3.351 \(\int \frac {A+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx\)

Optimal. Leaf size=116 \[ \frac {A b \sin (x)-A c \cos (x)+b C}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}-\frac {A \tanh ^{-1}\left (\frac {c \cos (x)-b \sin (x)}{\sqrt {b^2+c^2}}\right )}{2 \left (b^2+c^2\right )^{3/2}}-\frac {c^2 C \cos (x)-b c C \sin (x)}{\left (b^2+c^2\right )^2 (b \cos (x)+c \sin (x))} \]

[Out]

-1/2*A*arctanh((c*cos(x)-b*sin(x))/(b^2+c^2)^(1/2))/(b^2+c^2)^(3/2)+1/2*(b*C-A*c*cos(x)+A*b*sin(x))/(b^2+c^2)/
(b*cos(x)+c*sin(x))^2+(-c^2*C*cos(x)+b*c*C*sin(x))/(b^2+c^2)^2/(b*cos(x)+c*sin(x))

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Rubi [A]  time = 0.11, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3157, 3153, 3074, 206} \[ \frac {A b \sin (x)-A c \cos (x)+b C}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}-\frac {A \tanh ^{-1}\left (\frac {c \cos (x)-b \sin (x)}{\sqrt {b^2+c^2}}\right )}{2 \left (b^2+c^2\right )^{3/2}}-\frac {c^2 C \cos (x)-b c C \sin (x)}{\left (b^2+c^2\right )^2 (b \cos (x)+c \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Sin[x])/(b*Cos[x] + c*Sin[x])^3,x]

[Out]

-(A*ArcTanh[(c*Cos[x] - b*Sin[x])/Sqrt[b^2 + c^2]])/(2*(b^2 + c^2)^(3/2)) + (b*C - A*c*Cos[x] + A*b*Sin[x])/(2
*(b^2 + c^2)*(b*Cos[x] + c*Sin[x])^2) - (c^2*C*Cos[x] - b*c*C*Sin[x])/((b^2 + c^2)^2*(b*Cos[x] + c*Sin[x]))

Rule 206

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

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3153

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.) + (C_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + cos[(d_.) + (e_.)*(x_)]*(
b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^2, x_Symbol] :> Simp[(c*B - b*C - (a*C - c*A)*Cos[d + e*x] + (a*B - b*A)
*Sin[d + e*x])/(e*(a^2 - b^2 - c^2)*(a + b*Cos[d + e*x] + c*Sin[d + e*x])), x] + Dist[(a*A - b*B - c*C)/(a^2 -
b^2 - c^2), Int[1/(a + b*Cos[d + e*x] + c*Sin[d + e*x]), x], x] /; FreeQ[{a, b, c, d, e, A, B, C}, x] && NeQ[
a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B - c*C, 0]

Rule 3157

Int[((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(d_.) + (e
_.)*(x_)]), x_Symbol] :> Simp[((b*C + (a*C - c*A)*Cos[d + e*x] + b*A*Sin[d + e*x])*(a + b*Cos[d + e*x] + c*Sin
[d + e*x])^(n + 1))/(e*(n + 1)*(a^2 - b^2 - c^2)), x] + Dist[1/((n + 1)*(a^2 - b^2 - c^2)), Int[(a + b*Cos[d +
e*x] + c*Sin[d + e*x])^(n + 1)*Simp[(n + 1)*(a*A - c*C) - (n + 2)*b*A*Cos[d + e*x] + (n + 2)*(a*C - c*A)*Sin[
d + e*x], x], x], x] /; FreeQ[{a, b, c, d, e, A, C}, x] && LtQ[n, -1] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[n, -2]

Rubi steps

\begin {align*} \int \frac {A+C \sin (x)}{(b \cos (x)+c \sin (x))^3} \, dx &=\frac {b C-A c \cos (x)+A b \sin (x)}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}+\frac {\int \frac {2 c C+A b \cos (x)+A c \sin (x)}{(b \cos (x)+c \sin (x))^2} \, dx}{2 \left (b^2+c^2\right )}\\ &=\frac {b C-A c \cos (x)+A b \sin (x)}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}-\frac {c^2 C \cos (x)-b c C \sin (x)}{\left (b^2+c^2\right )^2 (b \cos (x)+c \sin (x))}+\frac {A \int \frac {1}{b \cos (x)+c \sin (x)} \, dx}{2 \left (b^2+c^2\right )}\\ &=\frac {b C-A c \cos (x)+A b \sin (x)}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}-\frac {c^2 C \cos (x)-b c C \sin (x)}{\left (b^2+c^2\right )^2 (b \cos (x)+c \sin (x))}-\frac {A \operatorname {Subst}\left (\int \frac {1}{b^2+c^2-x^2} \, dx,x,c \cos (x)-b \sin (x)\right )}{2 \left (b^2+c^2\right )}\\ &=-\frac {A \tanh ^{-1}\left (\frac {c \cos (x)-b \sin (x)}{\sqrt {b^2+c^2}}\right )}{2 \left (b^2+c^2\right )^{3/2}}+\frac {b C-A c \cos (x)+A b \sin (x)}{2 \left (b^2+c^2\right ) (b \cos (x)+c \sin (x))^2}-\frac {c^2 C \cos (x)-b c C \sin (x)}{\left (b^2+c^2\right )^2 (b \cos (x)+c \sin (x))}\\ \end {align*}

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Mathematica [C]  time = 0.39, size = 132, normalized size = 1.14 \[ \frac {\left (b^2+c^2\right ) \left (A b^2 \sin (x)-A b c \cos (x)+b C (b+c \sin (2 x))+2 c^2 C \sin ^2(x)\right )+2 A b \sqrt {b^2+c^2} (b \cos (x)+c \sin (x))^2 \tanh ^{-1}\left (\frac {b \tan \left (\frac {x}{2}\right )-c}{\sqrt {b^2+c^2}}\right )}{2 b (b-i c)^2 (b+i c)^2 (b \cos (x)+c \sin (x))^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Sin[x])/(b*Cos[x] + c*Sin[x])^3,x]

[Out]

(2*A*b*Sqrt[b^2 + c^2]*ArcTanh[(-c + b*Tan[x/2])/Sqrt[b^2 + c^2]]*(b*Cos[x] + c*Sin[x])^2 + (b^2 + c^2)*(-(A*b
*c*Cos[x]) + A*b^2*Sin[x] + 2*c^2*C*Sin[x]^2 + b*C*(b + c*Sin[2*x])))/(2*b*(b - I*c)^2*(b + I*c)^2*(b*Cos[x] +
c*Sin[x])^2)

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fricas [B]  time = 0.92, size = 279, normalized size = 2.41 \[ -\frac {8 \, C b c^{2} \cos \relax (x)^{2} - 2 \, C b^{3} - 6 \, C b c^{2} - {\left (2 \, A b c \cos \relax (x) \sin \relax (x) + A c^{2} + {\left (A b^{2} - A c^{2}\right )} \cos \relax (x)^{2}\right )} \sqrt {b^{2} + c^{2}} \log \left (-\frac {2 \, b c \cos \relax (x) \sin \relax (x) + {\left (b^{2} - c^{2}\right )} \cos \relax (x)^{2} - 2 \, b^{2} - c^{2} + 2 \, \sqrt {b^{2} + c^{2}} {\left (c \cos \relax (x) - b \sin \relax (x)\right )}}{2 \, b c \cos \relax (x) \sin \relax (x) + {\left (b^{2} - c^{2}\right )} \cos \relax (x)^{2} + c^{2}}\right ) + 2 \, {\left (A b^{2} c + A c^{3}\right )} \cos \relax (x) - 2 \, {\left (A b^{3} + A b c^{2} + 2 \, {\left (C b^{2} c - C c^{3}\right )} \cos \relax (x)\right )} \sin \relax (x)}{4 \, {\left (b^{4} c^{2} + 2 \, b^{2} c^{4} + c^{6} + {\left (b^{6} + b^{4} c^{2} - b^{2} c^{4} - c^{6}\right )} \cos \relax (x)^{2} + 2 \, {\left (b^{5} c + 2 \, b^{3} c^{3} + b c^{5}\right )} \cos \relax (x) \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sin(x))/(b*cos(x)+c*sin(x))^3,x, algorithm="fricas")

[Out]

-1/4*(8*C*b*c^2*cos(x)^2 - 2*C*b^3 - 6*C*b*c^2 - (2*A*b*c*cos(x)*sin(x) + A*c^2 + (A*b^2 - A*c^2)*cos(x)^2)*sq
rt(b^2 + c^2)*log(-(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*cos(x)^2 - 2*b^2 - c^2 + 2*sqrt(b^2 + c^2)*(c*cos(x) - b
*sin(x)))/(2*b*c*cos(x)*sin(x) + (b^2 - c^2)*cos(x)^2 + c^2)) + 2*(A*b^2*c + A*c^3)*cos(x) - 2*(A*b^3 + A*b*c^
2 + 2*(C*b^2*c - C*c^3)*cos(x))*sin(x))/(b^4*c^2 + 2*b^2*c^4 + c^6 + (b^6 + b^4*c^2 - b^2*c^4 - c^6)*cos(x)^2
+ 2*(b^5*c + 2*b^3*c^3 + b*c^5)*cos(x)*sin(x))

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giac [A]  time = 0.24, size = 199, normalized size = 1.72 \[ \frac {A \log \left (\frac {{\left | -2 \, b \tan \left (\frac {1}{2} \, x\right ) + 2 \, c - 2 \, \sqrt {b^{2} + c^{2}} \right |}}{{\left | -2 \, b \tan \left (\frac {1}{2} \, x\right ) + 2 \, c + 2 \, \sqrt {b^{2} + c^{2}} \right |}}\right )}{2 \, {\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}} + \frac {A b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, A b c^{2} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, C b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + A b^{2} c \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, C b c^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, A c^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + A b^{3} \tan \left (\frac {1}{2} \, x\right ) - 2 \, A b c^{2} \tan \left (\frac {1}{2} \, x\right ) - A b^{2} c}{{\left (b^{4} + b^{2} c^{2}\right )} {\left (b \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, c \tan \left (\frac {1}{2} \, x\right ) - b\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sin(x))/(b*cos(x)+c*sin(x))^3,x, algorithm="giac")

[Out]

1/2*A*log(abs(-2*b*tan(1/2*x) + 2*c - 2*sqrt(b^2 + c^2))/abs(-2*b*tan(1/2*x) + 2*c + 2*sqrt(b^2 + c^2)))/(b^2
+ c^2)^(3/2) + (A*b^3*tan(1/2*x)^3 + 2*A*b*c^2*tan(1/2*x)^3 + 2*C*b^3*tan(1/2*x)^2 + A*b^2*c*tan(1/2*x)^2 + 2*
C*b*c^2*tan(1/2*x)^2 - 2*A*c^3*tan(1/2*x)^2 + A*b^3*tan(1/2*x) - 2*A*b*c^2*tan(1/2*x) - A*b^2*c)/((b^4 + b^2*c
^2)*(b*tan(1/2*x)^2 - 2*c*tan(1/2*x) - b)^2)

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maple [A]  time = 0.19, size = 177, normalized size = 1.53 \[ -\frac {2 \left (-\frac {A \left (b^{2}+2 c^{2}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2 \left (b^{2}+c^{2}\right ) b}-\frac {\left (A \,b^{2} c -2 A \,c^{3}+2 C \,b^{3}+2 C b \,c^{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2 \left (b^{2}+c^{2}\right ) b^{2}}-\frac {A \left (b^{2}-2 c^{2}\right ) \tan \left (\frac {x}{2}\right )}{2 \left (b^{2}+c^{2}\right ) b}+\frac {A c}{2 b^{2}+2 c^{2}}\right )}{\left (b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 c \tan \left (\frac {x}{2}\right )-b \right )^{2}}+\frac {A \arctanh \left (\frac {2 b \tan \left (\frac {x}{2}\right )-2 c}{2 \sqrt {b^{2}+c^{2}}}\right )}{\left (b^{2}+c^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*sin(x))/(b*cos(x)+c*sin(x))^3,x)

[Out]

-2*(-1/2*A*(b^2+2*c^2)/(b^2+c^2)/b*tan(1/2*x)^3-1/2*(A*b^2*c-2*A*c^3+2*C*b^3+2*C*b*c^2)/(b^2+c^2)/b^2*tan(1/2*
x)^2-1/2*A*(b^2-2*c^2)/(b^2+c^2)/b*tan(1/2*x)+1/2*A*c/(b^2+c^2))/(b*tan(1/2*x)^2-2*c*tan(1/2*x)-b)^2+A/(b^2+c^
2)^(3/2)*arctanh(1/2*(2*b*tan(1/2*x)-2*c)/(b^2+c^2)^(1/2))

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maxima [B]  time = 0.44, size = 338, normalized size = 2.91 \[ -\frac {1}{2} \, A {\left (\frac {2 \, {\left (b^{2} c - \frac {{\left (b^{3} - 2 \, b c^{2}\right )} \sin \relax (x)}{\cos \relax (x) + 1} - \frac {{\left (b^{2} c - 2 \, c^{3}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {{\left (b^{3} + 2 \, b c^{2}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}\right )}}{b^{6} + b^{4} c^{2} + \frac {4 \, {\left (b^{5} c + b^{3} c^{3}\right )} \sin \relax (x)}{\cos \relax (x) + 1} - \frac {2 \, {\left (b^{6} - b^{4} c^{2} - 2 \, b^{2} c^{4}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {4 \, {\left (b^{5} c + b^{3} c^{3}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {{\left (b^{6} + b^{4} c^{2}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}} + \frac {\log \left (\frac {c - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {b^{2} + c^{2}}}{c - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {b^{2} + c^{2}}}\right )}{{\left (b^{2} + c^{2}\right )}^{\frac {3}{2}}}\right )} + \frac {2 \, C \sin \relax (x)^{2}}{{\left (b^{3} + \frac {4 \, b^{2} c \sin \relax (x)}{\cos \relax (x) + 1} - \frac {4 \, b^{2} c \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {b^{3} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {2 \, {\left (b^{3} - 2 \, b c^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}}\right )} {\left (\cos \relax (x) + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*sin(x))/(b*cos(x)+c*sin(x))^3,x, algorithm="maxima")

[Out]

-1/2*A*(2*(b^2*c - (b^3 - 2*b*c^2)*sin(x)/(cos(x) + 1) - (b^2*c - 2*c^3)*sin(x)^2/(cos(x) + 1)^2 - (b^3 + 2*b*
c^2)*sin(x)^3/(cos(x) + 1)^3)/(b^6 + b^4*c^2 + 4*(b^5*c + b^3*c^3)*sin(x)/(cos(x) + 1) - 2*(b^6 - b^4*c^2 - 2*
b^2*c^4)*sin(x)^2/(cos(x) + 1)^2 - 4*(b^5*c + b^3*c^3)*sin(x)^3/(cos(x) + 1)^3 + (b^6 + b^4*c^2)*sin(x)^4/(cos
(x) + 1)^4) + log((c - b*sin(x)/(cos(x) + 1) + sqrt(b^2 + c^2))/(c - b*sin(x)/(cos(x) + 1) - sqrt(b^2 + c^2)))
/(b^2 + c^2)^(3/2)) + 2*C*sin(x)^2/((b^3 + 4*b^2*c*sin(x)/(cos(x) + 1) - 4*b^2*c*sin(x)^3/(cos(x) + 1)^3 + b^3
*sin(x)^4/(cos(x) + 1)^4 - 2*(b^3 - 2*b*c^2)*sin(x)^2/(cos(x) + 1)^2)*(cos(x) + 1)^2)

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mupad [B]  time = 2.86, size = 227, normalized size = 1.96 \[ \frac {\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (A\,b^2+2\,A\,c^2\right )}{b\,\left (b^2+c^2\right )}-\frac {A\,c}{b^2+c^2}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,C\,b^3+A\,b^2\,c+2\,C\,b\,c^2-2\,A\,c^3\right )}{b^2\,\left (b^2+c^2\right )}+\frac {\mathrm {tan}\left (\frac {x}{2}\right )\,\left (A\,b^2-2\,A\,c^2\right )}{b\,\left (b^2+c^2\right )}}{b^2-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,b^2-4\,c^2\right )+b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,b\,c\,\mathrm {tan}\left (\frac {x}{2}\right )-4\,b\,c\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}+\frac {A\,\mathrm {atan}\left (\frac {b^2\,c\,1{}\mathrm {i}+c^3\,1{}\mathrm {i}-b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (b^2+c^2\right )\,1{}\mathrm {i}}{{\left (b^2+c^2\right )}^{3/2}}\right )\,1{}\mathrm {i}}{{\left (b^2+c^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*sin(x))/(b*cos(x) + c*sin(x))^3,x)

[Out]

((tan(x/2)^3*(A*b^2 + 2*A*c^2))/(b*(b^2 + c^2)) - (A*c)/(b^2 + c^2) + (tan(x/2)^2*(2*C*b^3 - 2*A*c^3 + A*b^2*c
+ 2*C*b*c^2))/(b^2*(b^2 + c^2)) + (tan(x/2)*(A*b^2 - 2*A*c^2))/(b*(b^2 + c^2)))/(b^2 - tan(x/2)^2*(2*b^2 - 4*
c^2) + b^2*tan(x/2)^4 + 4*b*c*tan(x/2) - 4*b*c*tan(x/2)^3) + (A*atan((b^2*c*1i + c^3*1i - b*tan(x/2)*(b^2 + c^
2)*1i)/(b^2 + c^2)^(3/2))*1i)/(b^2 + c^2)^(3/2)

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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((A+C*sin(x))/(b*cos(x)+c*sin(x))**3,x)

[Out]

Timed out

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