.399145, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {3056, 3001, 3770, 2659, 205} \[ -\frac{2 \left (a^2-2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^3 d \sqrt{a-b} \sqrt{a+b}}-\frac{2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{2 \tan (c+d x)}{a^2 d}-\frac{\tan (c+d x)}{a d (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^2,x]

[Out]

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

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + 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] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[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{\left (1-\cos ^2(c+d x)\right ) \sec ^2(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=-\frac{\tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (2 \left (a^2-b^2\right )-\left (a^2-b^2\right ) \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{2 \tan (c+d x)}{a^2 d}-\frac{\tan (c+d x)}{a d (a+b \cos (c+d x))}+\frac{\int \frac{\left (-2 b \left (a^2-b^2\right )-a \left (a^2-b^2\right ) \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{2 \tan (c+d x)}{a^2 d}-\frac{\tan (c+d x)}{a d (a+b \cos (c+d x))}-\frac{(2 b) \int \sec (c+d x) \, dx}{a^3}-\frac{\left (a^2-2 b^2\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^3}\\ &=-\frac{2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{2 \tan (c+d x)}{a^2 d}-\frac{\tan (c+d x)}{a d (a+b \cos (c+d x))}-\frac{\left (2 \left (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 d}\\ &=-\frac{2 \left (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 \sqrt{a-b} \sqrt{a+b} d}-\frac{2 b \tanh ^{-1}(\sin (c+d x))}{a^3 d}+\frac{2 \tan (c+d x)}{a^2 d}-\frac{\tan (c+d x)}{a d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.622995, size = 143, normalized size = 1.21 \[ \frac{\frac{2 \left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+\frac{a b \sin (c+d x)}{a+b \cos (c+d x)}+a \tan (c+d x)+2 b \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-2 b \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - Cos[c + d*x]^2)*Sec[c + d*x]^2)/(a + b*Cos[c + d*x])^2,x]

[Out]

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

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Maple [B]  time = 0.058, size = 231, normalized size = 2. \begin{align*} 2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) b}{d{a}^{2} \left ( a \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}-2\,{\frac{1}{da\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+4\,{\frac{{b}^{2}}{d{a}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{d{a}^{2}} \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}}}-{\frac{1}{d{a}^{2}} \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((1-cos(d*x+c)^2)*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 = 2.17619, size = 1403, normalized size = 11.89 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)**2)*sec(d*x+c)**2/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.59419, size = 317, normalized size = 2.69 \begin{align*} -\frac{2 \,{\left (\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}} - \frac{{\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^{2} - 2 \, b^{2}\right )}}{\sqrt{a^{2} - b^{2}} a^{3}} + \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, 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 ) + 2 \, b \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 )} a^{2}}\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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