\(\int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\) [97]

Optimal result

Integrand size = 25, antiderivative size = 87 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 (a+b)^{3/2} f}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 (a+b) f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4219, 482, 12, 385, 213} \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {a \text {arctanh}\left (\frac {\sqrt {a+b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f (a+b)^{3/2}}-\frac {\cot (e+f x) \csc (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f (a+b)} \]

[In]

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

[Out]

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

Rule 12

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 385

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 482

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 4219

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

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.61 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=-\frac {a \sqrt {a+2 b+a \cos (2 e+2 f x)} \sec (e+f x) \sqrt {a+b-a \sin ^2(e+f x)} \left (\frac {(a+b) \csc ^2(e+f x)}{a}+\frac {\text {arctanh}\left (\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}\right )}{\sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}}}\right )}{2 \sqrt {2} (a+b)^2 f \sqrt {a+b \sec ^2(e+f x)}} \]

[In]

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

[Out]

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

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(1214\) vs. \(2(75)=150\).

Time = 1.28 (sec) , antiderivative size = 1215, normalized size of antiderivative = 13.97

method	result	size
default	\(\text {Expression too large to display}\)	\(1215\)

[In]

int(csc(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8/f/(a+b)^(5/2)*(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*
x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)/((a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f
*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)/((1-cos(f*x+e))^2*csc(f*x+e)^
2-1)^2)^(1/2)/((1-cos(f*x+e))^2*csc(f*x+e)^2-1)/(1-cos(f*x+e))^2*((a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*
x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(3/2
)*(1-cos(f*x+e))^2+2*ln((a*(1-cos(f*x+e))^2*csc(f*x+e)^2+b*(1-cos(f*x+e))^2*csc(f*x+e)^2+(a*(1-cos(f*x+e))^4*c
sc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^
2+a+b)^(1/2)*(a+b)^(1/2)-a+b)/(a+b)^(1/2))*a^2*(1-cos(f*x+e))^2+2*ln((a*(1-cos(f*x+e))^2*csc(f*x+e)^2+b*(1-cos
(f*x+e))^2*csc(f*x+e)^2+(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*
csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(1/2)-a+b)/(a+b)^(1/2))*(1-cos(f*x+e))^2*a*b+2
*ln(2/(1-cos(f*x+e))^2*(-a*(1-cos(f*x+e))^2+b*(1-cos(f*x+e))^2+(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e
))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(1/2)*s
in(f*x+e)^2+a*sin(f*x+e)^2+b*sin(f*x+e)^2))*a^2*(1-cos(f*x+e))^2+2*ln(2/(1-cos(f*x+e))^2*(-a*(1-cos(f*x+e))^2+
b*(1-cos(f*x+e))^2+(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc(f
*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(1/2)*sin(f*x+e)^2+a*sin(f*x+e)^2+b*sin(f*x+e)^2))*
(1-cos(f*x+e))^2*a*b+(a*(1-cos(f*x+e))^4*csc(f*x+e)^4+b*(1-cos(f*x+e))^4*csc(f*x+e)^4-2*a*(1-cos(f*x+e))^2*csc
(f*x+e)^2+2*b*(1-cos(f*x+e))^2*csc(f*x+e)^2+a+b)^(1/2)*(a+b)^(3/2)*sin(f*x+e)^2)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 305, normalized size of antiderivative = 3.51 \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\left [\frac {2 \, {\left (a + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {a + b} \log \left (\frac {2 \, {\left (a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a + b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + a + 2 \, b\right )}}{\cos \left (f x + e\right )^{2} - 1}\right )}{4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}, \frac {{\left (a \cos \left (f x + e\right )^{2} - a\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-a - b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a + b}\right ) + {\left (a + b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} + 2 \, a b + b^{2}\right )} f\right )}}\right ] \]

[In]

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

[Out]

[1/4*(2*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + (a*cos(f*x + e)^2 - a)*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)/(c
os(f*x + e)^2 - 1)))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 - (a^2 + 2*a*b + b^2)*f), 1/2*((a*cos(f*x + e)^2 -
a)*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)) + (a + b
)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^2 - (a^2 + 2*a
*b + b^2)*f)]

Sympy [F]

\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {\csc ^{3}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

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

[Out]

integrate(csc(f*x + e)^3/sqrt(b*sec(f*x + e)^2 + a), x)

Giac [F]

\[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int { \frac {\csc \left (f x + e\right )^{3}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx=\int \frac {1}{{\sin \left (e+f\,x\right )}^3\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]

[In]

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

[Out]

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