\(\int \frac {\sin ^3(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\) [69]

Optimal result

Integrand size = 23, antiderivative size = 133 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 (a-b)^{7/2} f}-\frac {(a+b) \cos (e+f x)}{(a-b)^3 f}+\frac {\cos ^3(e+f x)}{3 (a-b)^2 f}-\frac {a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )} \]

[Out]

-(a+b)*cos(f*x+e)/(a-b)^3/f+1/3*cos(f*x+e)^3/(a-b)^2/f-1/2*a*b*sec(f*x+e)/(a-b)^3/f/(a-b+b*sec(f*x+e)^2)-1/2*(
3*a+2*b)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/(a-b)^(7/2)/f

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3745, 467, 1275, 211} \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=-\frac {\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 f (a-b)^{7/2}}+\frac {\cos ^3(e+f x)}{3 f (a-b)^2}-\frac {(a+b) \cos (e+f x)}{f (a-b)^3}-\frac {a b \sec (e+f x)}{2 f (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )} \]

[In]

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

[Out]

-1/2*(Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/((a - b)^(7/2)*f) - ((a + b)*Cos[e + f*x
])/((a - b)^3*f) + Cos[e + f*x]^3/(3*(a - b)^2*f) - (a*b*Sec[e + f*x])/(2*(a - b)^3*f*(a - b + b*Sec[e + f*x]^
2))

Rule 211

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

Rule 467

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[x^m*(a + b*x^2)^(p +
1)*ExpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)]
- ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &
& ILtQ[m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 3745

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m
+ 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {-1+x^2}{x^4 \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \text {Subst}\left (\int \frac {\frac {2}{(a-b) b}-\frac {2 a x^2}{(a-b)^2 b}+\frac {a x^4}{(a-b)^3}}{x^4 \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{2 f} \\ & = -\frac {a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {b \text {Subst}\left (\int \left (\frac {2}{(a-b)^2 b x^4}+\frac {2 (a+b)}{b (-a+b)^3 x^2}+\frac {3 a+2 b}{(a-b)^3 \left (a-b+b x^2\right )}\right ) \, dx,x,\sec (e+f x)\right )}{2 f} \\ & = -\frac {(a+b) \cos (e+f x)}{(a-b)^3 f}+\frac {\cos ^3(e+f x)}{3 (a-b)^2 f}-\frac {a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )}-\frac {(b (3 a+2 b)) \text {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{2 (a-b)^3 f} \\ & = -\frac {\sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{2 (a-b)^{7/2} f}-\frac {(a+b) \cos (e+f x)}{(a-b)^3 f}+\frac {\cos ^3(e+f x)}{3 (a-b)^2 f}-\frac {a b \sec (e+f x)}{2 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.19 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.37 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {6 \sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{7/2}}+\frac {6 \sqrt {b} (3 a+2 b) \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{7/2}}-\frac {\cos (e+f x) \left (9 a+15 b+\frac {12 a b}{a+b+(a-b) \cos (2 (e+f x))}\right )+(-a+b) \cos (3 (e+f x))}{(a-b)^3}}{12 f} \]

[In]

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

[Out]

((6*Sqrt[b]*(3*a + 2*b)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(7/2) + (6*Sqrt[b]*(
3*a + 2*b)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(7/2) - (Cos[e + f*x]*(9*a + 15*b
+ (12*a*b)/(a + b + (a - b)*Cos[2*(e + f*x)])) + (-a + b)*Cos[3*(e + f*x)])/(a - b)^3)/(12*f)

Maple [A] (verified)

Time = 7.74 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.14

method	result	size
derivativedivides	\(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\frac {b \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a -b \cos \left (f x +e \right )}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}+\frac {b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a +2 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{3}}}{f}\)	\(152\)
default	\(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\frac {b \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a -b \cos \left (f x +e \right )}{\left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}+\frac {b \left (-\frac {a \cos \left (f x +e \right )}{2 \left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )}+\frac {\left (3 a +2 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{2 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{3}}}{f}\)	\(152\)
risch	\(\frac {{\mathrm e}^{3 i \left (f x +e \right )}}{24 \left (a^{2}-2 a b +b^{2}\right ) f}-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )} a}{8 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )} b}{8 f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )} a}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )} b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}+\frac {{\mathrm e}^{-3 i \left (f x +e \right )}}{24 \left (a^{2}-2 a b +b^{2}\right ) f}-\frac {a b \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{f \left (a -b \right )^{3} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}+\frac {3 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) a}{4 \left (a -b \right )^{4} f}+\frac {i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{2 \left (a -b \right )^{4} f}-\frac {3 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) a}{4 \left (a -b \right )^{4} f}-\frac {i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{2 \left (a -b \right )^{4} f}\)	\(541\)

[In]

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

[Out]

1/f*(1/(a^2-2*a*b+b^2)/(a-b)*(1/3*a*cos(f*x+e)^3-1/3*b*cos(f*x+e)^3-cos(f*x+e)*a-b*cos(f*x+e))+b/(a-b)^3*(-1/2
*a*cos(f*x+e)/(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)+1/2*(3*a+2*b)/(b*(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b)
)^(1/2))))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 456, normalized size of antiderivative = 3.43 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [\frac {4 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 4 \, {\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 6 \, {\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )}{12 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}, \frac {2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left ({\left (3 \, a^{2} - a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a b + 2 \, b^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - 3 \, {\left (3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )}{6 \, {\left ({\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} f\right )}}\right ] \]

[In]

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

[Out]

[1/12*(4*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - 4*(3*a^2 - a*b - 2*b^2)*cos(f*x + e)^3 - 3*((3*a^2 - a*b - 2*b^2
)*cos(f*x + e)^2 + 3*a*b + 2*b^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b))*c
os(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) - 6*(3*a*b + 2*b^2)*cos(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 -
4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f), 1/6*(2*(a^2 - 2*a*b + b^2)*cos(f*x
+ e)^5 - 2*(3*a^2 - a*b - 2*b^2)*cos(f*x + e)^3 - 3*((3*a^2 - a*b - 2*b^2)*cos(f*x + e)^2 + 3*a*b + 2*b^2)*sqr
t(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) - 3*(3*a*b + 2*b^2)*cos(f*x + e))/((a^4 - 4*a^3*b
+ 6*a^2*b^2 - 4*a*b^3 + b^4)*f*cos(f*x + e)^2 + (a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*f)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate(sin(f*x+e)^3/(a+b*tan(f*x+e)^2)^2,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(b-a>0)', see `assume?` for mor
e details)Is

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (119) = 238\).

Time = 0.64 (sec) , antiderivative size = 354, normalized size of antiderivative = 2.66 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {a^{4} f^{11} \cos \left (f x + e\right )^{3} - 4 \, a^{3} b f^{11} \cos \left (f x + e\right )^{3} + 6 \, a^{2} b^{2} f^{11} \cos \left (f x + e\right )^{3} - 4 \, a b^{3} f^{11} \cos \left (f x + e\right )^{3} + b^{4} f^{11} \cos \left (f x + e\right )^{3} - 3 \, a^{4} f^{11} \cos \left (f x + e\right ) + 6 \, a^{3} b f^{11} \cos \left (f x + e\right ) - 6 \, a b^{3} f^{11} \cos \left (f x + e\right ) + 3 \, b^{4} f^{11} \cos \left (f x + e\right )}{3 \, {\left (a^{6} f^{12} - 6 \, a^{5} b f^{12} + 15 \, a^{4} b^{2} f^{12} - 20 \, a^{3} b^{3} f^{12} + 15 \, a^{2} b^{4} f^{12} - 6 \, a b^{5} f^{12} + b^{6} f^{12}\right )}} - \frac {a b \cos \left (f x + e\right )}{2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )} f} + \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt {a b - b^{2}}}\right )}{2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \sqrt {a b - b^{2}} f} \]

[In]

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

[Out]

1/3*(a^4*f^11*cos(f*x + e)^3 - 4*a^3*b*f^11*cos(f*x + e)^3 + 6*a^2*b^2*f^11*cos(f*x + e)^3 - 4*a*b^3*f^11*cos(
f*x + e)^3 + b^4*f^11*cos(f*x + e)^3 - 3*a^4*f^11*cos(f*x + e) + 6*a^3*b*f^11*cos(f*x + e) - 6*a*b^3*f^11*cos(
f*x + e) + 3*b^4*f^11*cos(f*x + e))/(a^6*f^12 - 6*a^5*b*f^12 + 15*a^4*b^2*f^12 - 20*a^3*b^3*f^12 + 15*a^2*b^4*
f^12 - 6*a*b^5*f^12 + b^6*f^12) - 1/2*a*b*cos(f*x + e)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(a*cos(f*x + e)^2 - b*
cos(f*x + e)^2 + b)*f) + 1/2*(3*a*b + 2*b^2)*arctan((a*cos(f*x + e) - b*cos(f*x + e))/sqrt(a*b - b^2))/((a^3 -
3*a^2*b + 3*a*b^2 - b^3)*sqrt(a*b - b^2)*f)

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 737, normalized size of antiderivative = 5.54 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {\sqrt {b}\,\left (3\,a+2\,b\right )\,\left (24\,a^9\,b-128\,a^8\,b^2+264\,a^7\,b^3-240\,a^6\,b^4+40\,a^5\,b^5+96\,a^4\,b^6-72\,a^3\,b^7+16\,a^2\,b^8\right )}{4\,a\,{\left (a-b\right )}^{13/2}}+\frac {\sqrt {b}\,\left (a-2\,b\right )\,{\left (3\,a+2\,b\right )}^2\,\left (16\,a^{12}-176\,a^{11}\,b+864\,a^{10}\,b^2-2496\,a^9\,b^3+4704\,a^8\,b^4-6048\,a^7\,b^5+5376\,a^6\,b^6-3264\,a^5\,b^7+1296\,a^4\,b^8-304\,a^3\,b^9+32\,a^2\,b^{10}\right )}{32\,a\,{\left (a-b\right )}^{21/2}}\right )+\frac {\sqrt {b}\,\left (a-2\,b\right )\,{\left (3\,a+2\,b\right )}^2\,\left (-16\,a^{12}+144\,a^{11}\,b-576\,a^{10}\,b^2+1344\,a^9\,b^3-2016\,a^8\,b^4+2016\,a^7\,b^5-1344\,a^6\,b^6+576\,a^5\,b^7-144\,a^4\,b^8+16\,a^3\,b^9\right )}{32\,a\,{\left (a-b\right )}^{21/2}}\right )\,{\left (a-b\right )}^7}{-9\,a^{10}\,b+42\,a^9\,b^2-67\,a^8\,b^3+24\,a^7\,b^4+45\,a^6\,b^5-46\,a^5\,b^6+3\,a^4\,b^7+12\,a^3\,b^8-4\,a^2\,b^9}\right )\,\left (3\,a+2\,b\right )}{2\,f\,{\left (a-b\right )}^{7/2}}-\frac {\frac {4\,a^2+11\,b\,a}{3\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (2\,b^2+3\,a\,b\right )}{\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (2\,a^2-3\,a\,b+11\,b^2\right )}{\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,a^2+9\,a\,b+19\,b^2\right )}{3\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}+\frac {2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (-10\,a^2+22\,a\,b+33\,b^2\right )}{3\,\left (a-b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+\left (a+4\,b\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+\left (12\,b-2\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+\left (12\,b-2\,a\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+\left (a+4\,b\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a\right )} \]

[In]

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

[Out]

(b^(1/2)*atan(((tan(e/2 + (f*x)/2)^2*((b^(1/2)*(3*a + 2*b)*(24*a^9*b + 16*a^2*b^8 - 72*a^3*b^7 + 96*a^4*b^6 +
40*a^5*b^5 - 240*a^6*b^4 + 264*a^7*b^3 - 128*a^8*b^2))/(4*a*(a - b)^(13/2)) + (b^(1/2)*(a - 2*b)*(3*a + 2*b)^2
*(16*a^12 - 176*a^11*b + 32*a^2*b^10 - 304*a^3*b^9 + 1296*a^4*b^8 - 3264*a^5*b^7 + 5376*a^6*b^6 - 6048*a^7*b^5
+ 4704*a^8*b^4 - 2496*a^9*b^3 + 864*a^10*b^2))/(32*a*(a - b)^(21/2))) + (b^(1/2)*(a - 2*b)*(3*a + 2*b)^2*(144
*a^11*b - 16*a^12 + 16*a^3*b^9 - 144*a^4*b^8 + 576*a^5*b^7 - 1344*a^6*b^6 + 2016*a^7*b^5 - 2016*a^8*b^4 + 1344
*a^9*b^3 - 576*a^10*b^2))/(32*a*(a - b)^(21/2)))*(a - b)^7)/(12*a^3*b^8 - 4*a^2*b^9 - 9*a^10*b + 3*a^4*b^7 - 4
6*a^5*b^6 + 45*a^6*b^5 + 24*a^7*b^4 - 67*a^8*b^3 + 42*a^9*b^2))*(3*a + 2*b))/(2*f*(a - b)^(7/2)) - ((11*a*b +
4*a^2)/(3*(a - b)*(a^2 - 2*a*b + b^2)) + (tan(e/2 + (f*x)/2)^8*(3*a*b + 2*b^2))/((a - b)*(a^2 - 2*a*b + b^2))
+ (2*tan(e/2 + (f*x)/2)^6*(2*a^2 - 3*a*b + 11*b^2))/((a - b)*(a^2 - 2*a*b + b^2)) + (2*tan(e/2 + (f*x)/2)^2*(9
*a*b + 2*a^2 + 19*b^2))/(3*(a - b)*(a^2 - 2*a*b + b^2)) + (2*tan(e/2 + (f*x)/2)^4*(22*a*b - 10*a^2 + 33*b^2))/
(3*(a - b)*(a^2 - 2*a*b + b^2)))/(f*(a + tan(e/2 + (f*x)/2)^2*(a + 4*b) + tan(e/2 + (f*x)/2)^8*(a + 4*b) - tan
(e/2 + (f*x)/2)^4*(2*a - 12*b) - tan(e/2 + (f*x)/2)^6*(2*a - 12*b) + a*tan(e/2 + (f*x)/2)^10))