3.5.67 \(\int \frac {1}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx\) [467]

Optimal. Leaf size=131 \[ \frac {2 d^2 \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{a^2 (c-d)^2 \sqrt {c^2-d^2} f}-\frac {(c-4 d) \cos (e+f x)}{3 a^2 (c-d)^2 f (1+\sin (e+f x))}-\frac {\cos (e+f x)}{3 (c-d) f (a+a \sin (e+f x))^2} \]

[Out]

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

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Rubi [A]
time = 0.20, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2845, 3057, 12, 2739, 632, 210} \begin {gather*} \frac {2 d^2 \text {ArcTan}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{a^2 f (c-d)^2 \sqrt {c^2-d^2}}-\frac {(c-4 d) \cos (e+f x)}{3 a^2 f (c-d)^2 (\sin (e+f x)+1)}-\frac {\cos (e+f x)}{3 f (c-d) (a \sin (e+f x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])),x]

[Out]

(2*d^2*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(a^2*(c - d)^2*Sqrt[c^2 - d^2]*f) - ((c - 4*d)*Cos[e
+ f*x])/(3*a^2*(c - d)^2*f*(1 + Sin[e + f*x])) - Cos[e + f*x]/(3*(c - d)*f*(a + a*Sin[e + f*x])^2)

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 2845

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

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

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Mathematica [A]
time = 0.24, size = 204, normalized size = 1.56 \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 (c-d) \sin \left (\frac {1}{2} (e+f x)\right )-(c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (c-4 d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+\frac {6 d^2 \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 )^3}{\sqrt {c^2-d^2}}\right )}{3 a^2 (c-d)^2 f (1+\sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c + d*Sin[e + f*x])),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*(c - d)*Sin[(e + f*x)/2] - (c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)
/2]) + 2*(c - 4*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + (6*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])^3)/Sqrt[c^2 - d^2]))/(3*a^2*(c - d)^2*f*(1 + Sin
[e + f*x])^2)

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Maple [A]
time = 0.48, size = 132, normalized size = 1.01

method	result	size
derivativedivides	\(\frac {\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2} \sqrt {c^{2}-d^{2}}}-\frac {2 \left (c -2 d \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {4}{3 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {2}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{2} f}\)	\(132\)
default	\(\frac {\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (c -d \right )^{2} \sqrt {c^{2}-d^{2}}}-\frac {2 \left (c -2 d \right )}{\left (c -d \right )^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {4}{3 \left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {2}{\left (c -d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}}{a^{2} f}\)	\(132\)
risch	\(\frac {2 d \,{\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 c}{3}-\frac {8 d}{3}-2 i c \,{\mathrm e}^{i \left (f x +e \right )}+6 i d \,{\mathrm e}^{i \left (f x +e \right )}}{\left ({\mathrm e}^{i \left (f x +e \right )}+i\right )^{3} \left (c -d \right )^{2} f \,a^{2}}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (c -d \right )^{2} f \,a^{2}}\)	\(231\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

2/f/a^2*(d^2/(c-d)^2/(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))-(c-2*d)/(c-d)^2/
(tan(1/2*f*x+1/2*e)+1)-2/3/(c-d)/(tan(1/2*f*x+1/2*e)+1)^3+1/(c-d)/(tan(1/2*f*x+1/2*e)+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(1/(a+a*sin(f*x+e))^2/(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 [B] Leaf count of result is larger than twice the leaf count of optimal. 464 vs. \(2 (127) = 254\).
time = 0.39, size = 1022, normalized size = 7.80 \begin {gather*} \left [\frac {2 \, c^{3} - 2 \, c^{2} d - 2 \, c d^{2} + 2 \, d^{3} + 2 \, {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} - d^{2} \cos \left (f x + e\right ) - 2 \, d^{2} - {\left (d^{2} \cos \left (f x + e\right ) + 2 \, d^{2}\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 (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right ) - 2 \, {\left (c^{3} - c^{2} d - c d^{2} + d^{3} - {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \, {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) - 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f - {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) + 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}}, \frac {c^{3} - c^{2} d - c d^{2} + d^{3} + {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - 3 \, {\left (d^{2} \cos \left (f x + e\right )^{2} - d^{2} \cos \left (f x + e\right ) - 2 \, d^{2} - {\left (d^{2} \cos \left (f x + e\right ) + 2 \, d^{2}\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 (2 \, c^{3} - 5 \, c^{2} d - 2 \, c d^{2} + 5 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (c^{3} - c^{2} d - c d^{2} + d^{3} - {\left (c^{3} - 4 \, c^{2} d - c d^{2} + 4 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \, {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) - 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f - {\left ({\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f \cos \left (f x + e\right ) + 2 \, {\left (a^{2} c^{4} - 2 \, a^{2} c^{3} d + 2 \, a^{2} c d^{3} - a^{2} d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/6*(2*c^3 - 2*c^2*d - 2*c*d^2 + 2*d^3 + 2*(c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e)^2 - 3*(d^2*cos(f*x +
e)^2 - d^2*cos(f*x + e) - 2*d^2 - (d^2*cos(f*x + e) + 2*d^2)*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*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*(2*c^3 - 5*c^2*d - 2*c*d^2 + 5*d^3)*cos(f*x
+ e) - 2*(c^3 - c^2*d - c*d^2 + d^3 - (c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*c^4 -
2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e)^2 - (a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos
(f*x + e) - 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f - ((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*
d^4)*f*cos(f*x + e) + 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f)*sin(f*x + e)), 1/3*(c^3 - c^2*d - c
*d^2 + d^3 + (c^3 - 4*c^2*d - c*d^2 + 4*d^3)*cos(f*x + e)^2 - 3*(d^2*cos(f*x + e)^2 - d^2*cos(f*x + e) - 2*d^2
- (d^2*cos(f*x + e) + 2*d^2)*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))) + (2*c^3 - 5*c^2*d - 2*c*d^2 + 5*d^3)*cos(f*x + e) - (c^3 - c^2*d - c*d^2 + d^3 - (c^3 - 4*c^2*d -
c*d^2 + 4*d^3)*cos(f*x + e))*sin(f*x + e))/((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e)^2 -
(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) - 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2
*d^4)*f - ((a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c*d^3 - a^2*d^4)*f*cos(f*x + e) + 2*(a^2*c^4 - 2*a^2*c^3*d + 2*a^2*c
*d^3 - a^2*d^4)*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(1/(a+a*sin(f*x+e))**2/(c+d*sin(f*x+e)),x)

[Out]

Timed out

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Giac [A]
time = 0.46, size = 195, normalized size = 1.49 \begin {gather*} \frac {2 \, {\left (\frac {3 \, {\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 )} d^{2}}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} \sqrt {c^{2} - d^{2}}} - \frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 6 \, d \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 ) - 9 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, c - 5 \, d}{{\left (a^{2} c^{2} - 2 \, a^{2} c d + a^{2} d^{2}\right )} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )}}{3 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

2/3*(3*(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^2/((
a^2*c^2 - 2*a^2*c*d + a^2*d^2)*sqrt(c^2 - d^2)) - (3*c*tan(1/2*f*x + 1/2*e)^2 - 6*d*tan(1/2*f*x + 1/2*e)^2 + 3
*c*tan(1/2*f*x + 1/2*e) - 9*d*tan(1/2*f*x + 1/2*e) + 2*c - 5*d)/((a^2*c^2 - 2*a^2*c*d + a^2*d^2)*(tan(1/2*f*x
+ 1/2*e) + 1)^3))/f

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Mupad [B]
time = 8.02, size = 250, normalized size = 1.91 \begin {gather*} \frac {2\,d^2\,\mathrm {atan}\left (\frac {\frac {d^2\,\left (2\,a^2\,c^2\,d-4\,a^2\,c\,d^2+2\,a^2\,d^3\right )}{a^2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}+\frac {2\,c\,d^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^2\,c^2-2\,a^2\,c\,d+a^2\,d^2\right )}{a^2\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}}{2\,d^2}\right )}{a^2\,f\,\sqrt {c+d}\,{\left (c-d\right )}^{5/2}}-\frac {\frac {2\,\left (2\,c-5\,d\right )}{3\,{\left (c-d\right )}^2}+\frac {2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (c-3\,d\right )}{{\left (c-d\right )}^2}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c-2\,d\right )}{{\left (c-d\right )}^2}}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+3\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + a*sin(e + f*x))^2*(c + d*sin(e + f*x))),x)

[Out]

(2*d^2*atan(((d^2*(2*a^2*d^3 - 4*a^2*c*d^2 + 2*a^2*c^2*d))/(a^2*(c + d)^(1/2)*(c - d)^(5/2)) + (2*c*d^2*tan(e/
2 + (f*x)/2)*(a^2*c^2 + a^2*d^2 - 2*a^2*c*d))/(a^2*(c + d)^(1/2)*(c - d)^(5/2)))/(2*d^2)))/(a^2*f*(c + d)^(1/2
)*(c - d)^(5/2)) - ((2*(2*c - 5*d))/(3*(c - d)^2) + (2*tan(e/2 + (f*x)/2)*(c - 3*d))/(c - d)^2 + (2*tan(e/2 +
(f*x)/2)^2*(c - 2*d))/(c - d)^2)/(f*(3*a^2*tan(e/2 + (f*x)/2)^2 + a^2*tan(e/2 + (f*x)/2)^3 + a^2 + 3*a^2*tan(e
/2 + (f*x)/2)))

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