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]

(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

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Rubi [A]  time = 0.237838, antiderivative size = 72, normalized size of antiderivative = 1., 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 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]

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 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 3770

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

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 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 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])

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*}

Mathematica [A]  time = 0.518584, 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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Maple [A]  time = 0.038, size = 120, normalized size = 1.7 \begin{align*}{\frac{b}{2\,c}\tan \left ({\frac{x}{2}} \right ) }+2\,{\frac{a}{\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( x/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }+2\,{\frac{b{d}^{2}}{{c}^{2}\sqrt{{c}^{2}-{d}^{2}}}\arctan \left ( 1/2\,{\frac{2\,c\tan \left ( x/2 \right ) +2\,d}{\sqrt{{c}^{2}-{d}^{2}}}} \right ) }-{\frac{b}{2\,c} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{db}{{c}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

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

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

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

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Fricas [B]  time = 9.01471, size = 817, normalized size = 11.35 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \csc ^{2}{\left (x \right )}}{c + d \sin{\left (x \right )}}\, dx \end{align*}

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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Giac [A]  time = 1.16477, size = 149, normalized size = 2.07 \begin{align*} -\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 )} \end{align*}

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))