3.5.0 ∫ (3+3 sin(e+fx))2 _________________________

Integrand size = 25, antiderivative size = 271 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx=\frac {9 \left (12 c^2-16 c d+7 d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{4 (c-d)^2 (c+d)^4 \sqrt {c^2-d^2} f}+\frac {9 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {3 (c+8 d) \cos (e+f x)}{4 d (c+d)^2 f (c+d \sin (e+f x))^3}-\frac {3 \left (2 c^2+16 c d-21 d^2\right ) \cos (e+f x)}{8 (c-d) d (c+d)^3 f (c+d \sin (e+f x))^2}-\frac {3 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \cos (e+f x)}{8 (c-d)^2 d (c+d)^4 f (c+d \sin (e+f x))} \]

1/4*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^4-1/12*a^2*(c+8*d)*cos(f*x+e)/d/(c+d)^2/f/(c+d*sin(f*x+e))

^3-1/24*a^2*(2*c^2+16*c*d-21*d^2)*cos(f*x+e)/(c-d)/d/(c+d)^3/f/(c+d*sin(f*x+e))^2-1/24*a^2*(2*c^3+16*c^2*d-59*

c*d^2+32*d^3)*cos(f*x+e)/(c-d)^2/d/(c+d)^4/f/(c+d*sin(f*x+e))+1/4*a^2*(12*c^2-16*c*d+7*d^2)*arctan((d+c*tan(1/

2*f*x+1/2*e))/(c^2-d^2)^(1/2))/(c-d)^2/(c+d)^4/f/(c^2-d^2)^(1/2)

Rubi [A] (veriﬁed)

Time = 0.34 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2841, 2833, 12, 2739, 632, 210} \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx=\frac {a^2 \left (12 c^2-16 c d+7 d^2\right ) \arctan \left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{4 f (c-d)^2 (c+d)^4 \sqrt {c^2-d^2}}-\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \cos (e+f x)}{24 d f (c-d) (c+d)^3 (c+d \sin (e+f x))^2}-\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \cos (e+f x)}{24 d f (c-d)^2 (c+d)^4 (c+d \sin (e+f x))}-\frac {a^2 (c+8 d) \cos (e+f x)}{12 d f (c+d)^2 (c+d \sin (e+f x))^3}+\frac {a^2 (c-d) \cos (e+f x)}{4 d f (c+d) (c+d \sin (e+f x))^4} \]

Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^5,x]

(a^2*(12*c^2 - 16*c*d + 7*d^2)*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/(4*(c - d)^2*(c + d)^4*Sqrt[c

^2 - d^2]*f) + (a^2*(c - d)*Cos[e + f*x])/(4*d*(c + d)*f*(c + d*Sin[e + f*x])^4) - (a^2*(c + 8*d)*Cos[e + f*x]

)/(12*d*(c + d)^2*f*(c + d*Sin[e + f*x])^3) - (a^2*(2*c^2 + 16*c*d - 21*d^2)*Cos[e + f*x])/(24*(c - d)*d*(c +

d)^3*f*(c + d*Sin[e + f*x])^2) - (a^2*(2*c^3 + 16*c^2*d - 59*c*d^2 + 32*d^3)*Cos[e + f*x])/(24*(c - d)^2*d*(c

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

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])

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]

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] &&

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]

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim

p[(-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*} \text {integral}& = \frac {a^2 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {a \int \frac {-8 a d-a (c+7 d) \sin (e+f x)}{(c+d \sin (e+f x))^4} \, dx}{4 d (c+d)} \\ & = \frac {a^2 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {a^2 (c+8 d) \cos (e+f x)}{12 d (c+d)^2 f (c+d \sin (e+f x))^3}+\frac {a \int \frac {21 a (c-d) d+2 a (c-d) (c+8 d) \sin (e+f x)}{(c+d \sin (e+f x))^3} \, dx}{12 (c-d) d (c+d)^2} \\ & = \frac {a^2 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {a^2 (c+8 d) \cos (e+f x)}{12 d (c+d)^2 f (c+d \sin (e+f x))^3}-\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \cos (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sin (e+f x))^2}-\frac {a \int \frac {-2 a (19 c-16 d) (c-d) d+a (c-d) \left (21 d^2-2 c (c+8 d)\right ) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{24 (c-d)^2 d (c+d)^3} \\ & = \frac {a^2 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {a^2 (c+8 d) \cos (e+f x)}{12 d (c+d)^2 f (c+d \sin (e+f x))^3}-\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \cos (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \cos (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sin (e+f x))}+\frac {a \int \frac {3 a (c-d) d \left (12 c^2-16 c d+7 d^2\right )}{c+d \sin (e+f x)} \, dx}{24 (c-d)^3 d (c+d)^4} \\ & = \frac {a^2 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {a^2 (c+8 d) \cos (e+f x)}{12 d (c+d)^2 f (c+d \sin (e+f x))^3}-\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \cos (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \cos (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (12 c^2-16 c d+7 d^2\right )\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{8 (c-d)^2 (c+d)^4} \\ & = \frac {a^2 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {a^2 (c+8 d) \cos (e+f x)}{12 d (c+d)^2 f (c+d \sin (e+f x))^3}-\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \cos (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \cos (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sin (e+f x))}+\frac {\left (a^2 \left (12 c^2-16 c d+7 d^2\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 )}{4 (c-d)^2 (c+d)^4 f} \\ & = \frac {a^2 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {a^2 (c+8 d) \cos (e+f x)}{12 d (c+d)^2 f (c+d \sin (e+f x))^3}-\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \cos (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \cos (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sin (e+f x))}-\frac {\left (a^2 \left (12 c^2-16 c d+7 d^2\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 )}{2 (c-d)^2 (c+d)^4 f} \\ & = \frac {a^2 \left (12 c^2-16 c d+7 d^2\right ) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{4 (c-d)^2 (c+d)^4 \sqrt {c^2-d^2} f}+\frac {a^2 (c-d) \cos (e+f x)}{4 d (c+d) f (c+d \sin (e+f x))^4}-\frac {a^2 (c+8 d) \cos (e+f x)}{12 d (c+d)^2 f (c+d \sin (e+f x))^3}-\frac {a^2 \left (2 c^2+16 c d-21 d^2\right ) \cos (e+f x)}{24 (c-d) d (c+d)^3 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (2 c^3+16 c^2 d-59 c d^2+32 d^3\right ) \cos (e+f x)}{24 (c-d)^2 d (c+d)^4 f (c+d \sin (e+f x))} \\ \end{align*}

Mathematica [A] (veriﬁed)

Time = 3.26 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.98 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx=-\frac {3 \cos (e+f x) \left (6 d (1+\sin (e+f x))^2+\frac {2 (5 c-2 d) d (1+\sin (e+f x))^2 (c+d \sin (e+f x))}{(c-d) (c+d)}+\frac {\left (12 c^2-16 c d+7 d^2\right ) (c+d \sin (e+f x))^2 \left (-6 \text {arctanh}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {1+\sin (e+f x)}}\right ) (c+d \sin (e+f x))^2+\sqrt {-c-d} \sqrt {c-d} \sqrt {\cos ^2(e+f x)} (4 c+d+(c+4 d) \sin (e+f x))\right )}{(-c-d)^{7/2} (c-d)^{3/2} \sqrt {\cos ^2(e+f x)}}\right )}{8 (-c+d) (c+d) f (c+d \sin (e+f x))^4} \]

Integrate[(3 + 3*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^5,x]

(-3*Cos[e + f*x]*(6*d*(1 + Sin[e + f*x])^2 + (2*(5*c - 2*d)*d*(1 + Sin[e + f*x])^2*(c + d*Sin[e + f*x]))/((c -

d)*(c + d)) + ((12*c^2 - 16*c*d + 7*d^2)*(c + d*Sin[e + f*x])^2*(-6*ArcTanh[(Sqrt[c - d]*Sqrt[1 - Sin[e + f*x

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

^2]*(4*c + d + (c + 4*d)*Sin[e + f*x])))/((-c - d)^(7/2)*(c - d)^(3/2)*Sqrt[Cos[e + f*x]^2])))/(8*(-c + d)*(c

Maple [B] (veriﬁed)

Leaf count of result is larger than twice the leaf count of optimal. \(1055\) vs. \(2(271)=542\).

Time = 3.22 (sec) , antiderivative size = 1056, normalized size of antiderivative = 3.90


method	result	size

derivativedivides	\(\text {Expression too large to display}\)	\(1056\)
default	\(\text {Expression too large to display}\)	\(1056\)
risch	\(\text {Expression too large to display}\)	\(1232\)

int((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^5,x,method=_RETURNVERBOSE)

2/f*a^2*((1/8*(4*c^6-32*c^5*d+15*c^4*d^2+32*c^3*d^3+8*c^2*d^4-16*c*d^5-8*d^6)/c/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3

-c^2*d^4+2*c*d^5+d^6)*tan(1/2*f*x+1/2*e)^7-1/8*(16*c^7-28*c^6*d+144*c^5*d^2-137*c^4*d^3-112*c^3*d^4+8*c^2*d^5+

64*c*d^6+24*d^7)/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3-c^2*d^4+2*c*d^5+d^6)/c^2*tan(1/2*f*x+1/2*e)^6+1/24/c^3*(12*c^8

-480*c^7*d+597*c^6*d^2-480*c^5*d^3+836*c^4*d^4+208*c^3*d^5-152*c^2*d^6-256*c*d^7-96*d^8)/(c^6+2*c^5*d-c^4*d^2-

4*c^3*d^3-c^2*d^4+2*c*d^5+d^6)*tan(1/2*f*x+1/2*e)^5-1/24/c^4*(144*c^9-204*c^8*d+1104*c^7*d^2-1617*c^6*d^3+48*c

^5*d^4-406*c^4*d^5+256*c^3*d^6+184*c^2*d^7+128*c*d^8+48*d^9)/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3-c^2*d^4+2*c*d^5+d^

6)*tan(1/2*f*x+1/2*e)^4-1/24/c^3*(12*c^8+672*c^7*d-1035*c^6*d^2+672*c^5*d^3-1220*c^4*d^4+80*c^3*d^5+152*c^2*d^

6+256*c*d^7+96*d^8)/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3-c^2*d^4+2*c*d^5+d^6)*tan(1/2*f*x+1/2*e)^3-1/24*(144*c^7-188

*c^6*d+656*c^5*d^2-1201*c^4*d^3+16*c^3*d^4+120*c^2*d^5+192*c*d^6+72*d^7)/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3-c^2*d^

4+2*c*d^5+d^6)/c^2*tan(1/2*f*x+1/2*e)^2-1/24*(12*c^6+288*c^5*d-499*c^4*d^2-32*c^3*d^3+64*c^2*d^4+80*c*d^5+24*d

^6)/c/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3-c^2*d^4+2*c*d^5+d^6)*tan(1/2*f*x+1/2*e)-1/24*(48*c^5-68*c^4*d-16*c^3*d^2+

5*c^2*d^3+16*c*d^4+6*d^5)/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3-c^2*d^4+2*c*d^5+d^6))/(tan(1/2*f*x+1/2*e)^2*c+2*d*tan

(1/2*f*x+1/2*e)+c)^4+1/8*(12*c^2-16*c*d+7*d^2)/(c^6+2*c^5*d-c^4*d^2-4*c^3*d^3-c^2*d^4+2*c*d^5+d^6)/(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)))

Fricas [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1033 vs. \(2 (271) = 542\).

Time = 0.37 (sec) , antiderivative size = 2151, normalized size of antiderivative = 7.94 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx=\text {Too large to display} \]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^5,x, algorithm="fricas")

[1/48*(2*(8*a^2*c^6*d + 64*a^2*c^5*d^2 - 208*a^2*c^4*d^3 + 16*a^2*c^3*d^4 + 221*a^2*c^2*d^5 - 80*a^2*c*d^6 - 2

1*a^2*d^7)*cos(f*x + e)^3 - 3*(12*a^2*c^6 - 16*a^2*c^5*d + 79*a^2*c^4*d^2 - 96*a^2*c^3*d^3 + 54*a^2*c^2*d^4 -

16*a^2*c*d^5 + 7*a^2*d^6 + (12*a^2*c^2*d^4 - 16*a^2*c*d^5 + 7*a^2*d^6)*cos(f*x + e)^4 - 2*(36*a^2*c^4*d^2 - 48

*a^2*c^3*d^3 + 33*a^2*c^2*d^4 - 16*a^2*c*d^5 + 7*a^2*d^6)*cos(f*x + e)^2 + 4*(12*a^2*c^5*d - 16*a^2*c^4*d^2 +

19*a^2*c^3*d^3 - 16*a^2*c^2*d^4 + 7*a^2*c*d^5 - (12*a^2*c^3*d^3 - 16*a^2*c^2*d^4 + 7*a^2*c*d^5)*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)) - 6*(16*a^2*c^7 - 20*a^2*c^6*d - 45*a^2*c^4*d^3 + 16*a^2*c^3*d^4 + 74*a^2*c^2*d^5 - 32*a^2*c*d^6 - 9*a^2*d

^7)*cos(f*x + e) + 2*((2*a^2*c^5*d^2 + 16*a^2*c^4*d^3 - 61*a^2*c^3*d^4 + 16*a^2*c^2*d^5 + 59*a^2*c*d^6 - 32*a^

2*d^7)*cos(f*x + e)^3 - 3*(4*a^2*c^7 + 32*a^2*c^6*d - 79*a^2*c^5*d^2 - 16*a^2*c^4*d^3 + 70*a^2*c^3*d^4 + 5*a^2

*c*d^6 - 16*a^2*d^7)*cos(f*x + e))*sin(f*x + e))/((c^8*d^4 + 2*c^7*d^5 - 2*c^6*d^6 - 6*c^5*d^7 + 6*c^3*d^9 + 2

*c^2*d^10 - 2*c*d^11 - d^12)*f*cos(f*x + e)^4 - 2*(3*c^10*d^2 + 6*c^9*d^3 - 5*c^8*d^4 - 16*c^7*d^5 - 2*c^6*d^6

+ 12*c^5*d^7 + 6*c^4*d^8 - c^2*d^10 - 2*c*d^11 - d^12)*f*cos(f*x + e)^2 + (c^12 + 2*c^11*d + 4*c^10*d^2 + 6*c

^9*d^3 - 11*c^8*d^4 - 28*c^7*d^5 + 28*c^5*d^7 + 11*c^4*d^8 - 6*c^3*d^9 - 4*c^2*d^10 - 2*c*d^11 - d^12)*f - 4*(

(c^9*d^3 + 2*c^8*d^4 - 2*c^7*d^5 - 6*c^6*d^6 + 6*c^4*d^8 + 2*c^3*d^9 - 2*c^2*d^10 - c*d^11)*f*cos(f*x + e)^2 -

(c^11*d + 2*c^10*d^2 - c^9*d^3 - 4*c^8*d^4 - 2*c^7*d^5 + 2*c^5*d^7 + 4*c^4*d^8 + c^3*d^9 - 2*c^2*d^10 - c*d^1

1)*f)*sin(f*x + e)), 1/24*((8*a^2*c^6*d + 64*a^2*c^5*d^2 - 208*a^2*c^4*d^3 + 16*a^2*c^3*d^4 + 221*a^2*c^2*d^5

- 80*a^2*c*d^6 - 21*a^2*d^7)*cos(f*x + e)^3 - 3*(12*a^2*c^6 - 16*a^2*c^5*d + 79*a^2*c^4*d^2 - 96*a^2*c^3*d^3 +

54*a^2*c^2*d^4 - 16*a^2*c*d^5 + 7*a^2*d^6 + (12*a^2*c^2*d^4 - 16*a^2*c*d^5 + 7*a^2*d^6)*cos(f*x + e)^4 - 2*(3

6*a^2*c^4*d^2 - 48*a^2*c^3*d^3 + 33*a^2*c^2*d^4 - 16*a^2*c*d^5 + 7*a^2*d^6)*cos(f*x + e)^2 + 4*(12*a^2*c^5*d -

16*a^2*c^4*d^2 + 19*a^2*c^3*d^3 - 16*a^2*c^2*d^4 + 7*a^2*c*d^5 - (12*a^2*c^3*d^3 - 16*a^2*c^2*d^4 + 7*a^2*c*d

^5)*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)))

- 3*(16*a^2*c^7 - 20*a^2*c^6*d - 45*a^2*c^4*d^3 + 16*a^2*c^3*d^4 + 74*a^2*c^2*d^5 - 32*a^2*c*d^6 - 9*a^2*d^7)

*cos(f*x + e) + ((2*a^2*c^5*d^2 + 16*a^2*c^4*d^3 - 61*a^2*c^3*d^4 + 16*a^2*c^2*d^5 + 59*a^2*c*d^6 - 32*a^2*d^7

)*cos(f*x + e)^3 - 3*(4*a^2*c^7 + 32*a^2*c^6*d - 79*a^2*c^5*d^2 - 16*a^2*c^4*d^3 + 70*a^2*c^3*d^4 + 5*a^2*c*d^

6 - 16*a^2*d^7)*cos(f*x + e))*sin(f*x + e))/((c^8*d^4 + 2*c^7*d^5 - 2*c^6*d^6 - 6*c^5*d^7 + 6*c^3*d^9 + 2*c^2*

d^10 - 2*c*d^11 - d^12)*f*cos(f*x + e)^4 - 2*(3*c^10*d^2 + 6*c^9*d^3 - 5*c^8*d^4 - 16*c^7*d^5 - 2*c^6*d^6 + 12

*c^5*d^7 + 6*c^4*d^8 - c^2*d^10 - 2*c*d^11 - d^12)*f*cos(f*x + e)^2 + (c^12 + 2*c^11*d + 4*c^10*d^2 + 6*c^9*d^

3 - 11*c^8*d^4 - 28*c^7*d^5 + 28*c^5*d^7 + 11*c^4*d^8 - 6*c^3*d^9 - 4*c^2*d^10 - 2*c*d^11 - d^12)*f - 4*((c^9*

d^3 + 2*c^8*d^4 - 2*c^7*d^5 - 6*c^6*d^6 + 6*c^4*d^8 + 2*c^3*d^9 - 2*c^2*d^10 - c*d^11)*f*cos(f*x + e)^2 - (c^1

1*d + 2*c^10*d^2 - c^9*d^3 - 4*c^8*d^4 - 2*c^7*d^5 + 2*c^5*d^7 + 4*c^4*d^8 + c^3*d^9 - 2*c^2*d^10 - c*d^11)*f)

Sympy [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx=\text {Timed out} \]

integrate((a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**5,x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx=\text {Exception raised: ValueError} \]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^5,x, algorithm="maxima")

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?`

Giac [B] (veriﬁcation not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1493 vs. \(2 (271) = 542\).

Time = 0.43 (sec) , antiderivative size = 1493, normalized size of antiderivative = 5.51 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx=\text {Too large to display} \]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^5,x, algorithm="giac")

1/12*(3*(12*a^2*c^2 - 16*a^2*c*d + 7*a^2*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)))/((c^6 + 2*c^5*d - c^4*d^2 - 4*c^3*d^3 - c^2*d^4 + 2*c*d^5 + d^6)*sqrt(c^2 - d

^2)) + (12*a^2*c^9*tan(1/2*f*x + 1/2*e)^7 - 96*a^2*c^8*d*tan(1/2*f*x + 1/2*e)^7 + 45*a^2*c^7*d^2*tan(1/2*f*x +

1/2*e)^7 + 96*a^2*c^6*d^3*tan(1/2*f*x + 1/2*e)^7 + 24*a^2*c^5*d^4*tan(1/2*f*x + 1/2*e)^7 - 48*a^2*c^4*d^5*tan

(1/2*f*x + 1/2*e)^7 - 24*a^2*c^3*d^6*tan(1/2*f*x + 1/2*e)^7 - 48*a^2*c^9*tan(1/2*f*x + 1/2*e)^6 + 84*a^2*c^8*d

*tan(1/2*f*x + 1/2*e)^6 - 432*a^2*c^7*d^2*tan(1/2*f*x + 1/2*e)^6 + 411*a^2*c^6*d^3*tan(1/2*f*x + 1/2*e)^6 + 33

6*a^2*c^5*d^4*tan(1/2*f*x + 1/2*e)^6 - 24*a^2*c^4*d^5*tan(1/2*f*x + 1/2*e)^6 - 192*a^2*c^3*d^6*tan(1/2*f*x + 1

/2*e)^6 - 72*a^2*c^2*d^7*tan(1/2*f*x + 1/2*e)^6 + 12*a^2*c^9*tan(1/2*f*x + 1/2*e)^5 - 480*a^2*c^8*d*tan(1/2*f*

x + 1/2*e)^5 + 597*a^2*c^7*d^2*tan(1/2*f*x + 1/2*e)^5 - 480*a^2*c^6*d^3*tan(1/2*f*x + 1/2*e)^5 + 836*a^2*c^5*d

^4*tan(1/2*f*x + 1/2*e)^5 + 208*a^2*c^4*d^5*tan(1/2*f*x + 1/2*e)^5 - 152*a^2*c^3*d^6*tan(1/2*f*x + 1/2*e)^5 -

256*a^2*c^2*d^7*tan(1/2*f*x + 1/2*e)^5 - 96*a^2*c*d^8*tan(1/2*f*x + 1/2*e)^5 - 144*a^2*c^9*tan(1/2*f*x + 1/2*e

)^4 + 204*a^2*c^8*d*tan(1/2*f*x + 1/2*e)^4 - 1104*a^2*c^7*d^2*tan(1/2*f*x + 1/2*e)^4 + 1617*a^2*c^6*d^3*tan(1/

2*f*x + 1/2*e)^4 - 48*a^2*c^5*d^4*tan(1/2*f*x + 1/2*e)^4 + 406*a^2*c^4*d^5*tan(1/2*f*x + 1/2*e)^4 - 256*a^2*c^

3*d^6*tan(1/2*f*x + 1/2*e)^4 - 184*a^2*c^2*d^7*tan(1/2*f*x + 1/2*e)^4 - 128*a^2*c*d^8*tan(1/2*f*x + 1/2*e)^4 -

48*a^2*d^9*tan(1/2*f*x + 1/2*e)^4 - 12*a^2*c^9*tan(1/2*f*x + 1/2*e)^3 - 672*a^2*c^8*d*tan(1/2*f*x + 1/2*e)^3

+ 1035*a^2*c^7*d^2*tan(1/2*f*x + 1/2*e)^3 - 672*a^2*c^6*d^3*tan(1/2*f*x + 1/2*e)^3 + 1220*a^2*c^5*d^4*tan(1/2*

f*x + 1/2*e)^3 - 80*a^2*c^4*d^5*tan(1/2*f*x + 1/2*e)^3 - 152*a^2*c^3*d^6*tan(1/2*f*x + 1/2*e)^3 - 256*a^2*c^2*

d^7*tan(1/2*f*x + 1/2*e)^3 - 96*a^2*c*d^8*tan(1/2*f*x + 1/2*e)^3 - 144*a^2*c^9*tan(1/2*f*x + 1/2*e)^2 + 188*a^

2*c^8*d*tan(1/2*f*x + 1/2*e)^2 - 656*a^2*c^7*d^2*tan(1/2*f*x + 1/2*e)^2 + 1201*a^2*c^6*d^3*tan(1/2*f*x + 1/2*e

)^2 - 16*a^2*c^5*d^4*tan(1/2*f*x + 1/2*e)^2 - 120*a^2*c^4*d^5*tan(1/2*f*x + 1/2*e)^2 - 192*a^2*c^3*d^6*tan(1/2

*f*x + 1/2*e)^2 - 72*a^2*c^2*d^7*tan(1/2*f*x + 1/2*e)^2 - 12*a^2*c^9*tan(1/2*f*x + 1/2*e) - 288*a^2*c^8*d*tan(

1/2*f*x + 1/2*e) + 499*a^2*c^7*d^2*tan(1/2*f*x + 1/2*e) + 32*a^2*c^6*d^3*tan(1/2*f*x + 1/2*e) - 64*a^2*c^5*d^4

*tan(1/2*f*x + 1/2*e) - 80*a^2*c^4*d^5*tan(1/2*f*x + 1/2*e) - 24*a^2*c^3*d^6*tan(1/2*f*x + 1/2*e) - 48*a^2*c^9

+ 68*a^2*c^8*d + 16*a^2*c^7*d^2 - 5*a^2*c^6*d^3 - 16*a^2*c^5*d^4 - 6*a^2*c^4*d^5)/((c^10 + 2*c^9*d - c^8*d^2

- 4*c^7*d^3 - c^6*d^4 + 2*c^5*d^5 + c^4*d^6)*(c*tan(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^4))/f

Mupad [B] (veriﬁcation not implemented)

Time = 11.08 (sec) , antiderivative size = 1411, normalized size of antiderivative = 5.21 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx=\text {Too large to display} \]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^5,x)

(a^2*atan((4*((a^2*(12*c^2 - 16*c*d + 7*d^2)*(16*c*d^6 + 8*c^6*d + 8*d^7 - 8*c^2*d^5 - 32*c^3*d^4 - 8*c^4*d^3

+ 16*c^5*d^2))/(32*(c + d)^(9/2)*(c - d)^(5/2)*(2*c*d^5 + 2*c^5*d + c^6 + d^6 - c^2*d^4 - 4*c^3*d^3 - c^4*d^2)

) + (a^2*c*tan(e/2 + (f*x)/2)*(12*c^2 - 16*c*d + 7*d^2))/(4*(c + d)^(9/2)*(c - d)^(5/2)))*(2*c*d^5 + 2*c^5*d +

c^6 + d^6 - c^2*d^4 - 4*c^3*d^3 - c^4*d^2))/(12*a^2*c^2 + 7*a^2*d^2 - 16*a^2*c*d))*(12*c^2 - 16*c*d + 7*d^2))

/(4*f*(c + d)^(9/2)*(c - d)^(5/2)) - ((48*a^2*c^5 + 6*a^2*d^5 + 16*a^2*c*d^4 - 68*a^2*c^4*d + 5*a^2*c^2*d^3 -

16*a^2*c^3*d^2)/(12*(2*c*d^5 + 2*c^5*d + c^6 + d^6 - c^2*d^4 - 4*c^3*d^3 - c^4*d^2)) + (a^2*tan(e/2 + (f*x)/2)

*(80*c*d^5 + 288*c^5*d + 12*c^6 + 24*d^6 + 64*c^2*d^4 - 32*c^3*d^3 - 499*c^4*d^2))/(12*c*(2*c*d^5 + 2*c^5*d +

c^6 + d^6 - c^2*d^4 - 4*c^3*d^3 - c^4*d^2)) + (a^2*tan(e/2 + (f*x)/2)^5*(256*c*d^7 + 480*c^7*d - 12*c^8 + 96*d

^8 + 152*c^2*d^6 - 208*c^3*d^5 - 836*c^4*d^4 + 480*c^5*d^3 - 597*c^6*d^2))/(12*c^3*(2*c*d^5 + 2*c^5*d + c^6 +

d^6 - c^2*d^4 - 4*c^3*d^3 - c^4*d^2)) + (a^2*tan(e/2 + (f*x)/2)^3*(256*c*d^7 + 672*c^7*d + 12*c^8 + 96*d^8 + 1

52*c^2*d^6 + 80*c^3*d^5 - 1220*c^4*d^4 + 672*c^5*d^3 - 1035*c^6*d^2))/(12*c^3*(2*c*d^5 + 2*c^5*d + c^6 + d^6 -

c^2*d^4 - 4*c^3*d^3 - c^4*d^2)) + (a^2*tan(e/2 + (f*x)/2)^6*(64*c*d^6 - 28*c^6*d + 16*c^7 + 24*d^7 + 8*c^2*d^

5 - 112*c^3*d^4 - 137*c^4*d^3 + 144*c^5*d^2))/(4*c^2*(2*c*d^5 + 2*c^5*d + c^6 + d^6 - c^2*d^4 - 4*c^3*d^3 - c^

4*d^2)) + (a^2*tan(e/2 + (f*x)/2)^2*(192*c*d^6 - 188*c^6*d + 144*c^7 + 72*d^7 + 120*c^2*d^5 + 16*c^3*d^4 - 120

1*c^4*d^3 + 656*c^5*d^2))/(12*c^2*(2*c*d^5 + 2*c^5*d + c^6 + d^6 - c^2*d^4 - 4*c^3*d^3 - c^4*d^2)) - (a^2*tan(

e/2 + (f*x)/2)^7*(4*c^6 - 32*c^5*d - 16*c*d^5 - 8*d^6 + 8*c^2*d^4 + 32*c^3*d^3 + 15*c^4*d^2))/(4*c*(2*c*d^5 +

2*c^5*d + c^6 + d^6 - c^2*d^4 - 4*c^3*d^3 - c^4*d^2)) + (a^2*tan(e/2 + (f*x)/2)^4*(3*c^4 + 8*d^4 + 24*c^2*d^2)

*(16*c*d^4 - 68*c^4*d + 48*c^5 + 6*d^5 + 5*c^2*d^3 - 16*c^3*d^2))/(12*c^4*(2*c*d^5 + 2*c^5*d + c^6 + d^6 - c^2

*d^4 - 4*c^3*d^3 - c^4*d^2)))/(f*(tan(e/2 + (f*x)/2)^4*(6*c^4 + 16*d^4 + 48*c^2*d^2) + c^4*tan(e/2 + (f*x)/2)^

8 + c^4 + tan(e/2 + (f*x)/2)^2*(4*c^4 + 24*c^2*d^2) + tan(e/2 + (f*x)/2)^6*(4*c^4 + 24*c^2*d^2) + tan(e/2 + (f

*x)/2)^3*(32*c*d^3 + 24*c^3*d) + tan(e/2 + (f*x)/2)^5*(32*c*d^3 + 24*c^3*d) + 8*c^3*d*tan(e/2 + (f*x)/2) + 8*c

\(\int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^5} \, dx\) [443]

Optimal result