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

Optimal. Leaf size=70 \[ -\frac {2 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {a x}{b^2}+\frac {\cos (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 70, 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 = {2695, 2735, 2660, 618, 204} \[ -\frac {2 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {a x}{b^2}+\frac {\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2695

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 + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && 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)} \, dx &=\frac {\cos (c+d x)}{b d}+\frac {\int \frac {b+a \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=\frac {a x}{b^2}+\frac {\cos (c+d x)}{b d}-\frac {\left (a^2-b^2\right ) \int \frac {1}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=\frac {a x}{b^2}+\frac {\cos (c+d x)}{b d}-\frac {\left (2 \left (a^2-b^2\right )\right ) \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 {a x}{b^2}+\frac {\cos (c+d x)}{b d}+\frac {\left (4 \left (a^2-b^2\right )\right ) \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 {a x}{b^2}-\frac {2 \sqrt {a^2-b^2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 d}+\frac {\cos (c+d x)}{b d}\\ \end {align*}

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Mathematica [B]  time = 1.40, size = 361, normalized size = 5.16 \[ \frac {\cos (c+d x) \left (2 (a-b) \sqrt {1-\sin (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}}{\sqrt {a+b} \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}}}\right )+\sqrt {a+b} \left (\sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \left (\sqrt {a-b} \sqrt {1-\sin (c+d x)} \sqrt {\frac {b (\sin (c+d x)+1)}{b-a}}+2 \sqrt {b} \sinh ^{-1}\left (\frac {\sqrt {a-b} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}}{\sqrt {2} \sqrt {b}}\right )\right )-2 \sqrt {a-b} \sqrt {1-\sin (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {\frac {b (\sin (c+d x)+1)}{b-a}}}{\sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}}}\right )\right )\right )}{b d \sqrt {a-b} \sqrt {a+b} \sqrt {1-\sin (c+d x)} \sqrt {-\frac {b (\sin (c+d x)-1)}{a+b}} \sqrt {-\frac {b (\sin (c+d x)+1)}{a-b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 214, normalized size = 3.06 \[ \left [\frac {2 \, a d x + 2 \, b \cos \left (d x + c\right ) + \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{2 \, b^{2} d}, \frac {a d x + b \cos \left (d x + c\right ) + \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{b^{2} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 95, normalized size = 1.36 \[ \frac {\frac {{\left (d x + c\right )} a}{b^{2}} - \frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} \sqrt {a^{2} - b^{2}}}{b^{2}} + \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.00, size = 142, normalized size = 2.03 \[ -\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{d \,b^{2} \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \sqrt {a^{2}-b^{2}}}+\frac {2}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/b^2/(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))*a^2+2/d/(a^2-b^2)^(1/2)*arct
an(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+2/d/b/(1+tan(1/2*d*x+1/2*c)^2)+2/d/b^2*a*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(cos(d*x+c)^2/(a+b*sin(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 = 3.92, size = 318, normalized size = 4.54 \[ \frac {2}{b\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a\,\mathrm {atan}\left (\frac {64\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^2-\frac {64\,a^4}{b^2}}+\frac {64\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,a^4-64\,a^2\,b^2}\right )}{b^2\,d}+\frac {2\,\mathrm {atanh}\left (\frac {64\,a^2\,\sqrt {b^2-a^2}}{64\,a^2\,b-\frac {64\,a^4}{b}-128\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+128\,a\,b^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {128\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{64\,a^2-\frac {64\,a^4}{b^2}-\frac {128\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{b}+128\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}+\frac {64\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {b^2-a^2}}{64\,a^4+128\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b-64\,a^2\,b^2-128\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^3}\right )\,\sqrt {b^2-a^2}}{b^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)),x)

[Out]

Timed out

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