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

Optimal. Leaf size=77 \[ -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2622, 321, 329, 212, 206, 203} \[ -\frac {b^{3/2} \tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}+\frac {2 b \sqrt {b \sec (e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 212

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

Rule 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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 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])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.96, size = 85, normalized size = 1.10 \[ \frac {(b \sec (e+f x))^{3/2} \left (4 \sqrt {\sec (e+f x)}+\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (\sqrt {\sec (e+f x)}+1\right )-2 \tan ^{-1}\left (\sqrt {\sec (e+f x)}\right )\right )}{2 f \sec ^{\frac {3}{2}}(e+f x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.80, size = 278, normalized size = 3.61 \[ \left [\frac {2 \, \sqrt {-b} b \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + \sqrt {-b} b \log \left (\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, b \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{4 \, f}, \frac {2 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) + b^{\frac {3}{2}} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right ) + 8 \, b \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{4 \, f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(-b)*b*arctan(1/2*sqrt(-b)*sqrt(b/cos(f*x + e))*(cos(f*x + e) + 1)/b) + sqrt(-b)*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*b*sqrt(b/cos(f*x + e)))/f, 1/4*(2*b^(3/2)*arctan(1/2*sqrt(b/cos(f*x + e))*(cos(f
*x + e) - 1)/sqrt(b)) + b^(3/2)*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*b*sqrt(b/cos(f*x + e)))/f]

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)b*2*(-2*b/(sqrt(-b
)*tan(1/2*(f*x+exp(1)))^2-sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b)-sqrt(-b))-1/4*sqrt(-b)*ln(abs(-sqrt(-b)*tan(1/2*(
f*x+exp(1)))^2+sqrt(-b*tan(1/2*(f*x+exp(1)))^4+b)))+1/2*b*atan((-sqrt(-b)*tan(1/2*(f*x+exp(1)))^2+sqrt(-b*tan(
1/2*(f*x+exp(1)))^4+b))/sqrt(-b))/sqrt(-b))*sign(cos(f*x+exp(1)))/f

________________________________________________________________________________________

maple [B]  time = 0.21, size = 235, normalized size = 3.05 \[ \frac {\left (-1+\cos \left (f x +e \right )\right )^{3} \left (4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+\cos \left (f x +e \right ) \ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right )+\cos \left (f x +e \right ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (f x +e \right )\right )}{2 f \sin \left (f x +e \right )^{6} \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(-1+cos(f*x+e))^3*(4*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)+cos(f*x+e)*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)+cos(f*x+e)*arctan(1/2/(-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))*(b/
cos(f*x+e))^(3/2)*cos(f*x+e)^2/sin(f*x+e)^6/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 87, normalized size = 1.13 \[ -\frac {{\left (2 \, \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right ) - \sqrt {b} \log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right ) - 4 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}\right )} b}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*sqrt(b)*arctan(sqrt(b/cos(f*x + e))/sqrt(b)) - sqrt(b)*log(-(sqrt(b) - sqrt(b/cos(f*x + e)))/(sqrt(b)
+ sqrt(b/cos(f*x + e)))) - 4*sqrt(b/cos(f*x + e)))*b/f

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [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)*(b*sec(f*x+e))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________