3.442 \(\int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=207 \[ \frac {a^2 (3 c-2 d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c-d) (c+d)^3 \sqrt {c^2-d^2}}-\frac {a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 d f (c-d) (c+d)^3 (c+d \sin (e+f x))}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d f (c+d)^2 (c+d \sin (e+f x))^2}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3} \]

[Out]

1/3*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^3-1/6*a^2*(c+6*d)*cos(f*x+e)/d/(c+d)^2/f/(c+d*sin(f*x+e))^
2-1/6*a^2*(c^2+6*c*d-10*d^2)*cos(f*x+e)/(c-d)/d/(c+d)^3/f/(c+d*sin(f*x+e))+a^2*(3*c-2*d)*arctan((d+c*tan(1/2*f
*x+1/2*e))/(c^2-d^2)^(1/2))/(c-d)/(c+d)^3/f/(c^2-d^2)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2762, 2754, 12, 2660, 618, 204} \[ \frac {a^2 (3 c-2 d) \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c-d) (c+d)^3 \sqrt {c^2-d^2}}-\frac {a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 d f (c-d) (c+d)^3 (c+d \sin (e+f x))}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d f (c+d)^2 (c+d \sin (e+f x))^2}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^4,x]

[Out]

(a^2*(3*c - 2*d)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c - d)*(c + d)^3*Sqrt[c^2 - d^2]*f) + (a^
2*(c - d)*Cos[e + f*x])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^3) - (a^2*(c + 6*d)*Cos[e + f*x])/(6*d*(c + d)^2*f
*(c + d*Sin[e + f*x])^2) - (a^2*(c^2 + 6*c*d - 10*d^2)*Cos[e + f*x])/(6*(c - d)*d*(c + d)^3*f*(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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

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 2660

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 2754

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 2762

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c
+ a*d)), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*
Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
&& (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx &=\frac {a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a \int \frac {-6 a d-a (c+5 d) \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx}{3 d (c+d)}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}+\frac {a \int \frac {10 a (c-d) d+a (c-d) (c+6 d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{6 (c-d) d (c+d)^2}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}-\frac {a \int -\frac {3 a (3 c-2 d) (c-d) d}{c+d \sin (e+f x)} \, dx}{6 (c-d)^2 d (c+d)^3}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}+\frac {\left (a^2 (3 c-2 d)\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 (c-d) (c+d)^3}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}+\frac {\left (a^2 (3 c-2 d)\right ) \operatorname {Subst}\left (\int \frac {1}{c+2 d x+c x^2} \, dx,x,\tan \left (\frac {1}{2} (e+f x)\right )\right )}{(c-d) (c+d)^3 f}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}-\frac {\left (2 a^2 (3 c-2 d)\right ) \operatorname {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 )}{(c-d) (c+d)^3 f}\\ &=\frac {a^2 (3 c-2 d) \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c-d) (c+d)^3 \sqrt {c^2-d^2} f}+\frac {a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 2.56, size = 196, normalized size = 0.95 \[ \frac {a^2 \cos (e+f x) \left (-\frac {d (\sin (e+f x)+1)^2}{(c+d \sin (e+f x))^3}-\frac {(3 c-2 d) \left (\frac {6 \tanh ^{-1}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {\sin (e+f x)+1}}\right )}{\sqrt {-c-d} \sqrt {c-d}}-\frac {\sqrt {\cos ^2(e+f x)} ((c+4 d) \sin (e+f x)+4 c+d)}{(c+d \sin (e+f x))^2}\right )}{2 (c+d)^2 \sqrt {\cos ^2(e+f x)}}\right )}{3 f (d-c) (c+d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^4,x]

[Out]

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

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fricas [B]  time = 0.56, size = 1366, normalized size = 6.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

[-1/12*(2*(a^2*c^4*d + 6*a^2*c^3*d^2 - 11*a^2*c^2*d^3 - 6*a^2*c*d^4 + 10*a^2*d^5)*cos(f*x + e)^3 - 6*(a^2*c^5
+ 6*a^2*c^4*d - 8*a^2*c^3*d^2 - 8*a^2*c^2*d^3 + 7*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x + e)*sin(f*x + e) - 3*(3*a^2*
c^4 - 2*a^2*c^3*d + 9*a^2*c^2*d^2 - 6*a^2*c*d^3 - 3*(3*a^2*c^2*d^2 - 2*a^2*c*d^3)*cos(f*x + e)^2 + (9*a^2*c^3*
d - 6*a^2*c^2*d^2 + 3*a^2*c*d^3 - 2*a^2*d^4 - (3*a^2*c*d^3 - 2*a^2*d^4)*cos(f*x + e)^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)) - 12*(2*a^2*c^5 - a^
2*c^4*d - 2*a^2*c^3*d^2 - a^2*c^2*d^3 + 2*a^2*d^5)*cos(f*x + e))/(3*(c^7*d^2 + 2*c^6*d^3 - c^5*d^4 - 4*c^4*d^5
- c^3*d^6 + 2*c^2*d^7 + c*d^8)*f*cos(f*x + e)^2 - (c^9 + 2*c^8*d + 2*c^7*d^2 + 2*c^6*d^3 - 4*c^5*d^4 - 10*c^4
*d^5 - 2*c^3*d^6 + 6*c^2*d^7 + 3*c*d^8)*f + ((c^6*d^3 + 2*c^5*d^4 - c^4*d^5 - 4*c^3*d^6 - c^2*d^7 + 2*c*d^8 +
d^9)*f*cos(f*x + e)^2 - (3*c^8*d + 6*c^7*d^2 - 2*c^6*d^3 - 10*c^5*d^4 - 4*c^4*d^5 + 2*c^3*d^6 + 2*c^2*d^7 + 2*
c*d^8 + d^9)*f)*sin(f*x + e)), -1/6*((a^2*c^4*d + 6*a^2*c^3*d^2 - 11*a^2*c^2*d^3 - 6*a^2*c*d^4 + 10*a^2*d^5)*c
os(f*x + e)^3 - 3*(a^2*c^5 + 6*a^2*c^4*d - 8*a^2*c^3*d^2 - 8*a^2*c^2*d^3 + 7*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x +
e)*sin(f*x + e) - 3*(3*a^2*c^4 - 2*a^2*c^3*d + 9*a^2*c^2*d^2 - 6*a^2*c*d^3 - 3*(3*a^2*c^2*d^2 - 2*a^2*c*d^3)*c
os(f*x + e)^2 + (9*a^2*c^3*d - 6*a^2*c^2*d^2 + 3*a^2*c*d^3 - 2*a^2*d^4 - (3*a^2*c*d^3 - 2*a^2*d^4)*cos(f*x + e
)^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))) - 6*(2*a^2*c^5
- a^2*c^4*d - 2*a^2*c^3*d^2 - a^2*c^2*d^3 + 2*a^2*d^5)*cos(f*x + e))/(3*(c^7*d^2 + 2*c^6*d^3 - c^5*d^4 - 4*c^
4*d^5 - c^3*d^6 + 2*c^2*d^7 + c*d^8)*f*cos(f*x + e)^2 - (c^9 + 2*c^8*d + 2*c^7*d^2 + 2*c^6*d^3 - 4*c^5*d^4 - 1
0*c^4*d^5 - 2*c^3*d^6 + 6*c^2*d^7 + 3*c*d^8)*f + ((c^6*d^3 + 2*c^5*d^4 - c^4*d^5 - 4*c^3*d^6 - c^2*d^7 + 2*c*d
^8 + d^9)*f*cos(f*x + e)^2 - (3*c^8*d + 6*c^7*d^2 - 2*c^6*d^3 - 10*c^5*d^4 - 4*c^4*d^5 + 2*c^3*d^6 + 2*c^2*d^7
+ 2*c*d^8 + d^9)*f)*sin(f*x + e))]

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giac [B]  time = 0.59, size = 781, normalized size = 3.77 \[ \frac {\frac {3 \, {\left (3 \, a^{2} c - 2 \, a^{2} d\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \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 (c^{4} + 2 \, c^{3} d - 2 \, c d^{3} - d^{4}\right )} \sqrt {c^{2} - d^{2}}} + \frac {3 \, a^{2} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 18 \, a^{2} c^{5} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 12 \, a^{2} c^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 6 \, a^{2} c^{2} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 12 \, a^{2} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 9 \, a^{2} c^{5} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 54 \, a^{2} c^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 24 \, a^{2} c^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 36 \, a^{2} c^{2} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 12 \, a^{2} c d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 72 \, a^{2} c^{5} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 42 \, a^{2} c^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 12 \, a^{2} c^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 40 \, a^{2} c^{2} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 24 \, a^{2} c d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 8 \, a^{2} d^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 24 \, a^{2} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, a^{2} c^{5} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 72 \, a^{2} c^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 66 \, a^{2} c^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 36 \, a^{2} c^{2} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, a^{2} c d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 3 \, a^{2} c^{6} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 54 \, a^{2} c^{5} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 42 \, a^{2} c^{4} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 24 \, a^{2} c^{3} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, a^{2} c^{2} d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, a^{2} c^{6} + 7 \, a^{2} c^{5} d + 6 \, a^{2} c^{4} d^{2} + 2 \, a^{2} c^{3} d^{3}}{{\left (c^{7} + 2 \, c^{6} d - 2 \, c^{4} d^{3} - c^{3} d^{4}\right )} {\left (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 ) + c\right )}^{3}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x, algorithm="giac")

[Out]

1/3*(3*(3*a^2*c - 2*a^2*d)*(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)))/((c^4 + 2*c^3*d - 2*c*d^3 - d^4)*sqrt(c^2 - d^2)) + (3*a^2*c^6*tan(1/2*f*x + 1/2*e)^5 - 18*a^2*c
^5*d*tan(1/2*f*x + 1/2*e)^5 + 12*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^5 + 6*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^5 - 1
2*a^2*c^6*tan(1/2*f*x + 1/2*e)^4 + 9*a^2*c^5*d*tan(1/2*f*x + 1/2*e)^4 - 54*a^2*c^4*d^2*tan(1/2*f*x + 1/2*e)^4
+ 24*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^4 + 36*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^4 + 12*a^2*c*d^5*tan(1/2*f*x + 1
/2*e)^4 - 72*a^2*c^5*d*tan(1/2*f*x + 1/2*e)^3 + 42*a^2*c^4*d^2*tan(1/2*f*x + 1/2*e)^3 - 12*a^2*c^3*d^3*tan(1/2
*f*x + 1/2*e)^3 + 40*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 + 24*a^2*c*d^5*tan(1/2*f*x + 1/2*e)^3 + 8*a^2*d^6*tan(
1/2*f*x + 1/2*e)^3 - 24*a^2*c^6*tan(1/2*f*x + 1/2*e)^2 + 12*a^2*c^5*d*tan(1/2*f*x + 1/2*e)^2 - 72*a^2*c^4*d^2*
tan(1/2*f*x + 1/2*e)^2 + 66*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^2 + 36*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^2 + 12*a^
2*c*d^5*tan(1/2*f*x + 1/2*e)^2 - 3*a^2*c^6*tan(1/2*f*x + 1/2*e) - 54*a^2*c^5*d*tan(1/2*f*x + 1/2*e) + 42*a^2*c
^4*d^2*tan(1/2*f*x + 1/2*e) + 24*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e) + 6*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e) - 12*a^
2*c^6 + 7*a^2*c^5*d + 6*a^2*c^4*d^2 + 2*a^2*c^3*d^3)/((c^7 + 2*c^6*d - 2*c^4*d^3 - c^3*d^4)*(c*tan(1/2*f*x + 1
/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^3))/f

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maple [B]  time = 0.36, size = 2425, normalized size = 11.71 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x)

[Out]

-2*a^2/f/(c^4+2*c^3*d-2*c*d^3-d^4)/(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))*d-
4*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4
-8*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^
2+a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^5
+8*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4*d^
3+4*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^5*d
^3-a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)+
7/3*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*c^2*d+8*a^2/f/(tan(1/2
*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)*d^3+22*a^2/f/(tan(1/2
*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^2*d^3-24*a^2/f/(tan(1
/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2*d/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+12*a^2/f/(t
an(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4*d^4+4*a^2/f
/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4*d^5-6*
a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^5*d
+2*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^5*
d^4+3*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*
e)^4*d-18*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/
2*e)^4*d^2+2*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x
+1/2*e)*d^4+14*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f
*x+1/2*e)*d^2-18*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1
/2*f*x+1/2*e)*d-24*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1
/2*f*x+1/2*e)^2*d^2+12*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c/(c^4+2*c^3*d-2*c*d^3-d^4)*t
an(1/2*f*x+1/2*e)^2*d^4+4*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2/(c^4+2*c^3*d-2*c*d^3-d
^4)*tan(1/2*f*x+1/2*e)^2*d^5+14*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3*c*d^2/(c^4+2*c^3*d-2
*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+40/3*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c*d^4/(c^4+2*c
^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+8*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^2*d^5/(c^
4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+8/3*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/c^3*
d^6/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+4*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3
*c^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^2*d-4*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c
)^3*d^3/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^3+2*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+
c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*c*d^2+3*a^2/f/(c^4+2*c^3*d-2*c*d^3-d^4)/(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-4*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2
*c*d^3-d^4)*c^3+2/3*a^2/f/(tan(1/2*f*x+1/2*e)^2*c+2*tan(1/2*f*x+1/2*e)*d+c)^3/(c^4+2*c^3*d-2*c*d^3-d^4)*d^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,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 details)Is 4*d^2-4*c^2 positive or negative?

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mupad [B]  time = 10.32, size = 735, normalized size = 3.55 \[ -\frac {\frac {-12\,a^2\,c^3+7\,a^2\,c^2\,d+6\,a^2\,c\,d^2+2\,a^2\,d^3}{3\,\left (-c^4-2\,c^3\,d+2\,c\,d^3+d^4\right )}+\frac {a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (c^4-6\,c^3\,d+4\,c\,d^3+2\,d^4\right )}{c\,\left (-c^4-2\,c^3\,d+2\,c\,d^3+d^4\right )}+\frac {2\,a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (-4\,c^5+2\,c^4\,d-12\,c^3\,d^2+11\,c^2\,d^3+6\,c\,d^4+2\,d^5\right )}{c^2\,\left (-c^4-2\,c^3\,d+2\,c\,d^3+d^4\right )}+\frac {a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (-4\,c^5+3\,c^4\,d-18\,c^3\,d^2+8\,c^2\,d^3+12\,c\,d^4+4\,d^5\right )}{c^2\,\left (-c^4-2\,c^3\,d+2\,c\,d^3+d^4\right )}+\frac {a^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (-c^4-18\,c^3\,d+14\,c^2\,d^2+8\,c\,d^3+2\,d^4\right )}{c\,\left (-c^4-2\,c^3\,d+2\,c\,d^3+d^4\right )}+\frac {2\,a^2\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,c^2+2\,d^2\right )\,\left (-12\,c^3+7\,c^2\,d+6\,c\,d^2+2\,d^3\right )}{3\,c^3\,\left (-c^4-2\,c^3\,d+2\,c\,d^3+d^4\right )}}{f\,\left (c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (3\,c^3+12\,c\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (3\,c^3+12\,c\,d^2\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (12\,c^2\,d+8\,d^3\right )+c^3+6\,c^2\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+6\,c^2\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\right )}-\frac {a^2\,\mathrm {atan}\left (\frac {\left (\frac {a^2\,\left (3\,c-2\,d\right )\,\left (-2\,c^4\,d-4\,c^3\,d^2+4\,c\,d^4+2\,d^5\right )}{2\,{\left (c+d\right )}^{7/2}\,{\left (c-d\right )}^{3/2}\,\left (-c^4-2\,c^3\,d+2\,c\,d^3+d^4\right )}+\frac {a^2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (3\,c-2\,d\right )}{{\left (c+d\right )}^{7/2}\,{\left (c-d\right )}^{3/2}}\right )\,\left (-c^4-2\,c^3\,d+2\,c\,d^3+d^4\right )}{3\,a^2\,c-2\,a^2\,d}\right )\,\left (3\,c-2\,d\right )}{f\,{\left (c+d\right )}^{7/2}\,{\left (c-d\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^4,x)

[Out]

- ((2*a^2*d^3 - 12*a^2*c^3 + 6*a^2*c*d^2 + 7*a^2*c^2*d)/(3*(2*c*d^3 - 2*c^3*d - c^4 + d^4)) + (a^2*tan(e/2 + (
f*x)/2)^5*(4*c*d^3 - 6*c^3*d + c^4 + 2*d^4))/(c*(2*c*d^3 - 2*c^3*d - c^4 + d^4)) + (2*a^2*tan(e/2 + (f*x)/2)^2
*(6*c*d^4 + 2*c^4*d - 4*c^5 + 2*d^5 + 11*c^2*d^3 - 12*c^3*d^2))/(c^2*(2*c*d^3 - 2*c^3*d - c^4 + d^4)) + (a^2*t
an(e/2 + (f*x)/2)^4*(12*c*d^4 + 3*c^4*d - 4*c^5 + 4*d^5 + 8*c^2*d^3 - 18*c^3*d^2))/(c^2*(2*c*d^3 - 2*c^3*d - c
^4 + d^4)) + (a^2*tan(e/2 + (f*x)/2)*(8*c*d^3 - 18*c^3*d - c^4 + 2*d^4 + 14*c^2*d^2))/(c*(2*c*d^3 - 2*c^3*d -
c^4 + d^4)) + (2*a^2*d*tan(e/2 + (f*x)/2)^3*(3*c^2 + 2*d^2)*(6*c*d^2 + 7*c^2*d - 12*c^3 + 2*d^3))/(3*c^3*(2*c*
d^3 - 2*c^3*d - c^4 + d^4)))/(f*(c^3*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^2*(12*c*d^2 + 3*c^3) + tan(e/2
+ (f*x)/2)^4*(12*c*d^2 + 3*c^3) + tan(e/2 + (f*x)/2)^3*(12*c^2*d + 8*d^3) + c^3 + 6*c^2*d*tan(e/2 + (f*x)/2) +
6*c^2*d*tan(e/2 + (f*x)/2)^5)) - (a^2*atan((((a^2*(3*c - 2*d)*(4*c*d^4 - 2*c^4*d + 2*d^5 - 4*c^3*d^2))/(2*(c
+ d)^(7/2)*(c - d)^(3/2)*(2*c*d^3 - 2*c^3*d - c^4 + d^4)) + (a^2*c*tan(e/2 + (f*x)/2)*(3*c - 2*d))/((c + d)^(7
/2)*(c - d)^(3/2)))*(2*c*d^3 - 2*c^3*d - c^4 + d^4))/(3*a^2*c - 2*a^2*d))*(3*c - 2*d))/(f*(c + d)^(7/2)*(c - d
)^(3/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**4,x)

[Out]

Timed out

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