∫ csc3 (e+fx)___ (a+b tan2(e+fx))3∕2 dx

3.132 \(\int \frac {\csc ^3(e+f x)}{(a+b \tan ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=127 \[ -\frac {(a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a^{5/2} f}-\frac {3 b \sec (e+f x)}{2 a^2 f \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b \sec ^2(e+f x)-b}} \]

[Out]

-1/2*(a-3*b)*arctanh(sec(f*x+e)*a^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))/a^(5/2)/f-1/2*cot(f*x+e)*csc(f*x+e)/a/f/(a

-b+b*sec(f*x+e)^2)^(1/2)-3/2*b*sec(f*x+e)/a^2/f/(a-b+b*sec(f*x+e)^2)^(1/2)

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Rubi [A] time = 0.16, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3664, 471, 527, 12, 377, 207} \[ -\frac {3 b \sec (e+f x)}{2 a^2 f \sqrt {a+b \sec ^2(e+f x)-b}}-\frac {(a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{2 a^{5/2} f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a+b \sec ^2(e+f x)-b}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((a - 3*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*a^(5/2)*f) - (Cot[e + f*x]*Csc[

e + f*x])/(2*a*f*Sqrt[a - b + b*Sec[e + f*x]^2]) - (3*b*Sec[e + f*x])/(2*a^2*f*Sqrt[a - b + b*Sec[e + f*x]^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 471

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 + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -

a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)

+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n

, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[

((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d

)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)

*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

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 \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {a-b-2 b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{2 a f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {3 b \sec (e+f x)}{2 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {(a-3 b) (a-b)}{\left (-1+x^2\right ) \sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 a^2 (a-b) f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {3 b \sec (e+f x)}{2 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {(a-3 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 a^2 f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {3 b \sec (e+f x)}{2 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}+\frac {(a-3 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 a^2 f}\\ &=-\frac {(a-3 b) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 a^{5/2} f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \sqrt {a-b+b \sec ^2(e+f x)}}-\frac {3 b \sec (e+f x)}{2 a^2 f \sqrt {a-b+b \sec ^2(e+f x)}}\\ \end {align*}

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Mathematica [B] time = 4.61, size = 308, normalized size = 2.43 \[ -\frac {\frac {(a-3 b) \cos (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\tanh ^{-1}\left (\frac {a-(a-2 b) \tan ^2\left (\frac {1}{2} (e+f x)\right )}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )+\tanh ^{-1}\left (\frac {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )+2 b}{\sqrt {a} \sqrt {a \left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-1\right )^2+4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )}}\right )\right )}{2 a^{5/2} \sqrt {\sec ^4\left (\frac {1}{2} (e+f x)\right ) ((a-b) \cos (2 (e+f x))+a+b)}}+\frac {\csc ^2(e+f x) \sec (e+f x) ((a-3 b) \cos (2 (e+f x))+a+3 b)}{\sqrt {2} a^2 \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}}}{2 f} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/2*(((a + 3*b + (a - 3*b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2*Sec[e + f*x])/(Sqrt[2]*a^2*Sqrt[(a + b + (a - b)*

Cos[2*(e + f*x)])*Sec[e + f*x]^2]) + ((a - 3*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])] + ArcTanh[(2*b + a*(-1 + Tan[(e + f*x)/2]^2))/(Sqrt[a]*S

qrt[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*a^(5/2)*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^

4]))/f

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fricas [A] time = 0.77, size = 455, normalized size = 3.58 \[ \left [-\frac {{\left ({\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 5 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, b^{2}\right )} \sqrt {a} \log \left (-\frac {2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right ) - 2 \, {\left ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, a b \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}, \frac {{\left ({\left (a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - {\left (a^{2} - 5 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - a b + 3 \, b^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a}\right ) + {\left ({\left (a^{2} - 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, a b \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a^{4} - a^{3} b\right )} f \cos \left (f x + e\right )^{4} - a^{3} b f - {\left (a^{4} - 2 \, a^{3} b\right )} f \cos \left (f x + e\right )^{2}\right )}}\right ] \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(((a^2 - 4*a*b + 3*b^2)*cos(f*x + e)^4 - (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^2 - a*b + 3*b^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)) - 2*((a^2 - 3*a*b)*cos(f*x + e)^3 + 3*a*b*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 +

b)/cos(f*x + e)^2))/((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^4 - 2*a^3*b)*f*cos(f*x + e)^2), 1/2*(((a^2

- 4*a*b + 3*b^2)*cos(f*x + e)^4 - (a^2 - 5*a*b + 6*b^2)*cos(f*x + e)^2 - a*b + 3*b^2)*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^2 - 3*a*b)*cos(f*x + e)^3 + 3*a*b*co

s(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a^4 - a^3*b)*f*cos(f*x + e)^4 - a^3*b*f - (a^

4 - 2*a^3*b)*f*cos(f*x + e)^2)]

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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)^(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)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/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/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)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)Warning, integration of abs or sign assumes consta

nt 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 checkedWarning, integration of abs or sign assumes constant sign by intervals (correct if

the argument is real):Check [abs(t_nostep^2-1)]Warning, need to choose a branch for the root of a polynomial

with parameters. This might be wrong.The choice was done assuming [a,b]=[85,53]Warning, need to choose a branc

h for the root of a polynomial with parameters. This might be wrong.The choice was done assuming [a,b]=[-33,71

]Discontinuities at zeroes of t_nostep^2-1 were not checkedUnable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_no

step/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*p

i/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Warning, integration of abs or sign ass

umes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep^2-1)]Evaluation time: 5.

61index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [B] time = 1.19, size = 5633, normalized size = 44.35 \[ \text {output too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [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)^3/(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

Timed out

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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