3.5.30 \(\int \frac {a+a \sin (e+f x)}{c+d \sin (e+f x)} \, dx\) [430]

Optimal. Leaf size=63 \[ \frac {a x}{d}-\frac {2 a (c-d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{d \sqrt {c^2-d^2} f} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2814, 2739, 632, 210} \begin {gather*} \frac {a x}{d}-\frac {2 a (c-d) \text {ArcTan}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{d f \sqrt {c^2-d^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 210

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 632

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 2739

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 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(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]

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.21, size = 182, normalized size = 2.89 \begin {gather*} \frac {a \left (-2 (c-d) \tan ^{-1}\left (\frac {\sec \left (\frac {f x}{2}\right ) (\cos (e)-i \sin (e)) \left (d \cos \left (e+\frac {f x}{2}\right )+c \sin \left (\frac {f x}{2}\right )\right )}{\sqrt {c^2-d^2} \sqrt {(\cos (e)-i \sin (e))^2}}\right ) (\cos (e)-i \sin (e))+\sqrt {c^2-d^2} f x \sqrt {(\cos (e)-i \sin (e))^2}\right ) (1+\sin (e+f x))}{d \sqrt {c^2-d^2} f \sqrt {(\cos (e)-i \sin (e))^2} \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-2*(c - d)*ArcTan[(Sec[(f*x)/2]*(Cos[e] - I*Sin[e])*(d*Cos[e + (f*x)/2] + c*Sin[(f*x)/2]))/(Sqrt[c^2 - d^2
]*Sqrt[(Cos[e] - I*Sin[e])^2])]*(Cos[e] - I*Sin[e]) + Sqrt[c^2 - d^2]*f*x*Sqrt[(Cos[e] - I*Sin[e])^2])*(1 + Si
n[e + f*x]))/(d*Sqrt[c^2 - d^2]*f*Sqrt[(Cos[e] - I*Sin[e])^2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)

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Maple [A]
time = 0.26, size = 72, normalized size = 1.14

method	result	size
derivativedivides	\(\frac {2 a \left (\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d}+\frac {\left (-c +d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(72\)
default	\(\frac {2 a \left (\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{d}+\frac {\left (-c +d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{d \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(72\)
risch	\(\frac {a x}{d}+\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-\frac {-i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right )}{\left (c +d \right ) f d}-\frac {\sqrt {-\left (c +d \right ) \left (c -d \right )}\, a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c +\sqrt {-\left (c +d \right ) \left (c -d \right )}}{d}\right )}{\left (c +d \right ) f d}\)	\(124\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*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 de

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Fricas [A]
time = 0.37, size = 237, normalized size = 3.76 \begin {gather*} \left [\frac {2 \, a f x + a \sqrt {-\frac {c - d}{c + d}} \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 ({\left (c^{2} + c d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left (c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {c - d}{c + d}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right )}{2 \, d f}, \frac {a f x + a \sqrt {\frac {c - d}{c + d}} \arctan \left (-\frac {{\left (c \sin \left (f x + e\right ) + d\right )} \sqrt {\frac {c - d}{c + d}}}{{\left (c - d\right )} \cos \left (f x + e\right )}\right )}{d f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*a*f*x + a*sqrt(-(c - d)/(c + 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^2 + c*d)*cos(f*x + e)*sin(f*x + e) + (c*d + d^2)*cos(f*x + e))*sqrt(-(c - d)/(c + d)))/(d^2*cos(f*x + e)
^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)))/(d*f), (a*f*x + a*sqrt((c - d)/(c + d))*arctan(-(c*sin(f*x + e) + d)*sq
rt((c - d)/(c + d))/((c - d)*cos(f*x + e))))/(d*f)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 537 vs. \(2 (49) = 98\).
time = 40.18, size = 537, normalized size = 8.52 \begin {gather*} \begin {cases} \frac {\tilde {\infty } x \left (a \sin {\left (e \right )} + a\right )}{\sin {\left (e \right )}} & \text {for}\: c = 0 \wedge d = 0 \wedge f = 0 \\\frac {x \left (a \sin {\left (e \right )} + a\right )}{c + d \sin {\left (e \right )}} & \text {for}\: f = 0 \\\frac {a x - \frac {a \cos {\left (e + f x \right )}}{f}}{c} & \text {for}\: d = 0 \\\frac {a x + \frac {a \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} \right )}}{f}}{d} & \text {for}\: c = 0 \\\frac {a d^{2} f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{d^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - f \left (d^{2}\right )^{\frac {3}{2}}} + \frac {2 a d^{2}}{d^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - f \left (d^{2}\right )^{\frac {3}{2}}} - \frac {a d f x \sqrt {d^{2}}}{d^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - f \left (d^{2}\right )^{\frac {3}{2}}} + \frac {2 a d \sqrt {d^{2}}}{d^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} - f \left (d^{2}\right )^{\frac {3}{2}}} & \text {for}\: c = - \sqrt {d^{2}} \\\frac {a d^{2} f x \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{d^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + f \left (d^{2}\right )^{\frac {3}{2}}} + \frac {2 a d^{2}}{d^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + f \left (d^{2}\right )^{\frac {3}{2}}} + \frac {a d f x \sqrt {d^{2}}}{d^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + f \left (d^{2}\right )^{\frac {3}{2}}} - \frac {2 a d \sqrt {d^{2}}}{d^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + f \left (d^{2}\right )^{\frac {3}{2}}} & \text {for}\: c = \sqrt {d^{2}} \\- \frac {a c \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {d}{c} - \frac {\sqrt {- c^{2} + d^{2}}}{c} \right )}}{d f \sqrt {- c^{2} + d^{2}}} + \frac {a c \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {d}{c} + \frac {\sqrt {- c^{2} + d^{2}}}{c} \right )}}{d f \sqrt {- c^{2} + d^{2}}} + \frac {a \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {d}{c} - \frac {\sqrt {- c^{2} + d^{2}}}{c} \right )}}{f \sqrt {- c^{2} + d^{2}}} - \frac {a \log {\left (\tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + \frac {d}{c} + \frac {\sqrt {- c^{2} + d^{2}}}{c} \right )}}{f \sqrt {- c^{2} + d^{2}}} + \frac {a x}{d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(a*sin(e) + a)/sin(e), Eq(c, 0) & Eq(d, 0) & Eq(f, 0)), (x*(a*sin(e) + a)/(c + d*sin(e)), Eq(
f, 0)), ((a*x - a*cos(e + f*x)/f)/c, Eq(d, 0)), ((a*x + a*log(tan(e/2 + f*x/2))/f)/d, Eq(c, 0)), (a*d**2*f*x*t
an(e/2 + f*x/2)/(d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)) + 2*a*d**2/(d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3
/2)) - a*d*f*x*sqrt(d**2)/(d**3*f*tan(e/2 + f*x/2) - f*(d**2)**(3/2)) + 2*a*d*sqrt(d**2)/(d**3*f*tan(e/2 + f*x
/2) - f*(d**2)**(3/2)), Eq(c, -sqrt(d**2))), (a*d**2*f*x*tan(e/2 + f*x/2)/(d**3*f*tan(e/2 + f*x/2) + f*(d**2)*
*(3/2)) + 2*a*d**2/(d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)) + a*d*f*x*sqrt(d**2)/(d**3*f*tan(e/2 + f*x/2) +
f*(d**2)**(3/2)) - 2*a*d*sqrt(d**2)/(d**3*f*tan(e/2 + f*x/2) + f*(d**2)**(3/2)), Eq(c, sqrt(d**2))), (-a*c*lo
g(tan(e/2 + f*x/2) + d/c - sqrt(-c**2 + d**2)/c)/(d*f*sqrt(-c**2 + d**2)) + a*c*log(tan(e/2 + f*x/2) + d/c + s
qrt(-c**2 + d**2)/c)/(d*f*sqrt(-c**2 + d**2)) + a*log(tan(e/2 + f*x/2) + d/c - sqrt(-c**2 + d**2)/c)/(f*sqrt(-
c**2 + d**2)) - a*log(tan(e/2 + f*x/2) + d/c + sqrt(-c**2 + d**2)/c)/(f*sqrt(-c**2 + d**2)) + a*x/d, True))

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Giac [A]
time = 0.48, size = 86, normalized size = 1.37 \begin {gather*} \frac {\frac {{\left (f x + e\right )} a}{d} - \frac {2 \, {\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 (a c - a d\right )}}{\sqrt {c^{2} - d^{2}} d}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((f*x + e)*a/d - 2*(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)))*(a*c - a*d)/(sqrt(c^2 - d^2)*d))/f

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Mupad [B]
time = 7.39, size = 449, normalized size = 7.13 \begin {gather*} \frac {2\,a\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f\,\left (c+d\right )}-\frac {2\,a\,\mathrm {atanh}\left (\frac {3\,d^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (d^2-c^2\right )}^{3/2}-2\,c^4\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}-2\,c^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (d^2-c^2\right )}^{3/2}+d^4\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}+2\,c^2\,d^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}+3\,c^2\,d^2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}+c\,d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (d^2-c^2\right )}^{3/2}+c\,d^3\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}+c^3\,d\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}+4\,c\,d^3\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}-2\,c^3\,d\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {d^2-c^2}}{2\,\left (d^2+c\,d\right )\,\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^3+2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c^2\,d-\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,c\,d^2-2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,d^3\right )}\right )\,\sqrt {d^2-c^2}}{d\,f\,\left (c+d\right )}+\frac {2\,a\,c\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{d\,f\,\left (c+d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*atan(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(f*(c + d)) - (2*a*atanh((3*d^2*sin(e/2 + (f*x)/2)*(d^2 - c^
2)^(3/2) - 2*c^4*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2) - 2*c^2*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2) + d^4*sin(e
/2 + (f*x)/2)*(d^2 - c^2)^(1/2) + 2*c^2*d^2*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2) + 3*c^2*d^2*sin(e/2 + (f*x)/2
)*(d^2 - c^2)^(1/2) + c*d*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(3/2) + c*d^3*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2) +
c^3*d*cos(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2) + 4*c*d^3*sin(e/2 + (f*x)/2)*(d^2 - c^2)^(1/2) - 2*c^3*d*sin(e/2 +
(f*x)/2)*(d^2 - c^2)^(1/2))/(2*(c*d + d^2)*(c^3*cos(e/2 + (f*x)/2) - 2*d^3*sin(e/2 + (f*x)/2) - c*d^2*cos(e/2
+ (f*x)/2) + 2*c^2*d*sin(e/2 + (f*x)/2))))*(d^2 - c^2)^(1/2))/(d*f*(c + d)) + (2*a*c*atan(sin(e/2 + (f*x)/2)/c
os(e/2 + (f*x)/2)))/(d*f*(c + d))

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