3.96 \(\int \csc ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=127 \[ -\frac {(a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 \sqrt {a} f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3664, 467, 523, 217, 206, 377, 207} \[ -\frac {(a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 \sqrt {a} f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-((a + b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*Sqrt[a]*f) + (Sqrt[b]*ArcTanh[(Sq
rt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/f - (Cot[e + f*x]*Csc[e + f*x]*Sqrt[a - b + 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 467

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, 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 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 ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2 \sqrt {a-b+b x^2}}{\left (-1+x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f}+\frac {\operatorname {Subst}\left (\int \frac {a-b+2 b x^2}{\left (-1+x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 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+b) \operatorname {Subst}\left (\int \frac {1}{\left (-1+x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 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 {(a+b) \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 )}{2 f}\\ &=-\frac {(a+b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 \sqrt {a} 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}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{2 f}\\ \end {align*}

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Mathematica [B]  time = 3.74, size = 460, normalized size = 3.62 \[ -\frac {\cot (e+f x) \csc (e+f x) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt {2} \sqrt {a} \sqrt {\sec ^4\left (\frac {1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}-16 \sqrt {a} \sqrt {b} \sin ^2\left (\frac {1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )+1\right )}{\sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )+4 (a+b) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac {a-(a-2 b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )+4 a \sin ^2\left (\frac {1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )+4 b \sin ^2\left (\frac {1}{2} (e+f x)\right ) \tanh ^{-1}\left (\frac {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )\right )}{4 \sqrt {a} f \sqrt {\sec ^4\left (\frac {1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-1/4*(Cot[e + f*x]*Csc[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(Sqrt[2]*Sqrt[a]*Sqrt[
(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4] - 16*Sqrt[a]*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]]*Sin[(e + f*x)/2]^2 + 4*(a + b)*ArcTanh[
(a - (a - 2*b)*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])]*Sin
[(e + f*x)/2]^2 + 4*a*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])]*Sin[(e + f*x)/2]^2 + 4*b*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])]*Sin[(e + f*x)/2]^2))/(Sqrt[a]*f*Sqrt[(a + b + (a -
b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4])

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fricas [A]  time = 1.38, size = 849, normalized size = 6.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(2*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 - a - b)*sq
rt(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)) + 2*(a*cos(f*x + e)^2 - a)*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*s
qrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2))/(a*f*cos(f*x + e)^2 - a*
f), 1/2*(((a + b)*cos(f*x + e)^2 - a - b)*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x +
e)^2)*cos(f*x + e)/a) + a*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + (a*cos(f*x + e)^2
- 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))/(a*f*cos(f*x + e)^2 - a*f), -1/4*(4*(a*cos(f*x + e)^2 - 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) - 2*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 - a - b)*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sq
rt(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)))/(a*f*cos(
f*x + e)^2 - a*f), 1/2*(((a + b)*cos(f*x + e)^2 - a - b)*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2
+ b)/cos(f*x + e)^2)*cos(f*x + e)/a) - 2*(a*cos(f*x + e)^2 - 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) + a*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x +
e))/(a*f*cos(f*x + e)^2 - a*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)^3*(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)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: (4*pi/t_nostep/2
)>(-4*pi/t_nostep/2)Warning, integration of abs or sign assumes constant sign by intervals (correct if the arg
ument is real):Check [abs(t_nostep^2-1)]Evaluation time: 1.22Error: Bad Argument Type

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maple [B]  time = 1.24, size = 2075, normalized size = 16.34 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

1/8/f*(-1+cos(f*x+e))*(4*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2*b^(3/2)*a^(1/
2)*4^(1/2)-4*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2*b^(1/2)*a^(3/2)*4^(1/2)+3
*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))*
cos(f*x+e)^2*b^(3/2)*4^(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)))*cos(f*x+e)^2*b^(3/2)*4^(1/2)*a-4*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))*cos(f*x+e)^2*b^(3/2)*4^(1/2)*a+4*arctanh(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))*cos(f*x+e)^2*a^(3/2)*4^(1/2)*b+8*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(3/2)*cos
(f*x+e)^2*b^(1/2)*a^(1/2)+2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*b^(1/2)*a^(3
/2)*4^(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))*cos(f*x+e)^2*b^(1/2)*4^(1/2)*a^2-ln(-4*(a^(1/2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+co
s(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)))*cos(f*x+e)^2*b^(1/2)*4^(1/2)*a^2+16*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)
^(3/2)*cos(f*x+e)*b^(1/2)*a^(1/2)-4*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)*a^(1/2)
*4^(1/2)+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)*a^(3/2)*4^(1/2)-3*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)*4^(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^(3/2)*4^(1/2)*a+4
*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))*
b^(3/2)*4^(1/2)*a-4*arctanh(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*c
os(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))*a^(3/2)*4^(1/2)*b+8*((a*cos(f*x+e)^2-co
s(f*x+e)^2*b+b)/(1+cos(f*x+e))^2)^(3/2)*b^(1/2)*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))*b^(1/2)*4^(1/2)*a^2+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)*4^(1/2)*a^2)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/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)/sin(f*x+e)^4/a^(3/2)/
b^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \tan \left (f x + e\right )^{2} + a} \csc \left (f x + e\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3*(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)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}}{{\sin \left (e+f\,x\right )}^3} \,d x \]

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)^3,x)

[Out]

int((a + b*tan(e + f*x)^2)^(1/2)/sin(e + f*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \tan ^{2}{\left (e + f x \right )}} \csc ^{3}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**3*(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(a + b*tan(e + f*x)**2)*csc(e + f*x)**3, x)

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