\(\int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^3} \, dx\) [717]

Optimal result

Integrand size = 25, antiderivative size = 179 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^3} \, dx=-\frac {\left (18 b c d-9 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{\left (9-b^2\right )^{5/2} f}+\frac {(b c-3 d)^2 \cos (e+f x)}{2 b \left (9-b^2\right ) f (3+b \sin (e+f x))^2}+\frac {(b c-3 d) \left (9 b c+9 d-4 b^2 d\right ) \cos (e+f x)}{2 b \left (9-b^2\right )^2 f (3+b \sin (e+f x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.09, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2869, 2833, 12, 2739, 632, 210} \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^3} \, dx=-\frac {\left (-\left (a^2 \left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \arctan \left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac {(b c-a d)^2 \cos (e+f x)}{2 b f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2}+\frac {\left (a^2 d+3 a b c-4 b^2 d\right ) (b c-a d) \cos (e+f x)}{2 b f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))} \]

[In]

Int[(c + d*Sin[e + f*x])^2/(a + b*Sin[e + f*x])^3,x]

[Out]

-(((6*a*b*c*d - a^2*(2*c^2 + d^2) - b^2*(c^2 + 2*d^2))*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2
- b^2)^(5/2)*f)) + ((b*c - a*d)^2*Cos[e + f*x])/(2*b*(a^2 - b^2)*f*(a + b*Sin[e + f*x])^2) + ((b*c - a*d)*(3*
a*b*c + a^2*d - 4*b^2*d)*Cos[e + f*x])/(2*b*(a^2 - b^2)^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]

Rule 2869

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(
-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] -
Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(2*b*c*d - a*(c^2 + d^2)) + (a
^2*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1) + c^2*(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]

Rubi steps \begin{align*} \text {integral}& = \frac {(b c-a d)^2 \cos (e+f x)}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {\int \frac {-2 b \left (2 b c d-a \left (c^2+d^2\right )\right )+\left (2 a b c d+a^2 d^2-b^2 \left (c^2+2 d^2\right )\right ) \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx}{2 b \left (a^2-b^2\right )} \\ & = \frac {(b c-a d)^2 \cos (e+f x)}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {(b c-a d) \left (3 a b c+a^2 d-4 b^2 d\right ) \cos (e+f x)}{2 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}-\frac {\int \frac {b \left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right )}{a+b \sin (e+f x)} \, dx}{2 b \left (a^2-b^2\right )^2} \\ & = \frac {(b c-a d)^2 \cos (e+f x)}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {(b c-a d) \left (3 a b c+a^2 d-4 b^2 d\right ) \cos (e+f x)}{2 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}-\frac {\left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \int \frac {1}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2} \\ & = \frac {(b c-a d)^2 \cos (e+f x)}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {(b c-a d) \left (3 a b c+a^2 d-4 b^2 d\right ) \cos (e+f x)}{2 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}-\frac {\left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \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 )^2 f} \\ & = \frac {(b c-a d)^2 \cos (e+f x)}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {(b c-a d) \left (3 a b c+a^2 d-4 b^2 d\right ) \cos (e+f x)}{2 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac {\left (2 \left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right )\right ) \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 )^2 f} \\ & = -\frac {\left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \arctan \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 )^{5/2} f}+\frac {(b c-a d)^2 \cos (e+f x)}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {(b c-a d) \left (3 a b c+a^2 d-4 b^2 d\right ) \cos (e+f x)}{2 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.98 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^3} \, dx=\frac {\frac {2 \left (\left (18+b^2\right ) c^2-18 b c d+\left (9+2 b^2\right ) d^2\right ) \arctan \left (\frac {b+3 \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {9-b^2}}\right )}{\left (9-b^2\right )^{5/2}}-\frac {(b c-3 d)^2 \cos (e+f x)}{b \left (-9+b^2\right ) (3+b \sin (e+f x))^2}+\frac {\left (-18 b c d-4 b^3 c d-27 d^2+3 b^2 \left (3 c^2+4 d^2\right )\right ) \cos (e+f x)}{b \left (-9+b^2\right )^2 (3+b \sin (e+f x))}}{2 f} \]

[In]

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

[Out]

((2*((18 + b^2)*c^2 - 18*b*c*d + (9 + 2*b^2)*d^2)*ArcTan[(b + 3*Tan[(e + f*x)/2])/Sqrt[9 - b^2]])/(9 - b^2)^(5
/2) - ((b*c - 3*d)^2*Cos[e + f*x])/(b*(-9 + b^2)*(3 + b*Sin[e + f*x])^2) + ((-18*b*c*d - 4*b^3*c*d - 27*d^2 +
3*b^2*(3*c^2 + 4*d^2))*Cos[e + f*x])/(b*(-9 + b^2)^2*(3 + b*Sin[e + f*x])))/(2*f)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(464\) vs. \(2(187)=374\).

Time = 1.35 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.60

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (a^{4} d^{2}-6 a^{3} b c d +5 a^{2} b^{2} c^{2}+2 a^{2} b^{2} d^{2}-2 b^{4} c^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {\left (4 a^{5} c d -4 a^{4} b \,c^{2}-3 a^{4} b \,d^{2}+10 a^{3} b^{2} c d -7 a^{2} b^{3} c^{2}-6 a^{2} b^{3} d^{2}+4 a \,b^{4} c d +2 b^{5} c^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}-\frac {\left (a^{4} d^{2}+10 a^{3} b c d -11 a^{2} b^{2} c^{2}-10 a^{2} b^{2} d^{2}+8 a \,b^{3} c d +2 b^{4} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}-\frac {4 a^{3} c d -4 a^{2} b \,c^{2}-3 a^{2} b \,d^{2}+2 a \,b^{2} c d +b^{3} c^{2}}{a^{4}-2 a^{2} b^{2}+b^{4}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a \right )}^{2}}+\frac {\left (2 a^{2} c^{2}+d^{2} a^{2}-6 a b c d +b^{2} c^{2}+2 d^{2} b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}}{f}\)	\(465\)
default	\(\frac {\frac {\frac {\left (a^{4} d^{2}-6 a^{3} b c d +5 a^{2} b^{2} c^{2}+2 a^{2} b^{2} d^{2}-2 b^{4} c^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {\left (4 a^{5} c d -4 a^{4} b \,c^{2}-3 a^{4} b \,d^{2}+10 a^{3} b^{2} c d -7 a^{2} b^{3} c^{2}-6 a^{2} b^{3} d^{2}+4 a \,b^{4} c d +2 b^{5} c^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}-\frac {\left (a^{4} d^{2}+10 a^{3} b c d -11 a^{2} b^{2} c^{2}-10 a^{2} b^{2} d^{2}+8 a \,b^{3} c d +2 b^{4} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}-\frac {4 a^{3} c d -4 a^{2} b \,c^{2}-3 a^{2} b \,d^{2}+2 a \,b^{2} c d +b^{3} c^{2}}{a^{4}-2 a^{2} b^{2}+b^{4}}}{{\left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a \right )}^{2}}+\frac {\left (2 a^{2} c^{2}+d^{2} a^{2}-6 a b c d +b^{2} c^{2}+2 d^{2} b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}}{f}\)	\(465\)
risch	\(\text {Expression too large to display}\)	\(1294\)

[In]

int((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(2*(1/2*(a^4*d^2-6*a^3*b*c*d+5*a^2*b^2*c^2+2*a^2*b^2*d^2-2*b^4*c^2)/a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*
e)^3-1/2*(4*a^5*c*d-4*a^4*b*c^2-3*a^4*b*d^2+10*a^3*b^2*c*d-7*a^2*b^3*c^2-6*a^2*b^3*d^2+4*a*b^4*c*d+2*b^5*c^2)/
(a^4-2*a^2*b^2+b^4)/a^2*tan(1/2*f*x+1/2*e)^2-1/2*(a^4*d^2+10*a^3*b*c*d-11*a^2*b^2*c^2-10*a^2*b^2*d^2+8*a*b^3*c
*d+2*b^4*c^2)/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*f*x+1/2*e)-1/2*(4*a^3*c*d-4*a^2*b*c^2-3*a^2*b*d^2+2*a*b^2*c*d+b^3*
c^2)/(a^4-2*a^2*b^2+b^4))/(tan(1/2*f*x+1/2*e)^2*a+2*b*tan(1/2*f*x+1/2*e)+a)^2+(2*a^2*c^2+a^2*d^2-6*a*b*c*d+b^2
*c^2+2*b^2*d^2)/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (187) = 374\).

Time = 0.33 (sec) , antiderivative size = 1025, normalized size of antiderivative = 5.73 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/4*(2*(3*(a^3*b^2 - a*b^4)*c^2 - 2*(a^4*b + a^2*b^3 - 2*b^5)*c*d - (a^5 - 5*a^3*b^2 + 4*a*b^4)*d^2)*cos(f*x
+ e)*sin(f*x + e) - ((2*a^4 + 3*a^2*b^2 + b^4)*c^2 - 6*(a^3*b + a*b^3)*c*d + (a^4 + 3*a^2*b^2 + 2*b^4)*d^2 +
(6*a*b^3*c*d - (2*a^2*b^2 + b^4)*c^2 - (a^2*b^2 + 2*b^4)*d^2)*cos(f*x + e)^2 - 2*(6*a^2*b^2*c*d - (2*a^3*b + a
*b^3)*c^2 - (a^3*b + 2*a*b^3)*d^2)*sin(f*x + e))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*si
n(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*((4*a^4*b - 5*a^2*b^3 + b^5)*c^2 - 2*(2*a^5 - a^3*b^2 - a*b^4)*c*d +
3*(a^4*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*f*cos(f*x + e)^2 - 2*(a^7*b -
3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*f*sin(f*x + e) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*f), -1/2*((3*(a^3*b^2 - a*
b^4)*c^2 - 2*(a^4*b + a^2*b^3 - 2*b^5)*c*d - (a^5 - 5*a^3*b^2 + 4*a*b^4)*d^2)*cos(f*x + e)*sin(f*x + e) - ((2*
a^4 + 3*a^2*b^2 + b^4)*c^2 - 6*(a^3*b + a*b^3)*c*d + (a^4 + 3*a^2*b^2 + 2*b^4)*d^2 + (6*a*b^3*c*d - (2*a^2*b^2
+ b^4)*c^2 - (a^2*b^2 + 2*b^4)*d^2)*cos(f*x + e)^2 - 2*(6*a^2*b^2*c*d - (2*a^3*b + a*b^3)*c^2 - (a^3*b + 2*a*
b^3)*d^2)*sin(f*x + e))*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) + ((4*a^4
*b - 5*a^2*b^3 + b^5)*c^2 - 2*(2*a^5 - a^3*b^2 - a*b^4)*c*d + 3*(a^4*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^6*b^2
- 3*a^4*b^4 + 3*a^2*b^6 - b^8)*f*cos(f*x + e)^2 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*f*sin(f*x + e) -
(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*f)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 586 vs. \(2 (187) = 374\).

Time = 0.32 (sec) , antiderivative size = 586, normalized size of antiderivative = 3.27 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^3} \, dx=\frac {\frac {{\left (2 \, a^{2} c^{2} + b^{2} c^{2} - 6 \, a b c d + a^{2} d^{2} + 2 \, b^{2} d^{2}\right )} {\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^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {5 \, a^{3} b^{2} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a b^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, a^{4} b c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a^{5} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a^{3} b^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 4 \, a^{4} b c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, a^{2} b^{3} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, b^{5} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, a^{5} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 10 \, a^{3} b^{2} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 4 \, a b^{4} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a^{4} b d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 6 \, a^{2} b^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 11 \, a^{3} b^{2} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a b^{4} c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 10 \, a^{4} b c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 8 \, a^{2} b^{3} c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a^{5} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 10 \, a^{3} b^{2} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a^{4} b c^{2} - a^{2} b^{3} c^{2} - 4 \, a^{5} c d - 2 \, a^{3} b^{2} c d + 3 \, a^{4} b d^{2}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\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 )}^{2}}}{f} \]

[In]

integrate((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

((2*a^2*c^2 + b^2*c^2 - 6*a*b*c*d + a^2*d^2 + 2*b^2*d^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^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (5*a^3*b^2*c^2*tan(1/2
*f*x + 1/2*e)^3 - 2*a*b^4*c^2*tan(1/2*f*x + 1/2*e)^3 - 6*a^4*b*c*d*tan(1/2*f*x + 1/2*e)^3 + a^5*d^2*tan(1/2*f*
x + 1/2*e)^3 + 2*a^3*b^2*d^2*tan(1/2*f*x + 1/2*e)^3 + 4*a^4*b*c^2*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*b^3*c^2*tan(1
/2*f*x + 1/2*e)^2 - 2*b^5*c^2*tan(1/2*f*x + 1/2*e)^2 - 4*a^5*c*d*tan(1/2*f*x + 1/2*e)^2 - 10*a^3*b^2*c*d*tan(1
/2*f*x + 1/2*e)^2 - 4*a*b^4*c*d*tan(1/2*f*x + 1/2*e)^2 + 3*a^4*b*d^2*tan(1/2*f*x + 1/2*e)^2 + 6*a^2*b^3*d^2*ta
n(1/2*f*x + 1/2*e)^2 + 11*a^3*b^2*c^2*tan(1/2*f*x + 1/2*e) - 2*a*b^4*c^2*tan(1/2*f*x + 1/2*e) - 10*a^4*b*c*d*t
an(1/2*f*x + 1/2*e) - 8*a^2*b^3*c*d*tan(1/2*f*x + 1/2*e) - a^5*d^2*tan(1/2*f*x + 1/2*e) + 10*a^3*b^2*d^2*tan(1
/2*f*x + 1/2*e) + 4*a^4*b*c^2 - a^2*b^3*c^2 - 4*a^5*c*d - 2*a^3*b^2*c*d + 3*a^4*b*d^2)/((a^6 - 2*a^4*b^2 + a^2
*b^4)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)^2))/f

Mupad [B] (verification not implemented)

Time = 11.34 (sec) , antiderivative size = 641, normalized size of antiderivative = 3.58 \[ \int \frac {(c+d \sin (e+f x))^2}{(3+b \sin (e+f x))^3} \, dx=\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2}\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{f\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {4\,a^3\,c\,d-4\,a^2\,b\,c^2-3\,a^2\,b\,d^2+2\,a\,b^2\,c\,d+b^3\,c^2}{a^4-2\,a^2\,b^2+b^4}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^4\,d^2+10\,a^3\,b\,c\,d-11\,a^2\,b^2\,c^2-10\,a^2\,b^2\,d^2+8\,a\,b^3\,c\,d+2\,b^4\,c^2\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (a^4\,d^2-6\,a^3\,b\,c\,d+5\,a^2\,b^2\,c^2+2\,a^2\,b^2\,d^2-2\,b^4\,c^2\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a^2+2\,b^2\right )\,\left (4\,a^3\,c\,d-4\,a^2\,b\,c^2-3\,a^2\,b\,d^2+2\,a\,b^2\,c\,d+b^3\,c^2\right )}{a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \]

[In]

int((c + d*sin(e + f*x))^2/(a + b*sin(e + f*x))^3,x)

[Out]

(atan(((((2*a^4*b + 2*b^5 - 4*a^2*b^3)*(2*a^2*c^2 + a^2*d^2 + b^2*c^2 + 2*b^2*d^2 - 6*a*b*c*d))/(2*(a + b)^(5/
2)*(a - b)^(5/2)*(a^4 + b^4 - 2*a^2*b^2)) + (a*tan(e/2 + (f*x)/2)*(2*a^2*c^2 + a^2*d^2 + b^2*c^2 + 2*b^2*d^2 -
6*a*b*c*d))/((a + b)^(5/2)*(a - b)^(5/2)))*(a^4 + b^4 - 2*a^2*b^2))/(2*a^2*c^2 + a^2*d^2 + b^2*c^2 + 2*b^2*d^
2 - 6*a*b*c*d))*(2*a^2*c^2 + a^2*d^2 + b^2*c^2 + 2*b^2*d^2 - 6*a*b*c*d))/(f*(a + b)^(5/2)*(a - b)^(5/2)) - ((b
^3*c^2 - 4*a^2*b*c^2 - 3*a^2*b*d^2 + 4*a^3*c*d + 2*a*b^2*c*d)/(a^4 + b^4 - 2*a^2*b^2) + (tan(e/2 + (f*x)/2)*(a
^4*d^2 + 2*b^4*c^2 - 11*a^2*b^2*c^2 - 10*a^2*b^2*d^2 + 8*a*b^3*c*d + 10*a^3*b*c*d))/(a*(a^4 + b^4 - 2*a^2*b^2)
) - (tan(e/2 + (f*x)/2)^3*(a^4*d^2 - 2*b^4*c^2 + 5*a^2*b^2*c^2 + 2*a^2*b^2*d^2 - 6*a^3*b*c*d))/(a*(a^4 + b^4 -
2*a^2*b^2)) + (tan(e/2 + (f*x)/2)^2*(a^2 + 2*b^2)*(b^3*c^2 - 4*a^2*b*c^2 - 3*a^2*b*d^2 + 4*a^3*c*d + 2*a*b^2*
c*d))/(a^2*(a^4 + b^4 - 2*a^2*b^2)))/(f*(tan(e/2 + (f*x)/2)^2*(2*a^2 + 4*b^2) + a^2*tan(e/2 + (f*x)/2)^4 + a^2
+ 4*a*b*tan(e/2 + (f*x)/2)^3 + 4*a*b*tan(e/2 + (f*x)/2)))