3.2.40 \(\int \csc ^2(a+b x) \sec (a+b x) \, dx\) [140]

Optimal. Leaf size=23 \[ \frac {\tanh ^{-1}(\sin (a+b x))}{b}-\frac {\csc (a+b x)}{b} \]

[Out]

arctanh(sin(b*x+a))/b-csc(b*x+a)/b

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Rubi [A]
time = 0.01, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2701, 327, 213} \begin {gather*} \frac {\tanh ^{-1}(\sin (a+b x))}{b}-\frac {\csc (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sec[a + b*x],x]

[Out]

ArcTanh[Sin[a + b*x]]/b - Csc[a + b*x]/b

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2701

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \csc ^2(a+b x) \sec (a+b x) \, dx &=-\frac {\text {Subst}\left (\int \frac {x^2}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{b}\\ &=-\frac {\csc (a+b x)}{b}-\frac {\text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\csc (a+b x)\right )}{b}\\ &=\frac {\tanh ^{-1}(\sin (a+b x))}{b}-\frac {\csc (a+b x)}{b}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.01, size = 27, normalized size = 1.17 \begin {gather*} -\frac {\csc (a+b x) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\sin ^2(a+b x)\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Csc[a + b*x]*Hypergeometric2F1[-1/2, 1, 1/2, Sin[a + b*x]^2])/b)

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Maple [A]
time = 0.04, size = 30, normalized size = 1.30

method	result	size
derivativedivides	\(\frac {-\frac {1}{\sin \left (b x +a \right )}+\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}\)	\(30\)
default	\(\frac {-\frac {1}{\sin \left (b x +a \right )}+\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}\)	\(30\)
risch	\(-\frac {2 i {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{b}\)	\(65\)
norman	\(\frac {-\frac {1}{2 b}-\frac {\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}{2 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b*(-1/sin(b*x+a)+ln(sec(b*x+a)+tan(b*x+a)))

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Maxima [A]
time = 0.30, size = 36, normalized size = 1.57 \begin {gather*} -\frac {\frac {2}{\sin \left (b x + a\right )} - \log \left (\sin \left (b x + a\right ) + 1\right ) + \log \left (\sin \left (b x + a\right ) - 1\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (23) = 46\).
time = 0.40, size = 50, normalized size = 2.17 \begin {gather*} \frac {\log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 2}{2 \, b \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec {\left (a + b x \right )}}{\sin ^{2}{\left (a + b x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)/sin(b*x+a)**2,x)

[Out]

Integral(sec(a + b*x)/sin(a + b*x)**2, x)

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Giac [A]
time = 7.08, size = 38, normalized size = 1.65 \begin {gather*} -\frac {\frac {2}{\sin \left (b x + a\right )} - \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{2 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2/sin(b*x + a) - log(abs(sin(b*x + a) + 1)) + log(abs(sin(b*x + a) - 1)))/b

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Mupad [B]
time = 0.02, size = 22, normalized size = 0.96 \begin {gather*} \frac {\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )-\frac {1}{\sin \left (a+b\,x\right )}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(a + b*x)*sin(a + b*x)^2),x)

[Out]

(atanh(sin(a + b*x)) - 1/sin(a + b*x))/b

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