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

Optimal. Leaf size=84 \[ \frac {2 a \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \sqrt {a^2-b^2}}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}-\frac {x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2693, 2735, 2660, 618, 204} \[ \frac {2 a \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d \sqrt {a^2-b^2}}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}-\frac {x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 2693

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2735

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x)}{(a+b \sin (c+d x))^2} \, dx &=-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}-\frac {\int \frac {\sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=-\frac {x}{b^2}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac {a \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=-\frac {x}{b^2}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 d}\\ &=-\frac {x}{b^2}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}-\frac {(4 a) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{b^2 d}\\ &=-\frac {x}{b^2}+\frac {2 a \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}-\frac {\cos (c+d x)}{b d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 2.69, size = 414, normalized size = 4.93 \[ \frac {\cos (c+d x) \left (\sqrt {a+b} \left ((b-a) \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \left (\sqrt {a-b} (a+b) \sqrt {1-\sin (c+d x)} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}+2 \sqrt {b} (a+b \sin (c+d x)) \sinh ^{-1}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}}{\sqrt {2} \sqrt {b}}\right )\right )+2 a \sqrt {a-b} \sqrt {1-\sin (c+d x)} (a+b \sin (c+d x)) \tanh ^{-1}\left (\frac {\sqrt {\frac {b (\sin (c+d x)+1)}{b-a}}}{\sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}}}\right )\right )-2 a (a-b) \sqrt {1-\sin (c+d x)} (a+b \sin (c+d x)) \tanh ^{-1}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}}{\sqrt {a+b} \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}}}\right )\right )}{b d (a-b)^{3/2} (a+b)^{3/2} \sqrt {1-\sin (c+d x)} \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}} (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]*(-2*a*(a - b)*ArcTanh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])/(Sqrt[a + b]*Sqrt[-(
(b*(-1 + Sin[c + d*x]))/(a + b))])]*Sqrt[1 - Sin[c + d*x]]*(a + b*Sin[c + d*x]) + Sqrt[a + b]*(2*a*Sqrt[a - b]
*ArcTanh[Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)]/Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]]*Sqrt[1 - Sin[c + d*x]
]*(a + b*Sin[c + d*x]) + (-a + b)*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*(Sqrt[a - b]*(a + b)*Sqrt[1 - Sin[c
+ d*x]]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))] + 2*Sqrt[b]*ArcSinh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[c + d*x])
)/(a - b))])/(Sqrt[2]*Sqrt[b])]*(a + b*Sin[c + d*x])))))/((a - b)^(3/2)*b*(a + b)^(3/2)*d*Sqrt[1 - Sin[c + d*x
]]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))]*(a + b*Sin[c + d*x]))

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fricas [A]  time = 0.50, size = 388, normalized size = 4.62 \[ \left [-\frac {2 \, {\left (a^{2} b - b^{3}\right )} d x \sin \left (d x + c\right ) + 2 \, {\left (a^{3} - a b^{2}\right )} d x + {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{3} - b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} d\right )}}, -\frac {{\left (a^{2} b - b^{3}\right )} d x \sin \left (d x + c\right ) + {\left (a^{3} - a b^{2}\right )} d x + {\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{{\left (a^{2} b^{3} - b^{5}\right )} d \sin \left (d x + c\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.38, size = 126, normalized size = 1.50 \[ \frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} a}{\sqrt {a^{2} - b^{2}} b^{2}} - \frac {d x + c}{b^{2}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )} a b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*a/(sqrt(a^
2 - b^2)*b^2) - (d*x + c)/b^2 - 2*(b*tan(1/2*d*x + 1/2*c) + a)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x +
1/2*c) + a)*a*b))/d

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maple [A]  time = 0.23, size = 153, normalized size = 1.82 \[ -\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) a}-\frac {2}{d b \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right )}+\frac {2 a \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,b^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(cos(d*x+c)^2/(a+b*sin(d*x+c))^2,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 = 5.56, size = 329, normalized size = 3.92 \[ -\frac {b^2\,\sin \left (c+d\,x\right )+\frac {\left (2\,a^3\,\mathrm {atan}\left (\frac {\left (-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )-a^2\,\sqrt {b^2-a^2}\,\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}}-\frac {b\,\left (a\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+a\,\cos \left (c+d\,x\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}-2\,a^2\,\mathrm {atan}\left (\frac {\left (-\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2+\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^2\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}\,\left (a\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+2\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}\right )\,\sin \left (c+d\,x\right )+a\,\sin \left (c+d\,x\right )\,\sqrt {b^2-a^2}\,\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{\sqrt {b^2-a^2}}}{a\,b^2\,d\,\left (a+b\,\sin \left (c+d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b^2*sin(c + d*x) + ((2*a^3*atan(((2*b^2*sin(c/2 + (d*x)/2) - a^2*sin(c/2 + (d*x)/2) + a*b*cos(c/2 + (d*x)/2)
)*1i)/((b^2 - a^2)^(1/2)*(a*cos(c/2 + (d*x)/2) + 2*b*sin(c/2 + (d*x)/2)))) - a^2*(b^2 - a^2)^(1/2)*(c + d*x)*1
i)*1i)/(b^2 - a^2)^(1/2) - (b*(a*(b^2 - a^2)^(1/2)*1i + a*cos(c + d*x)*(b^2 - a^2)^(1/2)*1i - 2*a^2*atan(((2*b
^2*sin(c/2 + (d*x)/2) - a^2*sin(c/2 + (d*x)/2) + a*b*cos(c/2 + (d*x)/2))*1i)/((b^2 - a^2)^(1/2)*(a*cos(c/2 + (
d*x)/2) + 2*b*sin(c/2 + (d*x)/2))))*sin(c + d*x) + a*sin(c + d*x)*(b^2 - a^2)^(1/2)*(c + d*x)*1i)*1i)/(b^2 - a
^2)^(1/2))/(a*b^2*d*(a + b*sin(c + d*x)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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