\(\int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx\) [165]

Optimal result

Integrand size = 21, antiderivative size = 110 \[ \int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx=-\frac {\left (3 a^2+4 b^2\right ) \text {arctanh}(\cos (e+f x))}{8 f}-\frac {2 a b \cot (e+f x)}{f}-\frac {2 a b \cot ^3(e+f x)}{3 f}-\frac {\left (3 a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac {a^2 \cot (e+f x) \csc ^3(e+f x)}{4 f} \]

[Out]

-1/8*(3*a^2+4*b^2)*arctanh(cos(f*x+e))/f-2*a*b*cot(f*x+e)/f-2/3*a*b*cot(f*x+e)^3/f-1/8*(3*a^2+4*b^2)*cot(f*x+e
)*csc(f*x+e)/f-1/4*a^2*cot(f*x+e)*csc(f*x+e)^3/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2868, 3852, 3091, 3853, 3855} \[ \int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx=-\frac {\left (3 a^2+4 b^2\right ) \text {arctanh}(\cos (e+f x))}{8 f}-\frac {\left (3 a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac {a^2 \cot (e+f x) \csc ^3(e+f x)}{4 f}-\frac {2 a b \cot ^3(e+f x)}{3 f}-\frac {2 a b \cot (e+f x)}{f} \]

[In]

Int[Csc[e + f*x]^5*(a + b*Sin[e + f*x])^2,x]

[Out]

-1/8*((3*a^2 + 4*b^2)*ArcTanh[Cos[e + f*x]])/f - (2*a*b*Cot[e + f*x])/f - (2*a*b*Cot[e + f*x]^3)/(3*f) - ((3*a
^2 + 4*b^2)*Cot[e + f*x]*Csc[e + f*x])/(8*f) - (a^2*Cot[e + f*x]*Csc[e + f*x]^3)/(4*f)

Rule 2868

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[2*c*(d/b)
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), x] /; FreeQ[{b, c,
d, e, f, m}, x]

Rule 3091

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A*Cos[e +
f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = (2 a b) \int \csc ^4(e+f x) \, dx+\int \csc ^5(e+f x) \left (a^2+b^2 \sin ^2(e+f x)\right ) \, dx \\ & = -\frac {a^2 \cot (e+f x) \csc ^3(e+f x)}{4 f}+\frac {1}{4} \left (3 a^2+4 b^2\right ) \int \csc ^3(e+f x) \, dx-\frac {(2 a b) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (e+f x)\right )}{f} \\ & = -\frac {2 a b \cot (e+f x)}{f}-\frac {2 a b \cot ^3(e+f x)}{3 f}-\frac {\left (3 a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac {a^2 \cot (e+f x) \csc ^3(e+f x)}{4 f}+\frac {1}{8} \left (3 a^2+4 b^2\right ) \int \csc (e+f x) \, dx \\ & = -\frac {\left (3 a^2+4 b^2\right ) \text {arctanh}(\cos (e+f x))}{8 f}-\frac {2 a b \cot (e+f x)}{f}-\frac {2 a b \cot ^3(e+f x)}{3 f}-\frac {\left (3 a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac {a^2 \cot (e+f x) \csc ^3(e+f x)}{4 f} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(255\) vs. \(2(110)=220\).

Time = 0.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.32 \[ \int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx=-\frac {4 a b \cot (e+f x)}{3 f}-\frac {3 a^2 \csc ^2\left (\frac {1}{2} (e+f x)\right )}{32 f}-\frac {b^2 \csc ^2\left (\frac {1}{2} (e+f x)\right )}{8 f}-\frac {a^2 \csc ^4\left (\frac {1}{2} (e+f x)\right )}{64 f}-\frac {2 a b \cot (e+f x) \csc ^2(e+f x)}{3 f}-\frac {3 a^2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{8 f}-\frac {b^2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 f}+\frac {3 a^2 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{8 f}+\frac {b^2 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 f}+\frac {3 a^2 \sec ^2\left (\frac {1}{2} (e+f x)\right )}{32 f}+\frac {b^2 \sec ^2\left (\frac {1}{2} (e+f x)\right )}{8 f}+\frac {a^2 \sec ^4\left (\frac {1}{2} (e+f x)\right )}{64 f} \]

[In]

Integrate[Csc[e + f*x]^5*(a + b*Sin[e + f*x])^2,x]

[Out]

(-4*a*b*Cot[e + f*x])/(3*f) - (3*a^2*Csc[(e + f*x)/2]^2)/(32*f) - (b^2*Csc[(e + f*x)/2]^2)/(8*f) - (a^2*Csc[(e
+ f*x)/2]^4)/(64*f) - (2*a*b*Cot[e + f*x]*Csc[e + f*x]^2)/(3*f) - (3*a^2*Log[Cos[(e + f*x)/2]])/(8*f) - (b^2*
Log[Cos[(e + f*x)/2]])/(2*f) + (3*a^2*Log[Sin[(e + f*x)/2]])/(8*f) + (b^2*Log[Sin[(e + f*x)/2]])/(2*f) + (3*a^
2*Sec[(e + f*x)/2]^2)/(32*f) + (b^2*Sec[(e + f*x)/2]^2)/(8*f) + (a^2*Sec[(e + f*x)/2]^4)/(64*f)

Maple [A] (verified)

Time = 1.98 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {a^{2} \left (\left (-\frac {\left (\csc ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \csc \left (f x +e \right )}{8}\right ) \cot \left (f x +e \right )+\frac {3 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{8}\right )+2 a b \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (f x +e \right )\right )}{3}\right ) \cot \left (f x +e \right )+b^{2} \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{2}\right )}{f}\)	\(114\)
default	\(\frac {a^{2} \left (\left (-\frac {\left (\csc ^{3}\left (f x +e \right )\right )}{4}-\frac {3 \csc \left (f x +e \right )}{8}\right ) \cot \left (f x +e \right )+\frac {3 \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{8}\right )+2 a b \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (f x +e \right )\right )}{3}\right ) \cot \left (f x +e \right )+b^{2} \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{2}\right )}{f}\)	\(114\)
parallelrisch	\(\frac {-3 \left (\cot ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2}+3 \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2}-16 \left (\cot ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a b +16 a b \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-24 \left (\cot ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2}-24 \left (\cot ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b^{2}+24 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2}+24 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b^{2}-144 \cot \left (\frac {f x}{2}+\frac {e}{2}\right ) a b +72 a^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )+96 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b^{2}+144 a b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{192 f}\)	\(189\)
risch	\(\frac {9 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}+12 b^{2} {\mathrm e}^{7 i \left (f x +e \right )}+96 i a b \,{\mathrm e}^{4 i \left (f x +e \right )}-33 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}-12 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}-128 i a b \,{\mathrm e}^{2 i \left (f x +e \right )}-33 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-12 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}+32 i a b +9 a^{2} {\mathrm e}^{i \left (f x +e \right )}+12 b^{2} {\mathrm e}^{i \left (f x +e \right )}}{12 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{8 f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b^{2}}{2 f}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{8 f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b^{2}}{2 f}\)	\(246\)
norman	\(\frac {-\frac {a^{2}}{64 f}+\frac {a^{2} \left (\tan ^{12}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{64 f}-\frac {\left (5 a^{2}+4 b^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{32 f}+\frac {\left (5 a^{2}+4 b^{2}\right ) \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{32 f}-\frac {\left (17 a^{2}+16 b^{2}\right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{32 f}-\frac {\left (17 a^{2}+16 b^{2}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{32 f}-\frac {a b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{12 f}-\frac {11 a b \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}-\frac {5 a b \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{6 f}+\frac {5 a b \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{6 f}+\frac {11 a b \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}+\frac {a b \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{2}}+\frac {\left (3 a^{2}+4 b^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}\)	\(297\)

[In]

int(csc(f*x+e)^5*(a+b*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(a^2*((-1/4*csc(f*x+e)^3-3/8*csc(f*x+e))*cot(f*x+e)+3/8*ln(-cot(f*x+e)+csc(f*x+e)))+2*a*b*(-2/3-1/3*csc(f*
x+e)^2)*cot(f*x+e)+b^2*(-1/2*csc(f*x+e)*cot(f*x+e)+1/2*ln(-cot(f*x+e)+csc(f*x+e))))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 229 vs. \(2 (102) = 204\).

Time = 0.29 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.08 \[ \int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {6 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 6 \, {\left (5 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right ) - 3 \, {\left ({\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 4 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 3 \, {\left ({\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 4 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 32 \, {\left (2 \, a b \cos \left (f x + e\right )^{3} - 3 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}} \]

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

1/48*(6*(3*a^2 + 4*b^2)*cos(f*x + e)^3 - 6*(5*a^2 + 4*b^2)*cos(f*x + e) - 3*((3*a^2 + 4*b^2)*cos(f*x + e)^4 -
2*(3*a^2 + 4*b^2)*cos(f*x + e)^2 + 3*a^2 + 4*b^2)*log(1/2*cos(f*x + e) + 1/2) + 3*((3*a^2 + 4*b^2)*cos(f*x + e
)^4 - 2*(3*a^2 + 4*b^2)*cos(f*x + e)^2 + 3*a^2 + 4*b^2)*log(-1/2*cos(f*x + e) + 1/2) + 32*(2*a*b*cos(f*x + e)^
3 - 3*a*b*cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f)

Sympy [F]

\[ \int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx=\int \left (a + b \sin {\left (e + f x \right )}\right )^{2} \csc ^{5}{\left (e + f x \right )}\, dx \]

[In]

integrate(csc(f*x+e)**5*(a+b*sin(f*x+e))**2,x)

[Out]

Integral((a + b*sin(e + f*x))**2*csc(e + f*x)**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.34 \[ \int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {3 \, a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\cos \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\cos \left (f x + e\right ) - 1\right )\right )} + 12 \, b^{2} {\left (\frac {2 \, \cos \left (f x + e\right )}{\cos \left (f x + e\right )^{2} - 1} - \log \left (\cos \left (f x + e\right ) + 1\right ) + \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - \frac {32 \, {\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a b}{\tan \left (f x + e\right )^{3}}}{48 \, f} \]

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

1/48*(3*a^2*(2*(3*cos(f*x + e)^3 - 5*cos(f*x + e))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1) - 3*log(cos(f*x + e
) + 1) + 3*log(cos(f*x + e) - 1)) + 12*b^2*(2*cos(f*x + e)/(cos(f*x + e)^2 - 1) - log(cos(f*x + e) + 1) + log(
cos(f*x + e) - 1)) - 32*(3*tan(f*x + e)^2 + 1)*a*b/tan(f*x + e)^3)/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (102) = 204\).

Time = 0.33 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.97 \[ \int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {3 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 16 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 24 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 144 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 24 \, {\left (3 \, a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) - \frac {150 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 200 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 144 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 24 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 24 \, b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 16 \, a b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a^{2}}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}}{192 \, f} \]

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e))^2,x, algorithm="giac")

[Out]

1/192*(3*a^2*tan(1/2*f*x + 1/2*e)^4 + 16*a*b*tan(1/2*f*x + 1/2*e)^3 + 24*a^2*tan(1/2*f*x + 1/2*e)^2 + 24*b^2*t
an(1/2*f*x + 1/2*e)^2 + 144*a*b*tan(1/2*f*x + 1/2*e) + 24*(3*a^2 + 4*b^2)*log(abs(tan(1/2*f*x + 1/2*e))) - (15
0*a^2*tan(1/2*f*x + 1/2*e)^4 + 200*b^2*tan(1/2*f*x + 1/2*e)^4 + 144*a*b*tan(1/2*f*x + 1/2*e)^3 + 24*a^2*tan(1/
2*f*x + 1/2*e)^2 + 24*b^2*tan(1/2*f*x + 1/2*e)^2 + 16*a*b*tan(1/2*f*x + 1/2*e) + 3*a^2)/tan(1/2*f*x + 1/2*e)^4
)/f

Mupad [B] (verification not implemented)

Time = 6.31 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.62 \[ \int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx=\frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (\frac {3\,a^2}{8}+\frac {b^2}{2}\right )}{f}+\frac {a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4}{64\,f}-\frac {{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,a^2+2\,b^2\right )+\frac {a^2}{4}+12\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+\frac {4\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{3}\right )}{16\,f}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a^2}{8}+\frac {b^2}{8}\right )}{f}+\frac {a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12\,f}+\frac {3\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{4\,f} \]

[In]

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

[Out]

(log(tan(e/2 + (f*x)/2))*((3*a^2)/8 + b^2/2))/f + (a^2*tan(e/2 + (f*x)/2)^4)/(64*f) - (cot(e/2 + (f*x)/2)^4*(t
an(e/2 + (f*x)/2)^2*(2*a^2 + 2*b^2) + a^2/4 + 12*a*b*tan(e/2 + (f*x)/2)^3 + (4*a*b*tan(e/2 + (f*x)/2))/3))/(16
*f) + (tan(e/2 + (f*x)/2)^2*(a^2/8 + b^2/8))/f + (a*b*tan(e/2 + (f*x)/2)^3)/(12*f) + (3*a*b*tan(e/2 + (f*x)/2)
)/(4*f)