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

Optimal. Leaf size=76 \[ \frac {2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d}+\frac {\tanh ^{-1}(\sin (c+d x))}{a d}-\frac {x}{b} \]

[Out]

-x/b+arctanh(sin(d*x+c))/a/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/b/
d

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Rubi [A]  time = 0.12, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3058, 2659, 205, 3770} \[ \frac {2 \sqrt {a-b} \sqrt {a+b} \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a b d}+\frac {\tanh ^{-1}(\sin (c+d x))}{a d}-\frac {x}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/b) + (2*Sqrt[a - b]*Sqrt[a + b]*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a*b*d) + ArcTanh[Sin[
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 3058

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

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 115, normalized size = 1.51 \[ -\frac {-2 \sqrt {b^2-a^2} \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )+a c+a d x+b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.60, size = 243, normalized size = 3.20 \[ \left [-\frac {2 \, a d x - b \log \left (\sin \left (d x + c\right ) + 1\right ) + b \log \left (-\sin \left (d x + c\right ) + 1\right ) - \sqrt {-a^{2} + b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) + {\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right )}{2 \, a b d}, -\frac {2 \, a d x - b \log \left (\sin \left (d x + c\right ) + 1\right ) + b \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right )}{2 \, a b d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.43, size = 130, normalized size = 1.71 \[ -\frac {\frac {d x + c}{b} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )} \sqrt {a^{2} - b^{2}}}{a b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.13, size = 153, normalized size = 2.01 \[ \frac {2 a \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d b \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b \arctan \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}-\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*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))-2/d/a*b/((a-b)*(a+b))^(1/2)*a
rctan(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))-1/d/a*ln(tan(1/2*d*x+1/2*c)-1)+1/d/a*ln(tan(1/2*d*x+1/2*c)
+1)-2/d/b*arctan(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((1-cos(d*x+c)^2)*sec(d*x+c)/(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 = 1.67, size = 121, normalized size = 1.59 \[ \frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a\,d}-\frac {2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d}-\frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )}\right )\,\sqrt {b^2-a^2}}{a\,b\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a*d) - (2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b*d)
- (2*atanh((sin(c/2 + (d*x)/2)*(b^2 - a^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a + b)))*(b^2 - a^2)^(1/2))/(a*b*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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