\(\int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx\) [171]

Optimal result

Integrand size = 21, antiderivative size = 68 \[ \int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=3 a b^2 x-\frac {3 a^2 b \text {arctanh}(\cos (e+f x))}{f}+\frac {b \left (a^2-b^2\right ) \cos (e+f x)}{f}-\frac {a^2 \cot (e+f x) (a+b \sin (e+f x))}{f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2871, 3102, 2814, 3855} \[ \int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=-\frac {3 a^2 b \text {arctanh}(\cos (e+f x))}{f}+\frac {b \left (a^2-b^2\right ) \cos (e+f x)}{f}-\frac {a^2 \cot (e+f x) (a+b \sin (e+f x))}{f}+3 a b^2 x \]

[In]

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

[Out]

3*a*b^2*x - (3*a^2*b*ArcTanh[Cos[e + f*x]])/f + (b*(a^2 - b^2)*Cos[e + f*x])/f - (a^2*Cot[e + f*x]*(a + b*Sin[
e + f*x]))/f

Rule 2814

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[b*(x/d)
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2871

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/
(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e
+ f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 +
c*d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 +
d*n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3102

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

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

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.28 \[ \int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {-2 b^3 \cos (e+f x)-a^3 \cot \left (\frac {1}{2} (e+f x)\right )+6 a b \left (b (e+f x)-a \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+a \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+a^3 \tan \left (\frac {1}{2} (e+f x)\right )}{2 f} \]

[In]

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

[Out]

(-2*b^3*Cos[e + f*x] - a^3*Cot[(e + f*x)/2] + 6*a*b*(b*(e + f*x) - a*Log[Cos[(e + f*x)/2]] + a*Log[Sin[(e + f*
x)/2]]) + a^3*Tan[(e + f*x)/2])/(2*f)

Maple [A] (verified)

Time = 1.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {-a^{3} \cot \left (f x +e \right )+3 a^{2} b \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+3 a \,b^{2} \left (f x +e \right )-\cos \left (f x +e \right ) b^{3}}{f}\)	\(61\)
default	\(\frac {-a^{3} \cot \left (f x +e \right )+3 a^{2} b \ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )+3 a \,b^{2} \left (f x +e \right )-\cos \left (f x +e \right ) b^{3}}{f}\)	\(61\)
parallelrisch	\(\frac {6 a \,b^{2} f x +2 b^{3}+6 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a^{2} b +\sec \left (\frac {f x}{2}+\frac {e}{2}\right ) \csc \left (\frac {f x}{2}+\frac {e}{2}\right ) a^{3}-2 \cot \left (\frac {f x}{2}+\frac {e}{2}\right ) a^{3}-2 \cos \left (f x +e \right ) b^{3}}{2 f}\)	\(83\)
risch	\(3 a \,b^{2} x -\frac {b^{3} {\mathrm e}^{i \left (f x +e \right )}}{2 f}-\frac {b^{3} {\mathrm e}^{-i \left (f x +e \right )}}{2 f}-\frac {2 i a^{3}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}+\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{f}-\frac {3 a^{2} b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{f}\)	\(107\)
norman	\(\frac {\frac {a^{3} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {2 b^{3} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {2 b^{3} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {4 b^{3} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {a^{3}}{2 f}-\frac {a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {a^{3} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}+3 a \,b^{2} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+9 a \,b^{2} x \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+9 a \,b^{2} x \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+3 a \,b^{2} x \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}+\frac {3 a^{2} b \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}\)	\(240\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.46 \[ \int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=-\frac {3 \, a^{2} b \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) - 3 \, a^{2} b \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) + 2 \, a^{3} \cos \left (f x + e\right ) - 2 \, {\left (3 \, a b^{2} f x - b^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{2 \, f \sin \left (f x + e\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\int \left (a + b \sin {\left (e + f x \right )}\right )^{3} \csc ^{2}{\left (e + f x \right )}\, dx \]

[In]

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

[Out]

Integral((a + b*sin(e + f*x))**3*csc(e + f*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00 \[ \int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {6 \, {\left (f x + e\right )} a b^{2} - 3 \, a^{2} b {\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - 2 \, b^{3} \cos \left (f x + e\right ) - \frac {2 \, a^{3}}{\tan \left (f x + e\right )}}{2 \, f} \]

[In]

integrate(csc(f*x+e)^2*(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

1/2*(6*(f*x + e)*a*b^2 - 3*a^2*b*(log(cos(f*x + e) + 1) - log(cos(f*x + e) - 1)) - 2*b^3*cos(f*x + e) - 2*a^3/
tan(f*x + e))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (68) = 136\).

Time = 0.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 2.01 \[ \int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {6 \, {\left (f x + e\right )} a b^{2} + 6 \, a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) + a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \frac {2 \, a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a^{3}}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}}{2 \, f} \]

[In]

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

[Out]

1/2*(6*(f*x + e)*a*b^2 + 6*a^2*b*log(abs(tan(1/2*f*x + 1/2*e))) + a^3*tan(1/2*f*x + 1/2*e) - (2*a^2*b*tan(1/2*
f*x + 1/2*e)^3 + a^3*tan(1/2*f*x + 1/2*e)^2 + 2*a^2*b*tan(1/2*f*x + 1/2*e) + 4*b^3*tan(1/2*f*x + 1/2*e) + a^3)
/(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e)))/f

Mupad [B] (verification not implemented)

Time = 6.55 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.85 \[ \int \csc ^2(e+f x) (a+b \sin (e+f x))^3 \, dx=\frac {a^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2\,f}-\frac {a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+a^3+4\,b^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}+\frac {6\,a\,b^2\,\mathrm {atan}\left (\frac {36\,a^2\,b^4}{36\,a^3\,b^3-36\,a^2\,b^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}+\frac {36\,a^3\,b^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{36\,a^3\,b^3-36\,a^2\,b^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f}+\frac {3\,a^2\,b\,\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f} \]

[In]

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

[Out]

(a^3*tan(e/2 + (f*x)/2))/(2*f) - (a^3*tan(e/2 + (f*x)/2)^2 + a^3 + 4*b^3*tan(e/2 + (f*x)/2))/(f*(2*tan(e/2 + (
f*x)/2) + 2*tan(e/2 + (f*x)/2)^3)) + (6*a*b^2*atan((36*a^2*b^4)/(36*a^3*b^3 - 36*a^2*b^4*tan(e/2 + (f*x)/2)) +
(36*a^3*b^3*tan(e/2 + (f*x)/2))/(36*a^3*b^3 - 36*a^2*b^4*tan(e/2 + (f*x)/2))))/f + (3*a^2*b*log(tan(e/2 + (f*
x)/2)))/f