Integrand size = 23, antiderivative size = 131 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {(4 a+b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{8 a^{3/2} (a+b)^{5/2} d}-\frac {\cos (c+d x) \sin (c+d x)}{4 (a+b) d \left (a+b \sin ^2(c+d x)\right )^2}-\frac {(2 a-b) \cos (c+d x) \sin (c+d x)}{8 a (a+b)^2 d \left (a+b \sin ^2(c+d x)\right )} \]
1/8*(4*a+b)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/a^(3/2)/(a+b)^(5/2)/d-1 /4*cos(d*x+c)*sin(d*x+c)/(a+b)/d/(a+b*sin(d*x+c)^2)^2-1/8*(2*a-b)*cos(d*x+ c)*sin(d*x+c)/a/(a+b)^2/d/(a+b*sin(d*x+c)^2)
Time = 11.64 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\frac {(4 a+b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{a^{3/2} (a+b)^{5/2}}-\frac {\left (8 a^2+4 a b-b^2+b (-2 a+b) \cos (2 (c+d x))\right ) \sin (2 (c+d x))}{a (a+b)^2 (2 a+b-b \cos (2 (c+d x)))^2}}{8 d} \]
(((4*a + b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a^(3/2)*(a + b)^( 5/2)) - ((8*a^2 + 4*a*b - b^2 + b*(-2*a + b)*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(a*(a + b)^2*(2*a + b - b*Cos[2*(c + d*x)])^2))/(8*d)
Time = 0.51 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3652, 3042, 3652, 27, 3042, 3660, 218}
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.
| \(\displaystyle \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{\left (a+b \sin (c+d x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 3652 |
| \(\displaystyle \frac {\int \frac {2 a \sin ^2(c+d x)+a}{\left (b \sin ^2(c+d x)+a\right )^2}dx}{4 a (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a \sin (c+d x)^2+a}{\left (b \sin (c+d x)^2+a\right )^2}dx}{4 a (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 3652 |
| \(\displaystyle \frac {\frac {\int \frac {a (4 a+b)}{b \sin ^2(c+d x)+a}dx}{2 a (a+b)}-\frac {(2 a-b) \sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {(4 a+b) \int \frac {1}{b \sin ^2(c+d x)+a}dx}{2 (a+b)}-\frac {(2 a-b) \sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(4 a+b) \int \frac {1}{b \sin (c+d x)^2+a}dx}{2 (a+b)}-\frac {(2 a-b) \sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 3660 |
| \(\displaystyle \frac {\frac {(4 a+b) \int \frac {1}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{2 d (a+b)}-\frac {(2 a-b) \sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {(4 a+b) \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{2 \sqrt {a} d (a+b)^{3/2}}-\frac {(2 a-b) \sin (c+d x) \cos (c+d x)}{2 d (a+b) \left (a+b \sin ^2(c+d x)\right )}}{4 a (a+b)}-\frac {\sin (c+d x) \cos (c+d x)}{4 d (a+b) \left (a+b \sin ^2(c+d x)\right )^2}\) |
-1/4*(Cos[c + d*x]*Sin[c + d*x])/((a + b)*d*(a + b*Sin[c + d*x]^2)^2) + (( (4*a + b)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(2*Sqrt[a]*(a + b)^( 3/2)*d) - ((2*a - b)*Cos[c + d*x]*Sin[c + d*x])/(2*(a + b)*d*(a + b*Sin[c + d*x]^2)))/(4*a*(a + b))
3.2.8.3.1 Defintions of rubi rules used
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[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b - a*B))*Cos[e + f*x]*Sin[e + f*x ]*((a + b*Sin[e + f*x]^2)^(p + 1)/(2*a*f*(a + b)*(p + 1))), x] - Simp[1/(2* a*(a + b)*(p + 1)) Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*( p + 1) + b*(2*p + 3)) + 2*(A*b - a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[1/(a + (a + b)*ff^2*x^ 2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]
Time = 1.05 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00
| method | result | size |
| derivativedivides | \(\frac {\frac {-\frac {\left (4 a -b \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{8 a \left (a +b \right )}-\frac {\left (4 a +b \right ) \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}^{2}}+\frac {\left (4 a +b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) a \sqrt {a \left (a +b \right )}}}{d}\) | \(131\) |
| default | \(\frac {\frac {-\frac {\left (4 a -b \right ) \left (\tan ^{3}\left (d x +c \right )\right )}{8 a \left (a +b \right )}-\frac {\left (4 a +b \right ) \tan \left (d x +c \right )}{8 \left (a^{2}+2 a b +b^{2}\right )}}{{\left (a \left (\tan ^{2}\left (d x +c \right )\right )+\left (\tan ^{2}\left (d x +c \right )\right ) b +a \right )}^{2}}+\frac {\left (4 a +b \right ) \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) a \sqrt {a \left (a +b \right )}}}{d}\) | \(131\) |
| risch | \(\frac {i \left (4 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+16 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+8 a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-2 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-3 b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-16 a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+2 a \,b^{2}-b^{3}\right )}{4 b a \left (a +b \right )^{2} d \left (-b \,{\mathrm e}^{4 i \left (d x +c \right )}+4 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {-2 i a^{2}-2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d a}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right )}{4 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a^{2}+2 i a b +2 a \sqrt {-a^{2}-a b}+b \sqrt {-a^{2}-a b}}{b \sqrt {-a^{2}-a b}}\right ) b}{16 \sqrt {-a^{2}-a b}\, \left (a +b \right )^{2} d a}\) | \(575\) |
1/d*((-1/8*(4*a-b)/a/(a+b)*tan(d*x+c)^3-1/8*(4*a+b)/(a^2+2*a*b+b^2)*tan(d* x+c))/(a*tan(d*x+c)^2+tan(d*x+c)^2*b+a)^2+1/8*(4*a+b)/(a^2+2*a*b+b^2)/a/(a *(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (117) = 234\).
Time = 0.32 (sec) , antiderivative size = 771, normalized size of antiderivative = 5.89 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\left [-\frac {{\left ({\left (4 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} + 4 \, a^{3} + 9 \, a^{2} b + 6 \, a b^{2} + b^{3} - 2 \, {\left (4 \, a^{2} b + 5 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} - a b} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{3} - {\left (a + b\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} - a b} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \, {\left ({\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{4} + 7 \, a^{3} b + 2 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{32 \, {\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} d\right )}}, -\frac {{\left ({\left (4 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{4} + 4 \, a^{3} + 9 \, a^{2} b + 6 \, a b^{2} + b^{3} - 2 \, {\left (4 \, a^{2} b + 5 \, a b^{2} + b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} + a b} \arctan \left (\frac {{\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b}{2 \, \sqrt {a^{2} + a b} \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) - 2 \, {\left ({\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )^{3} - {\left (4 \, a^{4} + 7 \, a^{3} b + 2 \, a^{2} b^{2} - a b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{16 \, {\left ({\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{6} b + 4 \, a^{5} b^{2} + 6 \, a^{4} b^{3} + 4 \, a^{3} b^{4} + a^{2} b^{5}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{7} + 5 \, a^{6} b + 10 \, a^{5} b^{2} + 10 \, a^{4} b^{3} + 5 \, a^{3} b^{4} + a^{2} b^{5}\right )} d\right )}}\right ] \]
[-1/32*(((4*a*b^2 + b^3)*cos(d*x + c)^4 + 4*a^3 + 9*a^2*b + 6*a*b^2 + b^3 - 2*(4*a^2*b + 5*a*b^2 + b^3)*cos(d*x + c)^2)*sqrt(-a^2 - a*b)*log(((8*a^2 + 8*a*b + b^2)*cos(d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 + 4*((2*a + b)*cos(d*x + c)^3 - (a + b)*cos(d*x + c))*sqrt(-a^2 - a*b)*sin(d *x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)) - 4*((2*a^3*b + a^2*b^2 - a*b^3)*cos(d*x + c)^ 3 - (4*a^4 + 7*a^3*b + 2*a^2*b^2 - a*b^3)*cos(d*x + c))*sin(d*x + c))/((a^ 5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*d*cos(d*x + c)^4 - 2*(a^6*b + 4*a ^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*d*cos(d*x + c)^2 + (a^7 + 5*a^6* b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a^2*b^5)*d), -1/16*(((4*a*b^2 + b^3)*cos(d*x + c)^4 + 4*a^3 + 9*a^2*b + 6*a*b^2 + b^3 - 2*(4*a^2*b + 5*a*b ^2 + b^3)*cos(d*x + c)^2)*sqrt(a^2 + a*b)*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)/(sqrt(a^2 + a*b)*cos(d*x + c)*sin(d*x + c))) - 2*((2*a^3*b + a^2*b^2 - a*b^3)*cos(d*x + c)^3 - (4*a^4 + 7*a^3*b + 2*a^2*b^2 - a*b^3)*c os(d*x + c))*sin(d*x + c))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*d* cos(d*x + c)^4 - 2*(a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*d *cos(d*x + c)^2 + (a^7 + 5*a^6*b + 10*a^5*b^2 + 10*a^4*b^3 + 5*a^3*b^4 + a ^2*b^5)*d)]
Timed out. \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.33 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.46 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\frac {{\left (4 \, a + b\right )} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {{\left (4 \, a^{2} + 3 \, a b - b^{2}\right )} \tan \left (d x + c\right )^{3} + {\left (4 \, a^{2} + a b\right )} \tan \left (d x + c\right )}{a^{5} + 2 \, a^{4} b + a^{3} b^{2} + {\left (a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4}\right )} \tan \left (d x + c\right )^{4} + 2 \, {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \tan \left (d x + c\right )^{2}}}{8 \, d} \]
1/8*((4*a + b)*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/((a^3 + 2*a^2* b + a*b^2)*sqrt((a + b)*a)) - ((4*a^2 + 3*a*b - b^2)*tan(d*x + c)^3 + (4*a ^2 + a*b)*tan(d*x + c))/(a^5 + 2*a^4*b + a^3*b^2 + (a^5 + 4*a^4*b + 6*a^3* b^2 + 4*a^2*b^3 + a*b^4)*tan(d*x + c)^4 + 2*(a^5 + 3*a^4*b + 3*a^3*b^2 + a ^2*b^3)*tan(d*x + c)^2))/d
Time = 0.38 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.46 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} {\left (4 \, a + b\right )}}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {a^{2} + a b}} - \frac {4 \, a^{2} \tan \left (d x + c\right )^{3} + 3 \, a b \tan \left (d x + c\right )^{3} - b^{2} \tan \left (d x + c\right )^{3} + 4 \, a^{2} \tan \left (d x + c\right ) + a b \tan \left (d x + c\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} {\left (a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right )^{2} + a\right )}^{2}}}{8 \, d} \]
1/8*((pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)))*(4*a + b)/((a^3 + 2*a^2*b + a*b^2)*sq rt(a^2 + a*b)) - (4*a^2*tan(d*x + c)^3 + 3*a*b*tan(d*x + c)^3 - b^2*tan(d* x + c)^3 + 4*a^2*tan(d*x + c) + a*b*tan(d*x + c))/((a^3 + 2*a^2*b + a*b^2) *(a*tan(d*x + c)^2 + b*tan(d*x + c)^2 + a)^2))/d
Time = 14.11 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.21 \[ \int \frac {\sin ^2(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )\,\left (4\,a+b\right )}{8\,a^{3/2}\,d\,{\left (a+b\right )}^{5/2}}-\frac {\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (4\,a+b\right )}{8\,\left (a^2+2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,\left (4\,a-b\right )}{8\,a\,\left (a+b\right )}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^4\,\left (a^2+2\,a\,b+b^2\right )+a^2+{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )\right )} \]
(atan((tan(c + d*x)*(2*a + 2*b)*(2*a*b + a^2 + b^2))/(2*a^(1/2)*(a + b)^(5 /2)))*(4*a + b))/(8*a^(3/2)*d*(a + b)^(5/2)) - ((tan(c + d*x)*(4*a + b))/( 8*(2*a*b + a^2 + b^2)) + (tan(c + d*x)^3*(4*a - b))/(8*a*(a + b)))/(d*(tan (c + d*x)^4*(2*a*b + a^2 + b^2) + a^2 + tan(c + d*x)^2*(2*a*b + 2*a^2)))