3.172 \(\int \csc ^3(e+f x) (a+b \sin (e+f x))^3 \, dx\)

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

[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 = 0.13, 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 = {2792, 3021, 2735, 3770} \[ -\frac {a \left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac {5 a^2 b \cot (e+f x)}{2 f}-\frac {a^2 \cot (e+f x) \csc (e+f x) (a+b \sin (e+f x))}{2 f}+b^3 x \]

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 2735

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 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((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 3021

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 3770

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 = 0.68, size = 152, normalized size = 1.92 \[ \frac {a^3 \left (-\csc ^2\left (\frac {1}{2} (e+f x)\right )\right )+a^3 \sec ^2\left (\frac {1}{2} (e+f x)\right )+4 a^3 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-4 a^3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+12 a^2 b \tan \left (\frac {1}{2} (e+f x)\right )-12 a^2 b \cot \left (\frac {1}{2} (e+f x)\right )+24 a b^2 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-24 a b^2 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+8 b^3 e+8 b^3 f x}{8 f} \]

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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fricas [B]  time = 0.50, size = 155, normalized size = 1.96 \[ \frac {4 \, b^{3} f x \cos \left (f x + e\right )^{2} - 4 \, b^{3} f x + 12 \, a^{2} b \cos \left (f x + e\right ) \sin \left (f x + e\right ) + 2 \, a^{3} \cos \left (f x + e\right ) + {\left (a^{3} + 6 \, a b^{2} - {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - {\left (a^{3} + 6 \, a b^{2} - {\left (a^{3} + 6 \, a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{4 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}} \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)2/f*((4*tan((f*x+exp(1))/2)^2*a^3+48*tan((f*x+exp(1))/2)*b*a^2)/64+(-36*tan((f*x+exp(1))/2)^2*b^2*a-6*tan(
(f*x+exp(1))/2)^2*a^3-12*tan((f*x+exp(1))/2)*b*a^2-a^3)*1/16/tan((f*x+exp(1))/2)^2+2*b^3/2*(f*x+exp(1))/2+(6*b
^2*a+a^3)/4*ln(abs(tan((f*x+exp(1))/2))))

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maple [A]  time = 0.36, size = 99, normalized size = 1.25 \[ -\frac {a^{3} \csc \left (f x +e \right ) \cot \left (f x +e \right )}{2 f}+\frac {a^{3} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2 f}-\frac {3 a^{2} b \cot \left (f x +e \right )}{f}+\frac {3 a \,b^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}+b^{3} x +\frac {b^{3} e}{f} \]

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)

[Out]

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

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maxima [A]  time = 1.28, size = 102, normalized size = 1.29 \[ \frac {4 \, {\left (f x + e\right )} b^{3} + a^{3} {\left (\frac {2 \, \cos \left (f x + e\right )}{\cos \left (f x + e\right )^{2} - 1} - \log \left (\cos \left (f x + e\right ) + 1\right ) + \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - 6 \, a b^{2} {\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - \frac {12 \, a^{2} b}{\tan \left (f x + e\right )}}{4 \, f} \]

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

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

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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