3.5 \(\int \csc (e+f x) (a+b \sec ^2(e+f x)) \, dx\)

Optimal. Leaf size=27 \[ \frac{b \sec (e+f x)}{f}-\frac{(a+b) \tanh ^{-1}(\cos (e+f x))}{f} \]

[Out]

-(((a + b)*ArcTanh[Cos[e + f*x]])/f) + (b*Sec[e + f*x])/f

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Rubi [A]  time = 0.0308487, antiderivative size = 27, normalized size of antiderivative = 1., 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 = {4133, 453, 206} \[ \frac{b \sec (e+f x)}{f}-\frac{(a+b) \tanh ^{-1}(\cos (e+f x))}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b)*ArcTanh[Cos[e + f*x]])/f) + (b*Sec[e + f*x])/f

Rule 4133

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*(ff*x)^n)^p)/(ff*x)^(n*
p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p
]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 206

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

Rubi steps

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

Mathematica [B]  time = 0.039863, size = 84, normalized size = 3.11 \[ \frac{a \log \left (\sin \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{f}-\frac{a \log \left (\cos \left (\frac{e}{2}+\frac{f x}{2}\right )\right )}{f}+\frac{b \sec (e+f x)}{f}+\frac{b \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )}{f}-\frac{b \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.04, size = 57, normalized size = 2.1 \begin{align*}{\frac{a\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}}+{\frac{b}{f\cos \left ( fx+e \right ) }}+{\frac{b\ln \left ( \csc \left ( fx+e \right ) -\cot \left ( fx+e \right ) \right ) }{f}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02554, size = 59, normalized size = 2.19 \begin{align*} -\frac{{\left (a + b\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) -{\left (a + b\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac{2 \, b}{\cos \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a + b)*log(cos(f*x + e) + 1) - (a + b)*log(cos(f*x + e) - 1) - 2*b/cos(f*x + e))/f

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Fricas [B]  time = 0.836694, size = 178, normalized size = 6.59 \begin{align*} -\frac{{\left (a + b\right )} \cos \left (f x + e\right ) \log \left (\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) -{\left (a + b\right )} \cos \left (f x + e\right ) \log \left (-\frac{1}{2} \, \cos \left (f x + e\right ) + \frac{1}{2}\right ) - 2 \, b}{2 \, f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a + b)*cos(f*x + e)*log(1/2*cos(f*x + e) + 1/2) - (a + b)*cos(f*x + e)*log(-1/2*cos(f*x + e) + 1/2) - 2
*b)/(f*cos(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \csc{\left (e + f x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.3427, size = 82, normalized size = 3.04 \begin{align*} \frac{{\left (a + b\right )} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) + \frac{4 \, b}{\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a + b)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1)) + 4*b/((cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1))/f