\(\int \csc (e+f x) (a+b \sec ^2(e+f x))^2 \, dx\) [18]

Optimal result

Integrand size = 21, antiderivative size = 52 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {(a+b)^2 \text {arctanh}(\cos (e+f x))}{f}+\frac {b (2 a+b) \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

[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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4218, 472, 213} \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {(a+b)^2 \text {arctanh}(\cos (e+f x))}{f}+\frac {b (2 a+b) \sec (e+f x)}{f}+\frac {b^2 \sec ^3(e+f x)}{3 f} \]

[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 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 472

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 4218

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*} \text {integral}& = -\frac {\text {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 {\text {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 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {(a+b)^2 \text {arctanh}(\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*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(108\) vs. \(2(52)=104\).

Time = 0.91 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.08 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {4 \left (b+a \cos ^2(e+f x)\right )^2 \left (-b^2-3 b (2 a+b) \cos ^2(e+f x)+3 (a+b)^2 \cos ^3(e+f x) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )\right ) \sec ^3(e+f x)}{3 f (a+2 b+a \cos (2 (e+f x)))^2} \]

[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)

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.81

method	result	size
derivativedivides	\(\frac {a^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )+2 a b \left (\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )+b^{2} \left (\frac {1}{3 \cos \left (f x +e \right )^{3}}+\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{f}\)	\(94\)
default	\(\frac {a^{2} \ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )+2 a b \left (\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )+b^{2} \left (\frac {1}{3 \cos \left (f x +e \right )^{3}}+\frac {1}{\cos \left (f x +e \right )}+\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )\right )}{f}\)	\(94\)
norman	\(\frac {\frac {\left (8 a b +4 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{f}-\frac {12 a b +8 b^{2}}{3 f}-\frac {\left (4 a b +4 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}\)	\(110\)
parallelrisch	\(\frac {3 \left (\frac {\cos \left (3 f x +3 e \right )}{3}+\cos \left (f x +e \right )\right ) \left (a +b \right )^{2} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )+6 b \left (\frac {\left (a +\frac {2 b}{3}\right ) \cos \left (3 f x +3 e \right )}{3}+\frac {\left (2 a +b \right ) \cos \left (2 f x +2 e \right )}{3}+\left (a +\frac {2 b}{3}\right ) \cos \left (f x +e \right )+\frac {2 a}{3}+\frac {5 b}{9}\right )}{f \left (\cos \left (3 f x +3 e \right )+3 \cos \left (f x +e \right )\right )}\)	\(115\)
risch	\(\frac {2 b \,{\mathrm e}^{i \left (f x +e \right )} \left (6 a \,{\mathrm e}^{4 i \left (f x +e \right )}+3 b \,{\mathrm e}^{4 i \left (f x +e \right )}+12 a \,{\mathrm e}^{2 i \left (f x +e \right )}+10 b \,{\mathrm e}^{2 i \left (f x +e \right )}+6 a +3 b \right )}{3 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a^{2}}{f}+\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a b}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b^{2}}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a^{2}}{f}-\frac {2 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a b}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b^{2}}{f}\)	\(211\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (50) = 100\).

Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.94 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - 6 \, {\left (2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, b^{2}}{6 \, f \cos \left (f x + e\right )^{3}} \]

[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)

Sympy [F]

\[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \csc {\left (e + f x \right )}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.58 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) - 3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )}}{\cos \left (f x + e\right )^{3}}}{6 \, f} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (50) = 100\).

Time = 0.32 (sec) , antiderivative size = 172, normalized size of antiderivative = 3.31 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right ) + \frac {8 \, {\left (3 \, a b + 2 \, b^{2} + \frac {6 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {3 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {3 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{{\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right )}^{3}}}{6 \, f} \]

[In]

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

[Out]

1/6*(3*(a^2 + 2*a*b + b^2)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1)) + 8*(3*a*b + 2*b^2 + 6*a*b*(cos(f
*x + e) - 1)/(cos(f*x + e) + 1) + 3*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 3*a*b*(cos(f*x + e) - 1)^2/(co
s(f*x + e) + 1)^2 + 3*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/((cos(f*x + e) - 1)/(cos(f*x + e) + 1) +
1)^3)/f

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.02 \[ \int \csc (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\cos \left (e+f\,x\right )}^2\,\left (b^2+2\,a\,b\right )+\frac {b^2}{3}}{f\,{\cos \left (e+f\,x\right )}^3}-\frac {\mathrm {atanh}\left (\cos \left (e+f\,x\right )\right )\,{\left (a+b\right )}^2}{f} \]

[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