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

Optimal. Leaf size=72 \[ \frac {2 \left (a c^2+b d^2\right ) \tan ^{-1}\left (\frac {c \tan \left (\frac {x}{2}\right )+d}{\sqrt {c^2-d^2}}\right )}{c^2 \sqrt {c^2-d^2}}+\frac {b d \tanh ^{-1}(\cos (x))}{c^2}-\frac {b \cot (x)}{c} \]

[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
)

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Rubi [A]  time = 0.24, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {4233, 3056, 3001, 3770, 2660, 618, 204} \[ \frac {2 \left (a c^2+b d^2\right ) \tan ^{-1}\left (\frac {c \tan \left (\frac {x}{2}\right )+d}{\sqrt {c^2-d^2}}\right )}{c^2 \sqrt {c^2-d^2}}+\frac {b d \tanh ^{-1}(\cos (x))}{c^2}-\frac {b \cot (x)}{c} \]

Antiderivative was successfully verified.

[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 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 3001

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 3056

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)), I
nt[(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] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

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

Rule 4233

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

Rubi steps

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

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Mathematica [A]  time = 0.53, size = 102, normalized size = 1.42 \[ \frac {\csc \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right ) \left (\frac {2 \sin (x) \left (a c^2+b d^2\right ) \tan ^{-1}\left (\frac {c \tan \left (\frac {x}{2}\right )+d}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}-b \left (c \cos (x)+d \sin (x) \left (\log \left (\sin \left (\frac {x}{2}\right )\right )-\log \left (\cos \left (\frac {x}{2}\right )\right )\right )\right )\right )}{2 c^2} \]

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 3.95, size = 332, normalized size = 4.61 \[ \left [-\frac {{\left (a c^{2} + b d^{2}\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \relax (x)^{2} - 2 \, c d \sin \relax (x) - c^{2} - d^{2} + 2 \, {\left (c \cos \relax (x) \sin \relax (x) + d \cos \relax (x)\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \relax (x)^{2} - 2 \, c d \sin \relax (x) - c^{2} - d^{2}}\right ) \sin \relax (x) - {\left (b c^{2} d - b d^{3}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + {\left (b c^{2} d - b d^{3}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + 2 \, {\left (b c^{3} - b c d^{2}\right )} \cos \relax (x)}{2 \, {\left (c^{4} - c^{2} d^{2}\right )} \sin \relax (x)}, -\frac {2 \, {\left (a c^{2} + b d^{2}\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \relax (x) + d}{\sqrt {c^{2} - d^{2}} \cos \relax (x)}\right ) \sin \relax (x) - {\left (b c^{2} d - b d^{3}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + {\left (b c^{2} d - b d^{3}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + 2 \, {\left (b c^{3} - b c d^{2}\right )} \cos \relax (x)}{2 \, {\left (c^{4} - c^{2} d^{2}\right )} \sin \relax (x)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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giac [A]  time = 0.17, size = 110, normalized size = 1.53 \[ -\frac {b d \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{c^{2}} + \frac {b \tan \left (\frac {1}{2} \, x\right )}{2 \, c} + \frac {2 \, {\left (a c^{2} + b d^{2}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, x\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{\sqrt {c^{2} - d^{2}} c^{2}} + \frac {2 \, b d \tan \left (\frac {1}{2} \, x\right ) - b c}{2 \, c^{2} \tan \left (\frac {1}{2} \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.11, size = 120, normalized size = 1.67 \[ \frac {b \tan \left (\frac {x}{2}\right )}{2 c}-\frac {b}{2 c \tan \left (\frac {x}{2}\right )}-\frac {d b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{c^{2}}+\frac {2 \arctan \left (\frac {2 c \tan \left (\frac {x}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) a}{\sqrt {c^{2}-d^{2}}}+\frac {2 \arctan \left (\frac {2 c \tan \left (\frac {x}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right ) b \,d^{2}}{c^{2} \sqrt {c^{2}-d^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b/c*tan(1/2*x)-1/2*b/c/tan(1/2*x)-1/c^2*d*b*ln(tan(1/2*x))+2/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*x)+2*
d)/(c^2-d^2)^(1/2))*a+2/c^2/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*x)+2*d)/(c^2-d^2)^(1/2))*b*d^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((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 details)Is 4*d^2-4*c^2 positive or negative?

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mupad [B]  time = 2.83, size = 463, normalized size = 6.43 \[ \frac {b\,d^3\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )-b\,c^2\,d\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )+a\,c^2\,\mathrm {atan}\left (\frac {a\,c^3\,\sqrt {d^2-c^2}\,1{}\mathrm {i}+b\,d^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,4{}\mathrm {i}+b\,c\,d^2\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+a\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}-b\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,1{}\mathrm {i}}{4\,b\,d^4\,\mathrm {tan}\left (\frac {x}{2}\right )-a\,c^4\,\mathrm {tan}\left (\frac {x}{2}\right )+a\,c^3\,d+2\,b\,c\,d^3-b\,c^3\,d+2\,a\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )-3\,b\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+b\,d^2\,\mathrm {atan}\left (\frac {a\,c^3\,\sqrt {d^2-c^2}\,1{}\mathrm {i}+b\,d^3\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,4{}\mathrm {i}+b\,c\,d^2\,\sqrt {d^2-c^2}\,2{}\mathrm {i}+a\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}-b\,c^2\,d\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {d^2-c^2}\,1{}\mathrm {i}}{4\,b\,d^4\,\mathrm {tan}\left (\frac {x}{2}\right )-a\,c^4\,\mathrm {tan}\left (\frac {x}{2}\right )+a\,c^3\,d+2\,b\,c\,d^3-b\,c^3\,d+2\,a\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )-3\,b\,c^2\,d^2\,\mathrm {tan}\left (\frac {x}{2}\right )}\right )\,\sqrt {d^2-c^2}\,2{}\mathrm {i}}{c^4-c^2\,d^2}-\frac {b\,c^3-b\,c\,d^2}{c^4\,\mathrm {tan}\relax (x)-c^2\,d^2\,\mathrm {tan}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \csc ^{2}{\relax (x )}}{c + d \sin {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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