3.3.27 \(\int \frac {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) [227]

Optimal. Leaf size=75 \[ -\frac {a x}{b^2}+\frac {2 a^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{b^2 \sqrt {a^2-b^2} d}-\frac {\cos (c+d x)}{b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2825, 12, 2814, 2739, 632, 210} \begin {gather*} \frac {2 a^2 \text {ArcTan}\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 {a x}{b^2}-\frac {\cos (c+d x)}{b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*x)/b^2) + (2*a^2*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)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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

Rule 632

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 2739

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 2814

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]

Rule 2825

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b^2
)*(Cos[e + f*x]/(d*f)), x] + Dist[1/d, Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(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 {\sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac {\cos (c+d x)}{b d}-\frac {\int \frac {a \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=-\frac {\cos (c+d x)}{b d}-\frac {a \int \frac {\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 {a^2 \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 a^2\right ) \text {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 a^2\right ) \text {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 a^2 \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}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.13, size = 71, normalized size = 0.95 \begin {gather*} -\frac {a (c+d x)-\frac {2 a^2 \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+b \cos (c+d x)}{b^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.10, size = 90, normalized size = 1.20

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{b^{2} \sqrt {a^{2}-b^{2}}}-\frac {2 \left (\frac {b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}}{d}\)	\(90\)
default	\(\frac {\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{b^{2} \sqrt {a^{2}-b^{2}}}-\frac {2 \left (\frac {b}{1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}+a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{b^{2}}}{d}\)	\(90\)
risch	\(-\frac {a x}{b^{2}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, d \,b^{2}}\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(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 de

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 283, normalized size = 3.77 \begin {gather*} \left [-\frac {\sqrt {-a^{2} + b^{2}} a^{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^{3} - a b^{2}\right )} d x + 2 \, {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d}, -\frac {\sqrt {a^{2} - b^{2}} a^{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^{3} - a b^{2}\right )} d x + {\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{{\left (a^{2} b^{2} - b^{4}\right )} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1690 vs. \(2 (61) = 122\).
time = 175.77, size = 1690, normalized size = 22.53 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*sin(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (-cos(c + d*x)/(b*d), Eq(a, 0)), (-b*d*x*tan(c/2 + d
*x/2)**3/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 +
d*x/2)) - b*d*x*tan(c/2 + d*x/2)/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b
*d*sqrt(b**2)*tan(c/2 + d*x/2)) - 2*b*tan(c/2 + d*x/2)**2/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2
)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)) - 4*b/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt
(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)) + d*x*sqrt(b**2)*tan(c/2 + d*x/2)**2/(b**2*d*tan
(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/2)) + d*x*sqrt(b
**2)/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 - b*d*sqrt(b**2)*tan(c/2 + d*x/
2)) + 2*sqrt(b**2)*tan(c/2 + d*x/2)/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d - b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3
- b*d*sqrt(b**2)*tan(c/2 + d*x/2)), Eq(a, -sqrt(b**2))), (-b*d*x*tan(c/2 + d*x/2)**3/(b**2*d*tan(c/2 + d*x/2)*
*2 + b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 + b*d*sqrt(b**2)*tan(c/2 + d*x/2)) - b*d*x*tan(c/2 + d*x/2)/(
b**2*d*tan(c/2 + d*x/2)**2 + b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 + b*d*sqrt(b**2)*tan(c/2 + d*x/2)) -
2*b*tan(c/2 + d*x/2)**2/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 + b*d*sqrt(b
**2)*tan(c/2 + d*x/2)) - 4*b/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 + b*d*s
qrt(b**2)*tan(c/2 + d*x/2)) - d*x*sqrt(b**2)*tan(c/2 + d*x/2)**2/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d + b*d*sq
rt(b**2)*tan(c/2 + d*x/2)**3 + b*d*sqrt(b**2)*tan(c/2 + d*x/2)) - d*x*sqrt(b**2)/(b**2*d*tan(c/2 + d*x/2)**2 +
b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 + b*d*sqrt(b**2)*tan(c/2 + d*x/2)) - 2*sqrt(b**2)*tan(c/2 + d*x/2
)/(b**2*d*tan(c/2 + d*x/2)**2 + b**2*d + b*d*sqrt(b**2)*tan(c/2 + d*x/2)**3 + b*d*sqrt(b**2)*tan(c/2 + d*x/2))
, Eq(a, sqrt(b**2))), ((x*sin(c + d*x)**2/2 + x*cos(c + d*x)**2/2 - sin(c + d*x)*cos(c + d*x)/(2*d))/a, Eq(b,
0)), (x*sin(c)**2/(a + b*sin(c)), Eq(d, 0)), (a**2*log(tan(c/2 + d*x/2) + b/a - sqrt(-a**2 + b**2)/a)*tan(c/2
+ d*x/2)**2/(b**2*d*sqrt(-a**2 + b**2)*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a**2 + b**2)) + a**2*log(tan(c/2 + d
*x/2) + b/a - sqrt(-a**2 + b**2)/a)/(b**2*d*sqrt(-a**2 + b**2)*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a**2 + b**2)
) - a**2*log(tan(c/2 + d*x/2) + b/a + sqrt(-a**2 + b**2)/a)*tan(c/2 + d*x/2)**2/(b**2*d*sqrt(-a**2 + b**2)*tan
(c/2 + d*x/2)**2 + b**2*d*sqrt(-a**2 + b**2)) - a**2*log(tan(c/2 + d*x/2) + b/a + sqrt(-a**2 + b**2)/a)/(b**2*
d*sqrt(-a**2 + b**2)*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a**2 + b**2)) - a*d*x*sqrt(-a**2 + b**2)*tan(c/2 + d*x
/2)**2/(b**2*d*sqrt(-a**2 + b**2)*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a**2 + b**2)) - a*d*x*sqrt(-a**2 + b**2)/
(b**2*d*sqrt(-a**2 + b**2)*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a**2 + b**2)) - 2*b*sqrt(-a**2 + b**2)/(b**2*d*s
qrt(-a**2 + b**2)*tan(c/2 + d*x/2)**2 + b**2*d*sqrt(-a**2 + b**2)), True))

________________________________________________________________________________________

Giac [A]
time = 3.14, size = 99, normalized size = 1.32 \begin {gather*} \frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \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^{2}}{\sqrt {a^{2} - b^{2}} b^{2}} - \frac {{\left (d x + c\right )} a}{b^{2}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Mupad [B]
time = 2.50, size = 127, normalized size = 1.69 \begin {gather*} -\frac {\cos \left (c+d\,x\right )}{b\,d}-\frac {a\,x}{b^2}-\frac {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 )\,2{}\mathrm {i}}{b^2\,d\,\sqrt {b^2-a^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________