3.83 \(\int \csc (e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)
Optimal. Leaf size=122 \[ \frac {b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}-\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}+\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f} \]
[Out]
-(a+b)^(3/2)*arctanh(sec(f*x+e)*(a+b)^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))/f+1/2*(3*a+2*b)*arctanh(sec(f*x+e)*b^(1/
2)/(a+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f+1/2*b*sec(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/f
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Rubi [A] time = 0.14, antiderivative size = 122, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used =
{4134, 416, 523, 217, 206, 377, 207} \[ \frac {b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}-\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}+\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
(Sqrt[b]*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/(2*f) - ((a + b)^(3/2)*ArcTan
h[(Sqrt[a + b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f + (b*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(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 416
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] && !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
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) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac {b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}+\frac {\operatorname {Subst}\left (\int \frac {a (2 a+b)+b (3 a+2 b) x^2}{\left (-1+x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac {b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}+\frac {(a+b)^2 \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 (3 a+2 b)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac {b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}+\frac {(a+b)^2 \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 {(b (3 a+2 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 )}{2 f}\\ &=\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}-\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}+\frac {b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}\\ \end {align*}
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Mathematica [A] time = 0.55, size = 171, normalized size = 1.40 \[ \frac {\sec (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (\sqrt {2} b \sqrt {a \cos (2 (e+f x))+a+2 b}+2 \sqrt {b} (3 a+2 b) \cos ^2(e+f x) \tanh ^{-1}\left (\frac {\sqrt {-a \sin ^2(e+f x)+a+b}}{\sqrt {b}}\right )-4 (a+b)^{3/2} \cos ^2(e+f x) \tanh ^{-1}\left (\frac {\sqrt {-a \sin ^2(e+f x)+a+b}}{\sqrt {a+b}}\right )\right )}{2 \sqrt {2} f \sqrt {a \cos (2 (e+f x))+a+2 b}} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[e + f*x]*(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
((2*Sqrt[b]*(3*a + 2*b)*ArcTanh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[b]]*Cos[e + f*x]^2 - 4*(a + b)^(3/2)*ArcTa
nh[Sqrt[a + b - a*Sin[e + f*x]^2]/Sqrt[a + b]]*Cos[e + f*x]^2 + Sqrt[2]*b*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])*
Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2])/(2*Sqrt[2]*f*Sqrt[a + 2*b + a*Cos[2*(e + f*x)]])
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fricas [A] time = 0.86, size = 715, normalized size = 5.86 \[ \left [\frac {2 \, {\left (a + b\right )}^{\frac {3}{2}} \cos \left (f x + e\right ) \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 ) + {\left (3 \, a + 2 \, b\right )} \sqrt {b} \cos \left (f x + e\right ) \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 \, b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right )}, \frac {4 \, {\left (a + b\right )} \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 ) \cos \left (f x + e\right ) + {\left (3 \, a + 2 \, b\right )} \sqrt {b} \cos \left (f x + e\right ) \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 \, b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right )}, -\frac {{\left (3 \, a + 2 \, 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 ) \cos \left (f x + e\right ) - {\left (a + b\right )}^{\frac {3}{2}} \cos \left (f x + e\right ) \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 ) - b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, f \cos \left (f x + e\right )}, \frac {2 \, {\left (a + b\right )} \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 ) \cos \left (f x + e\right ) - {\left (3 \, a + 2 \, 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 ) \cos \left (f x + e\right ) + b \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, f \cos \left (f x + e\right )}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/4*(2*(a + b)^(3/2)*cos(f*x + e)*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)) + (3*a + 2*b)*sqrt(b)*cos(f*x + e)*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) + 2*b*sqrt((a*co
s(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), 1/4*(4*(a + b)*sqrt(-a - b)*arctan(sqrt(-a - b)*sqrt((a*c
os(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b))*cos(f*x + e) + (3*a + 2*b)*sqrt(b)*cos(f*x + e)*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) +
2*b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), -1/2*((3*a + 2*b)*sqrt(-b)*arctan(sqrt(-b)
*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)*cos(f*x + e) - (a + b)^(3/2)*cos(f*x + e)*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)) - b*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)), 1/2*(2*(a + b)*sqrt(-a - b)*ar
ctan(sqrt(-a - b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/(a + b))*cos(f*x + e) - (3*a + 2*b)
*sqrt(-b)*arctan(sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b)*cos(f*x + e) + b*sqrt((a
*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e))]
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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*sec(f*x+e)^2)^(3/2),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)Warning, integration of ab
s or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(cos(f*x+exp(1)))]Una
ble 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_n
ostep/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/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)Una
ble to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Evaluation time: 1.52Erro
r: Bad Argument Type
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maple [B] time = 1.58, size = 2563, normalized size = 21.01 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(f*x+e)*(a+b*sec(f*x+e)^2)^(3/2),x)
[Out]
-1/4/f*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)*4^(1/2)*cos(f*x+e)*(-1+cos(f*x+e))^3*(-cos(f*x+e)^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^(13/2)-15*cos(f*x+e)^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^(9/2)*a^2-6*cos(f*x+e)^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^(11/2)*a-cos(f*x+e
)^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^6-cos(f*x+e)^2*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^(1/2)*a^6-6*cos(f*x+e)^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^(3/2)*a^5-6*cos(f*x+e)^2*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)*a^5-6*cos(f*x+e)^2*ln(-4*(-1+cos(f*x+e))*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*c
os(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^(5/2)*a^4-15*cos(f*x+e)^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^(5/2)*a^4-9*cos(f
*x+e)^2*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^(5/2)*a^4-20*cos(f*x+e
)^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^(7/2)*a^3+cos(f*x+e)^2*b^(13/2)*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))-20*cos(f*x+e)^2*b^(7/2)*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))*a^3+cos(f*x+e)*b^(5/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+c
os(f*x+e))^2)^(1/2)+b^(3/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a+2*cos(f*x+e)^2*(a+b)^(7/
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))*b^3+6*cos(f*x+e)^2*b^(11/2)*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-12*cos(f*x+e)^2*b^(11/2)*ln(-4*(-1+cos(f*x+e))*(((b+a*cos(f*x+e)^2)/(1+co
s(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+9*cos(f*x+e)^2*b^(9/2)*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^2-24*cos(f*x+e)^2*b^(9/2)*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))*a^2-2*cos(f*x+e)^2*b^(13/2)*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))+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(5/2)*(a+b)^(7/2)+3*cos(f*x+e)^2*(a+b)^(7/2)*arct
anh(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^2*b+5*cos(f*x+e)^2*(a+b)^(7/2)*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^2+cos
(f*x+e)*b^(3/2)*(a+b)^(7/2)*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*a)/sin(f*x+e)^6/((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(3/2)/b^(1/2)/(a+b)^(9/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \csc \left (f x + e\right )\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e)^2 + a)^(3/2)*csc(f*x + e), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}}{\sin \left (e+f\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b/cos(e + f*x)^2)^(3/2)/sin(e + f*x),x)
[Out]
int((a + b/cos(e + f*x)^2)^(3/2)/sin(e + f*x), x)
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)*(a+b*sec(f*x+e)**2)**(3/2),x)
[Out]
Timed out
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