3.165 \(\int \csc ^5(e+f x) (a+b \sin (e+f x))^2 \, dx\)

Optimal. Leaf size=110 \[ -\frac{\left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f}-\frac{\left (3 a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac{a^2 \cot (e+f x) \csc ^3(e+f x)}{4 f}-\frac{2 a b \cot ^3(e+f x)}{3 f}-\frac{2 a b \cot (e+f x)}{f} \]

[Out]

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

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Rubi [A]  time = 0.0939383, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2789, 3767, 3012, 3768, 3770} \[ -\frac{\left (3 a^2+4 b^2\right ) \tanh ^{-1}(\cos (e+f x))}{8 f}-\frac{\left (3 a^2+4 b^2\right ) \cot (e+f x) \csc (e+f x)}{8 f}-\frac{a^2 \cot (e+f x) \csc ^3(e+f x)}{4 f}-\frac{2 a b \cot ^3(e+f x)}{3 f}-\frac{2 a b \cot (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5*(a + b*Sin[e + f*x])^2,x]

[Out]

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

Rule 2789

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Dist[(2*c*d)/b
, Int[(b*Sin[e + f*x])^(m + 1), x], x] + Int[(b*Sin[e + f*x])^m*(c^2 + d^2*Sin[e + f*x]^2), 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 3012

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

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]

Rubi steps

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

Mathematica [B]  time = 0.0412662, size = 255, normalized size = 2.32 \[ -\frac{a^2 \csc ^4\left (\frac{1}{2} (e+f x)\right )}{64 f}-\frac{3 a^2 \csc ^2\left (\frac{1}{2} (e+f x)\right )}{32 f}+\frac{a^2 \sec ^4\left (\frac{1}{2} (e+f x)\right )}{64 f}+\frac{3 a^2 \sec ^2\left (\frac{1}{2} (e+f x)\right )}{32 f}+\frac{3 a^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{8 f}-\frac{3 a^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{8 f}-\frac{4 a b \cot (e+f x)}{3 f}-\frac{2 a b \cot (e+f x) \csc ^2(e+f x)}{3 f}-\frac{b^2 \csc ^2\left (\frac{1}{2} (e+f x)\right )}{8 f}+\frac{b^2 \sec ^2\left (\frac{1}{2} (e+f x)\right )}{8 f}+\frac{b^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{2 f}-\frac{b^2 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{2 f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^5*(a + b*Sin[e + f*x])^2,x]

[Out]

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

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Maple [A]  time = 0.111, size = 146, normalized size = 1.3 \begin{align*} -{\frac{{a}^{2}\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{3}}{4\,f}}-{\frac{3\,{a}^{2}\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{8\,f}}+{\frac{3\,{a}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{8\,f}}-{\frac{4\,ab\cot \left ( fx+e \right ) }{3\,f}}-{\frac{2\,ab\cot \left ( fx+e \right ) \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{3\,f}}-{\frac{{b}^{2}\cot \left ( fx+e \right ) \csc \left ( fx+e \right ) }{2\,f}}+{\frac{{b}^{2}\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{2\,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))^2,x)

[Out]

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

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Maxima [A]  time = 1.99729, size = 198, normalized size = 1.8 \begin{align*} \frac{3 \, a^{2}{\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 \, 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{32 \,{\left (3 \, \tan \left (f x + e\right )^{2} + 1\right )} a b}{\tan \left (f x + e\right )^{3}}}{48 \, 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))^2,x, algorithm="maxima")

[Out]

1/48*(3*a^2*(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*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)) - 32*(3*tan(f*x + e)^2 + 1)*a*b/tan(f*x + e)^3)/f

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Fricas [B]  time = 1.66156, size = 558, normalized size = 5.07 \begin{align*} \frac{6 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 6 \,{\left (5 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right ) - 3 \,{\left ({\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 4 \, b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) + 3 \,{\left ({\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 4 \, b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) + 32 \,{\left (2 \, a b \cos \left (f x + e\right )^{3} - 3 \, a b \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \,{\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))^2,x, algorithm="fricas")

[Out]

1/48*(6*(3*a^2 + 4*b^2)*cos(f*x + e)^3 - 6*(5*a^2 + 4*b^2)*cos(f*x + e) - 3*((3*a^2 + 4*b^2)*cos(f*x + e)^4 -
2*(3*a^2 + 4*b^2)*cos(f*x + e)^2 + 3*a^2 + 4*b^2)*log(1/2*cos(f*x + e) + 1/2) + 3*((3*a^2 + 4*b^2)*cos(f*x + e
)^4 - 2*(3*a^2 + 4*b^2)*cos(f*x + e)^2 + 3*a^2 + 4*b^2)*log(-1/2*cos(f*x + e) + 1/2) + 32*(2*a*b*cos(f*x + e)^
3 - 3*a*b*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))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.82501, size = 311, normalized size = 2.83 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 16 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 24 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 24 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 144 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 24 \,{\left (3 \, a^{2} + 4 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) \right |}\right ) - \frac{150 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 200 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 144 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 24 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 24 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 16 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, a^{2}}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4}}}{192 \, 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))^2,x, algorithm="giac")

[Out]

1/192*(3*a^2*tan(1/2*f*x + 1/2*e)^4 + 16*a*b*tan(1/2*f*x + 1/2*e)^3 + 24*a^2*tan(1/2*f*x + 1/2*e)^2 + 24*b^2*t
an(1/2*f*x + 1/2*e)^2 + 144*a*b*tan(1/2*f*x + 1/2*e) + 24*(3*a^2 + 4*b^2)*log(abs(tan(1/2*f*x + 1/2*e))) - (15
0*a^2*tan(1/2*f*x + 1/2*e)^4 + 200*b^2*tan(1/2*f*x + 1/2*e)^4 + 144*a*b*tan(1/2*f*x + 1/2*e)^3 + 24*a^2*tan(1/
2*f*x + 1/2*e)^2 + 24*b^2*tan(1/2*f*x + 1/2*e)^2 + 16*a*b*tan(1/2*f*x + 1/2*e) + 3*a^2)/tan(1/2*f*x + 1/2*e)^4
)/f