3.174 \(\int \csc ^5(e+f x) (a+b \sin (e+f x))^3 \, dx\)

Optimal. Leaf size=134 \[ -\frac{b \left (2 a^2+b^2\right ) \cot (e+f x)}{f}-\frac{3 a \left (a^2+4 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f}-\frac{3 a \left (a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac{3 a^2 b \cot (e+f x) \csc ^2(e+f x)}{4 f}-\frac{a^2 \cot (e+f x) \csc ^3(e+f x) (a+b \sin (e+f x))}{4 f} \]

[Out]

(-3*a*(a^2 + 4*b^2)*ArcTanh[Cos[e + f*x]])/(8*f) - (b*(2*a^2 + b^2)*Cot[e + f*x])/f - (3*a*(a^2 + 4*b^2)*Cot[e
+ f*x]*Csc[e + f*x])/(8*f) - (3*a^2*b*Cot[e + f*x]*Csc[e + f*x]^2)/(4*f) - (a^2*Cot[e + f*x]*Csc[e + f*x]^3*(
a + b*Sin[e + f*x]))/(4*f)

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Rubi [A]  time = 0.205392, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2792, 3021, 2748, 3768, 3770, 3767, 8} \[ -\frac{b \left (2 a^2+b^2\right ) \cot (e+f x)}{f}-\frac{3 a \left (a^2+4 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f}-\frac{3 a \left (a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac{3 a^2 b \cot (e+f x) \csc ^2(e+f x)}{4 f}-\frac{a^2 \cot (e+f x) \csc ^3(e+f x) (a+b \sin (e+f x))}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5*(a + b*Sin[e + f*x])^3,x]

[Out]

(-3*a*(a^2 + 4*b^2)*ArcTanh[Cos[e + f*x]])/(8*f) - (b*(2*a^2 + b^2)*Cot[e + f*x])/f - (3*a*(a^2 + 4*b^2)*Cot[e
+ f*x]*Csc[e + f*x])/(8*f) - (3*a^2*b*Cot[e + f*x]*Csc[e + f*x]^2)/(4*f) - (a^2*Cot[e + f*x]*Csc[e + f*x]^3*(
a + b*Sin[e + f*x]))/(4*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 3768

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, 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]

Rubi steps

\begin{align*} \int \csc ^5(e+f x) (a+b \sin (e+f x))^3 \, dx &=-\frac{a^2 \cot (e+f x) \csc ^3(e+f x) (a+b \sin (e+f x))}{4 f}+\frac{1}{4} \int \csc ^4(e+f x) \left (9 a^2 b+3 a \left (a^2+4 b^2\right ) \sin (e+f x)+2 b \left (a^2+2 b^2\right ) \sin ^2(e+f x)\right ) \, dx\\ &=-\frac{3 a^2 b \cot (e+f x) \csc ^2(e+f x)}{4 f}-\frac{a^2 \cot (e+f x) \csc ^3(e+f x) (a+b \sin (e+f x))}{4 f}+\frac{1}{12} \int \csc ^3(e+f x) \left (9 a \left (a^2+4 b^2\right )+12 b \left (2 a^2+b^2\right ) \sin (e+f x)\right ) \, dx\\ &=-\frac{3 a^2 b \cot (e+f x) \csc ^2(e+f x)}{4 f}-\frac{a^2 \cot (e+f x) \csc ^3(e+f x) (a+b \sin (e+f x))}{4 f}+\left (b \left (2 a^2+b^2\right )\right ) \int \csc ^2(e+f x) \, dx+\frac{1}{4} \left (3 a \left (a^2+4 b^2\right )\right ) \int \csc ^3(e+f x) \, dx\\ &=-\frac{3 a \left (a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac{3 a^2 b \cot (e+f x) \csc ^2(e+f x)}{4 f}-\frac{a^2 \cot (e+f x) \csc ^3(e+f x) (a+b \sin (e+f x))}{4 f}+\frac{1}{8} \left (3 a \left (a^2+4 b^2\right )\right ) \int \csc (e+f x) \, dx-\frac{\left (b \left (2 a^2+b^2\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,\cot (e+f x))}{f}\\ &=-\frac{3 a \left (a^2+4 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f}-\frac{b \left (2 a^2+b^2\right ) \cot (e+f x)}{f}-\frac{3 a \left (a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac{3 a^2 b \cot (e+f x) \csc ^2(e+f x)}{4 f}-\frac{a^2 \cot (e+f x) \csc ^3(e+f x) (a+b \sin (e+f x))}{4 f}\\ \end{align*}

Mathematica [B]  time = 6.17099, size = 322, normalized size = 2.4 \[ -\frac{3 \left (a^3+4 a b^2\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )}{32 f}+\frac{3 \left (a^3+4 a b^2\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{32 f}+\frac{3 \left (a^3+4 a b^2\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{8 f}-\frac{3 \left (a^3+4 a b^2\right ) \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{8 f}+\frac{\csc \left (\frac{1}{2} (e+f x)\right ) \left (b^3 \left (-\cos \left (\frac{1}{2} (e+f x)\right )\right )-2 a^2 b \cos \left (\frac{1}{2} (e+f x)\right )\right )}{2 f}+\frac{\sec \left (\frac{1}{2} (e+f x)\right ) \left (2 a^2 b \sin \left (\frac{1}{2} (e+f x)\right )+b^3 \sin \left (\frac{1}{2} (e+f x)\right )\right )}{2 f}-\frac{a^2 b \cot \left (\frac{1}{2} (e+f x)\right ) \csc ^2\left (\frac{1}{2} (e+f x)\right )}{8 f}+\frac{a^2 b \tan \left (\frac{1}{2} (e+f x)\right ) \sec ^2\left (\frac{1}{2} (e+f x)\right )}{8 f}-\frac{a^3 \csc ^4\left (\frac{1}{2} (e+f x)\right )}{64 f}+\frac{a^3 \sec ^4\left (\frac{1}{2} (e+f x)\right )}{64 f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^5*(a + b*Sin[e + f*x])^3,x]

[Out]

((-2*a^2*b*Cos[(e + f*x)/2] - b^3*Cos[(e + f*x)/2])*Csc[(e + f*x)/2])/(2*f) - (3*(a^3 + 4*a*b^2)*Csc[(e + f*x)
/2]^2)/(32*f) - (a^2*b*Cot[(e + f*x)/2]*Csc[(e + f*x)/2]^2)/(8*f) - (a^3*Csc[(e + f*x)/2]^4)/(64*f) - (3*(a^3
+ 4*a*b^2)*Log[Cos[(e + f*x)/2]])/(8*f) + (3*(a^3 + 4*a*b^2)*Log[Sin[(e + f*x)/2]])/(8*f) + (3*(a^3 + 4*a*b^2)
*Sec[(e + f*x)/2]^2)/(32*f) + (a^3*Sec[(e + f*x)/2]^4)/(64*f) + (Sec[(e + f*x)/2]*(2*a^2*b*Sin[(e + f*x)/2] +
b^3*Sin[(e + f*x)/2]))/(2*f) + (a^2*b*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(8*f)

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Maple [A]  time = 0.067, size = 166, normalized size = 1.2 \begin{align*} -{\frac{{a}^{3}\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{3}}{4\,f}}-{\frac{3\,{a}^{3}\csc \left ( fx+e \right ) \cot \left ( fx+e \right ) }{8\,f}}+{\frac{3\,{a}^{3}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{8\,f}}-2\,{\frac{{a}^{2}b\cot \left ( fx+e \right ) }{f}}-{\frac{{a}^{2}b\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{f}}-{\frac{3\,a{b}^{2}\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{2\,f}}+{\frac{3\,a{b}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,f}}-{\frac{{b}^{3}\cot \left ( fx+e \right ) }{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5*(a+b*sin(f*x+e))^3,x)

[Out]

-1/4/f*a^3*cot(f*x+e)*csc(f*x+e)^3-3/8/f*a^3*csc(f*x+e)*cot(f*x+e)+3/8/f*a^3*ln(csc(f*x+e)-cot(f*x+e))-2*a^2*b
*cot(f*x+e)/f-a^2*b*cot(f*x+e)*csc(f*x+e)^2/f-3/2/f*a*b^2*cot(f*x+e)*csc(f*x+e)+3/2/f*a*b^2*ln(csc(f*x+e)-cot(
f*x+e))-1/f*b^3*cot(f*x+e)

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Maxima [A]  time = 1.65598, size = 219, normalized size = 1.63 \begin{align*} \frac{a^{3}{\left (\frac{2 \,{\left (3 \, \cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )\right )}}{\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1} - 3 \, \log \left (\cos \left (f x + e\right ) + 1\right ) + 3 \, \log \left (\cos \left (f x + e\right ) - 1\right )\right )} + 12 \, a b^{2}{\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{16 \, b^{3}}{\tan \left (f x + e\right )} - \frac{16 \,{\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a^{2} b}{\tan \left (f x + e\right )^{3}}}{16 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

1/16*(a^3*(2*(3*cos(f*x + e)^3 - 5*cos(f*x + e))/(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1) - 3*log(cos(f*x + e)
+ 1) + 3*log(cos(f*x + e) - 1)) + 12*a*b^2*(2*cos(f*x + e)/(cos(f*x + e)^2 - 1) - log(cos(f*x + e) + 1) + log(
cos(f*x + e) - 1)) - 16*b^3/tan(f*x + e) - 16*(3*tan(f*x + e)^2 + 1)*a^2*b/tan(f*x + e)^3)/f

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Fricas [A]  time = 1.79595, size = 589, normalized size = 4.4 \begin{align*} \frac{6 \,{\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 2 \,{\left (5 \, a^{3} + 12 \, a b^{2}\right )} \cos \left (f x + e\right ) - 3 \,{\left ({\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{3} + 4 \, a b^{2} - 2 \,{\left (a^{3} + 4 \, 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 ) + 3 \,{\left ({\left (a^{3} + 4 \, a b^{2}\right )} \cos \left (f x + e\right )^{4} + a^{3} + 4 \, a b^{2} - 2 \,{\left (a^{3} + 4 \, 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 ) + 16 \,{\left ({\left (2 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right )^{3} -{\left (3 \, a^{2} b + b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \,{\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/16*(6*(a^3 + 4*a*b^2)*cos(f*x + e)^3 - 2*(5*a^3 + 12*a*b^2)*cos(f*x + e) - 3*((a^3 + 4*a*b^2)*cos(f*x + e)^4
+ a^3 + 4*a*b^2 - 2*(a^3 + 4*a*b^2)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 3*((a^3 + 4*a*b^2)*cos(f*x
+ e)^4 + a^3 + 4*a*b^2 - 2*(a^3 + 4*a*b^2)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2) + 16*((2*a^2*b + b^3)*
cos(f*x + e)^3 - (3*a^2*b + b^3)*cos(f*x + e))*sin(f*x + e))/(f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f)

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

[Out]

Timed out

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Giac [B]  time = 1.90712, size = 363, normalized size = 2.71 \begin{align*} \frac{a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 8 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 8 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 24 \, a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 72 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 32 \, b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 24 \,{\left (a^{3} + 4 \, a b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) - \frac{50 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 200 \, a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 72 \, a^{2} b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 32 \, b^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 8 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 24 \, a b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 8 \, 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 )^{4}}}{64 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

1/64*(a^3*tan(1/2*f*x + 1/2*e)^4 + 8*a^2*b*tan(1/2*f*x + 1/2*e)^3 + 8*a^3*tan(1/2*f*x + 1/2*e)^2 + 24*a*b^2*ta
n(1/2*f*x + 1/2*e)^2 + 72*a^2*b*tan(1/2*f*x + 1/2*e) + 32*b^3*tan(1/2*f*x + 1/2*e) + 24*(a^3 + 4*a*b^2)*log(ab
s(tan(1/2*f*x + 1/2*e))) - (50*a^3*tan(1/2*f*x + 1/2*e)^4 + 200*a*b^2*tan(1/2*f*x + 1/2*e)^4 + 72*a^2*b*tan(1/
2*f*x + 1/2*e)^3 + 32*b^3*tan(1/2*f*x + 1/2*e)^3 + 8*a^3*tan(1/2*f*x + 1/2*e)^2 + 24*a*b^2*tan(1/2*f*x + 1/2*e
)^2 + 8*a^2*b*tan(1/2*f*x + 1/2*e) + a^3)/tan(1/2*f*x + 1/2*e)^4)/f