3.376 \(\int \csc ^3(e+f x) \sqrt{b \sec (e+f x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0758129, 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, 288, 329, 298, 203, 206} \[ -\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b f}+\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{3 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[b]*ArcTan[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(4*f) - (3*Sqrt[b]*ArcTanh[Sqrt[b*Sec[e + f*x]]/Sqrt[b]])/(4*
f) - (Cot[e + f*x]^2*(b*Sec[e + f*x])^(3/2))/(2*b*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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& 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 \csc ^3(e+f x) \sqrt{b \sec (e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^{5/2}}{\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 f}+\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{x}}{-1+\frac{x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{4 b f}\\ &=-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b f}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{-1+\frac{x^4}{b^2}} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{2 b f}\\ &=-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b f}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{4 f}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \sec (e+f x)}\right )}{4 f}\\ &=\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{3 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b \sec (e+f x)}}{\sqrt{b}}\right )}{4 f}-\frac{\cot ^2(e+f x) (b \sec (e+f x))^{3/2}}{2 b f}\\ \end{align*}

Mathematica [A]  time = 0.612232, size = 95, normalized size = 1.02 \[ -\frac{\sqrt{b \sec (e+f x)} \left (-3 \log \left (1-\sqrt{\sec (e+f x)}\right )+3 \log \left (\sqrt{\sec (e+f x)}+1\right )+\frac{4 \csc ^2(e+f x)}{\sqrt{\sec (e+f x)}}-6 \tan ^{-1}\left (\sqrt{\sec (e+f x)}\right )\right )}{8 f \sqrt{\sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((-6*ArcTan[Sqrt[Sec[e + f*x]]] - 3*Log[1 - Sqrt[Sec[e + f*x]]] + 3*Log[1 + Sqrt[Sec[e + f*x]]] + (4*Csc[e +
f*x]^2)/Sqrt[Sec[e + f*x]])*Sqrt[b*Sec[e + f*x]])/(8*f*Sqrt[Sec[e + f*x]])

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Maple [B]  time = 0.148, size = 603, normalized size = 6.5 \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))^(1/2),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)-3*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)/s
in(f*x+e)^2)+4*cos(f*x+e)^2*ln(-2*(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)+3*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)-4*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-co
s(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))*cos(f*x+e)*(b/cos(f*x+e))^(1/
2)/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/sin(f*x+e)^4

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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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.66575, size = 941, normalized size = 10.12 \begin{align*} \left [\frac{6 \,{\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 ) + 3 \,{\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 (f \cos \left (f x + e\right )^{2} - f\right )}}, -\frac{6 \,{\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 ) - 3 \,{\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 (f \cos \left (f x + e\right )^{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))^(1/2),x, algorithm="fricas")

[Out]

[1/16*(6*(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) + 3*(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))/(f*c
os(f*x + e)^2 - f), -1/16*(6*(cos(f*x + e)^2 - 1)*sqrt(b)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f*x + e) - 1)/s
qrt(b)) - 3*(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)*sqr
t(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))/(f*cos(f*x + e)^2 - f)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec{\left (e + f x \right )}} \csc ^{3}{\left (e + f x \right )}\, 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))**(1/2),x)

[Out]

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

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Giac [A]  time = 1.12906, size = 139, normalized size = 1.49 \begin{align*} \frac{b^{4}{\left (\frac{2 \, \sqrt{b \cos \left (f x + e\right )}}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} b^{2}} + \frac{3 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{-b}}\right )}{\sqrt{-b} b^{3}} - \frac{3 \, \arctan \left (\frac{\sqrt{b \cos \left (f x + e\right )}}{\sqrt{b}}\right )}{b^{\frac{7}{2}}}\right )} \mathrm{sgn}\left (\cos \left (f x + e\right )\right )}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^4*(2*sqrt(b*cos(f*x + e))/((b^2*cos(f*x + e)^2 - b^2)*b^2) + 3*arctan(sqrt(b*cos(f*x + e))/sqrt(-b))/(sq
rt(-b)*b^3) - 3*arctan(sqrt(b*cos(f*x + e))/sqrt(b))/b^(7/2))*sgn(cos(f*x + e))/f