Integrand size = 41, antiderivative size = 239 \[ \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \, dx=-\frac {\cos (d+e x) (b+a \sin (d+e x))}{2 e \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}-\frac {\text {arctanh}\left (\frac {a+b \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {a^2-b^2}}\right ) \left (a b+a^2 \sin (d+e x)\right )^3}{a^2 \left (a^2-b^2\right )^{3/2} e \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}}+\frac {b \cos (d+e x) \left (a b+a^2 \sin (d+e x)\right )^3}{2 \left (a^2-b^2\right ) e \left (a^3 b+a^4 \sin (d+e x)\right ) \left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \] Output:
-1/2*cos(e*x+d)*(b+a*sin(e*x+d))/e/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2) ^(3/2)-arctanh((a+b*tan(1/2*e*x+1/2*d))/(a^2-b^2)^(1/2))*(a*b+a^2*sin(e*x+ d))^3/a^2/(a^2-b^2)^(3/2)/e/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2)+ 1/2*b*cos(e*x+d)*(a*b+a^2*sin(e*x+d))^3/(a^2-b^2)/e/(a^3*b+a^4*sin(e*x+d)) /(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2)
Time = 0.60 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.60 \[ \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \, dx=\frac {-2 a \arctan \left (\frac {a+b \tan \left (\frac {1}{2} (d+e x)\right )}{\sqrt {-a^2+b^2}}\right ) (b+a \sin (d+e x))^2+\sqrt {-a^2+b^2} \cos (d+e x) \left (a^2-2 b^2-a b \sin (d+e x)\right )}{2 (-a+b) (a+b) \sqrt {-a^2+b^2} e (b+a \sin (d+e x)) \sqrt {(b+a \sin (d+e x))^2}} \] Input:
Integrate[(a + b*Sin[d + e*x])/(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x ]^2)^(3/2),x]
Output:
(-2*a*ArcTan[(a + b*Tan[(d + e*x)/2])/Sqrt[-a^2 + b^2]]*(b + a*Sin[d + e*x ])^2 + Sqrt[-a^2 + b^2]*Cos[d + e*x]*(a^2 - 2*b^2 - a*b*Sin[d + e*x]))/(2* (-a + b)*(a + b)*Sqrt[-a^2 + b^2]*e*(b + a*Sin[d + e*x])*Sqrt[(b + a*Sin[d + e*x])^2])
Time = 0.76 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.79, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 3771, 27, 3042, 3233, 27, 3042, 3233, 27, 3042, 3139, 1083, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {a+b \sin (d+e x)}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {a+b \sin (d+e x)}{\left (a^2 \sin (d+e x)^2+2 a b \sin (d+e x)+b^2\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3771 |
| \(\displaystyle \frac {8 \left (a^2 \sin (d+e x)+a b\right )^3 \int \frac {a+b \sin (d+e x)}{8 \left (\sin (d+e x) a^2+b a\right )^3}dx}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \int \frac {a+b \sin (d+e x)}{\left (\sin (d+e x) a^2+b a\right )^3}dx}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \int \frac {a+b \sin (d+e x)}{\left (\sin (d+e x) a^2+b a\right )^3}dx}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\int \frac {a \left (a^2-b^2\right ) \sin (d+e x)}{\left (\sin (d+e x) a^2+b a\right )^2}dx}{2 a^2 \left (a^2-b^2\right )}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\int \frac {\sin (d+e x)}{\left (\sin (d+e x) a^2+b a\right )^2}dx}{2 a}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\int \frac {\sin (d+e x)}{\left (\sin (d+e x) a^2+b a\right )^2}dx}{2 a}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 3233 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\frac {\int \frac {a^2}{\sin (d+e x) a^2+b a}dx}{a^2 \left (a^2-b^2\right )}+\frac {b \cos (d+e x)}{e \left (a^2-b^2\right ) \left (a^3 \sin (d+e x)+a^2 b\right )}}{2 a}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\frac {\int \frac {1}{\sin (d+e x) a^2+b a}dx}{a^2-b^2}+\frac {b \cos (d+e x)}{e \left (a^2-b^2\right ) \left (a^3 \sin (d+e x)+a^2 b\right )}}{2 a}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\frac {\int \frac {1}{\sin (d+e x) a^2+b a}dx}{a^2-b^2}+\frac {b \cos (d+e x)}{e \left (a^2-b^2\right ) \left (a^3 \sin (d+e x)+a^2 b\right )}}{2 a}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 3139 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\frac {2 \int \frac {1}{2 \tan \left (\frac {1}{2} (d+e x)\right ) a^2+b \tan ^2\left (\frac {1}{2} (d+e x)\right ) a+b a}d\tan \left (\frac {1}{2} (d+e x)\right )}{e \left (a^2-b^2\right )}+\frac {b \cos (d+e x)}{e \left (a^2-b^2\right ) \left (a^3 \sin (d+e x)+a^2 b\right )}}{2 a}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\frac {b \cos (d+e x)}{e \left (a^2-b^2\right ) \left (a^3 \sin (d+e x)+a^2 b\right )}-\frac {4 \int \frac {1}{4 a^2 \left (a^2-b^2\right )-\left (2 a^2+2 b \tan \left (\frac {1}{2} (d+e x)\right ) a\right )^2}d\left (2 a^2+2 b \tan \left (\frac {1}{2} (d+e x)\right ) a\right )}{e \left (a^2-b^2\right )}}{2 a}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\left (a^2 \sin (d+e x)+a b\right )^3 \left (\frac {\frac {b \cos (d+e x)}{e \left (a^2-b^2\right ) \left (a^3 \sin (d+e x)+a^2 b\right )}-\frac {2 \text {arctanh}\left (\frac {2 a^2+2 a b \tan \left (\frac {1}{2} (d+e x)\right )}{2 a \sqrt {a^2-b^2}}\right )}{a e \left (a^2-b^2\right )^{3/2}}}{2 a}-\frac {\cos (d+e x)}{2 a e \left (a^2 \sin (d+e x)+a b\right )^2}\right )}{\left (a^2 \sin ^2(d+e x)+2 a b \sin (d+e x)+b^2\right )^{3/2}}\) |
Input:
Int[(a + b*Sin[d + e*x])/(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^( 3/2),x]
Output:
((a*b + a^2*Sin[d + e*x])^3*(-1/2*Cos[d + e*x]/(a*e*(a*b + a^2*Sin[d + e*x ])^2) + ((-2*ArcTanh[(2*a^2 + 2*a*b*Tan[(d + e*x)/2])/(2*a*Sqrt[a^2 - b^2] )])/(a*(a^2 - b^2)^(3/2)*e) + (b*Cos[d + e*x])/((a^2 - b^2)*e*(a^2*b + a^3 *Sin[d + e*x])))/(2*a)))/(b^2 + 2*a*b*Sin[d + e*x] + a^2*Sin[d + e*x]^2)^( 3/2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[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]
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] + Simp[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]
Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])*((a_) + (b_.)*sin[(d_.) + (e_.)* (x_)] + (c_.)*sin[(d_.) + (e_.)*(x_)]^2)^(n_), x_Symbol] :> Simp[(a + b*Sin [d + e*x] + c*Sin[d + e*x]^2)^n/(b + 2*c*Sin[d + e*x])^(2*n) Int[(A + B*S in[d + e*x])*(b + 2*c*Sin[d + e*x])^(2*n), x], x] /; FreeQ[{a, b, c, d, e, A, B}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[n]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.10 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.46
| method | result | size |
| default | \(\frac {\left (-2 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a^{3} b^{2} \sin \left (e x +d \right )^{2}-4 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2} b^{3} \sin \left (e x +d \right )-2 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a \,b^{4}-\sqrt {-a^{2}+b^{2}}\, a^{4} \sin \left (e x +d \right )^{2}-2 \sqrt {-a^{2}+b^{2}}\, a^{3} b \sin \left (e x +d \right )+\left (-2 \cos \left (e x +d \right )^{2}-\cos \left (e x +d \right )+1\right ) b^{2} \sqrt {-a^{2}+b^{2}}\, a^{2}+\left (\cos \left (e x +d \right )+4\right ) \sin \left (e x +d \right ) \sqrt {-a^{2}+b^{2}}\, a \,b^{3}+\left (2 \cos \left (e x +d \right )+2\right ) b^{4} \sqrt {-a^{2}+b^{2}}\right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{2 e \sqrt {-a^{2}+b^{2}}\, \left (a^{2}-b^{2}\right ) b^{2} \left (b^{2}+2 a b \sin \left (e x +d \right )+a^{2} \sin \left (e x +d \right )^{2}\right )}\) | \(349\) |
| parts | \(-\frac {a \left (-2 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a^{4} b^{2} \sin \left (e x +d \right )^{2}-4 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a^{3} b^{3} \sin \left (e x +d \right )+\left (4 \cos \left (e x +d \right )^{2}-6\right ) b^{4} \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2}-8 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a \,b^{5} \sin \left (e x +d \right )-4 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) b^{6}-\sqrt {-a^{2}+b^{2}}\, a^{5} \sin \left (e x +d \right )^{2}-2 \sqrt {-a^{2}+b^{2}}\, a^{4} b \sin \left (e x +d \right )+\left (-4 \cos \left (e x +d \right )^{2}-\cos \left (e x +d \right )+3\right ) b^{2} \sqrt {-a^{2}+b^{2}}\, a^{3}+\left (3 \cos \left (e x +d \right )+8\right ) \sin \left (e x +d \right ) \sqrt {-a^{2}+b^{2}}\, a^{2} b^{3}+\left (4 \cos \left (e x +d \right )+4\right ) b^{4} \sqrt {-a^{2}+b^{2}}\, a \right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{2 e \sqrt {-a^{2}+b^{2}}\, b^{2} \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left (-b^{2}-2 a b \sin \left (e x +d \right )-a^{2} \sin \left (e x +d \right )^{2}\right )}+\frac {\left (-6 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a^{3} b^{2} \sin \left (e x +d \right )^{2}-12 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2} b^{3} \sin \left (e x +d \right )-6 \arctan \left (\frac {b \cot \left (e x +d \right )-a -b \csc \left (e x +d \right )}{\sqrt {-a^{2}+b^{2}}}\right ) a \,b^{4}+\sqrt {-a^{2}+b^{2}}\, a^{4} \sin \left (e x +d \right )^{2}+\left (2 \cos \left (e x +d \right )+2\right ) \sin \left (e x +d \right ) \sqrt {-a^{2}+b^{2}}\, a^{3} b +\left (-2 \cos \left (e x +d \right )^{2}+\cos \left (e x +d \right )+3\right ) b^{2} \sqrt {-a^{2}+b^{2}}\, a^{2}+\left (\cos \left (e x +d \right )+4\right ) \sin \left (e x +d \right ) \sqrt {-a^{2}+b^{2}}\, a \,b^{3}+\left (2 \cos \left (e x +d \right )+2\right ) b^{4} \sqrt {-a^{2}+b^{2}}\right ) \operatorname {csgn}\left (b +a \sin \left (e x +d \right )\right )}{2 e \sqrt {-a^{2}+b^{2}}\, \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \left (-b^{2}-2 a b \sin \left (e x +d \right )-a^{2} \sin \left (e x +d \right )^{2}\right )}\) | \(823\) |
Input:
int((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x,metho d=_RETURNVERBOSE)
Output:
1/2/e/(-a^2+b^2)^(1/2)/(a^2-b^2)/b^2*(-2*arctan((b*cot(e*x+d)-a-b*csc(e*x+ d))/(-a^2+b^2)^(1/2))*a^3*b^2*sin(e*x+d)^2-4*arctan((b*cot(e*x+d)-a-b*csc( e*x+d))/(-a^2+b^2)^(1/2))*a^2*b^3*sin(e*x+d)-2*arctan((b*cot(e*x+d)-a-b*cs c(e*x+d))/(-a^2+b^2)^(1/2))*a*b^4-(-a^2+b^2)^(1/2)*a^4*sin(e*x+d)^2-2*(-a^ 2+b^2)^(1/2)*a^3*b*sin(e*x+d)+(-2*cos(e*x+d)^2-cos(e*x+d)+1)*b^2*(-a^2+b^2 )^(1/2)*a^2+(cos(e*x+d)+4)*sin(e*x+d)*(-a^2+b^2)^(1/2)*a*b^3+(2*cos(e*x+d) +2)*b^4*(-a^2+b^2)^(1/2))*csgn(b+a*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*s in(e*x+d)^2)
Time = 0.10 (sec) , antiderivative size = 527, normalized size of antiderivative = 2.21 \[ \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \, dx=\left [-\frac {2 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) + {\left (a^{3} \cos \left (e x + d\right )^{2} - 2 \, a^{2} b \sin \left (e x + d\right ) - a^{3} - a b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (e x + d\right )^{2} + 2 \, a b \sin \left (e x + d\right ) + a^{2} + b^{2} + 2 \, {\left (b \cos \left (e x + d\right ) \sin \left (e x + d\right ) + a \cos \left (e x + d\right )\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \left (e x + d\right )^{2} - 2 \, a b \sin \left (e x + d\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{4} - 3 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (e x + d\right )}{4 \, {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} e \cos \left (e x + d\right )^{2} - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} e \sin \left (e x + d\right ) - {\left (a^{6} - a^{4} b^{2} - a^{2} b^{4} + b^{6}\right )} e\right )}}, -\frac {{\left (a^{3} b - a b^{3}\right )} \cos \left (e x + d\right ) \sin \left (e x + d\right ) + {\left (a^{3} \cos \left (e x + d\right )^{2} - 2 \, a^{2} b \sin \left (e x + d\right ) - a^{3} - a b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \left (e x + d\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (e x + d\right )}\right ) - {\left (a^{4} - 3 \, a^{2} b^{2} + 2 \, b^{4}\right )} \cos \left (e x + d\right )}{2 \, {\left ({\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} e \cos \left (e x + d\right )^{2} - 2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} e \sin \left (e x + d\right ) - {\left (a^{6} - a^{4} b^{2} - a^{2} b^{4} + b^{6}\right )} e\right )}}\right ] \] Input:
integrate((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x , algorithm="fricas")
Output:
[-1/4*(2*(a^3*b - a*b^3)*cos(e*x + d)*sin(e*x + d) + (a^3*cos(e*x + d)^2 - 2*a^2*b*sin(e*x + d) - a^3 - a*b^2)*sqrt(a^2 - b^2)*log(((a^2 - 2*b^2)*co s(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)) - 2*(a^4 - 3*a^2*b^2 + 2*b^4)*cos(e*x + d))/((a^6 - 2 *a^4*b^2 + a^2*b^4)*e*cos(e*x + d)^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*e*sin (e*x + d) - (a^6 - a^4*b^2 - a^2*b^4 + b^6)*e), -1/2*((a^3*b - a*b^3)*cos( e*x + d)*sin(e*x + d) + (a^3*cos(e*x + d)^2 - 2*a^2*b*sin(e*x + d) - a^3 - a*b^2)*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))) - (a^4 - 3*a^2*b^2 + 2*b^4)*cos(e*x + d))/((a^6 - 2*a^4*b^2 + a^2*b^4)*e*cos(e*x + d)^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*e*s in(e*x + d) - (a^6 - a^4*b^2 - a^2*b^4 + b^6)*e)]
Timed out. \[ \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \, dx=\text {Timed out} \] Input:
integrate((a+b*sin(e*x+d))/(b**2+2*a*b*sin(e*x+d)+a**2*sin(e*x+d)**2)**(3/ 2),x)
Output:
Timed out
Exception generated. \[ \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x , algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.37 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.22 \[ \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {e x + d}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a}{{\left (a^{2} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) - b^{2} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} \sqrt {-a^{2} + b^{2}}} - \frac {2 \, a^{3} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} - 3 \, a b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{3} + 2 \, a^{4} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 3 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} - 2 \, b^{4} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, a^{3} b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) - 5 \, a b^{3} \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + a^{2} b^{2} - 2 \, b^{4}}{{\left (a^{2} b^{2} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right ) - b^{4} \mathrm {sgn}\left (a \sin \left (e x + d\right ) + b\right )\right )} {\left (b \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right )^{2} + 2 \, a \tan \left (\frac {1}{2} \, e x + \frac {1}{2} \, d\right ) + b\right )}^{2}}}{e} \] Input:
integrate((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x , algorithm="giac")
Output:
((pi*floor(1/2*(e*x + d)/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*e*x + 1/2*d) + a)/sqrt(-a^2 + b^2)))*a/((a^2*sgn(a*sin(e*x + d) + b) - b^2*sgn(a*sin(e *x + d) + b))*sqrt(-a^2 + b^2)) - (2*a^3*b*tan(1/2*e*x + 1/2*d)^3 - 3*a*b^ 3*tan(1/2*e*x + 1/2*d)^3 + 2*a^4*tan(1/2*e*x + 1/2*d)^2 - 3*a^2*b^2*tan(1/ 2*e*x + 1/2*d)^2 - 2*b^4*tan(1/2*e*x + 1/2*d)^2 + 2*a^3*b*tan(1/2*e*x + 1/ 2*d) - 5*a*b^3*tan(1/2*e*x + 1/2*d) + a^2*b^2 - 2*b^4)/((a^2*b^2*sgn(a*sin (e*x + d) + b) - b^4*sgn(a*sin(e*x + d) + b))*(b*tan(1/2*e*x + 1/2*d)^2 + 2*a*tan(1/2*e*x + 1/2*d) + b)^2))/e
Timed out. \[ \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \, dx=\int \frac {a+b\,\sin \left (d+e\,x\right )}{{\left (a^2\,{\sin \left (d+e\,x\right )}^2+2\,a\,b\,\sin \left (d+e\,x\right )+b^2\right )}^{3/2}} \,d x \] Input:
int((a + b*sin(d + e*x))/(b^2 + a^2*sin(d + e*x)^2 + 2*a*b*sin(d + e*x))^( 3/2),x)
Output:
int((a + b*sin(d + e*x))/(b^2 + a^2*sin(d + e*x)^2 + 2*a*b*sin(d + e*x))^( 3/2), x)
Time = 0.16 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.62 \[ \int \frac {a+b \sin (d+e x)}{\left (b^2+2 a b \sin (d+e x)+a^2 \sin ^2(d+e x)\right )^{3/2}} \, dx=\frac {-4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right ) \sin \left (e x +d \right )^{2} a^{3}-8 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right ) \sin \left (e x +d \right ) a^{2} b -4 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {e x}{2}+\frac {d}{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right ) a \,b^{2}+2 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a^{3} b -2 \cos \left (e x +d \right ) \sin \left (e x +d \right ) a \,b^{3}-2 \cos \left (e x +d \right ) a^{4}+6 \cos \left (e x +d \right ) a^{2} b^{2}-4 \cos \left (e x +d \right ) b^{4}+\sin \left (e x +d \right )^{2} a^{4}-\sin \left (e x +d \right )^{2} a^{2} b^{2}+2 \sin \left (e x +d \right ) a^{3} b -2 \sin \left (e x +d \right ) a \,b^{3}+a^{2} b^{2}-b^{4}}{4 e \left (\sin \left (e x +d \right )^{2} a^{6}-2 \sin \left (e x +d \right )^{2} a^{4} b^{2}+\sin \left (e x +d \right )^{2} a^{2} b^{4}+2 \sin \left (e x +d \right ) a^{5} b -4 \sin \left (e x +d \right ) a^{3} b^{3}+2 \sin \left (e x +d \right ) a \,b^{5}+a^{4} b^{2}-2 a^{2} b^{4}+b^{6}\right )} \] Input:
int((a+b*sin(e*x+d))/(b^2+2*a*b*sin(e*x+d)+a^2*sin(e*x+d)^2)^(3/2),x)
Output:
( - 4*sqrt( - a**2 + b**2)*atan((tan((d + e*x)/2)*b + a)/sqrt( - a**2 + b* *2))*sin(d + e*x)**2*a**3 - 8*sqrt( - a**2 + b**2)*atan((tan((d + e*x)/2)* b + a)/sqrt( - a**2 + b**2))*sin(d + e*x)*a**2*b - 4*sqrt( - a**2 + b**2)* atan((tan((d + e*x)/2)*b + a)/sqrt( - a**2 + b**2))*a*b**2 + 2*cos(d + e*x )*sin(d + e*x)*a**3*b - 2*cos(d + e*x)*sin(d + e*x)*a*b**3 - 2*cos(d + e*x )*a**4 + 6*cos(d + e*x)*a**2*b**2 - 4*cos(d + e*x)*b**4 + sin(d + e*x)**2* a**4 - sin(d + e*x)**2*a**2*b**2 + 2*sin(d + e*x)*a**3*b - 2*sin(d + e*x)* a*b**3 + a**2*b**2 - b**4)/(4*e*(sin(d + e*x)**2*a**6 - 2*sin(d + e*x)**2* a**4*b**2 + sin(d + e*x)**2*a**2*b**4 + 2*sin(d + e*x)*a**5*b - 4*sin(d + e*x)*a**3*b**3 + 2*sin(d + e*x)*a*b**5 + a**4*b**2 - 2*a**2*b**4 + b**6))