3.3.70 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^2} \, dx\) [270]

Optimal. Leaf size=181 \[ -\frac {2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a (c-d) \left (c^2-d^2\right )^{3/2} f}+\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a (c-d)^2 (c+d) f (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) (c+d \sin (e+f x))} \]

[Out]

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

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Rubi [A]
time = 0.24, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3057, 2833, 12, 2739, 632, 210} \begin {gather*} -\frac {2 \left (A d (2 c+d)-B \left (c^2+c d+d^2\right )\right ) \text {ArcTan}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a f (c-d) \left (c^2-d^2\right )^{3/2}}+\frac {d (B (2 c+d)-A (c+2 d)) \cos (e+f x)}{a f (c-d)^2 (c+d) (c+d \sin (e+f x))}-\frac {(A-B) \cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) (c+d \sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3057

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

Rubi steps

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

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Mathematica [A]
time = 0.87, size = 209, normalized size = 1.15 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 (A-B) \sin \left (\frac {1}{2} (e+f x)\right )+\frac {2 \left (-A d (2 c+d)+B \left (c^2+c d+d^2\right )\right ) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) \sqrt {c^2-d^2}}+\frac {d (B c-A d) \cos (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))}\right )}{a (c-d)^2 f (1+\sin (e+f x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(A - B)*Sin[(e + f*x)/2] + (2*(-(A*d*(2*c + d)) + B*(c^2 + c*d + d^2
))*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/((c + d)*Sqrt[c^2 -
d^2]) + (d*(B*c - A*d)*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[e + f*x]))))/(
a*(c - d)^2*f*(1 + Sin[e + f*x]))

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Maple [A]
time = 0.53, size = 197, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A -B \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 \left (\frac {\frac {d^{2} \left (A d -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (A d -B c \right )}{c +d}}{c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 A c d +A \,d^{2}-B \,c^{2}-B c d -B \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2}}}{a f}\)	\(197\)
default	\(\frac {-\frac {2 \left (A -B \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {2 \left (\frac {\frac {d^{2} \left (A d -B c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (c +d \right ) c}+\frac {d \left (A d -B c \right )}{c +d}}{c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c}+\frac {\left (2 A c d +A \,d^{2}-B \,c^{2}-B c d -B \,d^{2}\right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c +d \right ) \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2}}}{a f}\)	\(197\)
risch	\(-\frac {2 i \left (-3 i A c d \,{\mathrm e}^{i \left (f x +e \right )}-i A \,d^{2} {\mathrm e}^{i \left (f x +e \right )}-2 A c d \,{\mathrm e}^{2 i \left (f x +e \right )}+3 i B \,c^{2} {\mathrm e}^{i \left (f x +e \right )}+3 i B c d \,{\mathrm e}^{i \left (f x +e \right )}+B \,c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+2 A \,d^{2}-2 B c d -2 i A \,c^{2} {\mathrm e}^{i \left (f x +e \right )}-A \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+B c d \,{\mathrm e}^{2 i \left (f x +e \right )}+B \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+A c d -B \,d^{2}\right )}{\left (c +d \right ) \left (i d -i d \,{\mathrm e}^{2 i \left (f x +e \right )}+2 c \,{\mathrm e}^{i \left (f x +e \right )}\right ) \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) \left (c -d \right )^{2} f a}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) A c d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) A \,d^{2}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B \,c^{2}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B c d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B \,d^{2}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) A c d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) A \,d^{2}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B \,c^{2}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B c d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) B \,d^{2}}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right ) \left (c -d \right )^{2} f a}\)	\(1081\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 739 vs. \(2 (182) = 364\).
time = 0.42, size = 1571, normalized size = 8.68 \begin {gather*} \left [\frac {2 \, {\left (A - B\right )} c^{4} - 4 \, {\left (A - B\right )} c^{2} d^{2} + 2 \, {\left (A - B\right )} d^{4} + 2 \, {\left ({\left (A - 2 \, B\right )} c^{3} d + {\left (2 \, A - B\right )} c^{2} d^{2} - {\left (A - 2 \, B\right )} c d^{3} - {\left (2 \, A - B\right )} d^{4}\right )} \cos \left (f x + e\right )^{2} + {\left (B c^{3} - 2 \, {\left (A - B\right )} c^{2} d - {\left (3 \, A - 2 \, B\right )} c d^{2} - {\left (A - B\right )} d^{3} - {\left (B c^{2} d - {\left (2 \, A - B\right )} c d^{2} - {\left (A - B\right )} d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (B c^{3} - {\left (2 \, A - B\right )} c^{2} d - {\left (A - B\right )} c d^{2}\right )} \cos \left (f x + e\right ) + {\left (B c^{3} - 2 \, {\left (A - B\right )} c^{2} d - {\left (3 \, A - 2 \, B\right )} c d^{2} - {\left (A - B\right )} d^{3} + {\left (B c^{2} d - {\left (2 \, A - B\right )} c d^{2} - {\left (A - B\right )} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}} \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 (c \cos \left (f x + e\right ) \sin \left (f x + e\right ) + d \cos \left (f x + e\right )\right )} \sqrt {-c^{2} + d^{2}}}{d^{2} \cos \left (f x + e\right )^{2} - 2 \, c d \sin \left (f x + e\right ) - c^{2} - d^{2}}\right ) + 2 \, {\left ({\left (A - B\right )} c^{4} + {\left (A - 2 \, B\right )} c^{3} d + B c^{2} d^{2} - {\left (A - 2 \, B\right )} c d^{3} - A d^{4}\right )} \cos \left (f x + e\right ) - 2 \, {\left ({\left (A - B\right )} c^{4} - 2 \, {\left (A - B\right )} c^{2} d^{2} + {\left (A - B\right )} d^{4} - {\left ({\left (A - 2 \, B\right )} c^{3} d + {\left (2 \, A - B\right )} c^{2} d^{2} - {\left (A - 2 \, B\right )} c d^{3} - {\left (2 \, A - B\right )} d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (a c^{5} d - a c^{4} d^{2} - 2 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + a c d^{5} - a d^{6}\right )} f \cos \left (f x + e\right )^{2} - {\left (a c^{6} - a c^{5} d - 2 \, a c^{4} d^{2} + 2 \, a c^{3} d^{3} + a c^{2} d^{4} - a c d^{5}\right )} f \cos \left (f x + e\right ) - {\left (a c^{6} - 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} - a d^{6}\right )} f - {\left ({\left (a c^{5} d - a c^{4} d^{2} - 2 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + a c d^{5} - a d^{6}\right )} f \cos \left (f x + e\right ) + {\left (a c^{6} - 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} - a d^{6}\right )} f\right )} \sin \left (f x + e\right )\right )}}, \frac {{\left (A - B\right )} c^{4} - 2 \, {\left (A - B\right )} c^{2} d^{2} + {\left (A - B\right )} d^{4} + {\left ({\left (A - 2 \, B\right )} c^{3} d + {\left (2 \, A - B\right )} c^{2} d^{2} - {\left (A - 2 \, B\right )} c d^{3} - {\left (2 \, A - B\right )} d^{4}\right )} \cos \left (f x + e\right )^{2} + {\left (B c^{3} - 2 \, {\left (A - B\right )} c^{2} d - {\left (3 \, A - 2 \, B\right )} c d^{2} - {\left (A - B\right )} d^{3} - {\left (B c^{2} d - {\left (2 \, A - B\right )} c d^{2} - {\left (A - B\right )} d^{3}\right )} \cos \left (f x + e\right )^{2} + {\left (B c^{3} - {\left (2 \, A - B\right )} c^{2} d - {\left (A - B\right )} c d^{2}\right )} \cos \left (f x + e\right ) + {\left (B c^{3} - 2 \, {\left (A - B\right )} c^{2} d - {\left (3 \, A - 2 \, B\right )} c d^{2} - {\left (A - B\right )} d^{3} + {\left (B c^{2} d - {\left (2 \, A - B\right )} c d^{2} - {\left (A - B\right )} d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {c^{2} - d^{2}} \arctan \left (-\frac {c \sin \left (f x + e\right ) + d}{\sqrt {c^{2} - d^{2}} \cos \left (f x + e\right )}\right ) + {\left ({\left (A - B\right )} c^{4} + {\left (A - 2 \, B\right )} c^{3} d + B c^{2} d^{2} - {\left (A - 2 \, B\right )} c d^{3} - A d^{4}\right )} \cos \left (f x + e\right ) - {\left ({\left (A - B\right )} c^{4} - 2 \, {\left (A - B\right )} c^{2} d^{2} + {\left (A - B\right )} d^{4} - {\left ({\left (A - 2 \, B\right )} c^{3} d + {\left (2 \, A - B\right )} c^{2} d^{2} - {\left (A - 2 \, B\right )} c d^{3} - {\left (2 \, A - B\right )} d^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{{\left (a c^{5} d - a c^{4} d^{2} - 2 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + a c d^{5} - a d^{6}\right )} f \cos \left (f x + e\right )^{2} - {\left (a c^{6} - a c^{5} d - 2 \, a c^{4} d^{2} + 2 \, a c^{3} d^{3} + a c^{2} d^{4} - a c d^{5}\right )} f \cos \left (f x + e\right ) - {\left (a c^{6} - 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} - a d^{6}\right )} f - {\left ({\left (a c^{5} d - a c^{4} d^{2} - 2 \, a c^{3} d^{3} + 2 \, a c^{2} d^{4} + a c d^{5} - a d^{6}\right )} f \cos \left (f x + e\right ) + {\left (a c^{6} - 3 \, a c^{4} d^{2} + 3 \, a c^{2} d^{4} - a d^{6}\right )} f\right )} \sin \left (f x + e\right )}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(A - B)*c^4 - 4*(A - B)*c^2*d^2 + 2*(A - B)*d^4 + 2*((A - 2*B)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c
*d^3 - (2*A - B)*d^4)*cos(f*x + e)^2 + (B*c^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 - (B*c^2*d -
(2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e)^2 + (B*c^3 - (2*A - B)*c^2*d - (A - B)*c*d^2)*cos(f*x + e) + (B*c
^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 + (B*c^2*d - (2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e
))*sin(f*x + e))*sqrt(-c^2 + d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 2*(c*co
s(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*((A - B)*c^4 + (A - 2*B)*c^3*d + B*c^2*d^2 - (A - 2*B)*c*d^3 - A*d^4)*cos(f*x + e) - 2*((A - B)*c^4 -
2*(A - B)*c^2*d^2 + (A - B)*d^4 - ((A - 2*B)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos
(f*x + e))*sin(f*x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f*cos(f*x + e)^2
- (a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x + e) - (a*c^6 - 3*a*c^4*d^2 +
3*a*c^2*d^4 - a*d^6)*f - ((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d^5 - a*d^6)*f*cos(f*x + e)
+ (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f)*sin(f*x + e)), ((A - B)*c^4 - 2*(A - B)*c^2*d^2 + (A - B)*d^4
+ ((A - 2*B)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e)^2 + (B*c^3 - 2*(A - B)
*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 - (B*c^2*d - (2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e)^2 + (B*c^3 -
(2*A - B)*c^2*d - (A - B)*c*d^2)*cos(f*x + e) + (B*c^3 - 2*(A - B)*c^2*d - (3*A - 2*B)*c*d^2 - (A - B)*d^3 +
(B*c^2*d - (2*A - B)*c*d^2 - (A - B)*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(c^2 - d^2)*arctan(-(c*sin(f*x + e)
+ d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + ((A - B)*c^4 + (A - 2*B)*c^3*d + B*c^2*d^2 - (A - 2*B)*c*d^3 - A*d^4)*c
os(f*x + e) - ((A - B)*c^4 - 2*(A - B)*c^2*d^2 + (A - B)*d^4 - ((A - 2*B)*c^3*d + (2*A - B)*c^2*d^2 - (A - 2*B
)*c*d^3 - (2*A - B)*d^4)*cos(f*x + e))*sin(f*x + e))/((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*d
^5 - a*d^6)*f*cos(f*x + e)^2 - (a*c^6 - a*c^5*d - 2*a*c^4*d^2 + 2*a*c^3*d^3 + a*c^2*d^4 - a*c*d^5)*f*cos(f*x +
e) - (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f - ((a*c^5*d - a*c^4*d^2 - 2*a*c^3*d^3 + 2*a*c^2*d^4 + a*c*
d^5 - a*d^6)*f*cos(f*x + e) + (a*c^6 - 3*a*c^4*d^2 + 3*a*c^2*d^4 - a*d^6)*f)*sin(f*x + e))]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (182) = 364\).
time = 0.48, size = 443, normalized size = 2.45 \begin {gather*} \frac {2 \, {\left (\frac {{\left (B c^{2} - 2 \, A c d + B c d - A d^{2} + B d^{2}\right )} {\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^{3} - a c^{2} d - a c d^{2} + a d^{3}\right )} \sqrt {c^{2} - d^{2}}} - \frac {A c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + A c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - B c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + A d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, A c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, B c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 3 \, B c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + A c^{3} - B c^{3} + A c^{2} d - 2 \, B c^{2} d + A c d^{2}}{{\left (a c^{4} - a c^{3} d - a c^{2} d^{2} + a c d^{3}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}}\right )}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((B*c^2 - 2*A*c*d + B*c*d - A*d^2 + B*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^3 - a*c^2*d - a*c*d^2 + a*d^3)*sqrt(c^2 - d^2)) - (A*c^3*tan(1/2*f*x + 1
/2*e)^2 - B*c^3*tan(1/2*f*x + 1/2*e)^2 + A*c^2*d*tan(1/2*f*x + 1/2*e)^2 - B*c^2*d*tan(1/2*f*x + 1/2*e)^2 - B*c
*d^2*tan(1/2*f*x + 1/2*e)^2 + A*d^3*tan(1/2*f*x + 1/2*e)^2 + 2*A*c^2*d*tan(1/2*f*x + 1/2*e) - 3*B*c^2*d*tan(1/
2*f*x + 1/2*e) + 3*A*c*d^2*tan(1/2*f*x + 1/2*e) - 3*B*c*d^2*tan(1/2*f*x + 1/2*e) + A*d^3*tan(1/2*f*x + 1/2*e)
+ A*c^3 - B*c^3 + A*c^2*d - 2*B*c^2*d + A*c*d^2)/((a*c^4 - a*c^3*d - a*c^2*d^2 + a*c*d^3)*(c*tan(1/2*f*x + 1/2
*e)^3 + c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e)^2 + c*tan(1/2*f*x + 1/2*e) + 2*d*tan(1/2*f*x + 1/2
*e) + c)))/f

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Mupad [B]
time = 15.53, size = 437, normalized size = 2.41 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {\frac {\left (2\,a\,c^3\,d-2\,a\,c^2\,d^2-2\,a\,c\,d^3+2\,a\,d^4\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c^3-a\,c^2\,d-a\,c\,d^2+a\,d^3\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}}{2\,B\,c^2-2\,A\,d^2+2\,B\,d^2-4\,A\,c\,d+2\,B\,c\,d}\right )\,\left (B\,c^2-A\,d^2+B\,d^2-2\,A\,c\,d+B\,c\,d\right )}{a\,f\,{\left (c+d\right )}^{3/2}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {2\,\left (A\,c^2+A\,d^2-B\,c^2+A\,c\,d-2\,B\,c\,d\right )}{\left (c+d\right )\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,d^2+2\,A\,c\,d-3\,B\,c\,d\right )}{c\,{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (A\,c^3+A\,d^3-B\,c^3+A\,c^2\,d-B\,c\,d^2-B\,c^2\,d\right )}{c\,\left (c+d\right )\,{\left (c-d\right )}^2}}{f\,\left (a\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\left (a\,c+2\,a\,d\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+\left (a\,c+2\,a\,d\right )\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\,c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atan((((2*a*d^4 - 2*a*c^2*d^2 - 2*a*c*d^3 + 2*a*c^3*d)*(B*c^2 - A*d^2 + B*d^2 - 2*A*c*d + B*c*d))/(a*(c + d
)^(3/2)*(c - d)^(5/2)) + (2*c*tan(e/2 + (f*x)/2)*(a*c^3 + a*d^3 - a*c*d^2 - a*c^2*d)*(B*c^2 - A*d^2 + B*d^2 -
2*A*c*d + B*c*d))/(a*(c + d)^(3/2)*(c - d)^(5/2)))/(2*B*c^2 - 2*A*d^2 + 2*B*d^2 - 4*A*c*d + 2*B*c*d))*(B*c^2 -
A*d^2 + B*d^2 - 2*A*c*d + B*c*d))/(a*f*(c + d)^(3/2)*(c - d)^(5/2)) - ((2*(A*c^2 + A*d^2 - B*c^2 + A*c*d - 2*
B*c*d))/((c + d)*(c - d)^2) + (2*tan(e/2 + (f*x)/2)*(A*d^2 + 2*A*c*d - 3*B*c*d))/(c*(c - d)^2) + (2*tan(e/2 +
(f*x)/2)^2*(A*c^3 + A*d^3 - B*c^3 + A*c^2*d - B*c*d^2 - B*c^2*d))/(c*(c + d)*(c - d)^2))/(f*(a*c + tan(e/2 + (
f*x)/2)^2*(a*c + 2*a*d) + tan(e/2 + (f*x)/2)*(a*c + 2*a*d) + a*c*tan(e/2 + (f*x)/2)^3))

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