3.502 \(\int \frac {a+b \sin (d+e x)}{(b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x))^2} \, dx\)

Optimal. Leaf size=157 \[ -\frac {\left (2 a^2+b^2\right ) \cos (d+e x)}{3 e \left (a^2-b^2\right )^2 (a \sin (d+e x)+b)}+\frac {b \cos (d+e x)}{3 e \left (a^2-b^2\right ) (a \sin (d+e x)+b)^2}+\frac {2 a b \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-b^2}}\right )}{e \left (a^2-b^2\right )^{5/2}}-\frac {\cos (d+e x)}{3 e (a \sin (d+e x)+b)^3} \]

[Out]

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

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Rubi [A]  time = 0.42, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3288, 2754, 12, 2660, 618, 206} \[ -\frac {\left (2 a^2+b^2\right ) \cos (d+e x)}{3 e \left (a^2-b^2\right )^2 (a \sin (d+e x)+b)}+\frac {b \cos (d+e x)}{3 e \left (a^2-b^2\right ) (a \sin (d+e x)+b)^2}+\frac {2 a b \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-b^2}}\right )}{e \left (a^2-b^2\right )^{5/2}}-\frac {\cos (d+e x)}{3 e (a \sin (d+e x)+b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 3288

Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sin[(d_.) + (e_.)*(x_)] + (c_.)*sin[(d_.) + (e_.)*(x_
)]^2)^(n_), x_Symbol] :> Dist[1/(4^n*c^n), Int[(A + B*Sin[d + e*x])*(b + 2*c*Sin[d + e*x])^(2*n), x], x] /; Fr
eeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^2} \, dx &=\left (16 a^4\right ) \int \frac {a+b \sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^4} \, dx\\ &=-\frac {\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac {\left (4 a^2\right ) \int \frac {4 a \left (a^2-b^2\right ) \sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^3} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac {1}{3} \left (16 a^3\right ) \int \frac {\sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^3} \, dx\\ &=-\frac {\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac {b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}+\frac {(2 a) \int \frac {4 a^2-2 a b \sin (d+e x)}{\left (2 a b+2 a^2 \sin (d+e x)\right )^2} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac {b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac {\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}+\frac {\int -\frac {12 a^3 b}{2 a b+2 a^2 \sin (d+e x)} \, dx}{6 a \left (a^2-b^2\right )^2}\\ &=-\frac {\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac {b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac {\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}-\frac {\left (2 a^2 b\right ) \int \frac {1}{2 a b+2 a^2 \sin (d+e x)} \, dx}{\left (a^2-b^2\right )^2}\\ &=-\frac {\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac {b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac {\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}-\frac {\left (4 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{2 a b+4 a^2 x+2 a b x^2} \, dx,x,\tan \left (\frac {1}{2} (d+e x)\right )\right )}{\left (a^2-b^2\right )^2 e}\\ &=-\frac {\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac {b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac {\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}+\frac {\left (8 a^2 b\right ) \operatorname {Subst}\left (\int \frac {1}{16 a^2 \left (a^2-b^2\right )-x^2} \, dx,x,4 a^2+4 a b \tan \left (\frac {1}{2} (d+e x)\right )\right )}{\left (a^2-b^2\right )^2 e}\\ &=\frac {2 a b \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} e}-\frac {\cos (d+e x)}{3 e (b+a \sin (d+e x))^3}+\frac {b \cos (d+e x)}{3 \left (a^2-b^2\right ) e (b+a \sin (d+e x))^2}-\frac {\left (2 a^2+b^2\right ) \cos (d+e x)}{3 \left (a^2-b^2\right )^2 e (b+a \sin (d+e x))}\\ \end {align*}

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Mathematica [A]  time = 0.96, size = 140, normalized size = 0.89 \[ -\frac {\frac {6 a b \tan ^{-1}\left (\frac {a+b \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac {\cos (d+e x) \left (a^4+a^2 \left (2 a^2+b^2\right ) \sin ^2(d+e x)+3 a b \left (a^2+b^2\right ) \sin (d+e x)-a^2 b^2+3 b^4\right )}{(a-b)^2 (a+b)^2 (a \sin (d+e x)+b)^3}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((6*a*b*ArcTan[(a + b*Tan[(d + e*x)/2])/Sqrt[-a^2 + b^2]])/(-a^2 + b^2)^(5/2) + (Cos[d + e*x]*(a^4 - a^2*
b^2 + 3*b^4 + 3*a*b*(a^2 + b^2)*Sin[d + e*x] + a^2*(2*a^2 + b^2)*Sin[d + e*x]^2))/((a - b)^2*(a + b)^2*(b + a*
Sin[d + e*x])^3))/e

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fricas [B]  time = 1.98, size = 795, normalized size = 5.06 \[ \left [-\frac {2 \, {\left (2 \, a^{6} - a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (e x + d\right )^{3} - 6 \, {\left (a^{5} b - a b^{5}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) - 3 \, {\left (3 \, a^{3} b^{2} \cos \left (e x + d\right )^{2} - 3 \, a^{3} b^{2} - a b^{4} + {\left (a^{4} b \cos \left (e x + d\right )^{2} - a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (e x + d\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, a b \sin \left (e x + d\right ) + a^{2} + b^{2} + 2 \, {\left (b \cos \left (e x + d\right ) \sin \left (e x + d\right ) + a \cos \left (e x + d\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (e x + d\right )^{2} - 2 \, a b \sin \left (e x + d\right ) - a^{2} - b^{2}}\right ) - 6 \, {\left (a^{6} - a^{4} b^{2} + a^{2} b^{4} - b^{6}\right )} \cos \left (e x + d\right )}{6 \, {\left (3 \, {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} e \cos \left (e x + d\right )^{2} - {\left (3 \, a^{8} b - 8 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - b^{9}\right )} e + {\left ({\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} e \cos \left (e x + d\right )^{2} - {\left (a^{9} - 6 \, a^{5} b^{4} + 8 \, a^{3} b^{6} - 3 \, a b^{8}\right )} e\right )} \sin \left (e x + d\right )\right )}}, -\frac {{\left (2 \, a^{6} - a^{4} b^{2} - a^{2} b^{4}\right )} \cos \left (e x + d\right )^{3} - 3 \, {\left (a^{5} b - a b^{5}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) - 3 \, {\left (3 \, a^{3} b^{2} \cos \left (e x + d\right )^{2} - 3 \, a^{3} b^{2} - a b^{4} + {\left (a^{4} b \cos \left (e x + d\right )^{2} - a^{4} b - 3 \, a^{2} b^{3}\right )} \sin \left (e x + d\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (e x + d\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (e x + d\right )}\right ) - 3 \, {\left (a^{6} - a^{4} b^{2} + a^{2} b^{4} - b^{6}\right )} \cos \left (e x + d\right )}{3 \, {\left (3 \, {\left (a^{8} b - 3 \, a^{6} b^{3} + 3 \, a^{4} b^{5} - a^{2} b^{7}\right )} e \cos \left (e x + d\right )^{2} - {\left (3 \, a^{8} b - 8 \, a^{6} b^{3} + 6 \, a^{4} b^{5} - b^{9}\right )} e + {\left ({\left (a^{9} - 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} - a^{3} b^{6}\right )} e \cos \left (e x + d\right )^{2} - {\left (a^{9} - 6 \, a^{5} b^{4} + 8 \, a^{3} b^{6} - 3 \, a b^{8}\right )} e\right )} \sin \left (e x + d\right )\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(2*(2*a^6 - a^4*b^2 - a^2*b^4)*cos(e*x + d)^3 - 6*(a^5*b - a*b^5)*cos(e*x + d)*sin(e*x + d) - 3*(3*a^3*b
^2*cos(e*x + d)^2 - 3*a^3*b^2 - a*b^4 + (a^4*b*cos(e*x + d)^2 - a^4*b - 3*a^2*b^3)*sin(e*x + d))*sqrt(a^2 - b^
2)*log(((a^2 - 2*b^2)*cos(e*x + d)^2 + 2*a*b*sin(e*x + d) + a^2 + b^2 + 2*(b*cos(e*x + d)*sin(e*x + d) + a*cos
(e*x + d))*sqrt(a^2 - b^2))/(a^2*cos(e*x + d)^2 - 2*a*b*sin(e*x + d) - a^2 - b^2)) - 6*(a^6 - a^4*b^2 + a^2*b^
4 - b^6)*cos(e*x + d))/(3*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*e*cos(e*x + d)^2 - (3*a^8*b - 8*a^6*b^3 +
6*a^4*b^5 - b^9)*e + ((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*e*cos(e*x + d)^2 - (a^9 - 6*a^5*b^4 + 8*a^3*b^6
- 3*a*b^8)*e)*sin(e*x + d)), -1/3*((2*a^6 - a^4*b^2 - a^2*b^4)*cos(e*x + d)^3 - 3*(a^5*b - a*b^5)*cos(e*x + d)
*sin(e*x + d) - 3*(3*a^3*b^2*cos(e*x + d)^2 - 3*a^3*b^2 - a*b^4 + (a^4*b*cos(e*x + d)^2 - a^4*b - 3*a^2*b^3)*s
in(e*x + d))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*sin(e*x + d) + a)/((a^2 - b^2)*cos(e*x + d))) - 3*(a
^6 - a^4*b^2 + a^2*b^4 - b^6)*cos(e*x + d))/(3*(a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*e*cos(e*x + d)^2 - (3
*a^8*b - 8*a^6*b^3 + 6*a^4*b^5 - b^9)*e + ((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*e*cos(e*x + d)^2 - (a^9 - 6
*a^5*b^4 + 8*a^3*b^6 - 3*a*b^8)*e)*sin(e*x + d))]

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giac [B]  time = 0.25, size = 454, normalized size = 2.89 \[ -\frac {2}{3} \, {\left (\frac {3 \, {\left (\pi \left \lfloor \frac {x e + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a b}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {3 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{5} - 6 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{5} + 6 \, a b^{6} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{5} + 6 \, a^{6} b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{4} - 9 \, a^{4} b^{3} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{4} + 15 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{4} + 3 \, b^{7} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{4} + 4 \, a^{7} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 2 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 6 \, a^{3} b^{4} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 18 \, a b^{6} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{3} + 6 \, a^{6} b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 18 \, a^{2} b^{5} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 6 \, b^{7} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 3 \, a^{5} b^{2} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + 12 \, a b^{6} \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + a^{4} b^{3} - a^{2} b^{5} + 3 \, b^{7}}{{\left (a^{4} b^{3} - 2 \, a^{2} b^{5} + b^{7}\right )} {\left (b \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, x e + \frac {1}{2} \, d\right ) + b\right )}^{3}}\right )} e^{\left (-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*(pi*floor(1/2*(x*e + d)/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x*e + 1/2*d) + a)/sqrt(-a^2 + b^2)))*a*b/
((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + (3*a^5*b^2*tan(1/2*x*e + 1/2*d)^5 - 6*a^3*b^4*tan(1/2*x*e + 1/2*d
)^5 + 6*a*b^6*tan(1/2*x*e + 1/2*d)^5 + 6*a^6*b*tan(1/2*x*e + 1/2*d)^4 - 9*a^4*b^3*tan(1/2*x*e + 1/2*d)^4 + 15*
a^2*b^5*tan(1/2*x*e + 1/2*d)^4 + 3*b^7*tan(1/2*x*e + 1/2*d)^4 + 4*a^7*tan(1/2*x*e + 1/2*d)^3 + 2*a^5*b^2*tan(1
/2*x*e + 1/2*d)^3 + 6*a^3*b^4*tan(1/2*x*e + 1/2*d)^3 + 18*a*b^6*tan(1/2*x*e + 1/2*d)^3 + 6*a^6*b*tan(1/2*x*e +
1/2*d)^2 + 18*a^2*b^5*tan(1/2*x*e + 1/2*d)^2 + 6*b^7*tan(1/2*x*e + 1/2*d)^2 + 3*a^5*b^2*tan(1/2*x*e + 1/2*d)
+ 12*a*b^6*tan(1/2*x*e + 1/2*d) + a^4*b^3 - a^2*b^5 + 3*b^7)/((a^4*b^3 - 2*a^2*b^5 + b^7)*(b*tan(1/2*x*e + 1/2
*d)^2 + 2*a*tan(1/2*x*e + 1/2*d) + b)^3))*e^(-1)

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maple [B]  time = 0.33, size = 1297, normalized size = 8.26 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3*a^5/b/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^5+4/e/(b
*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3*a^3*b/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^5-4/e/(b*tan(1/
2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3*a*b^3/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^5-4/e/(b*tan(1/2*d+1/2
*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3/b^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^4*a^6+6/e/(b*tan(1/2*d+1/2*e*x)
^2+2*a*tan(1/2*d+1/2*e*x)+b)^3/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^4*a^4-10/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*t
an(1/2*d+1/2*e*x)+b)^3*b^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^4*a^2-2/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/
2*d+1/2*e*x)+b)^3*b^4/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^4-8/3/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2
*e*x)+b)^3*a^7/b^3/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^3-4/3/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*
x)+b)^3*a^5/b/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^3-4/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3
*a^3*b/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^3-12/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3*a*b^3
/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^3-4/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3/b^2/(a^4-2*a
^2*b^2+b^4)*tan(1/2*d+1/2*e*x)^2*a^6-12/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3*b^2/(a^4-2*a^2*b
^2+b^4)*tan(1/2*d+1/2*e*x)^2*a^2-4/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3*b^4/(a^4-2*a^2*b^2+b^
4)*tan(1/2*d+1/2*e*x)^2-2/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3*a^5/b/(a^4-2*a^2*b^2+b^4)*tan(
1/2*d+1/2*e*x)-8/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3*a*b^3/(a^4-2*a^2*b^2+b^4)*tan(1/2*d+1/2
*e*x)-2/3/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3/(a^4-2*a^2*b^2+b^4)*a^4+2/3/e/(b*tan(1/2*d+1/2
*e*x)^2+2*a*tan(1/2*d+1/2*e*x)+b)^3/(a^4-2*a^2*b^2+b^4)*a^2*b^2-2/e/(b*tan(1/2*d+1/2*e*x)^2+2*a*tan(1/2*d+1/2*
e*x)+b)^3/(a^4-2*a^2*b^2+b^4)*b^4-2/e*a*b/(a^4-2*a^2*b^2+b^4)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*d+1/2*e
*x)+2*a)/(-a^2+b^2)^(1/2))

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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*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^2,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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 6.06, size = 497, normalized size = 3.17 \[ \frac {2\,a\,b\,\mathrm {atanh}\left (\frac {\left (2\,a+2\,b\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )}{e\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {2\,\left (a^4-a^2\,b^2+3\,b^4\right )}{3\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (a^6+3\,a^2\,b^4+b^6\right )}{b^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (2\,a^6-3\,a^4\,b^2+5\,a^2\,b^4+b^6\right )}{b^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )\,\left (a^4+4\,b^4\right )}{b\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {2\,a\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\,\left (a^4-2\,a^2\,b^2+2\,b^4\right )}{b\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {4\,a\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (2\,a^2+3\,b^2\right )\,\left (a^4-a^2\,b^2+3\,b^4\right )}{3\,b^3\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{e\,\left (b^3\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^3\,\left (8\,a^3+12\,a\,b^2\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^2\,\left (12\,a^2\,b+3\,b^3\right )+{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^4\,\left (12\,a^2\,b+3\,b^3\right )+b^3+6\,a\,b^2\,\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )+6\,a\,b^2\,{\mathrm {tan}\left (\frac {d}{2}+\frac {e\,x}{2}\right )}^5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*b*atanh(((2*a + 2*b*tan(d/2 + (e*x)/2))*(a^4 + b^4 - 2*a^2*b^2))/(2*(a + b)^(5/2)*(a - b)^(5/2))))/(e*(a
+ b)^(5/2)*(a - b)^(5/2)) - ((2*(a^4 + 3*b^4 - a^2*b^2))/(3*(a^4 + b^4 - 2*a^2*b^2)) + (4*tan(d/2 + (e*x)/2)^2
*(a^6 + b^6 + 3*a^2*b^4))/(b^2*(a^4 + b^4 - 2*a^2*b^2)) + (2*tan(d/2 + (e*x)/2)^4*(2*a^6 + b^6 + 5*a^2*b^4 - 3
*a^4*b^2))/(b^2*(a^4 + b^4 - 2*a^2*b^2)) + (2*a*tan(d/2 + (e*x)/2)*(a^4 + 4*b^4))/(b*(a^4 + b^4 - 2*a^2*b^2))
+ (2*a*tan(d/2 + (e*x)/2)^5*(a^4 + 2*b^4 - 2*a^2*b^2))/(b*(a^4 + b^4 - 2*a^2*b^2)) + (4*a*tan(d/2 + (e*x)/2)^3
*(2*a^2 + 3*b^2)*(a^4 + 3*b^4 - a^2*b^2))/(3*b^3*(a^4 + b^4 - 2*a^2*b^2)))/(e*(b^3*tan(d/2 + (e*x)/2)^6 + tan(
d/2 + (e*x)/2)^3*(12*a*b^2 + 8*a^3) + tan(d/2 + (e*x)/2)^2*(12*a^2*b + 3*b^3) + tan(d/2 + (e*x)/2)^4*(12*a^2*b
+ 3*b^3) + b^3 + 6*a*b^2*tan(d/2 + (e*x)/2) + 6*a*b^2*tan(d/2 + (e*x)/2)^5))

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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+b*sin(e*x+d))/(b**2+2*a*b*sin(e*x+d)+a**2*sin(e*x+d)**2)**2,x)

[Out]

Timed out

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