\(\int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx\) [216]

Optimal result

Integrand size = 17, antiderivative size = 72 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\frac {2 \left (a c^2+b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {x}{2}\right )}{\sqrt {c^2-d^2}}\right )}{c^2 \sqrt {c^2-d^2}}+\frac {b d \text {arctanh}(\cos (x))}{c^2}-\frac {b \cot (x)}{c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {4318, 3135, 3080, 3855, 2739, 632, 210} \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\frac {2 \left (a c^2+b d^2\right ) \arctan \left (\frac {c \tan \left (\frac {x}{2}\right )+d}{\sqrt {c^2-d^2}}\right )}{c^2 \sqrt {c^2-d^2}}+\frac {b d \text {arctanh}(\cos (x))}{c^2}-\frac {b \cot (x)}{c} \]

[In]

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

[Out]

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

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 3080

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 3135

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)),
Int[(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] && L
tQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3855

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

Rule 4318

Int[(csc[(a_.) + (b_.)*(x_)]^2*(C_.) + (A_))*(u_), x_Symbol] :> Int[ActivateTrig[u]*((C + A*Sin[a + b*x]^2)/Si
n[a + b*x]^2), x] /; FreeQ[{a, b, A, C}, x] && KnownSineIntegrandQ[u, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\csc ^2(x) \left (b+a \sin ^2(x)\right )}{c+d \sin (x)} \, dx \\ & = -\frac {b \cot (x)}{c}+\frac {\int \frac {\csc (x) (-b d+a c \sin (x))}{c+d \sin (x)} \, dx}{c} \\ & = -\frac {b \cot (x)}{c}-\frac {(b d) \int \csc (x) \, dx}{c^2}+\left (a+\frac {b d^2}{c^2}\right ) \int \frac {1}{c+d \sin (x)} \, dx \\ & = \frac {b d \text {arctanh}(\cos (x))}{c^2}-\frac {b \cot (x)}{c}+\left (2 \left (a+\frac {b d^2}{c^2}\right )\right ) \text {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right ) \\ & = \frac {b d \text {arctanh}(\cos (x))}{c^2}-\frac {b \cot (x)}{c}-\left (4 \left (a+\frac {b d^2}{c^2}\right )\right ) \text {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {x}{2}\right )\right ) \\ & = \frac {2 \left (a+\frac {b d^2}{c^2}\right ) \arctan \left (\frac {d+c \tan \left (\frac {x}{2}\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+\frac {b d \text {arctanh}(\cos (x))}{c^2}-\frac {b \cot (x)}{c} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.42 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\frac {\csc \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right ) \left (\frac {2 \left (a c^2+b d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {x}{2}\right )}{\sqrt {c^2-d^2}}\right ) \sin (x)}{\sqrt {c^2-d^2}}-b \left (c \cos (x)+d \left (-\log \left (\cos \left (\frac {x}{2}\right )\right )+\log \left (\sin \left (\frac {x}{2}\right )\right )\right ) \sin (x)\right )\right )}{2 c^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {\tan \left (\frac {x}{2}\right ) b}{2 c}+\frac {\left (4 a \,c^{2}+4 b \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {x}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 c^{2} \sqrt {c^{2}-d^{2}}}-\frac {b}{2 c \tan \left (\frac {x}{2}\right )}-\frac {b d \ln \left (\tan \left (\frac {x}{2}\right )\right )}{c^{2}}\)	\(90\)
parts	\(\frac {2 a \arctan \left (\frac {2 c \tan \left (\frac {x}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\sqrt {c^{2}-d^{2}}}+b \left (\frac {\tan \left (\frac {x}{2}\right )}{2 c}-\frac {1}{2 c \tan \left (\frac {x}{2}\right )}-\frac {d \ln \left (\tan \left (\frac {x}{2}\right )\right )}{c^{2}}+\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {x}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{c^{2} \sqrt {c^{2}-d^{2}}}\right )\)	\(119\)
risch	\(-\frac {2 i b}{c \left ({\mathrm e}^{2 i x}-1\right )}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{d \sqrt {-c^{2}+d^{2}}}\right ) a}{\sqrt {-c^{2}+d^{2}}}-\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{d \sqrt {-c^{2}+d^{2}}}\right ) b \,d^{2}}{\sqrt {-c^{2}+d^{2}}\, c^{2}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{d \sqrt {-c^{2}+d^{2}}}\right ) a}{\sqrt {-c^{2}+d^{2}}}+\frac {\ln \left ({\mathrm e}^{i x}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{d \sqrt {-c^{2}+d^{2}}}\right ) b \,d^{2}}{\sqrt {-c^{2}+d^{2}}\, c^{2}}-\frac {b d \ln \left ({\mathrm e}^{i x}-1\right )}{c^{2}}+\frac {b d \ln \left ({\mathrm e}^{i x}+1\right )}{c^{2}}\)	\(297\)

[In]

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

[Out]

1/2*tan(1/2*x)*b/c+1/2/c^2*(4*a*c^2+4*b*d^2)/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*x)+2*d)/(c^2-d^2)^(1/2))-
1/2*b/c/tan(1/2*x)-b*d/c^2*ln(tan(1/2*x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (66) = 132\).

Time = 0.59 (sec) , antiderivative size = 332, normalized size of antiderivative = 4.61 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\left [-\frac {{\left (a c^{2} + b d^{2}\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (x\right )^{2} - 2 \, c d \sin \left (x\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (x\right ) \sin \left (x\right ) + d \cos \left (x\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (x\right )^{2} - 2 \, c d \sin \left (x\right ) - c^{2} - d^{2}}\right ) \sin \left (x\right ) - {\left (b c^{2} d - b d^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + {\left (b c^{2} d - b d^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + 2 \, {\left (b c^{3} - b c d^{2}\right )} \cos \left (x\right )}{2 \, {\left (c^{4} - c^{2} d^{2}\right )} \sin \left (x\right )}, -\frac {2 \, {\left (a c^{2} + b d^{2}\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (x\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (x\right )}\right ) \sin \left (x\right ) - {\left (b c^{2} d - b d^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + {\left (b c^{2} d - b d^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) + 2 \, {\left (b c^{3} - b c d^{2}\right )} \cos \left (x\right )}{2 \, {\left (c^{4} - c^{2} d^{2}\right )} \sin \left (x\right )}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\int \frac {a + b \csc ^{2}{\left (x \right )}}{c + d \sin {\left (x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((a+b*csc(x)^2)/(c+d*sin(x)),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*d^2-4*c^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.53 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=-\frac {b d \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{c^{2}} + \frac {b \tan \left (\frac {1}{2} \, x\right )}{2 \, c} + \frac {2 \, {\left (a c^{2} + b d^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, x\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} c^{2}} + \frac {2 \, b d \tan \left (\frac {1}{2} \, x\right ) - b c}{2 \, c^{2} \tan \left (\frac {1}{2} \, x\right )} \]

[In]

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

[Out]

-b*d*log(abs(tan(1/2*x)))/c^2 + 1/2*b*tan(1/2*x)/c + 2*(a*c^2 + b*d^2)*(pi*floor(1/2*x/pi + 1/2)*sgn(c) + arct
an((c*tan(1/2*x) + d)/sqrt(c^2 - d^2)))/(sqrt(c^2 - d^2)*c^2) + 1/2*(2*b*d*tan(1/2*x) - b*c)/(c^2*tan(1/2*x))

Mupad [B] (verification not implemented)

Time = 27.90 (sec) , antiderivative size = 463, normalized size of antiderivative = 6.43 \[ \int \frac {a+b \csc ^2(x)}{c+d \sin (x)} \, dx=\frac {b\,d^3\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )-b\,c^2\,d\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )+a\,c^2\,\mathrm {atan}\left (\frac {a\,c^3\,\sqrt {d^2-c^2}\,1{}\mathrm {i}+b\,d^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,4{}\mathrm {i}+b\,c\,d^2\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+a\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}-b\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,1{}\mathrm {i}}{4\,b\,d^4\,\mathrm {tan}\left (\frac {x}{2}\right )-a\,c^4\,\mathrm {tan}\left (\frac {x}{2}\right )+a\,c^3\,d+2\,b\,c\,d^3-b\,c^3\,d+2\,a\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )-3\,b\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+b\,d^2\,\mathrm {atan}\left (\frac {a\,c^3\,\sqrt {d^2-c^2}\,1{}\mathrm {i}+b\,d^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,4{}\mathrm {i}+b\,c\,d^2\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+a\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}-b\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,1{}\mathrm {i}}{4\,b\,d^4\,\mathrm {tan}\left (\frac {x}{2}\right )-a\,c^4\,\mathrm {tan}\left (\frac {x}{2}\right )+a\,c^3\,d+2\,b\,c\,d^3-b\,c^3\,d+2\,a\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )-3\,b\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}}{c^4-c^2\,d^2}-\frac {b\,c^3-b\,c\,d^2}{c^4\,\mathrm {tan}\left (x\right )-c^2\,d^2\,\mathrm {tan}\left (x\right )} \]

[In]

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

[Out]

(b*d^3*log(tan(x/2)) + a*c^2*atan((a*c^3*(d^2 - c^2)^(1/2)*1i + b*d^3*tan(x/2)*(d^2 - c^2)^(1/2)*4i + b*c*d^2*
(d^2 - c^2)^(1/2)*2i + a*c^2*d*tan(x/2)*(d^2 - c^2)^(1/2)*2i - b*c^2*d*tan(x/2)*(d^2 - c^2)^(1/2)*1i)/(4*b*d^4
*tan(x/2) - a*c^4*tan(x/2) + a*c^3*d + 2*b*c*d^3 - b*c^3*d + 2*a*c^2*d^2*tan(x/2) - 3*b*c^2*d^2*tan(x/2)))*(d^
2 - c^2)^(1/2)*2i + b*d^2*atan((a*c^3*(d^2 - c^2)^(1/2)*1i + b*d^3*tan(x/2)*(d^2 - c^2)^(1/2)*4i + b*c*d^2*(d^
2 - c^2)^(1/2)*2i + a*c^2*d*tan(x/2)*(d^2 - c^2)^(1/2)*2i - b*c^2*d*tan(x/2)*(d^2 - c^2)^(1/2)*1i)/(4*b*d^4*ta
n(x/2) - a*c^4*tan(x/2) + a*c^3*d + 2*b*c*d^3 - b*c^3*d + 2*a*c^2*d^2*tan(x/2) - 3*b*c^2*d^2*tan(x/2)))*(d^2 -
c^2)^(1/2)*2i - b*c^2*d*log(tan(x/2)))/(c^4 - c^2*d^2) - (b*c^3 - b*c*d^2)/(c^4*tan(x) - c^2*d^2*tan(x))