\(\int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx\) [690]

Optimal result

Integrand size = 25, antiderivative size = 129 \[ \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx=\frac {b^2 x}{d^2}-\frac {2 (b c-a d) \left (a c d+b \left (c^2-2 d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d^2 \left (c^2-d^2\right )^{3/2} f}+\frac {(b c-a d)^2 \cos (e+f x)}{d \left (c^2-d^2\right ) f (c+d \sin (e+f x))} \] Output:

b^2*x/d^2-2*(-a*d+b*c)*(a*c*d+b*(c^2-2*d^2))*arctan((d+c*tan(1/2*f*x+1/2*e 
))/(c^2-d^2)^(1/2))/d^2/(c^2-d^2)^(3/2)/f+(-a*d+b*c)^2*cos(f*x+e)/d/(c^2-d 
^2)/f/(c+d*sin(f*x+e))
 

Mathematica [A] (verified)

Time = 0.91 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx=\frac {b^2 (e+f x)-\frac {2 \left (-a^2 c d^2+2 a b d^3+b^2 \left (c^3-2 c d^2\right )\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\left (c^2-d^2\right )^{3/2}}+\frac {d (b c-a d)^2 \cos (e+f x)}{(c-d) (c+d) (c+d \sin (e+f x))}}{d^2 f} \] Input:

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

Output:

(b^2*(e + f*x) - (2*(-(a^2*c*d^2) + 2*a*b*d^3 + b^2*(c^3 - 2*c*d^2))*ArcTa 
n[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(c^2 - d^2)^(3/2) + (d*(b*c - 
 a*d)^2*Cos[e + f*x])/((c - d)*(c + d)*(c + d*Sin[e + f*x])))/(d^2*f)
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.24, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3269, 3042, 3214, 3042, 3139, 1083, 217}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2}dx\)

\(\Big \downarrow \) 3269

\(\displaystyle \frac {\int \frac {\left (c^2-d^2\right ) \sin (e+f x) b^2+d \left (\left (a^2+b^2\right ) c-2 a b d\right )}{c+d \sin (e+f x)}dx}{d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\left (c^2-d^2\right ) \sin (e+f x) b^2+d \left (\left (a^2+b^2\right ) c-2 a b d\right )}{c+d \sin (e+f x)}dx}{d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3214

\(\displaystyle \frac {\frac {b^2 x \left (c^2-d^2\right )}{d}-\frac {(b c-a d) \left (a c d+b c^2-2 b d^2\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{d}}{d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {b^2 x \left (c^2-d^2\right )}{d}-\frac {(b c-a d) \left (a c d+b c^2-2 b d^2\right ) \int \frac {1}{c+d \sin (e+f x)}dx}{d}}{d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {\frac {b^2 x \left (c^2-d^2\right )}{d}-\frac {2 (b c-a d) \left (a c d+b c^2-2 b d^2\right ) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{d f}}{d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {4 (b c-a d) \left (a c d+b c^2-2 b d^2\right ) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{d f}+\frac {b^2 x \left (c^2-d^2\right )}{d}}{d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {b^2 x \left (c^2-d^2\right )}{d}-\frac {2 (b c-a d) \left (a c d+b c^2-2 b d^2\right ) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}}}{d \left (c^2-d^2\right )}+\frac {(b c-a d)^2 \cos (e+f x)}{d f \left (c^2-d^2\right ) (c+d \sin (e+f x))}\)

Input:

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

Output:

((b^2*(c^2 - d^2)*x)/d - (2*(b*c - a*d)*(b*c^2 + a*c*d - 2*b*d^2)*ArcTan[( 
2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/(d*Sqrt[c^2 - d^2]*f))/( 
d*(c^2 - d^2)) + ((b*c - a*d)^2*Cos[e + f*x])/(d*(c^2 - d^2)*f*(c + d*Sin[ 
e + f*x]))
 

Defintions of rubi rules used

rule 217

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

rule 1083

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

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]
 

rule 3214

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

rule 3269

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^2, x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e 
+ f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[ 
1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1) 
*(2*b*c*d - a*(c^2 + d^2)) + (a^2*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1 
) + c^2*(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]
 

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.69

method	result	size
derivativedivides	\(\frac {\frac {2 b^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{2}}+\frac {\frac {2 \left (\frac {d^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{c^{2}-d^{2}}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a^{2} c \,d^{2}-2 a b \,d^{3}-b^{2} c^{3}+2 b^{2} c \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{d^{2}}}{f}\)	\(218\)
default	\(\frac {\frac {2 b^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d^{2}}+\frac {\frac {2 \left (\frac {d^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c^{2}-d^{2}\right ) c}+\frac {d \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{c^{2}-d^{2}}\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {2 \left (a^{2} c \,d^{2}-2 a b \,d^{3}-b^{2} c^{3}+2 b^{2} c \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c^{2}-d^{2}\right )^{\frac {3}{2}}}}{d^{2}}}{f}\)	\(218\)
risch	\(\frac {b^{2} x}{d^{2}}-\frac {2 i \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (i d +c \,{\mathrm e}^{i \left (f x +e \right )}\right )}{d^{2} \left (c^{2}-d^{2}\right ) f \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+2 c \,{\mathrm e}^{i \left (f x +e \right )}+i d \right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a^{2} c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}+\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a b}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b^{2} c^{3}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f \,d^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b^{2} c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a^{2} c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}-\frac {2 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) a b}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b^{2} c^{3}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f \,d^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) b^{2} c}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right ) f}\)	\(761\)

Input:

int((a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

1/f*(2*b^2/d^2*arctan(tan(1/2*f*x+1/2*e))+2/d^2*((d^2*(a^2*d^2-2*a*b*c*d+b 
^2*c^2)/(c^2-d^2)/c*tan(1/2*f*x+1/2*e)+d*(a^2*d^2-2*a*b*c*d+b^2*c^2)/(c^2- 
d^2))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan(1/2*f*x+1/2*e)+c)+(a^2*c*d^2-2*a*b*d 
^3-b^2*c^3+2*b^2*c*d^2)/(c^2-d^2)^(3/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e) 
+2*d)/(c^2-d^2)^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (124) = 248\).

Time = 0.12 (sec) , antiderivative size = 675, normalized size of antiderivative = 5.23 \[ \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx=\left [\frac {2 \, {\left (b^{2} c^{4} d - 2 \, b^{2} c^{2} d^{3} + b^{2} d^{5}\right )} f x \sin \left (f x + e\right ) + 2 \, {\left (b^{2} c^{5} - 2 \, b^{2} c^{3} d^{2} + b^{2} c d^{4}\right )} f x + {\left (b^{2} c^{4} + 2 \, a b c d^{3} - {\left (a^{2} + 2 \, b^{2}\right )} c^{2} d^{2} + {\left (b^{2} c^{3} d + 2 \, a b d^{4} - {\left (a^{2} + 2 \, b^{2}\right )} c d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \log \left (\frac {{\left (2 \, c^{2} - d^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2} + 2 \, {\left (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + 2 \, a b c d^{4} - a^{2} d^{5} + {\left (a^{2} - b^{2}\right )} c^{2} d^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{4} d^{3} - 2 \, c^{2} d^{5} + d^{7}\right )} f \sin \left (f x + e\right ) + {\left (c^{5} d^{2} - 2 \, c^{3} d^{4} + c d^{6}\right )} f\right )}}, \frac {{\left (b^{2} c^{4} d - 2 \, b^{2} c^{2} d^{3} + b^{2} d^{5}\right )} f x \sin \left (f x + e\right ) + {\left (b^{2} c^{5} - 2 \, b^{2} c^{3} d^{2} + b^{2} c d^{4}\right )} f x + {\left (b^{2} c^{4} + 2 \, a b c d^{3} - {\left (a^{2} + 2 \, b^{2}\right )} c^{2} d^{2} + {\left (b^{2} c^{3} d + 2 \, a b d^{4} - {\left (a^{2} + 2 \, b^{2}\right )} c d^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left (b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + 2 \, a b c d^{4} - a^{2} d^{5} + {\left (a^{2} - b^{2}\right )} c^{2} d^{3}\right )} \cos \left (f x + e\right )}{{\left (c^{4} d^{3} - 2 \, c^{2} d^{5} + d^{7}\right )} f \sin \left (f x + e\right ) + {\left (c^{5} d^{2} - 2 \, c^{3} d^{4} + c d^{6}\right )} f}\right ] \] Input:

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

Output:

[1/2*(2*(b^2*c^4*d - 2*b^2*c^2*d^3 + b^2*d^5)*f*x*sin(f*x + e) + 2*(b^2*c^ 
5 - 2*b^2*c^3*d^2 + b^2*c*d^4)*f*x + (b^2*c^4 + 2*a*b*c*d^3 - (a^2 + 2*b^2 
)*c^2*d^2 + (b^2*c^3*d + 2*a*b*d^4 - (a^2 + 2*b^2)*c*d^3)*sin(f*x + e))*sq 
rt(-c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^ 
2 - d^2 + 2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2 + d^2 
))/(d^2*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) + 2*(b^2*c^4*d - 
 2*a*b*c^3*d^2 + 2*a*b*c*d^4 - a^2*d^5 + (a^2 - b^2)*c^2*d^3)*cos(f*x + e) 
)/((c^4*d^3 - 2*c^2*d^5 + d^7)*f*sin(f*x + e) + (c^5*d^2 - 2*c^3*d^4 + c*d 
^6)*f), ((b^2*c^4*d - 2*b^2*c^2*d^3 + b^2*d^5)*f*x*sin(f*x + e) + (b^2*c^5 
 - 2*b^2*c^3*d^2 + b^2*c*d^4)*f*x + (b^2*c^4 + 2*a*b*c*d^3 - (a^2 + 2*b^2) 
*c^2*d^2 + (b^2*c^3*d + 2*a*b*d^4 - (a^2 + 2*b^2)*c*d^3)*sin(f*x + e))*sqr 
t(c^2 - d^2)*arctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) 
+ (b^2*c^4*d - 2*a*b*c^3*d^2 + 2*a*b*c*d^4 - a^2*d^5 + (a^2 - b^2)*c^2*d^3 
)*cos(f*x + e))/((c^4*d^3 - 2*c^2*d^5 + d^7)*f*sin(f*x + e) + (c^5*d^2 - 2 
*c^3*d^4 + c*d^6)*f)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+b*sin(f*x+e))^2/(c+d*sin(f*x+e))^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*d^2-4*c^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.87 \[ \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx=\frac {\frac {{\left (f x + e\right )} b^{2}}{d^{2}} - \frac {2 \, {\left (b^{2} c^{3} - a^{2} c d^{2} - 2 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )}}{{\left (c^{2} d^{2} - d^{4}\right )} \sqrt {c^{2} - d^{2}}} + \frac {2 \, {\left (b^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a b c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )}}{{\left (c^{3} d - c d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}}}{f} \] Input:

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

Output:

((f*x + e)*b^2/d^2 - 2*(b^2*c^3 - a^2*c*d^2 - 2*b^2*c*d^2 + 2*a*b*d^3)*(pi 
*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2*e) + d 
)/sqrt(c^2 - d^2)))/((c^2*d^2 - d^4)*sqrt(c^2 - d^2)) + 2*(b^2*c^2*d*tan(1 
/2*f*x + 1/2*e) - 2*a*b*c*d^2*tan(1/2*f*x + 1/2*e) + a^2*d^3*tan(1/2*f*x + 
 1/2*e) + b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)/((c^3*d - c*d^3)*(c*tan(1/2*f 
*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)))/f
 

Mupad [B] (verification not implemented)

Time = 23.36 (sec) , antiderivative size = 5776, normalized size of antiderivative = 44.78 \[ \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

((2*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(d*(c^2 - d^2)) + (2*tan(e/2 + (f*x)/ 
2)*(a^2*d^2 + b^2*c^2 - 2*a*b*c*d))/(c*(c^2 - d^2)))/(f*(c + 2*d*tan(e/2 + 
 (f*x)/2) + c*tan(e/2 + (f*x)/2)^2)) - (2*b^2*atan(((b^2*((b^2*((32*(b^2*c 
*d^8 + a^2*c^3*d^6 - a^2*c^5*d^4 - b^2*c^3*d^6 - 2*a*b*c^2*d^7 + 2*a*b*c^4 
*d^5))/(d^6 - 2*c^2*d^4 + c^4*d^2) + (32*tan(e/2 + (f*x)/2)*(2*a^2*c^2*d^8 
 - 2*a^2*c^4*d^6 + 4*b^2*c^2*d^8 - 6*b^2*c^4*d^6 + 2*b^2*c^6*d^4 - 4*a*b*c 
*d^9 + 4*a*b*c^3*d^7))/(d^7 - 2*c^2*d^5 + c^4*d^3) - (b^2*((32*(c^2*d^9 - 
2*c^4*d^7 + c^6*d^5))/(d^6 - 2*c^2*d^4 + c^4*d^2) + (32*tan(e/2 + (f*x)/2) 
*(3*c*d^11 - 8*c^3*d^9 + 7*c^5*d^7 - 2*c^7*d^5))/(d^7 - 2*c^2*d^5 + c^4*d^ 
3))*1i)/d^2)*1i)/d^2 - (32*(b^4*c^6*d + b^4*c^2*d^5 - 2*b^4*c^4*d^3))/(d^6 
 - 2*c^2*d^4 + c^4*d^2) + (32*tan(e/2 + (f*x)/2)*(2*b^4*c^7*d - 2*b^4*c*d^ 
7 + a^4*c^3*d^5 + 9*b^4*c^3*d^5 - 8*b^4*c^5*d^3 - 8*a*b^3*c^2*d^6 + 4*a*b^ 
3*c^4*d^4 + 4*a^2*b^2*c*d^7 - 4*a^3*b*c^2*d^6 + 4*a^2*b^2*c^3*d^5 - 2*a^2* 
b^2*c^5*d^3))/(d^7 - 2*c^2*d^5 + c^4*d^3)))/d^2 - (b^2*((32*(b^4*c^6*d + b 
^4*c^2*d^5 - 2*b^4*c^4*d^3))/(d^6 - 2*c^2*d^4 + c^4*d^2) + (b^2*((32*(b^2* 
c*d^8 + a^2*c^3*d^6 - a^2*c^5*d^4 - b^2*c^3*d^6 - 2*a*b*c^2*d^7 + 2*a*b*c^ 
4*d^5))/(d^6 - 2*c^2*d^4 + c^4*d^2) + (32*tan(e/2 + (f*x)/2)*(2*a^2*c^2*d^ 
8 - 2*a^2*c^4*d^6 + 4*b^2*c^2*d^8 - 6*b^2*c^4*d^6 + 2*b^2*c^6*d^4 - 4*a*b* 
c*d^9 + 4*a*b*c^3*d^7))/(d^7 - 2*c^2*d^5 + c^4*d^3) + (b^2*((32*(c^2*d^9 - 
 2*c^4*d^7 + c^6*d^5))/(d^6 - 2*c^2*d^4 + c^4*d^2) + (32*tan(e/2 + (f*x...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 627, normalized size of antiderivative = 4.86 \[ \int \frac {(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^2} \, dx=\frac {2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (f x +e \right ) a^{2} c \,d^{3}-4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (f x +e \right ) a b \,d^{4}-2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (f x +e \right ) b^{2} c^{3} d +4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) \sin \left (f x +e \right ) b^{2} c \,d^{3}+2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) a^{2} c^{2} d^{2}-4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) a b c \,d^{3}-2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) b^{2} c^{4}+4 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) b^{2} c^{2} d^{2}+\cos \left (f x +e \right ) a^{2} c^{2} d^{3}-\cos \left (f x +e \right ) a^{2} d^{5}-2 \cos \left (f x +e \right ) a b \,c^{3} d^{2}+2 \cos \left (f x +e \right ) a b c \,d^{4}+\cos \left (f x +e \right ) b^{2} c^{4} d -\cos \left (f x +e \right ) b^{2} c^{2} d^{3}+\sin \left (f x +e \right ) b^{2} c^{4} d f x -2 \sin \left (f x +e \right ) b^{2} c^{2} d^{3} f x +\sin \left (f x +e \right ) b^{2} d^{5} f x +b^{2} c^{5} f x -2 b^{2} c^{3} d^{2} f x +b^{2} c \,d^{4} f x}{d^{2} f \left (\sin \left (f x +e \right ) c^{4} d -2 \sin \left (f x +e \right ) c^{2} d^{3}+\sin \left (f x +e \right ) d^{5}+c^{5}-2 c^{3} d^{2}+c \,d^{4}\right )} \] Input:

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

Output:

(2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin( 
e + f*x)*a**2*c*d**3 - 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/s 
qrt(c**2 - d**2))*sin(e + f*x)*a*b*d**4 - 2*sqrt(c**2 - d**2)*atan((tan((e 
 + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*b**2*c**3*d + 4*sqrt(c** 
2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*b* 
*2*c*d**3 + 2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - 
d**2))*a**2*c**2*d**2 - 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/ 
sqrt(c**2 - d**2))*a*b*c*d**3 - 2*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2) 
*c + d)/sqrt(c**2 - d**2))*b**2*c**4 + 4*sqrt(c**2 - d**2)*atan((tan((e + 
f*x)/2)*c + d)/sqrt(c**2 - d**2))*b**2*c**2*d**2 + cos(e + f*x)*a**2*c**2* 
d**3 - cos(e + f*x)*a**2*d**5 - 2*cos(e + f*x)*a*b*c**3*d**2 + 2*cos(e + f 
*x)*a*b*c*d**4 + cos(e + f*x)*b**2*c**4*d - cos(e + f*x)*b**2*c**2*d**3 + 
sin(e + f*x)*b**2*c**4*d*f*x - 2*sin(e + f*x)*b**2*c**2*d**3*f*x + sin(e + 
 f*x)*b**2*d**5*f*x + b**2*c**5*f*x - 2*b**2*c**3*d**2*f*x + b**2*c*d**4*f 
*x)/(d**2*f*(sin(e + f*x)*c**4*d - 2*sin(e + f*x)*c**2*d**3 + sin(e + f*x) 
*d**5 + c**5 - 2*c**3*d**2 + c*d**4))