\(\int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx\) [156]

Optimal result

Integrand size = 19, antiderivative size = 64 \[ \int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx=-\frac {b \text {arctanh}(\cos (e+f x))}{2 f}-\frac {a \cot (e+f x)}{f}-\frac {a \cot ^3(e+f x)}{3 f}-\frac {b \cot (e+f x) \csc (e+f x)}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2827, 3852, 3853, 3855} \[ \int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx=-\frac {a \cot ^3(e+f x)}{3 f}-\frac {a \cot (e+f x)}{f}-\frac {b \text {arctanh}(\cos (e+f x))}{2 f}-\frac {b \cot (e+f x) \csc (e+f x)}{2 f} \]

[In]

Int[Csc[e + f*x]^4*(a + b*Sin[e + f*x]),x]

[Out]

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

Rule 2827

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.80 \[ \int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx=-\frac {2 a \cot (e+f x)}{3 f}-\frac {b \csc ^2\left (\frac {1}{2} (e+f x)\right )}{8 f}-\frac {a \cot (e+f x) \csc ^2(e+f x)}{3 f}-\frac {b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 f}+\frac {b \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 f}+\frac {b \sec ^2\left (\frac {1}{2} (e+f x)\right )}{8 f} \]

[In]

Integrate[Csc[e + f*x]^4*(a + b*Sin[e + f*x]),x]

[Out]

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

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.95

method	result	size
derivativedivides	\(\frac {a \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (f x +e \right )\right )}{3}\right ) \cot \left (f x +e \right )+b \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{2}\right )}{f}\)	\(61\)
default	\(\frac {a \left (-\frac {2}{3}-\frac {\left (\csc ^{2}\left (f x +e \right )\right )}{3}\right ) \cot \left (f x +e \right )+b \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right )}{2}\right )}{f}\)	\(61\)
risch	\(\frac {3 b \,{\mathrm e}^{5 i \left (f x +e \right )}+12 i a \,{\mathrm e}^{2 i \left (f x +e \right )}-4 i a -3 b \,{\mathrm e}^{i \left (f x +e \right )}}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{2 f}-\frac {b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{2 f}\)	\(98\)
parallelrisch	\(\frac {\left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a -\left (\cot ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) a +3 \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b -3 \left (\cot ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b +9 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-9 \cot \left (\frac {f x}{2}+\frac {e}{2}\right ) a +12 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) b}{24 f}\)	\(99\)
norman	\(\frac {-\frac {a}{24 f}-\frac {5 a \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}+\frac {5 a \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{12 f}+\frac {a \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{24 f}-\frac {b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{8 f}+\frac {b \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}-\frac {b \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}+\frac {b \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}\)	\(152\)

[In]

int(csc(f*x+e)^4*(a+b*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

1/f*(a*(-2/3-1/3*csc(f*x+e)^2)*cot(f*x+e)+b*(-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. 128 vs. \(2 (58) = 116\).

Time = 0.32 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.00 \[ \int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx=-\frac {8 \, a \cos \left (f x + e\right )^{3} - 6 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 3 \, {\left (b \cos \left (f x + e\right )^{2} - b\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) - 3 \, {\left (b \cos \left (f x + e\right )^{2} - b\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) \sin \left (f x + e\right ) - 12 \, a \cos \left (f x + e\right )}{12 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )} \sin \left (f x + e\right )} \]

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

-1/12*(8*a*cos(f*x + e)^3 - 6*b*cos(f*x + e)*sin(f*x + e) + 3*(b*cos(f*x + e)^2 - b)*log(1/2*cos(f*x + e) + 1/
2)*sin(f*x + e) - 3*(b*cos(f*x + e)^2 - b)*log(-1/2*cos(f*x + e) + 1/2)*sin(f*x + e) - 12*a*cos(f*x + e))/((f*
cos(f*x + e)^2 - f)*sin(f*x + e))

Sympy [F]

\[ \int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx=\int \left (a + b \sin {\left (e + f x \right )}\right ) \csc ^{4}{\left (e + f x \right )}\, dx \]

[In]

integrate(csc(f*x+e)**4*(a+b*sin(f*x+e)),x)

[Out]

Integral((a + b*sin(e + f*x))*csc(e + f*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx=\frac {3 \, b {\left (\frac {2 \, \cos \left (f x + e\right )}{\cos \left (f x + e\right )^{2} - 1} - \log \left (\cos \left (f x + e\right ) + 1\right ) + \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - \frac {4 \, {\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a}{\tan \left (f x + e\right )^{3}}}{12 \, f} \]

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.78 \[ \int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx=\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) + 9 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \frac {22 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 9 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3}}}{24 \, f} \]

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

1/24*(a*tan(1/2*f*x + 1/2*e)^3 + 3*b*tan(1/2*f*x + 1/2*e)^2 + 12*b*log(abs(tan(1/2*f*x + 1/2*e))) + 9*a*tan(1/
2*f*x + 1/2*e) - (22*b*tan(1/2*f*x + 1/2*e)^3 + 9*a*tan(1/2*f*x + 1/2*e)^2 + 3*b*tan(1/2*f*x + 1/2*e) + a)/tan
(1/2*f*x + 1/2*e)^3)/f

Mupad [B] (verification not implemented)

Time = 6.33 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.73 \[ \int \csc ^4(e+f x) (a+b \sin (e+f x)) \, dx=\frac {3\,a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8\,f}+\frac {a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{24\,f}+\frac {b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{8\,f}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{2\,f}-\frac {{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+\frac {a}{3}\right )}{8\,f} \]

[In]

int((a + b*sin(e + f*x))/sin(e + f*x)^4,x)

[Out]

(3*a*tan(e/2 + (f*x)/2))/(8*f) + (a*tan(e/2 + (f*x)/2)^3)/(24*f) + (b*tan(e/2 + (f*x)/2)^2)/(8*f) + (b*log(tan
(e/2 + (f*x)/2)))/(2*f) - (cot(e/2 + (f*x)/2)^3*(a/3 + b*tan(e/2 + (f*x)/2) + 3*a*tan(e/2 + (f*x)/2)^2))/(8*f)