3.429 \(\int \frac{\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=93 \[ -\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}-\frac{\tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 b^{3/2} f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 b^{3/2} f} \]

[Out]

-ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(4*b^(3/2)*f) + ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(4*b^(3/2)*f) - (C
ot[e + f*x]^2*(b*Sec[e + f*x])^(3/2))/(2*b^3*f)

________________________________________________________________________________________

Rubi [A]  time = 0.0738407, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2622, 290, 329, 298, 203, 206} \[ -\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}-\frac{\tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 b^{3/2} f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 b^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(4*b^(3/2)*f) + ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]]/(4*b^(3/2)*f) - (C
ot[e + f*x]^2*(b*Sec[e + f*x])^(3/2))/(2*b^3*f)

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 203

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

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])

Rubi steps

\begin{align*} \int \frac{\csc ^3(e+f x)}{(b \sec (e+f x))^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{\left (-1+\frac{x^2}{b^2}\right )^2} \, dx,x,b \sec (e+f x)\right )}{b^3 f}\\ &=-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+\frac{x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{4 b^3 f}\\ &=-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}-\frac{\operatorname{Subst}\left (\int \frac{x^2}{-1+\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{2 b^3 f}\\ &=-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}+\frac{\operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{4 b f}-\frac{\operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{4 b f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 b^{3/2} f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 b^{3/2} f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b^3 f}\\ \end{align*}

Mathematica [A]  time = 0.466018, size = 98, normalized size = 1.05 \[ \frac{-4 \csc ^2(e+f x)+\sqrt{\sec (e+f x)} \left (\log \left (\sqrt{\sec (e+f x)}+1\right )-\log \left (1-\sqrt{\sec (e+f x)}\right )\right )-2 \sqrt{\sec (e+f x)} \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )}{8 b f \sqrt{b \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Csc[e + f*x]^2 - 2*ArcTan[Sqrt[Sec[e + f*x]]]*Sqrt[Sec[e + f*x]] + (-Log[1 - Sqrt[Sec[e + f*x]]] + Log[1 +
Sqrt[Sec[e + f*x]]])*Sqrt[Sec[e + f*x]])/(8*b*f*Sqrt[b*Sec[e + f*x]])

________________________________________________________________________________________

Maple [B]  time = 0.125, size = 426, normalized size = 4.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 3.0209, size = 957, normalized size = 10.29 \begin{align*} \left [-\frac{2 \,{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} \sqrt{\frac{b}{\cos \left (f x + e\right )}}{\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) +{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt{-b} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt{-b} \sqrt{\frac{b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) - 8 \, \sqrt{\frac{b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{16 \,{\left (b^{2} f \cos \left (f x + e\right )^{2} - b^{2} f\right )}}, \frac{2 \,{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt{b} \arctan \left (\frac{\sqrt{\frac{b}{\cos \left (f x + e\right )}}{\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt{b}}\right ) +{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sqrt{b} \log \left (\frac{b \cos \left (f x + e\right )^{2} + 4 \,{\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt{b} \sqrt{\frac{b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, \sqrt{\frac{b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{16 \,{\left (b^{2} f \cos \left (f x + e\right )^{2} - b^{2} f\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(2*(cos(f*x + e)^2 - 1)*sqrt(-b)*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) + (cos(
f*x + e)^2 - 1)*sqrt(-b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - cos(f*x + e))*sqrt(-b)*sqrt(b/cos(f*x + e
)) - 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 8*sqrt(b/cos(f*x + e))*cos(f*x + e))/(b^2*
f*cos(f*x + e)^2 - b^2*f), 1/16*(2*(cos(f*x + e)^2 - 1)*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e)
- 1)/sqrt(b)) + (cos(f*x + e)^2 - 1)*sqrt(b)*log((b*cos(f*x + e)^2 + 4*(cos(f*x + e)^2 + cos(f*x + e))*sqrt(b)
*sqrt(b/cos(f*x + e)) + 6*b*cos(f*x + e) + b)/(cos(f*x + e)^2 - 2*cos(f*x + e) + 1)) + 8*sqrt(b/cos(f*x + e))*
cos(f*x + e))/(b^2*f*cos(f*x + e)^2 - b^2*f)]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{3}{\left (e + f x \right )}}{\left (b \sec{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{3}}{\left (b \sec \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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