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

Optimal. Leaf size=82 \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}-\frac {\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 82, 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 = {4134, 402, 217, 206, 377, 207} \[ \frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}-\frac {\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - (Sqrt[a + b]*ArcTanh[(Sqrt[a + b]*Sec
[e + f*x])/Sqrt[a + 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 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 119, normalized size = 1.45 \[ \frac {\sqrt {2} \cos (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {-a \sin ^2(e+f x)+a+b}}{\sqrt {b}}\right )-\sqrt {a+b} \tanh ^{-1}\left (\frac {\sqrt {-a \sin ^2(e+f x)+a+b}}{\sqrt {a+b}}\right )\right )}{f \sqrt {a \cos (2 e+2 f x)+a+2 b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.78, size = 496, normalized size = 6.05 \[ \left [\frac {\sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) + \sqrt {b} \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, \frac {2 \, \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right ) + \sqrt {b} \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right )}{2 \, f}, -\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) - \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{2 \, f}, \frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right ) - \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right )}{f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x
+ e) + a + 2*b)/(cos(f*x + e)^2 - 1)) + sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*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(-a - b)*arctan(sqrt(-a - b)*sqrt((a*cos(f*
x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) + sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*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
*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - sqrt(a + b)*log(2*(a*cos(f*x + e)^2 - 2*sqrt(a + b)*sqr
t((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + 2*b)/(cos(f*x + e)^2 - 1)))/f, (sqrt(-a - b)*arcta
n(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b)) - sqrt(-b)*arctan(sqrt(-b)*sq
rt((a*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 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)-2*(1/4*sqrt(a+b)*ln(abs(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sq
rt(a+b)+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+
exp(1)))^2+a+b)))-1/4*sqrt(a+b)*ln(abs(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2-sqrt(a+b)+sqrt(a*tan(1/2*(f*x+exp(1)
))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b)))+b*atan(1/2*(-sqr
t(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a+b)+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*
(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))/sqrt(-b))/sqrt(-b)+sqrt(a+b)*(a+b)*ln(abs(-(a+b)*(-sqrt(a+b)
*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+
2*b*tan(1/2*(f*x+exp(1)))^2+a+b))-sqrt(a+b)*(a-b)))/(-4*a-4*b))*sign(cos(f*x+exp(1)))/f

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maple [B]  time = 2.01, size = 688, normalized size = 8.39 \[ -\frac {\sqrt {4}\, \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right ) \left (\ln \left (-\frac {2 \left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}-a \cos \left (f x +e \right )+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a +b}}\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 {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}-a \cos \left (f x +e \right )+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a +b}}\right )+2 \arctanh \left (\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (\cos \left (f x +e \right ) \sqrt {4}-2 \cos \left (f x +e \right )-\sqrt {4}-2\right ) \sqrt {b}\, \sqrt {4}}{8 \sin \left (f x +e \right )^{2} \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}}\right ) \sqrt {a +b}\, b -\ln \left (-\frac {2 \left (-1+\cos \left (f x +e \right )\right ) \left (\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}-a \cos \left (f x +e \right )+b \right )}{\sin \left (f x +e \right )^{2} \sqrt {a +b}}\right ) a \sqrt {b}-\ln \left (-\frac {4 \left (\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+a \cos \left (f x +e \right )+\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}+b \right )}{-1+\cos \left (f x +e \right )}\right ) \sqrt {b}\, a -\ln \left (-\frac {4 \left (\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {a +b}+a \cos \left (f x +e \right )+\sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {a +b}+b \right )}{-1+\cos \left (f x +e \right )}\right ) b^{\frac {3}{2}}\right ) \left (-1+\cos \left (f x +e \right )\right )}{4 f \sin \left (f x +e \right )^{2} \sqrt {\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\left (1+\cos \left (f x +e \right )\right )^{2}}}\, \sqrt {b}\, \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sec(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 \[ \int \frac {\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}}{\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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