3.1.18 \(\int \csc (2 a+2 b x) \sin ^2(a+b x) \, dx\) [18]

Optimal. Leaf size=14 \[ -\frac {\log (\cos (a+b x))}{2 b} \]

[Out]

-1/2*ln(cos(b*x+a))/b

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Rubi [A]
time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4373, 3556} \begin {gather*} -\frac {\log (\cos (a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[2*a + 2*b*x]*Sin[a + b*x]^2,x]

[Out]

-1/2*Log[Cos[a + b*x]]/b

Rule 3556

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

Rule 4373

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/f^p, Int[Cos[a
+ b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rubi steps

\begin {align*} \int \csc (2 a+2 b x) \sin ^2(a+b x) \, dx &=\frac {1}{2} \int \tan (a+b x) \, dx\\ &=-\frac {\log (\cos (a+b x))}{2 b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} -\frac {\log (\cos (a+b x))}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[2*a + 2*b*x]*Sin[a + b*x]^2,x]

[Out]

-1/2*Log[Cos[a + b*x]]/b

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Maple [A]
time = 0.06, size = 13, normalized size = 0.93

method	result	size
default	\(-\frac {\ln \left (\cos \left (x b +a \right )\right )}{2 b}\)	\(13\)
risch	\(\frac {i x}{2}+\frac {i a}{b}-\frac {\ln \left ({\mathrm e}^{2 i \left (x b +a \right )}+1\right )}{2 b}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(2*b*x+2*a)*sin(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

-1/2*ln(cos(b*x+a))/b

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 55 vs. \(2 (12) = 24\).
time = 0.27, size = 55, normalized size = 3.93 \begin {gather*} -\frac {\log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, a\right ) + \sin \left (2 \, a\right )^{2}\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + cos(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2
)/b

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Fricas [A]
time = 2.92, size = 14, normalized size = 1.00 \begin {gather*} -\frac {\log \left (-\cos \left (b x + a\right )\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(-cos(b*x + a))/b

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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Giac [A]
time = 0.41, size = 18, normalized size = 1.29 \begin {gather*} -\frac {\log \left (-\sin \left (b x + a\right )^{2} + 1\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*log(-sin(b*x + a)^2 + 1)/b

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Mupad [B]
time = 0.08, size = 12, normalized size = 0.86 \begin {gather*} -\frac {\ln \left (\cos \left (a+b\,x\right )\right )}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(cos(a + b*x))/(2*b)

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