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

Optimal. Leaf size=82 \[ -\frac {2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d}-\frac {b \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {\tan (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ -\frac {2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d}-\frac {b \tanh ^{-1}(\sin (c+d x))}{a^2 d}+\frac {\tan (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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 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 3770

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

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

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Mathematica [A]  time = 0.28, size = 112, normalized size = 1.37 \[ \frac {-2 \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )+a \tan (c+d x)+b \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.64, size = 297, normalized size = 3.62 \[ \left [-\frac {b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) - \sqrt {-a^{2} + b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - 2 \, a \sin \left (d x + c\right )}{2 \, a^{2} d \cos \left (d x + c\right )}, -\frac {b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - 2 \, a \sin \left (d x + c\right )}{2 \, a^{2} d \cos \left (d x + c\right )}\right ] \]

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)),x, algorithm="fricas")

[Out]

[-1/2*(b*cos(d*x + c)*log(sin(d*x + c) + 1) - b*cos(d*x + c)*log(-sin(d*x + c) + 1) - sqrt(-a^2 + b^2)*cos(d*x
+ c)*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*sin(d*x + c))/(a^2*d*cos(d*x + c))
, -1/2*(b*cos(d*x + c)*log(sin(d*x + c) + 1) - b*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*sqrt(a^2 - b^2)*arcta
n(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^2)*sin(d*x + c)))*cos(d*x + c) - 2*a*sin(d*x + c))/(a^2*d*cos(d*x + c))]

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giac [B]  time = 0.42, size = 150, normalized size = 1.83 \[ -\frac {\frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2}} + \frac {2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a}}{d} \]

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)),x, algorithm="giac")

[Out]

-(b*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - b*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2 - 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)))*sq
rt(a^2 - b^2)/a^2 + 2*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*a))/d

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maple [B]  time = 0.15, size = 177, normalized size = 2.16 \[ -\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) b^{2}}{d \,a^{2} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {1}{d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}-\frac {1}{d a \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}} \]

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)),x)

[Out]

-2/d/((a-b)*(a+b))^(1/2)*arctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))+2/d/a^2/((a-b)*(a+b))^(1/2)*arct
an(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*b^2-1/d/a/(tan(1/2*d*x+1/2*c)-1)+1/d*b/a^2*ln(tan(1/2*d*x+1/2
*c)-1)-1/d/a/(tan(1/2*d*x+1/2*c)+1)-1/d*b/a^2*ln(tan(1/2*d*x+1/2*c)+1)

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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((1-cos(d*x+c)^2)*sec(d*x+c)^2/(a+b*cos(d*x+c)),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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 1.62, size = 436, normalized size = 5.32 \[ \frac {2\,b\,\mathrm {atanh}\left (\frac {64\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a\,b-64\,b^2-\frac {64\,b^3}{a}+\frac {64\,b^4}{a^2}}-\frac {64\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a\,b^2-64\,a^2\,b+64\,b^3-\frac {64\,b^4}{a}}+\frac {64\,b^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{-64\,a^3\,b+64\,a^2\,b^2+64\,a\,b^3-64\,b^4}-\frac {64\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a\,b-64\,b^2-\frac {64\,b^3}{a}+\frac {64\,b^4}{a^2}}\right )}{a^2\,d}-\frac {2\,\mathrm {atanh}\left (\frac {64\,b^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{64\,a^4-128\,a^3\,b+128\,a\,b^3-64\,b^4}+\frac {192\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{128\,a^2\,b-64\,a^3-128\,b^3+\frac {64\,b^4}{a}}+\frac {64\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{128\,a\,b-64\,a^2-\frac {128\,b^3}{a}+\frac {64\,b^4}{a^2}}-\frac {192\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{128\,a\,b-64\,a^2-\frac {128\,b^3}{a}+\frac {64\,b^4}{a^2}}\right )\,\sqrt {b^2-a^2}}{a^2\,d}-\frac {2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*atanh((64*b^2*tan(c/2 + (d*x)/2))/(64*a*b - 64*b^2 - (64*b^3)/a + (64*b^4)/a^2) - (64*b^3*tan(c/2 + (d*x)
/2))/(64*a*b^2 - 64*a^2*b + 64*b^3 - (64*b^4)/a) + (64*b^4*tan(c/2 + (d*x)/2))/(64*a*b^3 - 64*a^3*b - 64*b^4 +
64*a^2*b^2) - (64*a*b*tan(c/2 + (d*x)/2))/(64*a*b - 64*b^2 - (64*b^3)/a + (64*b^4)/a^2)))/(a^2*d) - (2*atanh(
(64*b^3*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(128*a*b^3 - 128*a^3*b + 64*a^4 - 64*b^4) + (192*b^2*tan(c/2 + (
d*x)/2)*(b^2 - a^2)^(1/2))/(128*a^2*b - 64*a^3 - 128*b^3 + (64*b^4)/a) + (64*a*tan(c/2 + (d*x)/2)*(b^2 - a^2)^
(1/2))/(128*a*b - 64*a^2 - (128*b^3)/a + (64*b^4)/a^2) - (192*b*tan(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(128*a*b
- 64*a^2 - (128*b^3)/a + (64*b^4)/a^2))*(b^2 - a^2)^(1/2))/(a^2*d) - (2*tan(c/2 + (d*x)/2))/(a*d*(tan(c/2 + (
d*x)/2)^2 - 1))

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

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)),x)

[Out]

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

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