3.8.9 \(\int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx\) [709]

Optimal. Leaf size=97 \[ \frac {2 (a c-b d) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {(b c-a d) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2833, 12, 2739, 632, 210} \begin {gather*} \frac {2 (a c-b d) \text {ArcTan}\left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}+\frac {(b c-a d) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 2833

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 + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(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] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx &=\frac {(b c-a d) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {\int \frac {-a c+b d}{a+b \sin (e+f x)} \, dx}{-a^2+b^2}\\ &=\frac {(b c-a d) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {(a c-b d) \int \frac {1}{a+b \sin (e+f x)} \, dx}{a^2-b^2}\\ &=\frac {(b c-a d) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac {(2 (a c-b d)) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{\left (a^2-b^2\right ) f}\\ &=\frac {(b c-a d) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac {(4 (a c-b d)) \text {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 )}{\left (a^2-b^2\right ) f}\\ &=\frac {2 (a c-b d) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {(b c-a d) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 96, normalized size = 0.99 \begin {gather*} \frac {\frac {2 (a c-b d) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {(b c-a d) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.21, size = 144, normalized size = 1.48

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 b \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) a}-\frac {2 \left (a d -b c \right )}{a^{2}-b^{2}}}{a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {2 \left (a c -b d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{f}\)	\(144\)
default	\(\frac {\frac {-\frac {2 b \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) a}-\frac {2 \left (a d -b c \right )}{a^{2}-b^{2}}}{a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {2 \left (a c -b d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{f}\)	\(144\)
risch	\(\frac {2 i \left (a d -b c \right ) \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{b \left (a^{2}-b^{2}\right ) f \left (-i b \,{\mathrm e}^{2 i \left (f x +e \right )}+i b +2 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a c}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b d}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a c}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b d}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}\)	\(396\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(2*(-b*(a*d-b*c)/(a^2-b^2)/a*tan(1/2*f*x+1/2*e)-(a*d-b*c)/(a^2-b^2))/(a*tan(1/2*f*x+1/2*e)^2+2*b*tan(1/2*f
*x+1/2*e)+a)+2*(a*c-b*d)/(a^2-b^2)^(3/2)*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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 0.38, size = 414, normalized size = 4.27 \begin {gather*} \left [-\frac {{\left (a^{2} c - a b d + {\left (a b c - b^{2} d\right )} \sin \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left ({\left (a^{2} b - b^{3}\right )} c - {\left (a^{3} - a b^{2}\right )} d\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f\right )}}, -\frac {{\left (a^{2} c - a b d + {\left (a b c - b^{2} d\right )} \sin \left (f x + e\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) - {\left ({\left (a^{2} b - b^{3}\right )} c - {\left (a^{3} - a b^{2}\right )} d\right )} \cos \left (f x + e\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*((a^2*c - a*b*d + (a*b*c - b^2*d)*sin(f*x + e))*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)) - 2*((a^2*b - b^3)*c - (a^3 - a*b^2)*d)*cos(f*x + e))/((a^4*b - 2*a^
2*b^3 + b^5)*f*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*b^4)*f), -((a^2*c - a*b*d + (a*b*c - b^2*d)*sin(f*x + e))*s
qrt(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 - (a^3 - a*b^2)
*d)*cos(f*x + e))/((a^4*b - 2*a^2*b^3 + b^5)*f*sin(f*x + e) + (a^5 - 2*a^3*b^2 + a*b^4)*f)]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.48, size = 157, normalized size = 1.62 \begin {gather*} \frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a c - b d\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {b^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a b d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a b c - a^{2} d}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}}\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Mupad [B]
time = 8.02, size = 215, normalized size = 2.22 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\left (\frac {2\,\left (a^2\,b-b^3\right )\,\left (a\,c-b\,d\right )}{{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c-b\,d\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (a^2-b^2\right )}{2\,\left (a\,c-b\,d\right )}\right )\,\left (a\,c-b\,d\right )}{f\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {\frac {2\,\left (a\,d-b\,c\right )}{a^2-b^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,d-b\,c\right )}{a\,\left (a^2-b^2\right )}}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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