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

Optimal. Leaf size=35 \[ -\frac {a^2 \tanh ^{-1}(\cos (e+f x))}{f}+2 a b x-\frac {b^2 \cos (e+f x)}{f} \]

[Out]

2*a*b*x-a^2*arctanh(cos(f*x+e))/f-b^2*cos(f*x+e)/f

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Rubi [A]  time = 0.06, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2746, 2735, 3770} \[ -\frac {a^2 \tanh ^{-1}(\cos (e+f x))}{f}+2 a b x-\frac {b^2 \cos (e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*b*x - (a^2*ArcTanh[Cos[e + f*x]])/f - (b^2*Cos[e + f*x])/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 2746

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b^2
*Cos[e + f*x])/(d*f), x] + Dist[1/d, Int[Simp[a^2*d - b*(b*c - 2*a*d)*Sin[e + f*x], x]/(c + d*Sin[e + f*x]), x
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 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 (e+f x) (a+b \sin (e+f x))^2 \, dx &=-\frac {b^2 \cos (e+f x)}{f}+\int \csc (e+f x) \left (a^2+2 a b \sin (e+f x)\right ) \, dx\\ &=2 a b x-\frac {b^2 \cos (e+f x)}{f}+a^2 \int \csc (e+f x) \, dx\\ &=2 a b x-\frac {a^2 \tanh ^{-1}(\cos (e+f x))}{f}-\frac {b^2 \cos (e+f x)}{f}\\ \end {align*}

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Mathematica [B]  time = 0.02, size = 76, normalized size = 2.17 \[ \frac {a^2 \log \left (\sin \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f}-\frac {a^2 \log \left (\cos \left (\frac {e}{2}+\frac {f x}{2}\right )\right )}{f}+2 a b x+\frac {b^2 \sin (e) \sin (f x)}{f}-\frac {b^2 \cos (e) \cos (f x)}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*a*b*x - (b^2*Cos[e]*Cos[f*x])/f - (a^2*Log[Cos[e/2 + (f*x)/2]])/f + (a^2*Log[Sin[e/2 + (f*x)/2]])/f + (b^2*S
in[e]*Sin[f*x])/f

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fricas [A]  time = 0.49, size = 54, normalized size = 1.54 \[ \frac {4 \, a b f x - 2 \, b^{2} \cos \left (f x + e\right ) - a^{2} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + a^{2} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right )}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

1/2*(4*a*b*f*x - 2*b^2*cos(f*x + e) - a^2*log(1/2*cos(f*x + e) + 1/2) + a^2*log(-1/2*cos(f*x + e) + 1/2))/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)*(a+b*sin(f*x+e))^2,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*(a^2/2*ln(abs(tan((f*x+exp(1))/2)))+4*a*b/2*(f*x+exp(1))/2-b^2/(tan((f*x+exp(1))/2)^2+1))

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maple [A]  time = 0.20, size = 52, normalized size = 1.49 \[ 2 a b x +\frac {a^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}-\frac {b^{2} \cos \left (f x +e \right )}{f}+\frac {2 a b e}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)*(a+b*sin(f*x+e))^2,x)

[Out]

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

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maxima [A]  time = 0.64, size = 44, normalized size = 1.26 \[ \frac {2 \, {\left (f x + e\right )} a b - b^{2} \cos \left (f x + e\right ) - a^{2} \log \left (\cot \left (f x + e\right ) + \csc \left (f x + e\right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

(2*(f*x + e)*a*b - b^2*cos(f*x + e) - a^2*log(cot(f*x + e) + csc(f*x + e)))/f

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mupad [B]  time = 6.48, size = 125, normalized size = 3.57 \[ \frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{f}-\frac {2\,b^2}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {4\,a\,b\,\mathrm {atan}\left (\frac {16\,a^2\,b^2}{8\,a^3\,b-16\,a^2\,b^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}+\frac {8\,a^3\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{8\,a^3\,b-16\,a^2\,b^2\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x))^2/sin(e + f*x),x)

[Out]

(a^2*log(tan(e/2 + (f*x)/2)))/f - (2*b^2)/(f*(tan(e/2 + (f*x)/2)^2 + 1)) + (4*a*b*atan((16*a^2*b^2)/(8*a^3*b -
16*a^2*b^2*tan(e/2 + (f*x)/2)) + (8*a^3*b*tan(e/2 + (f*x)/2))/(8*a^3*b - 16*a^2*b^2*tan(e/2 + (f*x)/2))))/f

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (e + f x \right )}\right )^{2} \csc {\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*sin(f*x+e))**2,x)

[Out]

Integral((a + b*sin(e + f*x))**2*csc(e + f*x), x)

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