3.441 \(\int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=138 \[ \frac {3 a^2 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d)^2 \sqrt {c^2-d^2}}-\frac {a^2 (c+4 d) \cos (e+f x)}{2 d f (c+d)^2 (c+d \sin (e+f x))}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 138, 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 = {2762, 2754, 12, 2660, 618, 204} \[ \frac {3 a^2 \tan ^{-1}\left (\frac {c \tan \left (\frac {1}{2} (e+f x)\right )+d}{\sqrt {c^2-d^2}}\right )}{f (c+d)^2 \sqrt {c^2-d^2}}-\frac {a^2 (c+4 d) \cos (e+f x)}{2 d f (c+d)^2 (c+d \sin (e+f x))}+\frac {a^2 (c-d) \cos (e+f x)}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a^2*ArcTan[(d + c*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/((c + d)^2*Sqrt[c^2 - d^2]*f) + (a^2*(c - d)*Cos[e +
f*x])/(2*d*(c + d)*f*(c + d*Sin[e + f*x])^2) - (a^2*(c + 4*d)*Cos[e + f*x])/(2*d*(c + d)^2*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))^3} \, dx &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a \int \frac {-4 a d-a (c+3 d) \sin (e+f x)}{(c+d \sin (e+f x))^2} \, dx}{2 d (c+d)}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}+\frac {a \int \frac {3 a (c-d) d}{c+d \sin (e+f x)} \, dx}{2 (c-d) d (c+d)^2}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}+\frac {\left (3 a^2\right ) \int \frac {1}{c+d \sin (e+f x)} \, dx}{2 (c+d)^2}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}+\frac {\left (3 a^2\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)^2 f}\\ &=\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}-\frac {\left (6 a^2\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)^2 f}\\ &=\frac {3 a^2 \tan ^{-1}\left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c+d)^2 \sqrt {c^2-d^2} f}+\frac {a^2 (c-d) \cos (e+f x)}{2 d (c+d) f (c+d \sin (e+f x))^2}-\frac {a^2 (c+4 d) \cos (e+f x)}{2 d (c+d)^2 f (c+d \sin (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 0.62, size = 140, normalized size = 1.01 \[ \frac {a^2 \cos (e+f x) \left (-\frac {(c+4 d) \sin (e+f x)+4 c+d}{(c+d) (c+d \sin (e+f x))^2}-\frac {6 \tanh ^{-1}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {\sin (e+f x)+1}}\right )}{(-c-d)^{3/2} \sqrt {c-d} \sqrt {\cos ^2(e+f x)}}\right )}{2 f (c+d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.50, size = 679, normalized size = 4.92 \[ \left [\frac {2 \, {\left (a^{2} c^{3} + 4 \, a^{2} c^{2} d - a^{2} c d^{2} - 4 \, a^{2} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, {\left (a^{2} d^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} c d \sin \left (f x + e\right ) - a^{2} c^{2} - a^{2} d^{2}\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 (4 \, a^{2} c^{3} + a^{2} c^{2} d - 4 \, a^{2} c d^{2} - a^{2} d^{3}\right )} \cos \left (f x + e\right )}{4 \, {\left ({\left (c^{4} d^{2} + 2 \, c^{3} d^{3} - 2 \, c d^{5} - d^{6}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{5} d + 2 \, c^{4} d^{2} - 2 \, c^{2} d^{4} - c d^{5}\right )} f \sin \left (f x + e\right ) - {\left (c^{6} + 2 \, c^{5} d + c^{4} d^{2} - c^{2} d^{4} - 2 \, c d^{5} - d^{6}\right )} f\right )}}, \frac {{\left (a^{2} c^{3} + 4 \, a^{2} c^{2} d - a^{2} c d^{2} - 4 \, a^{2} d^{3}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 3 \, {\left (a^{2} d^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} c d \sin \left (f x + e\right ) - a^{2} c^{2} - a^{2} d^{2}\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 (4 \, a^{2} c^{3} + a^{2} c^{2} d - 4 \, a^{2} c d^{2} - a^{2} d^{3}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (c^{4} d^{2} + 2 \, c^{3} d^{3} - 2 \, c d^{5} - d^{6}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{5} d + 2 \, c^{4} d^{2} - 2 \, c^{2} d^{4} - c d^{5}\right )} f \sin \left (f x + e\right ) - {\left (c^{6} + 2 \, c^{5} d + c^{4} d^{2} - c^{2} d^{4} - 2 \, c d^{5} - d^{6}\right )} f\right )}}\right ] \]

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))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.26, size = 348, normalized size = 2.52 \[ \frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (c) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} a^{2}}{{\left (c^{2} + 2 \, c d + d^{2}\right )} \sqrt {c^{2} - d^{2}}} + \frac {a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 4 \, a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 8 \, a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a^{2} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 12 \, a^{2} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a^{2} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a^{2} c^{3} - a^{2} c^{2} d}{{\left (c^{4} + 2 \, c^{3} d + c^{2} d^{2}\right )} {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + c\right )}^{2}}}{f} \]

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))^3,x, algorithm="giac")

[Out]

(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)))*a^2/((c^2
+ 2*c*d + d^2)*sqrt(c^2 - d^2)) + (a^2*c^3*tan(1/2*f*x + 1/2*e)^3 - 4*a^2*c^2*d*tan(1/2*f*x + 1/2*e)^3 - 2*a^2
*c*d^2*tan(1/2*f*x + 1/2*e)^3 - 4*a^2*c^3*tan(1/2*f*x + 1/2*e)^2 - a^2*c^2*d*tan(1/2*f*x + 1/2*e)^2 - 8*a^2*c*
d^2*tan(1/2*f*x + 1/2*e)^2 - 2*a^2*d^3*tan(1/2*f*x + 1/2*e)^2 - a^2*c^3*tan(1/2*f*x + 1/2*e) - 12*a^2*c^2*d*ta
n(1/2*f*x + 1/2*e) - 2*a^2*c*d^2*tan(1/2*f*x + 1/2*e) - 4*a^2*c^3 - a^2*c^2*d)/((c^4 + 2*c^3*d + c^2*d^2)*(c*t
an(1/2*f*x + 1/2*e)^2 + 2*d*tan(1/2*f*x + 1/2*e) + c)^2))/f

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maple [B]  time = 0.32, size = 799, normalized size = 5.79 \[ \frac {a^{2} c \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right )}-\frac {4 a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right )}-\frac {2 a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{2}}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right ) c}-\frac {4 a^{2} c \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right )}-\frac {a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right )}-\frac {8 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{2}}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right ) c}-\frac {2 a^{2} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) d^{3}}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right ) c^{2}}-\frac {a^{2} c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right )}-\frac {12 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right )}-\frac {2 a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d^{2}}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} c \left (c^{2}+2 c d +d^{2}\right )}-\frac {4 a^{2} c}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right )}-\frac {a^{2} d}{f \left (\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) c +2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d +c \right )^{2} \left (c^{2}+2 c d +d^{2}\right )}+\frac {3 a^{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 )}{f \left (c^{2}+2 c d +d^{2}\right ) \sqrt {c^{2}-d^{2}}} \]

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))^3,x)

[Out]

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

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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))^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*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 = 9.43, size = 362, normalized size = 2.62 \[ \frac {3\,a^2\,\mathrm {atan}\left (\frac {\left (\frac {3\,a^2\,\left (2\,c^2\,d+4\,c\,d^2+2\,d^3\right )}{2\,{\left (c+d\right )}^{5/2}\,\sqrt {c-d}\,\left (c^2+2\,c\,d+d^2\right )}+\frac {3\,a^2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{{\left (c+d\right )}^{5/2}\,\sqrt {c-d}}\right )\,\left (c^2+2\,c\,d+d^2\right )}{3\,a^2}\right )}{f\,{\left (c+d\right )}^{5/2}\,\sqrt {c-d}}-\frac {\frac {4\,a^2\,c+a^2\,d}{c^2+2\,c\,d+d^2}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^2\,c^2+12\,a^2\,c\,d+2\,a^2\,d^2\right )}{c\,\left (c^2+2\,c\,d+d^2\right )}+\frac {a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (-c^2+4\,c\,d+2\,d^2\right )}{c\,\left (c^2+2\,c\,d+d^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (c^2+2\,d^2\right )\,\left (4\,a^2\,c+a^2\,d\right )}{c^2\,\left (c^2+2\,c\,d+d^2\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,c^2+4\,d^2\right )+c^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+c^2+4\,c\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,c\,d\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \]

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))^3,x)

[Out]

(3*a^2*atan((((3*a^2*(4*c*d^2 + 2*c^2*d + 2*d^3))/(2*(c + d)^(5/2)*(c - d)^(1/2)*(2*c*d + c^2 + d^2)) + (3*a^2
*c*tan(e/2 + (f*x)/2))/((c + d)^(5/2)*(c - d)^(1/2)))*(2*c*d + c^2 + d^2))/(3*a^2)))/(f*(c + d)^(5/2)*(c - d)^
(1/2)) - ((4*a^2*c + a^2*d)/(2*c*d + c^2 + d^2) + (tan(e/2 + (f*x)/2)*(a^2*c^2 + 2*a^2*d^2 + 12*a^2*c*d))/(c*(
2*c*d + c^2 + d^2)) + (a^2*tan(e/2 + (f*x)/2)^3*(4*c*d - c^2 + 2*d^2))/(c*(2*c*d + c^2 + d^2)) + (tan(e/2 + (f
*x)/2)^2*(c^2 + 2*d^2)*(4*a^2*c + a^2*d))/(c^2*(2*c*d + c^2 + d^2)))/(f*(tan(e/2 + (f*x)/2)^2*(2*c^2 + 4*d^2)
+ c^2*tan(e/2 + (f*x)/2)^4 + c^2 + 4*c*d*tan(e/2 + (f*x)/2)^3 + 4*c*d*tan(e/2 + (f*x)/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))**3,x)

[Out]

Timed out

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