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3.95 \(\int \csc (e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=84 \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3664, 402, 217, 206, 377, 207} \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f}-\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

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

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])

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 3664

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

Rubi steps

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

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Mathematica [B]  time = 6.65, size = 295, normalized size = 3.51 \[ \frac {\cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )+1\right )}{\sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )-\sqrt {a} \left (\tanh ^{-1}\left (\frac {a-(a-2 b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )\right )\right )}{2 f \sqrt {\sec ^4\left (\frac {1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

((2*Sqrt[b]*ArcTanh[(Sqrt[b]*(1 + Tan[(e + f*x)/2]^2))/Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^
2)^2]] - Sqrt[a]*(ArcTanh[(a - (a - 2*b)*Tan[(e + f*x)/2]^2)/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Ta
n[(e + f*x)/2]^2)^2])] + ArcTanh[(2*b + a*(-1 + Tan[(e + f*x)/2]^2))/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*
(-1 + Tan[(e + f*x)/2]^2)^2])]))*Cos[e + f*x]*Sec[(e + f*x)/2]^2*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e
 + f*x]^2])/(2*f*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])

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fricas [A]  time = 0.68, size = 514, normalized size = 6.12 \[ \left [\frac {\sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, -\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) - \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, f}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) - \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right )}{f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*(sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos
(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) + sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(
f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2))/f, -1/2*(2*sqrt(-b)*arctan(sqrt(-b)*sqrt(
((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt
(a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)))/f, 1/2*(2*s
qrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) + sqrt(b)*log(-((a -
 b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x +
 e)^2))/f, (sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) - sqrt(
-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b))/f]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to ch
eck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable
 to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)
Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nos
tep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi
/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>
(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nost
ep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/
t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign:
(2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check
sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to
check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unab
le to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/
2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_n
ostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*
pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2
)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_no
step/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*p
i/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign
: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to chec
k sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable t
o check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Un
able to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_noste
p/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t
_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-
2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep
/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_
nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2
*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check si
gn: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs or sign assumes constant sign by intervals
 (correct if the argument is real):Check [abs(t_nostep^2-1)]Discontinuities at zeroes of t_nostep^2-1 were not
 checkedUnable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Warning, integration of abs or sign assumes
constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Warning, replacing 0 by
` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable s
hould perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning
, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substi
tution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps
be purged.Warning, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0
by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variabl
e should perhaps be purged.Warning, integration of abs or sign assumes constant sign by intervals (correct if
the argument is real):Check [abs(t_nostep)]Warning, need to choose a branch for the root of a polynomial with
parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[-25]Warning, n
eed to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular value [0]
 was discarded and replaced randomly by 0=[84]Warning, need to choose a branch for the root of a polynomial wi
th parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[-99]Warning
, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular value
[0] was discarded and replaced randomly by 0=[-54]Warning, need to choose a branch for the root of a polynomia
l with parameters. This might be wrong.Non regular value [0] was discarded and replaced randomly by 0=[-56]War
ning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regular va
lue [0] was discarded and replaced randomly by 0=[-32]Evaluation time: 1.39index.cc index_m operator + Error:
Bad Argument Value

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maple [B]  time = 1.19, size = 721, normalized size = 8.58 \[ -\frac {\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}}\, \sqrt {4}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \left (2 \sqrt {a}\, \arctanh \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {4}\, \cos \left (f x +e \right )-\sqrt {4}-2 \cos \left (f x +e \right )-2\right ) \sqrt {b}\, \sqrt {4}}{8 \sin \left (f x +e \right )^{2} \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\right ) b +2 \ln \left (-\frac {2 \left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}-a \cos \left (f x +e \right )+b \cos \left (f x +e \right )+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a}}\right ) b^{\frac {3}{2}}-2 b^{\frac {3}{2}} \ln \left (-\frac {4 \left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}-a \cos \left (f x +e \right )+b \cos \left (f x +e \right )+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a}}\right )-\ln \left (-\frac {2 \left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}-a \cos \left (f x +e \right )+b \cos \left (f x +e \right )+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a}}\right ) a \sqrt {b}-a \ln \left (-\frac {4 \left (\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}+\sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}+a \cos \left (f x +e \right )-b \cos \left (f x +e \right )+b \right )}{-1+\cos \left (f x +e \right )}\right ) \sqrt {b}\right )}{4 f \sin \left (f x +e \right )^{2} \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a}\, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

-1/4/f*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)*4^(1/2)*cos(f*x+e)*(-1+cos(f*x+e))*(2*a^(1/2)*ar
ctanh(1/8*(-1+cos(f*x+e))*(4^(1/2)*cos(f*x+e)-4^(1/2)-2*cos(f*x+e)-2)/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)
^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*4^(1/2))*b+2*ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^
2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*a^(1/2)
-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*b^(3/2)-2*b^(3/2)*ln(-4*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e
)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))
^2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))-ln(-2*(-1+cos(f*x+e))*(a^(1/2)*cos(f*x+e)
*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^
2)^(1/2)*a^(1/2)-a*cos(f*x+e)+b*cos(f*x+e)+b)/sin(f*x+e)^2/a^(1/2))*a*b^(1/2)-a*ln(-4*(a^(1/2)*cos(f*x+e)*((a*
cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1
/2)*a^(1/2)+a*cos(f*x+e)-b*cos(f*x+e)+b)/(-1+cos(f*x+e)))*b^(1/2))/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*
b+b)/(1+cos(f*x+e))^2)^(1/2)/a^(1/2)/b^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*tan(f*x + e)^2 + a)*csc(f*x + e), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x)^2)^(1/2)/sin(e + f*x),x)

[Out]

int((a + b*tan(e + f*x)^2)^(1/2)/sin(e + f*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)*(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

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

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