3.2.72 \(\int \csc ^3(e+f x) (a+b \sin (e+f x))^3 \, dx\) [172]

Optimal. Leaf size=79 \[ b^3 x-\frac {a \left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac {5 a^2 b \cot (e+f x)}{2 f}-\frac {a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f} \]

[Out]

b^3*x-1/2*a*(a^2+6*b^2)*arctanh(cos(f*x+e))/f-5/2*a^2*b*cot(f*x+e)/f-1/2*a^2*cot(f*x+e)*csc(f*x+e)*(a+b*sin(f*
x+e))/f

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Rubi [A]
time = 0.10, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2871, 3100, 2814, 3855} \begin {gather*} -\frac {a \left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac {5 a^2 b \cot (e+f x)}{2 f}-\frac {a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+b^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

b^3*x - (a*(a^2 + 6*b^2)*ArcTanh[Cos[e + f*x]])/(2*f) - (5*a^2*b*Cot[e + f*x])/(2*f) - (a^2*Cot[e + f*x]*Csc[e
+ f*x]*(a + b*Sin[e + f*x]))/(2*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 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

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*} \int \csc ^3(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac {a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+\frac {1}{2} \int \csc ^2(e+f x) \left (5 a^2 b+a \left (a^2+6 b^2\right ) \sin (e+f x)+2 b^3 \sin ^2(e+f x)\right ) \, dx\\ &=-\frac {5 a^2 b \cot (e+f x)}{2 f}-\frac {a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+\frac {1}{2} \int \csc (e+f x) \left (a \left (a^2+6 b^2\right )+2 b^3 \sin (e+f x)\right ) \, dx\\ &=b^3 x-\frac {5 a^2 b \cot (e+f x)}{2 f}-\frac {a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+\frac {1}{2} \left (a \left (a^2+6 b^2\right )\right ) \int \csc (e+f x) \, dx\\ &=b^3 x-\frac {a \left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac {5 a^2 b \cot (e+f x)}{2 f}-\frac {a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}\\ \end {align*}

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Mathematica [A]
time = 0.45, size = 152, normalized size = 1.92 \begin {gather*} \frac {8 b^3 e+8 b^3 f x-12 a^2 b \cot \left (\frac {1}{2} (e+f x)\right )-a^3 \csc ^2\left (\frac {1}{2} (e+f x)\right )-4 a^3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )-24 a b^2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+4 a^3 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )+24 a b^2 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )+a^3 \sec ^2\left (\frac {1}{2} (e+f x)\right )+12 a^2 b \tan \left (\frac {1}{2} (e+f x)\right )}{8 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b^3*e + 8*b^3*f*x - 12*a^2*b*Cot[(e + f*x)/2] - a^3*Csc[(e + f*x)/2]^2 - 4*a^3*Log[Cos[(e + f*x)/2]] - 24*a
*b^2*Log[Cos[(e + f*x)/2]] + 4*a^3*Log[Sin[(e + f*x)/2]] + 24*a*b^2*Log[Sin[(e + f*x)/2]] + a^3*Sec[(e + f*x)/
2]^2 + 12*a^2*b*Tan[(e + f*x)/2])/(8*f)

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Maple [A]
time = 0.38, size = 86, normalized size = 1.09

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )-3 a^{2} b \cot \left (f x +e \right )+3 a \,b^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )+b^{3} \left (f x +e \right )}{f}\)	\(86\)
default	\(\frac {a^{3} \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )-3 a^{2} b \cot \left (f x +e \right )+3 a \,b^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )+b^{3} \left (f x +e \right )}{f}\)	\(86\)
risch	\(b^{3} x -\frac {i a^{2} \left (i a \,{\mathrm e}^{3 i \left (f x +e \right )}+i a \,{\mathrm e}^{i \left (f x +e \right )}+6 b \,{\mathrm e}^{2 i \left (f x +e \right )}-6 b \right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{2 f}+\frac {3 a \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b^{2}}{f}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{2 f}-\frac {3 a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b^{2}}{f}\)	\(153\)
norman	\(\frac {b^{3} x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+b^{3} x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {a^{3}}{8 f}+\frac {a^{3} \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}+3 b^{3} x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+3 b^{3} x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {3 a^{3} \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4 f}-\frac {7 a^{3} \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}-\frac {11 a^{3} \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{8 f}-\frac {3 a^{2} b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 f}-\frac {3 a^{2} b \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a^{2} b \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}+\frac {3 a^{2} b \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3}}+\frac {a \left (a^{2}+6 b^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{2 f}\)	\(283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 110, normalized size = 1.39 \begin {gather*} \frac {4 \, {\left (f x + e\right )} b^{3} + a^{3} {\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 )} - 6 \, a b^{2} {\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - \frac {12 \, a^{2} b}{\tan \left (f x + e\right )}}{4 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (78) = 156\).
time = 0.52, size = 164, normalized size = 2.08 \begin {gather*} \frac {4 \, b^{3} f x \cos \left (f x + e\right )^{2} - 4 \, b^{3} f x + 12 \, a^{2} b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, a^{3} \cos \left (f x + e\right ) + {\left (a^{3} + 6 \, a b^{2} - {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - {\left (a^{3} + 6 \, a b^{2} - {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*b^3*f*x*cos(f*x + e)^2 - 4*b^3*f*x + 12*a^2*b*cos(f*x + e)*sin(f*x + e) + 2*a^3*cos(f*x + e) + (a^3 + 6
*a*b^2 - (a^3 + 6*a*b^2)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) - (a^3 + 6*a*b^2 - (a^3 + 6*a*b^2)*cos(f*
x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/(f*cos(f*x + e)^2 - f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (e + f x \right )}\right )^{3} \csc ^{3}{\left (e + f x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.46, size = 142, normalized size = 1.80 \begin {gather*} \frac {a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 8 \, {\left (f x + e\right )} b^{3} + 12 \, a^{2} b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, {\left (a^{3} + 6 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}\right ) - \frac {6 \, a^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 36 \, a b^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 12 \, a^{2} b \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 )^{2}}}{8 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a^3*tan(1/2*f*x + 1/2*e)^2 + 8*(f*x + e)*b^3 + 12*a^2*b*tan(1/2*f*x + 1/2*e) + 4*(a^3 + 6*a*b^2)*log(abs(
tan(1/2*f*x + 1/2*e))) - (6*a^3*tan(1/2*f*x + 1/2*e)^2 + 36*a*b^2*tan(1/2*f*x + 1/2*e)^2 + 12*a^2*b*tan(1/2*f*
x + 1/2*e) + a^3)/tan(1/2*f*x + 1/2*e)^2)/f

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Mupad [B]
time = 6.96, size = 234, normalized size = 2.96 \begin {gather*} \frac {2\,b^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^3+6\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a\,b^2+2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,b^3}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a^3+6\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,a\,b^2-2\,\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,b^3}\right )}{f}-\frac {a^3\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{8\,f}+\frac {a^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2}{8\,f}+\frac {a^3\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{2\,f}-\frac {3\,a^2\,b\,\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2\,f}+\frac {3\,a\,b^2\,\ln \left (\frac {\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f}+\frac {3\,a^2\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{2\,f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^3*atan((2*b^3*cos(e/2 + (f*x)/2) + a^3*sin(e/2 + (f*x)/2) + 6*a*b^2*sin(e/2 + (f*x)/2))/(a^3*cos(e/2 + (f
*x)/2) - 2*b^3*sin(e/2 + (f*x)/2) + 6*a*b^2*cos(e/2 + (f*x)/2))))/f - (a^3*cot(e/2 + (f*x)/2)^2)/(8*f) + (a^3*
tan(e/2 + (f*x)/2)^2)/(8*f) + (a^3*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/(2*f) - (3*a^2*b*cot(e/2 + (f*x
)/2))/(2*f) + (3*a*b^2*log(sin(e/2 + (f*x)/2)/cos(e/2 + (f*x)/2)))/f + (3*a^2*b*tan(e/2 + (f*x)/2))/(2*f)

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