3.703 \(\int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx\)

Optimal. Leaf size=117 \[ \frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2} (b c-a d)}-\frac {2 d \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f \sqrt {c^2-d^2} (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2747, 2660, 618, 204} \[ \frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{f \sqrt {a^2-b^2} (b c-a d)}-\frac {2 d \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f \sqrt {c^2-d^2} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])),x]

[Out]

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 2747

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

Rubi steps

\begin {align*} \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx &=\frac {b \int \frac {1}{a+b \sin (e+f x)} \, dx}{b c-a d}-\frac {d \int \frac {1}{c+d \sin (e+f x)} \, dx}{b c-a d}\\ &=\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(b c-a d) f}-\frac {(2 d) \operatorname {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(b c-a d) f}\\ &=-\frac {(4 b) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (e+f x)\right )\right )}{(b c-a d) f}+\frac {(4 d) \operatorname {Subst}\left (\int \frac {1}{-4 \left (c^2-d^2\right )-x^2} \, dx,x,2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{(b c-a d) f}\\ &=\frac {2 b \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} (b c-a d) f}-\frac {2 d \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(b c-a d) \sqrt {c^2-d^2} f}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 104, normalized size = 0.89 \[ \frac {\frac {2 b \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {2 d \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}}{b c f-a d f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Sin[e + f*x])*(c + d*Sin[e + f*x])),x]

[Out]

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

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fricas [A]  time = 1.85, size = 1057, normalized size = 9.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 145, normalized size = 1.24 \[ \frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{\sqrt {a^{2} - b^{2}} {\left (b c - a d\right )}} - \frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} d}{{\left (b c - a d\right )} \sqrt {c^{2} - d^{2}}}\right )}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.30, size = 116, normalized size = 0.99 \[ \frac {2 d \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{f \left (d a -c b \right ) \sqrt {c^{2}-d^{2}}}-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{f \left (d a -c b \right ) \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/(a*d-b*c)*d/(c^2-d^2)^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2))-2/f*b/(a*d-b*c)/(a^2-
b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(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(1/(a+b*sin(f*x+e))/(c+d*sin(f*x+e)),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*d^2-4*c^2>0)', see `assume?`
for more details)Is 4*d^2-4*c^2 positive or negative?

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mupad [B]  time = 9.99, size = 3281, normalized size = 28.04 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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