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

Optimal. Leaf size=52 \[ \frac {b (2 a+b) \sec (e+f x)}{f}-\frac {(a+b)^2 \tanh ^{-1}(\cos (e+f x))}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

[Out]

-(a+b)^2*arctanh(cos(f*x+e))/f+b*(2*a+b)*sec(f*x+e)/f+1/3*b^2*sec(f*x+e)^3/f

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Rubi [A]  time = 0.07, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4133, 461, 207} \[ \frac {b (2 a+b) \sec (e+f x)}{f}-\frac {(a+b)^2 \tanh ^{-1}(\cos (e+f x))}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b)^2*ArcTanh[Cos[e + f*x]])/f) + (b*(2*a + b)*Sec[e + f*x])/f + (b^2*Sec[e + f*x]^3)/(3*f)

Rule 207

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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
]

Rubi steps

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

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Mathematica [B]  time = 0.51, size = 108, normalized size = 2.08 \[ -\frac {4 \sec ^3(e+f x) \left (a \cos ^2(e+f x)+b\right )^2 \left (-3 b (2 a+b) \cos ^2(e+f x)+3 (a+b)^2 \cos ^3(e+f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )-b^2\right )}{3 f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.66, size = 101, normalized size = 1.94 \[ -\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - 6 \, {\left (2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}}{6 \, f \cos \left (f x + e\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(a^2 + 2*a*b + b^2)*cos(f*x + e)^3*log(1/2*cos(f*x + e) + 1/2) - 3*(a^2 + 2*a*b + b^2)*cos(f*x + e)^3*
log(-1/2*cos(f*x + e) + 1/2) - 6*(2*a*b + b^2)*cos(f*x + e)^2 - 2*b^2)/(f*cos(f*x + e)^3)

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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*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*((-6*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b-6*((1
-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b^2+12*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b+6*(1-cos(f*x+exp(1
)))/(1+cos(f*x+exp(1)))*b^2-6*a*b-4*b^2)*1/3/((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))-1)^3+(a^2+2*a*b+b^2)/4*l
n(abs(1-cos(f*x+exp(1)))/abs(1+cos(f*x+exp(1)))))

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maple [B]  time = 0.78, size = 117, normalized size = 2.25 \[ \frac {a^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}+\frac {2 a b}{f \cos \left (f x +e \right )}+\frac {2 a b \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f}+\frac {b^{2}}{3 f \cos \left (f x +e \right )^{3}}+\frac {b^{2}}{f \cos \left (f x +e \right )}+\frac {b^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 82, normalized size = 1.58 \[ -\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )}}{\cos \left (f x + e\right )^{3}}}{6 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(a^2 + 2*a*b + b^2)*log(cos(f*x + e) + 1) - 3*(a^2 + 2*a*b + b^2)*log(cos(f*x + e) - 1) - 2*(3*(2*a*b
+ b^2)*cos(f*x + e)^2 + b^2)/cos(f*x + e)^3)/f

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mupad [B]  time = 0.12, size = 53, normalized size = 1.02 \[ \frac {{\cos \left (e+f\,x\right )}^2\,\left (b^2+2\,a\,b\right )+\frac {b^2}{3}}{f\,{\cos \left (e+f\,x\right )}^3}-\frac {\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )\,{\left (a+b\right )}^2}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(e + f*x)^2*(2*a*b + b^2) + b^2/3)/(f*cos(e + f*x)^3) - (atanh(cos(e + f*x))*(a + b)^2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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