3.8.19 \(\int \frac {1}{(a+b \sin (e+f x))^3} \, dx\) [719]

Optimal. Leaf size=131 \[ \frac {\left (2 a^2+b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} f}+\frac {b \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {3 a b \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))} \]

[Out]

(2*a^2+b^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*b*cos(f*x+e)/(a^2-b^2)/f/(a
+b*sin(f*x+e))^2+3/2*a*b*cos(f*x+e)/(a^2-b^2)^2/f/(a+b*sin(f*x+e))

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Rubi [A]
time = 0.08, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2743, 2833, 12, 2739, 632, 210} \begin {gather*} \frac {\left (2 a^2+b^2\right ) \text {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 {3 a b \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac {b \cos (e+f x)}{2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x])^(-3),x]

[Out]

((2*a^2 + b^2)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*f) + (b*Cos[e + f*x])/(2*(
a^2 - b^2)*f*(a + b*Sin[e + f*x])^2) + (3*a*b*Cos[e + f*x])/(2*(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 2743

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((a + b*Sin[c + d*x])^(n
+ 1)/(d*(n + 1)*(a^2 - b^2))), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n +
1) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ
erQ[2*n]

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]

Rubi steps

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

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Mathematica [A]
time = 0.43, size = 114, normalized size = 0.87 \begin {gather*} \frac {\frac {2 \left (2 a^2+b^2\right ) \tan ^{-1}\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}}+\frac {b \cos (e+f x) \left (4 a^2-b^2+3 a b \sin (e+f x)\right )}{(a-b)^2 (a+b)^2 (a+b \sin (e+f x))^2}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x])^(-3),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(281\) vs. \(2(122)=244\).
time = 0.29, size = 282, normalized size = 2.15

method	result	size
derivativedivides	\(\frac {\frac {\frac {b^{2} \left (5 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{4}+7 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}-2 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 b \left (4 a^{2}-b^{2}\right )}{2 a^{4}-4 a^{2} b^{2}+2 b^{4}}}{\left (a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}}{f}\)	\(282\)
default	\(\frac {\frac {\frac {b^{2} \left (5 a^{2}-2 b^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}+\frac {b \left (4 a^{4}+7 a^{2} b^{2}-2 b^{4}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}+\frac {b^{2} \left (11 a^{2}-2 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}+\frac {2 b \left (4 a^{2}-b^{2}\right )}{2 a^{4}-4 a^{2} b^{2}+2 b^{4}}}{\left (a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2}+b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}}{f}\)	\(282\)
risch	\(-\frac {i \left (-2 i b \,a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-i {\mathrm e}^{3 i \left (f x +e \right )} b^{3}+10 i a^{2} b \,{\mathrm e}^{i \left (f x +e \right )}-i b^{3} {\mathrm e}^{i \left (f x +e \right )}+6 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+3 b^{2} a \,{\mathrm e}^{2 i \left (f x +e \right )}-3 a \,b^{2}\right )}{\left (-i b \,{\mathrm e}^{2 i \left (f x +e \right )}+i b +2 a \,{\mathrm e}^{i \left (f x +e \right )}\right )^{2} \left (a^{2}-b^{2}\right )^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{2}}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b^{2}}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} f}\)	\(472\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

1/f*(2*(1/2*b^2*(5*a^2-2*b^2)/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*f*x+1/2*e)^3+1/2*b*(4*a^4+7*a^2*b^2-2*b^4)/(a^4-2*
a^2*b^2+b^4)/a^2*tan(1/2*f*x+1/2*e)^2+1/2*b^2*(11*a^2-2*b^2)/a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)+1/2*b*(4
*a^2-b^2)/(a^4-2*a^2*b^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+(2*a^2+b^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)))

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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+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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (127) = 254\).
time = 0.37, size = 641, normalized size = 4.89 \begin {gather*} \left [-\frac {6 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (f x + e\right )}{4 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} f \sin \left (f x + e\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} f\right )}}, -\frac {3 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4} - {\left (2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (2 \, a^{3} b + a b^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) + {\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} f \sin \left (f x + e\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/4*(6*(a^3*b^2 - a*b^4)*cos(f*x + e)*sin(f*x + e) - (2*a^4 + 3*a^2*b^2 + b^4 - (2*a^2*b^2 + b^4)*cos(f*x +
e)^2 + 2*(2*a^3*b + a*b^3)*sin(f*x + e))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(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)*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)*cos(f*x + e)*sin(f*x + e) - (2*a^4 + 3*a^2*b^2 + b^4 - (2*a^2*b^2 + b^
4)*cos(f*x + e)^2 + 2*(2*a^3*b + a*b^3)*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)*cos(f*x + e))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*f*co
s(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)]

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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+b*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (127) = 254\).
time = 0.46, size = 284, normalized size = 2.17 \begin {gather*} \frac {\frac {{\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 (2 \, a^{2} + b^{2}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {5 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a b^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 4 \, a^{4} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, a^{2} b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, b^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 11 \, a^{3} b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a b^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a^{4} b - a^{2} b^{3}}{{\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

((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)))*(2*a^2 + b^2
)/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (5*a^3*b^2*tan(1/2*f*x + 1/2*e)^3 - 2*a*b^4*tan(1/2*f*x + 1/2*e)
^3 + 4*a^4*b*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*b^3*tan(1/2*f*x + 1/2*e)^2 - 2*b^5*tan(1/2*f*x + 1/2*e)^2 + 11*a^3
*b^2*tan(1/2*f*x + 1/2*e) - 2*a*b^4*tan(1/2*f*x + 1/2*e) + 4*a^4*b - a^2*b^3)/((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

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Mupad [B]
time = 10.15, size = 395, normalized size = 3.02 \begin {gather*} \frac {\frac {4\,a^2\,b-b^3}{a^4-2\,a^2\,b^2+b^4}+\frac {b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (11\,a^2\,b-2\,b^3\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 (4\,a^2\,b-b^3\right )\,\left (a^2+2\,b^2\right )}{a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (5\,a^2\,b-2\,b^3\right )}{a\,\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 )}+\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,a^2+b^2\right )\,\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\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+b^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+b^2}\right )\,\left (2\,a^2+b^2\right )}{f\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(e + f*x))^3,x)

[Out]

((4*a^2*b - b^3)/(a^4 + b^4 - 2*a^2*b^2) + (b*tan(e/2 + (f*x)/2)*(11*a^2*b - 2*b^3))/(a*(a^4 + b^4 - 2*a^2*b^2
)) + (tan(e/2 + (f*x)/2)^2*(4*a^2*b - b^3)*(a^2 + 2*b^2))/(a^2*(a^4 + b^4 - 2*a^2*b^2)) + (b*tan(e/2 + (f*x)/2
)^3*(5*a^2*b - 2*b^3))/(a*(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))) + (atan(((((2*a^2 + b^2)*(2*a^4*b +
2*b^5 - 4*a^2*b^3))/(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 + b
^2))/((a + b)^(5/2)*(a - b)^(5/2)))*(a^4 + b^4 - 2*a^2*b^2))/(2*a^2 + b^2))*(2*a^2 + b^2))/(f*(a + b)^(5/2)*(a
- b)^(5/2))

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