3.173 \(\int \csc ^4(e+f x) (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=109 \[ -\frac{a \left (2 a^2+9 b^2\right ) \cot (e+f x)}{3 f}-\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f} \]

[Out]

-(b*(3*a^2 + 2*b^2)*ArcTanh[Cos[e + f*x]])/(2*f) - (a*(2*a^2 + 9*b^2)*Cot[e + f*x])/(3*f) - (7*a^2*b*Cot[e + f
*x]*Csc[e + f*x])/(6*f) - (a^2*Cot[e + f*x]*Csc[e + f*x]^2*(a + b*Sin[e + f*x]))/(3*f)

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Rubi [A]  time = 0.181392, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2792, 3021, 2748, 3767, 8, 3770} \[ -\frac{a \left (2 a^2+9 b^2\right ) \cot (e+f x)}{3 f}-\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^4*(a + b*Sin[e + f*x])^3,x]

[Out]

-(b*(3*a^2 + 2*b^2)*ArcTanh[Cos[e + f*x]])/(2*f) - (a*(2*a^2 + 9*b^2)*Cot[e + f*x])/(3*f) - (7*a^2*b*Cot[e + f
*x]*Csc[e + f*x])/(6*f) - (a^2*Cot[e + f*x]*Csc[e + f*x]^2*(a + b*Sin[e + f*x]))/(3*f)

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 2748

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 3767

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 ^4(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac{a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}+\frac{1}{3} \int \csc ^3(e+f x) \left (7 a^2 b+a \left (2 a^2+9 b^2\right ) \sin (e+f x)+b \left (a^2+3 b^2\right ) \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}+\frac{1}{6} \int \csc ^2(e+f x) \left (2 a \left (2 a^2+9 b^2\right )+3 b \left (3 a^2+2 b^2\right ) \sin (e+f x)\right ) \, dx\\ &=-\frac{7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}+\frac{1}{2} \left (b \left (3 a^2+2 b^2\right )\right ) \int \csc (e+f x) \, dx+\frac{1}{3} \left (a \left (2 a^2+9 b^2\right )\right ) \int \csc ^2(e+f x) \, dx\\ &=-\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}-\frac{\left (a \left (2 a^2+9 b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (e+f x))}{3 f}\\ &=-\frac{b \left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{2 f}-\frac{a \left (2 a^2+9 b^2\right ) \cot (e+f x)}{3 f}-\frac{7 a^2 b \cot (e+f x) \csc (e+f x)}{6 f}-\frac{a^2 \cot (e+f x) \csc ^2(e+f x) (a+b \sin (e+f x))}{3 f}\\ \end{align*}

Mathematica [B]  time = 6.183, size = 525, normalized size = 4.82 \[ \frac{\sin ^3(e+f x) \csc \left (\frac{1}{2} (e+f x)\right ) \left (-2 a^3 \cos \left (\frac{1}{2} (e+f x)\right )-9 a b^2 \cos \left (\frac{1}{2} (e+f x)\right )\right ) (a \csc (e+f x)+b)^3}{6 f (a+b \sin (e+f x))^3}+\frac{\left (3 a^2 b+2 b^3\right ) \sin ^3(e+f x) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right ) (a \csc (e+f x)+b)^3}{2 f (a+b \sin (e+f x))^3}+\frac{\sin ^3(e+f x) \sec \left (\frac{1}{2} (e+f x)\right ) \left (2 a^3 \sin \left (\frac{1}{2} (e+f x)\right )+9 a b^2 \sin \left (\frac{1}{2} (e+f x)\right )\right ) (a \csc (e+f x)+b)^3}{6 f (a+b \sin (e+f x))^3}+\frac{\left (-3 a^2 b-2 b^3\right ) \sin ^3(e+f x) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right ) (a \csc (e+f x)+b)^3}{2 f (a+b \sin (e+f x))^3}-\frac{3 a^2 b \sin ^3(e+f x) \csc ^2\left (\frac{1}{2} (e+f x)\right ) (a \csc (e+f x)+b)^3}{8 f (a+b \sin (e+f x))^3}-\frac{a^3 \sin ^3(e+f x) \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right ) (a \csc (e+f x)+b)^3}{24 f (a+b \sin (e+f x))^3}+\frac{3 a^2 b \sin ^3(e+f x) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a \csc (e+f x)+b)^3}{8 f (a+b \sin (e+f x))^3}+\frac{a^3 \sin ^3(e+f x) \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right ) (a \csc (e+f x)+b)^3}{24 f (a+b \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^4*(a + b*Sin[e + f*x])^3,x]

[Out]

((-2*a^3*Cos[(e + f*x)/2] - 9*a*b^2*Cos[(e + f*x)/2])*Csc[(e + f*x)/2]*(b + a*Csc[e + f*x])^3*Sin[e + f*x]^3)/
(6*f*(a + b*Sin[e + f*x])^3) - (3*a^2*b*Csc[(e + f*x)/2]^2*(b + a*Csc[e + f*x])^3*Sin[e + f*x]^3)/(8*f*(a + b*
Sin[e + f*x])^3) - (a^3*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2*(b + a*Csc[e + f*x])^3*Sin[e + f*x]^3)/(24*f*(a +
b*Sin[e + f*x])^3) + ((-3*a^2*b - 2*b^3)*(b + a*Csc[e + f*x])^3*Log[Cos[(e + f*x)/2]]*Sin[e + f*x]^3)/(2*f*(a
+ b*Sin[e + f*x])^3) + ((3*a^2*b + 2*b^3)*(b + a*Csc[e + f*x])^3*Log[Sin[(e + f*x)/2]]*Sin[e + f*x]^3)/(2*f*(a
+ b*Sin[e + f*x])^3) + (3*a^2*b*(b + a*Csc[e + f*x])^3*Sec[(e + f*x)/2]^2*Sin[e + f*x]^3)/(8*f*(a + b*Sin[e +
f*x])^3) + ((b + a*Csc[e + f*x])^3*Sec[(e + f*x)/2]*(2*a^3*Sin[(e + f*x)/2] + 9*a*b^2*Sin[(e + f*x)/2])*Sin[e
+ f*x]^3)/(6*f*(a + b*Sin[e + f*x])^3) + (a^3*(b + a*Csc[e + f*x])^3*Sec[(e + f*x)/2]^2*Sin[e + f*x]^3*Tan[(e
+ f*x)/2])/(24*f*(a + b*Sin[e + f*x])^3)

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Maple [A]  time = 0.063, size = 122, normalized size = 1.1 \begin{align*} -{\frac{2\,{a}^{3}\cot \left ( fx+e \right ) }{3\,f}}-{\frac{{a}^{3}\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{3\,f}}-{\frac{3\,{a}^{2}b\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{2\,f}}+{\frac{3\,{a}^{2}b\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,f}}-3\,{\frac{a{b}^{2}\cot \left ( fx+e \right ) }{f}}+{\frac{{b}^{3}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^4*(a+b*sin(f*x+e))^3,x)

[Out]

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

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Maxima [A]  time = 2.64138, size = 159, normalized size = 1.46 \begin{align*} \frac{9 \, a^{2} 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 )} - 6 \, b^{3}{\left (\log \left (\cos \left (f x + e\right ) + 1\right ) - \log \left (\cos \left (f x + e\right ) - 1\right )\right )} - \frac{36 \, a b^{2}}{\tan \left (f x + e\right )} - \frac{4 \,{\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a^{3}}{\tan \left (f x + e\right )^{3}}}{12 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

1/12*(9*a^2*b*(2*cos(f*x + e)/(cos(f*x + e)^2 - 1) - log(cos(f*x + e) + 1) + log(cos(f*x + e) - 1)) - 6*b^3*(l
og(cos(f*x + e) + 1) - log(cos(f*x + e) - 1)) - 36*a*b^2/tan(f*x + e) - 4*(3*tan(f*x + e)^2 + 1)*a^3/tan(f*x +
e)^3)/f

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Fricas [A]  time = 1.72114, size = 471, normalized size = 4.32 \begin{align*} \frac{18 \, a^{2} b \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 4 \,{\left (2 \, a^{3} + 9 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (3 \, a^{2} b + 2 \, b^{3} -{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) - 3 \,{\left (3 \, a^{2} b + 2 \, b^{3} -{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) \sin \left (f x + e\right ) + 12 \,{\left (a^{3} + 3 \, a b^{2}\right )} \cos \left (f x + e\right )}{12 \,{\left (f \cos \left (f x + e\right )^{2} - f\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/12*(18*a^2*b*cos(f*x + e)*sin(f*x + e) - 4*(2*a^3 + 9*a*b^2)*cos(f*x + e)^3 + 3*(3*a^2*b + 2*b^3 - (3*a^2*b
+ 2*b^3)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2)*sin(f*x + e) - 3*(3*a^2*b + 2*b^3 - (3*a^2*b + 2*b^3)*cos
(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2)*sin(f*x + e) + 12*(a^3 + 3*a*b^2)*cos(f*x + e))/((f*cos(f*x + e)^2 -
f)*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**4*(a+b*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 2.11076, size = 271, normalized size = 2.49 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 9 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 9 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 36 \, a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 12 \,{\left (3 \, a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) - \frac{66 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 44 \, b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 9 \, 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} + 9 \, 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 )^{3}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^4*(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/24*(a^3*tan(1/2*f*x + 1/2*e)^3 + 9*a^2*b*tan(1/2*f*x + 1/2*e)^2 + 9*a^3*tan(1/2*f*x + 1/2*e) + 36*a*b^2*tan(
1/2*f*x + 1/2*e) + 12*(3*a^2*b + 2*b^3)*log(abs(tan(1/2*f*x + 1/2*e))) - (66*a^2*b*tan(1/2*f*x + 1/2*e)^3 + 44
*b^3*tan(1/2*f*x + 1/2*e)^3 + 9*a^3*tan(1/2*f*x + 1/2*e)^2 + 36*a*b^2*tan(1/2*f*x + 1/2*e)^2 + 9*a^2*b*tan(1/2
*f*x + 1/2*e) + a^3)/tan(1/2*f*x + 1/2*e)^3)/f