3.5.65 \(\int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx\) [465]

Optimal. Leaf size=65 \[ -\frac {(c-d) \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac {(c+2 d) \cos (e+f x)}{3 f \left (a^2+a^2 \sin (e+f x)\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2829, 2727} \begin {gather*} -\frac {(c+2 d) \cos (e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )}-\frac {(c-d) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((c - d)*Cos[e + f*x])/(f*(a + a*Sin[e + f*x])^2) - ((c + 2*d)*Cos[e + f*x])/(3*f*(a^2 + a^2*Sin[e + f*x]
))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2829

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*
c - a*d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {c+d \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx &=-\frac {(c-d) \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}+\frac {(c+2 d) \int \frac {1}{a+a \sin (e+f x)} \, dx}{3 a}\\ &=-\frac {(c-d) \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac {(c+2 d) \cos (e+f x)}{3 f \left (a^2+a^2 \sin (e+f x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.04, size = 43, normalized size = 0.66 \begin {gather*} -\frac {\cos (e+f x) (2 c+d+(c+2 d) \sin (e+f x))}{3 a^2 f (1+\sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(Cos[e + f*x]*(2*c + d + (c + 2*d)*Sin[e + f*x]))/(a^2*f*(1 + Sin[e + f*x])^2)

________________________________________________________________________________________

Maple [A]
time = 0.26, size = 70, normalized size = 1.08

method	result	size
risch	\(-\frac {2 \left (-c +3 i c \,{\mathrm e}^{i \left (f x +e \right )}+3 i d \,{\mathrm e}^{i \left (f x +e \right )}+3 d \,{\mathrm e}^{2 i \left (f x +e \right )}-2 d \right )}{3 f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3}}\)	\(68\)
derivativedivides	\(\frac {-\frac {-2 c +2 d}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (2 c -2 d \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 c}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a^{2} f}\)	\(70\)
default	\(\frac {-\frac {-2 c +2 d}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {2 \left (2 c -2 d \right )}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {2 c}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}}{a^{2} f}\)	\(70\)
norman	\(\frac {-\frac {2 c \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}+\frac {\left (-2 c -2 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {-4 c -2 d}{3 a f}+\frac {2 \left (-c -d \right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a f}+\frac {2 \left (-5 c -d \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 a f}}{a \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 232 vs. \(2 (65) = 130\).
time = 0.30, size = 232, normalized size = 3.57 \begin {gather*} -\frac {2 \, {\left (\frac {c {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac {d {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.32, size = 125, normalized size = 1.92 \begin {gather*} \frac {{\left (c + 2 \, d\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, c + d\right )} \cos \left (f x + e\right ) + {\left ({\left (c + 2 \, d\right )} \cos \left (f x + e\right ) - c + d\right )} \sin \left (f x + e\right ) + c - d}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f - {\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (56) = 112\).
time = 1.41, size = 372, normalized size = 5.72 \begin {gather*} \begin {cases} - \frac {6 c \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {6 c \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {4 c}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {6 d \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} - \frac {2 d}{3 a^{2} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 9 a^{2} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 3 a^{2} f} & \text {for}\: f \neq 0 \\\frac {x \left (c + d \sin {\left (e \right )}\right )}{\left (a \sin {\left (e \right )} + a\right )^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-6*c*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*ta
n(e/2 + f*x/2) + 3*a**2*f) - 6*c*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2
+ 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 4*c/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 +
9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 6*d*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2
+ f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 2*d/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f
*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f), Ne(f, 0)), (x*(c + d*sin(e))/(a*sin(e) + a)**2, True))

________________________________________________________________________________________

Giac [A]
time = 0.44, size = 68, normalized size = 1.05 \begin {gather*} -\frac {2 \, {\left (3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, c + d\right )}}{3 \, a^{2} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 7.21, size = 97, normalized size = 1.49 \begin {gather*} -\frac {2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {5\,c}{2}+\frac {d}{2}-\frac {c\,\cos \left (e+f\,x\right )}{2}+\frac {d\,\cos \left (e+f\,x\right )}{2}+\frac {3\,c\,\sin \left (e+f\,x\right )}{2}+\frac {3\,d\,\sin \left (e+f\,x\right )}{2}\right )}{3\,a^2\,f\,\left (\frac {3\,\sqrt {2}\,\cos \left (\frac {e}{2}-\frac {\pi }{4}+\frac {f\,x}{2}\right )}{2}-\frac {\sqrt {2}\,\cos \left (\frac {3\,e}{2}+\frac {\pi }{4}+\frac {3\,f\,x}{2}\right )}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*cos(e/2 + (f*x)/2)*((5*c)/2 + d/2 - (c*cos(e + f*x))/2 + (d*cos(e + f*x))/2 + (3*c*sin(e + f*x))/2 + (3*d*
sin(e + f*x))/2))/(3*a^2*f*((3*2^(1/2)*cos(e/2 - pi/4 + (f*x)/2))/2 - (2^(1/2)*cos((3*e)/2 + pi/4 + (3*f*x)/2)
)/2))

________________________________________________________________________________________