\(\int \frac {d+e \sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx\) [446]

Optimal result

Integrand size = 21, antiderivative size = 242 \[ \int \frac {d+e \sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {\sqrt {2} \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {\sqrt {2} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)+b \sqrt {b^2-4 a c}}} \] Output:

2^(1/2)*(e+(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))*arctan(1/2*(2*c+(b-(-4*a*c+b^2 
)^(1/2))*tan(1/2*x))*2^(1/2)/(b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1/2))^(1/2))/( 
b^2-2*c*(a+c)-b*(-4*a*c+b^2)^(1/2))^(1/2)+2^(1/2)*(e-(-b*e+2*c*d)/(-4*a*c+ 
b^2)^(1/2))*arctan(1/2*(2*c+(b+(-4*a*c+b^2)^(1/2))*tan(1/2*x))*2^(1/2)/(b^ 
2-2*c*(a+c)+b*(-4*a*c+b^2)^(1/2))^(1/2))/(b^2-2*c*(a+c)+b*(-4*a*c+b^2)^(1/ 
2))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.04 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.18 \[ \int \frac {d+e \sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\frac {\frac {\left (-2 i c d+\left (i b+\sqrt {-b^2+4 a c}\right ) e\right ) \arctan \left (\frac {2 c+\left (b-i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)-i b \sqrt {-b^2+4 a c}}}+\frac {\left (2 i c d+\left (-i b+\sqrt {-b^2+4 a c}\right ) e\right ) \arctan \left (\frac {2 c+\left (b+i \sqrt {-b^2+4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2} \sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}\right )}{\sqrt {b^2-2 c (a+c)+i b \sqrt {-b^2+4 a c}}}}{\sqrt {-\frac {b^2}{2}+2 a c}} \] Input:

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

Output:

((((-2*I)*c*d + (I*b + Sqrt[-b^2 + 4*a*c])*e)*ArcTan[(2*c + (b - I*Sqrt[-b 
^2 + 4*a*c])*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4 
*a*c]])])/Sqrt[b^2 - 2*c*(a + c) - I*b*Sqrt[-b^2 + 4*a*c]] + (((2*I)*c*d + 
 ((-I)*b + Sqrt[-b^2 + 4*a*c])*e)*ArcTan[(2*c + (b + I*Sqrt[-b^2 + 4*a*c]) 
*Tan[x/2])/(Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]])])/Sq 
rt[b^2 - 2*c*(a + c) + I*b*Sqrt[-b^2 + 4*a*c]])/Sqrt[-1/2*b^2 + 2*a*c]
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3773, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3773

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle -4 \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{4 \left (4 c^2-\left (b+\sqrt {b^2-4 a c}\right )^2\right )-\left (4 c+2 \left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )\right )^2}d\left (4 c+2 \left (b+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )\right )-4 \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \int \frac {1}{-\left (4 c+2 \left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )\right )^2-8 \left (b^2-\sqrt {b^2-4 a c} b-2 c (a+c)\right )}d\left (4 c+2 \left (b-\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\sqrt {2} \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) \left (b-\sqrt {b^2-4 a c}\right )+4 c}{2 \sqrt {2} \sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt {-b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}+\frac {\sqrt {2} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) \left (\sqrt {b^2-4 a c}+b\right )+4 c}{2 \sqrt {2} \sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\right )}{\sqrt {b \sqrt {b^2-4 a c}-2 c (a+c)+b^2}}\)

Input:

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

Output:

(Sqrt[2]*(e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(4*c + 2*(b - Sqrt[b 
^2 - 4*a*c])*Tan[x/2])/(2*Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4* 
a*c]])])/Sqrt[b^2 - 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]] + (Sqrt[2]*(e - (2* 
c*d - b*e)/Sqrt[b^2 - 4*a*c])*ArcTan[(4*c + 2*(b + Sqrt[b^2 - 4*a*c])*Tan[ 
x/2])/(2*Sqrt[2]*Sqrt[b^2 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]])])/Sqrt[b^2 
 - 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]
 

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 3773

Int[((A_) + (B_.)*sin[(d_.) + (e_.)*(x_)])/((a_.) + (b_.)*sin[(d_.) + (e_.) 
*(x_)] + (c_.)*sin[(d_.) + (e_.)*(x_)]^2), x_Symbol] :> Module[{q = Rt[b^2 
- 4*a*c, 2]}, Simp[(B + (b*B - 2*A*c)/q)   Int[1/(b + q + 2*c*Sin[d + e*x]) 
, x], x] + Simp[(B - (b*B - 2*A*c)/q)   Int[1/(b - q + 2*c*Sin[d + e*x]), x 
], x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[b^2 - 4*a*c, 0]
 

Maple [A] (verified)

Time = 1.50 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.09

method	result	size
default	\(2 a \left (-\frac {\sqrt {-4 a c +b^{2}}\, \left (\sqrt {-4 a c +b^{2}}\, d -2 a e +b d \right ) \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+b +\sqrt {-4 a c +b^{2}}}{\sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (4 a c -b^{2}\right ) a \sqrt {4 a c -2 b^{2}-2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}+\frac {\left (\sqrt {-4 a c +b^{2}}\, d +2 a e -b d \right ) \sqrt {-4 a c +b^{2}}\, \arctan \left (\frac {-2 a \tan \left (\frac {x}{2}\right )+\sqrt {-4 a c +b^{2}}-b}{\sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )}{\left (4 a c -b^{2}\right ) a \sqrt {4 a c -2 b^{2}+2 b \sqrt {-4 a c +b^{2}}+4 a^{2}}}\right )\)	\(263\)
risch	\(\text {Expression too large to display}\)	\(8331\)

Input:

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

Output:

2*a*(-(-4*a*c+b^2)^(1/2)*((-4*a*c+b^2)^(1/2)*d-2*a*e+b*d)/(4*a*c-b^2)/a/(4 
*a*c-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((2*a*tan(1/2*x)+b+(- 
4*a*c+b^2)^(1/2))/(4*a*c-2*b^2-2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2))+((-4*a 
*c+b^2)^(1/2)*d+2*a*e-b*d)*(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)/a/(4*a*c-2*b^2+2 
*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)*arctan((-2*a*tan(1/2*x)+(-4*a*c+b^2)^(1 
/2)-b)/(4*a*c-2*b^2+2*b*(-4*a*c+b^2)^(1/2)+4*a^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6695 vs. \(2 (208) = 416\).

Time = 5.51 (sec) , antiderivative size = 6695, normalized size of antiderivative = 27.67 \[ \int \frac {d+e \sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

\[ \int \frac {d+e \sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\int { \frac {e \sin \left (x\right ) + d}{c \sin \left (x\right )^{2} + b \sin \left (x\right ) + a} \,d x } \] Input:

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 30.50 (sec) , antiderivative size = 10465, normalized size of antiderivative = 43.24 \[ \int \frac {d+e \sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((-(b^4*d^2 - b^4*e^2 + 8*a*c^3*d^2 + b*d^2*(-(4*a*c - b^2)^3)^(1/2) 
- 8*a^3*c*e^2 + b*e^2*(-(4*a*c - b^2)^3)^(1/2) + 2*a^2*b^2*e^2 + 8*a^2*c^2 
*d^2 - 8*a^2*c^2*e^2 - 2*b^2*c^2*d^2 - 2*a*b^3*d*e - 2*a*d*e*(-(4*a*c - b^ 
2)^3)^(1/2) + 2*b^3*c*d*e - 2*c*d*e*(-(4*a*c - b^2)^3)^(1/2) - 6*a*b^2*c*d 
^2 + 6*a*b^2*c*e^2 - 8*a*b*c^2*d*e + 8*a^2*b*c*d*e)/(2*(a^2*b^4 - b^6 + 16 
*a^2*c^4 + 32*a^3*c^3 + 16*a^4*c^2 + b^4*c^2 - 8*a*b^2*c^3 - 8*a^3*b^2*c - 
 32*a^2*b^2*c^2 + 10*a*b^4*c)))^(1/2)*((-(b^4*d^2 - b^4*e^2 + 8*a*c^3*d^2 
+ b*d^2*(-(4*a*c - b^2)^3)^(1/2) - 8*a^3*c*e^2 + b*e^2*(-(4*a*c - b^2)^3)^ 
(1/2) + 2*a^2*b^2*e^2 + 8*a^2*c^2*d^2 - 8*a^2*c^2*e^2 - 2*b^2*c^2*d^2 - 2* 
a*b^3*d*e - 2*a*d*e*(-(4*a*c - b^2)^3)^(1/2) + 2*b^3*c*d*e - 2*c*d*e*(-(4* 
a*c - b^2)^3)^(1/2) - 6*a*b^2*c*d^2 + 6*a*b^2*c*e^2 - 8*a*b*c^2*d*e + 8*a^ 
2*b*c*d*e)/(2*(a^2*b^4 - b^6 + 16*a^2*c^4 + 32*a^3*c^3 + 16*a^4*c^2 + b^4* 
c^2 - 8*a*b^2*c^3 - 8*a^3*b^2*c - 32*a^2*b^2*c^2 + 10*a*b^4*c)))^(1/2)*((- 
(b^4*d^2 - b^4*e^2 + 8*a*c^3*d^2 + b*d^2*(-(4*a*c - b^2)^3)^(1/2) - 8*a^3* 
c*e^2 + b*e^2*(-(4*a*c - b^2)^3)^(1/2) + 2*a^2*b^2*e^2 + 8*a^2*c^2*d^2 - 8 
*a^2*c^2*e^2 - 2*b^2*c^2*d^2 - 2*a*b^3*d*e - 2*a*d*e*(-(4*a*c - b^2)^3)^(1 
/2) + 2*b^3*c*d*e - 2*c*d*e*(-(4*a*c - b^2)^3)^(1/2) - 6*a*b^2*c*d^2 + 6*a 
*b^2*c*e^2 - 8*a*b*c^2*d*e + 8*a^2*b*c*d*e)/(2*(a^2*b^4 - b^6 + 16*a^2*c^4 
 + 32*a^3*c^3 + 16*a^4*c^2 + b^4*c^2 - 8*a*b^2*c^3 - 8*a^3*b^2*c - 32*a^2* 
b^2*c^2 + 10*a*b^4*c)))^(1/2)*(tan(x/2)*(96*a*b^4 + 256*a^4*c - 64*a^3*...
 

Reduce [F]

\[ \int \frac {d+e \sin (x)}{a+b \sin (x)+c \sin ^2(x)} \, dx=\left (\int \frac {\sin \left (x \right )}{\sin \left (x \right )^{2} c +\sin \left (x \right ) b +a}d x \right ) e +\left (\int \frac {1}{\sin \left (x \right )^{2} c +\sin \left (x \right ) b +a}d x \right ) d \] Input:

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

Output:

int(sin(x)/(sin(x)**2*c + sin(x)*b + a),x)*e + int(1/(sin(x)**2*c + sin(x) 
*b + a),x)*d