3.23 \(\int \frac {\sin ^2(x)}{(a \cos (x)+b \sin (x))^3} \, dx\)

Optimal. Leaf size=92 \[ \frac {a \left (\left (a^2+4 b^2\right ) \sin (x)+3 a b \cos (x)\right )}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))^2}-\frac {\left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

[Out]

-(a^2-2*b^2)*arctanh((-b+a*tan(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2)+1/2*a*(3*a*b*cos(x)+(a^2+4*b^2)*sin(x)
)/(a^2+b^2)^2/(a*cos(x)+b*sin(x))^2

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Rubi [B]  time = 0.70, antiderivative size = 300, normalized size of antiderivative = 3.26, number of steps used = 13, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {4401, 1660, 12, 618, 206, 3155, 3074} \[ -\frac {a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )+3 a^2 b^2+4 a^4+2 b^4}{a b \left (a^2+b^2\right )^2 \left (-a \tan ^2\left (\frac {x}{2}\right )+a+2 b \tan \left (\frac {x}{2}\right )\right )}+\frac {2 \left (\left (a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )+a b\right )}{a \left (a^2+b^2\right ) \left (-a \tan ^2\left (\frac {x}{2}\right )+a+2 b \tan \left (\frac {x}{2}\right )\right )^2}+\frac {2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac {a^2 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{5/2}}+\frac {2 a^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a*Cos[x] + b*Sin[x])^3,x]

[Out]

(2*a^2*ArcTanh[(b*Cos[x] - a*Sin[x])/Sqrt[a^2 + b^2]])/(b^2*(a^2 + b^2)^(3/2)) - ArcTanh[(b*Cos[x] - a*Sin[x])
/Sqrt[a^2 + b^2]]/(b^2*Sqrt[a^2 + b^2]) - (a^2*(2*a^2 - b^2)*ArcTanh[(b - a*Tan[x/2])/Sqrt[a^2 + b^2]])/(b^2*(
a^2 + b^2)^(5/2)) + (2*a)/(b*(a^2 + b^2)*(a*Cos[x] + b*Sin[x])) + (2*(a*b + (a^2 + 2*b^2)*Tan[x/2]))/(a*(a^2 +
b^2)*(a + 2*b*Tan[x/2] - a*Tan[x/2]^2)^2) - (4*a^4 + 3*a^2*b^2 + 2*b^4 + a*b*(5*a^2 + 2*b^2)*Tan[x/2])/(a*b*(
a^2 + b^2)^2*(a + 2*b*Tan[x/2] - a*Tan[x/2]^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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 1660

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[((b*f - 2*a*g + (2*c*f - b*g)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

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 3155

Int[((A_.) + cos[(d_.) + (e_.)*(x_)]*(B_.))/((a_.) + cos[(d_.) + (e_.)*(x_)]*(b_.) + (c_.)*sin[(d_.) + (e_.)*(
x_)])^2, x_Symbol] :> Simp[(c*B + 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)/(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}, x] && NeQ[a^2 - b^2 - c^2, 0] && NeQ[a*A - b*B, 0]

Rule 4401

Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /;  !InertTrigFreeQ[u]

Rubi steps

\begin {align*} \int \frac {\sin ^2(x)}{(a \cos (x)+b \sin (x))^3} \, dx &=\int \left (\frac {a^2 \cos ^2(x)}{b^2 (a \cos (x)+b \sin (x))^3}-\frac {2 a \cos (x)}{b^2 (a \cos (x)+b \sin (x))^2}+\frac {1}{b^2 (a \cos (x)+b \sin (x))}\right ) \, dx\\ &=\frac {\int \frac {1}{a \cos (x)+b \sin (x)} \, dx}{b^2}-\frac {(2 a) \int \frac {\cos (x)}{(a \cos (x)+b \sin (x))^2} \, dx}{b^2}+\frac {a^2 \int \frac {\cos ^2(x)}{(a \cos (x)+b \sin (x))^3} \, dx}{b^2}\\ &=\frac {2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}-\frac {\operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{b^2}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{\left (a+2 b x-a x^2\right )^3} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b^2}-\frac {\left (2 a^2\right ) \int \frac {1}{a \cos (x)+b \sin (x)} \, dx}{b^2 \left (a^2+b^2\right )}\\ &=-\frac {\tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}+\frac {2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac {2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )^2}-\frac {a^2 \operatorname {Subst}\left (\int \frac {-\frac {8 \left (a^4+2 b^4\right )}{a^3}+16 b \left (1+\frac {b^2}{a^2}\right ) x+8 \left (a+\frac {b^2}{a}\right ) x^2}{\left (a+2 b x-a x^2\right )^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{4 b^2 \left (a^2+b^2\right )}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{b^2 \left (a^2+b^2\right )}\\ &=\frac {2 a^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}+\frac {2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac {2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )^2}-\frac {4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )}{a b \left (a^2+b^2\right )^2 \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )}+\frac {a^2 \operatorname {Subst}\left (\int \frac {16 \left (2 a^2-b^2\right )}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{16 b^2 \left (a^2+b^2\right )^2}\\ &=\frac {2 a^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}+\frac {2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac {2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )^2}-\frac {4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )}{a b \left (a^2+b^2\right )^2 \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )}+\frac {\left (a^2 \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{b^2 \left (a^2+b^2\right )^2}\\ &=\frac {2 a^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}+\frac {2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac {2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )^2}-\frac {4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )}{a b \left (a^2+b^2\right )^2 \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )}-\frac {\left (2 a^2 \left (2 a^2-b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tan \left (\frac {x}{2}\right )\right )}{b^2 \left (a^2+b^2\right )^2}\\ &=\frac {2 a^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{3/2}}-\frac {\tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{b^2 \sqrt {a^2+b^2}}-\frac {a^2 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac {b-a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{b^2 \left (a^2+b^2\right )^{5/2}}+\frac {2 a}{b \left (a^2+b^2\right ) (a \cos (x)+b \sin (x))}+\frac {2 \left (a b+\left (a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )\right )}{a \left (a^2+b^2\right ) \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )^2}-\frac {4 a^4+3 a^2 b^2+2 b^4+a b \left (5 a^2+2 b^2\right ) \tan \left (\frac {x}{2}\right )}{a b \left (a^2+b^2\right )^2 \left (a+2 b \tan \left (\frac {x}{2}\right )-a \tan ^2\left (\frac {x}{2}\right )\right )}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 92, normalized size = 1.00 \[ \frac {a \left (\left (a^2+4 b^2\right ) \sin (x)+3 a b \cos (x)\right )}{2 \left (a^2+b^2\right )^2 (a \cos (x)+b \sin (x))^2}-\frac {\left (a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a*Cos[x] + b*Sin[x])^3,x]

[Out]

-(((a^2 - 2*b^2)*ArcTanh[(-b + a*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2)) + (a*(3*a*b*Cos[x] + (a^2 + 4*
b^2)*Sin[x]))/(2*(a^2 + b^2)^2*(a*Cos[x] + b*Sin[x])^2)

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fricas [B]  time = 1.41, size = 282, normalized size = 3.07 \[ -\frac {{\left (a^{2} b^{2} - 2 \, b^{4} + {\left (a^{4} - 3 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \relax (x) \sin \relax (x)\right )} \sqrt {a^{2} + b^{2}} \log \left (-\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) - 6 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \relax (x) - 2 \, {\left (a^{5} + 5 \, a^{3} b^{2} + 4 \, a b^{4}\right )} \sin \relax (x)}{4 \, {\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8} + {\left (a^{8} + 2 \, a^{6} b^{2} - 2 \, a^{2} b^{6} - b^{8}\right )} \cos \relax (x)^{2} + 2 \, {\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \cos \relax (x) \sin \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a*cos(x)+b*sin(x))^3,x, algorithm="fricas")

[Out]

-1/4*((a^2*b^2 - 2*b^4 + (a^4 - 3*a^2*b^2 + 2*b^4)*cos(x)^2 + 2*(a^3*b - 2*a*b^3)*cos(x)*sin(x))*sqrt(a^2 + b^
2)*log(-(2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 - 2*a^2 - b^2 + 2*sqrt(a^2 + b^2)*(b*cos(x) - a*sin(x)))/(
2*a*b*cos(x)*sin(x) + (a^2 - b^2)*cos(x)^2 + b^2)) - 6*(a^4*b + a^2*b^3)*cos(x) - 2*(a^5 + 5*a^3*b^2 + 4*a*b^4
)*sin(x))/(a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8 + (a^8 + 2*a^6*b^2 - 2*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b + 3
*a^5*b^3 + 3*a^3*b^5 + a*b^7)*cos(x)*sin(x))

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giac [B]  time = 0.28, size = 197, normalized size = 2.14 \[ \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 2 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 6 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + a^{3} \tan \left (\frac {1}{2} \, x\right ) + 10 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) + 3 \, a^{2} b}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} - 2 \, b \tan \left (\frac {1}{2} \, x\right ) - a\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a*cos(x)+b*sin(x))^3,x, algorithm="giac")

[Out]

1/2*(a^2 - 2*b^2)*log(abs(2*a*tan(1/2*x) - 2*b - 2*sqrt(a^2 + b^2))/abs(2*a*tan(1/2*x) - 2*b + 2*sqrt(a^2 + b^
2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + (a^3*tan(1/2*x)^3 - 2*a*b^2*tan(1/2*x)^3 - 3*a^2*b*tan(1/2*x)
^2 + 6*b^3*tan(1/2*x)^2 + a^3*tan(1/2*x) + 10*a*b^2*tan(1/2*x) + 3*a^2*b)/((a^4 + 2*a^2*b^2 + b^4)*(a*tan(1/2*
x)^2 - 2*b*tan(1/2*x) - a)^2)

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maple [B]  time = 0.59, size = 212, normalized size = 2.30 \[ -\frac {8 \left (-\frac {a \left (a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{8 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}+\frac {3 b \left (a^{2}-2 b^{2}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{8 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {\left (a^{2}+10 b^{2}\right ) a \tan \left (\frac {x}{2}\right )}{8 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}-\frac {3 a^{2} b}{8 \left (a^{4}+2 a^{2} b^{2}+b^{4}\right )}\right )}{\left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 \tan \left (\frac {x}{2}\right ) b -a \right )^{2}}-\frac {\left (a^{2}-2 b^{2}\right ) \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}+b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a*cos(x)+b*sin(x))^3,x)

[Out]

-8*(-1/8*a*(a^2-2*b^2)/(a^4+2*a^2*b^2+b^4)*tan(1/2*x)^3+3/8*b*(a^2-2*b^2)/(a^4+2*a^2*b^2+b^4)*tan(1/2*x)^2-1/8
*(a^2+10*b^2)*a/(a^4+2*a^2*b^2+b^4)*tan(1/2*x)-3/8*a^2*b/(a^4+2*a^2*b^2+b^4))/(a*tan(1/2*x)^2-2*tan(1/2*x)*b-a
)^2-(a^2-2*b^2)/(a^4+2*a^2*b^2+b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tan(1/2*x)-2*b)/(a^2+b^2)^(1/2))

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maxima [B]  time = 0.43, size = 299, normalized size = 3.25 \[ \frac {{\left (a^{2} - 2 \, b^{2}\right )} \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {3 \, a^{2} b + \frac {{\left (a^{3} + 10 \, a b^{2}\right )} \sin \relax (x)}{\cos \relax (x) + 1} - \frac {3 \, {\left (a^{2} b - 2 \, b^{3}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {{\left (a^{3} - 2 \, a b^{2}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4} + \frac {4 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \relax (x)}{\cos \relax (x) + 1} - \frac {2 \, {\left (a^{6} - 3 \, a^{2} b^{4} - 2 \, b^{6}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {4 \, {\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^2/(a*cos(x)+b*sin(x))^3,x, algorithm="maxima")

[Out]

1/2*(a^2 - 2*b^2)*log((b - a*sin(x)/(cos(x) + 1) + sqrt(a^2 + b^2))/(b - a*sin(x)/(cos(x) + 1) - sqrt(a^2 + b^
2)))/((a^4 + 2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) + (3*a^2*b + (a^3 + 10*a*b^2)*sin(x)/(cos(x) + 1) - 3*(a^2*b -
2*b^3)*sin(x)^2/(cos(x) + 1)^2 + (a^3 - 2*a*b^2)*sin(x)^3/(cos(x) + 1)^3)/(a^6 + 2*a^4*b^2 + a^2*b^4 + 4*(a^5*
b + 2*a^3*b^3 + a*b^5)*sin(x)/(cos(x) + 1) - 2*(a^6 - 3*a^2*b^4 - 2*b^6)*sin(x)^2/(cos(x) + 1)^2 - 4*(a^5*b +
2*a^3*b^3 + a*b^5)*sin(x)^3/(cos(x) + 1)^3 + (a^6 + 2*a^4*b^2 + a^2*b^4)*sin(x)^4/(cos(x) + 1)^4)

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mupad [B]  time = 0.80, size = 263, normalized size = 2.86 \[ \frac {\frac {3\,a^2\,b}{a^4+2\,a^2\,b^2+b^4}+\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^2+10\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (a^2-2\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}-\frac {3\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^2-2\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}}{a^2-{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (2\,a^2-4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+4\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )-4\,a\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}+\frac {\mathrm {atanh}\left (\frac {2\,a^4\,b+4\,a^2\,b^3+2\,b^5}{2\,{\left (a^2+b^2\right )}^{5/2}}-\frac {a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{{\left (a^2+b^2\right )}^{5/2}}\right )\,\left (a^2-2\,b^2\right )}{{\left (a^2+b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^2/(a*cos(x) + b*sin(x))^3,x)

[Out]

((3*a^2*b)/(a^4 + b^4 + 2*a^2*b^2) + (a*tan(x/2)*(a^2 + 10*b^2))/(a^4 + b^4 + 2*a^2*b^2) + (a*tan(x/2)^3*(a^2
- 2*b^2))/(a^4 + b^4 + 2*a^2*b^2) - (3*b*tan(x/2)^2*(a^2 - 2*b^2))/(a^4 + b^4 + 2*a^2*b^2))/(a^2 - tan(x/2)^2*
(2*a^2 - 4*b^2) + a^2*tan(x/2)^4 + 4*a*b*tan(x/2) - 4*a*b*tan(x/2)^3) + (atanh((2*a^4*b + 2*b^5 + 4*a^2*b^3)/(
2*(a^2 + b^2)^(5/2)) - (a*tan(x/2)*(a^4 + b^4 + 2*a^2*b^2))/(a^2 + b^2)^(5/2))*(a^2 - 2*b^2))/(a^2 + b^2)^(5/2
)

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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(sin(x)**2/(a*cos(x)+b*sin(x))**3,x)

[Out]

Timed out

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