3.5.15 \(\int \frac {\csc ^3(e+f x)}{\sqrt {b \sec (e+f x)}} \, dx\) [415]

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

[Out]

-1/4*arctan((b*sec(f*x+e))^(1/2)/b^(1/2))/f/b^(1/2)-1/4*arctanh((b*sec(f*x+e))^(1/2)/b^(1/2))/f/b^(1/2)-1/2*co
t(f*x+e)^2*(b*sec(f*x+e))^(1/2)/b/f

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Rubi [A]
time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 = {2702, 294, 335, 218, 212, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f}-\frac {\cot ^2(e+f x) \sqrt {b \sec (e+f x)}}{2 b f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{4 \sqrt {b} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 209

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

Rule 212

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

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 294

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 335

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 2702

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

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Mathematica [A]
time = 0.86, size = 93, normalized size = 1.00 \begin {gather*} \frac {\left (-2 \tan ^{-1}\left (\sqrt {\sec (e+f x)}\right )+\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )-\frac {4 \csc ^2(e+f x)}{\sec ^{\frac {3}{2}}(e+f x)}\right ) \sqrt {\sec (e+f x)}}{8 f \sqrt {b \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(424\) vs. \(2(73)=146\).
time = 0.25, size = 425, normalized size = 4.57

method	result	size
default	\(-\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (8 \left (\cos ^{2}\left (f x +e \right )\right ) \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}+16 \cos \left (f x +e \right ) \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}+8 \left (-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}\right )^{\frac {3}{2}}-4 \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 ) \arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-\left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {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}{\sin \left (f x +e \right )^{2}}\right )+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}}}+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+\ln \left (-\frac {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}{\sin \left (f x +e \right )^{2}}\right )\right )}{8 f \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sin \left (f x +e \right )^{4} \sqrt {\frac {b}{\cos \left (f x +e \right )}}}\)	\(425\)

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,method=_RETURNVERBOSE)

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

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Maxima [A]
time = 0.52, size = 110, normalized size = 1.18 \begin {gather*} \frac {b {\left (\frac {4 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{b^{2} - \frac {b^{2}}{\cos \left (f x + e\right )^{2}}} - \frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}} + \frac {\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 )}{b^{\frac {3}{2}}}\right )}}{8 \, f} \end {gather*}

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]

1/8*b*(4*sqrt(b/cos(f*x + e))/(b^2 - b^2/cos(f*x + e)^2) - 2*arctan(sqrt(b/cos(f*x + e))/sqrt(b))/b^(3/2) + lo
g(-(sqrt(b) - sqrt(b/cos(f*x + e)))/(sqrt(b) + sqrt(b/cos(f*x + e))))/b^(3/2))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (77) = 154\).
time = 0.46, size = 389, normalized size = 4.18 \begin {gather*} \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 ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - {\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 )}{16 \, {\left (b f \cos \left (f x + e\right )^{2} - b 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 ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} + {\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 )}{16 \, {\left (b f \cos \left (f x + e\right )^{2} - b f\right )}}\right ] \end {gather*}

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*(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) + 8*sqrt
(b/cos(f*x + e))*cos(f*x + e)^2 - (cos(f*x + e)^2 - 1)*sqrt(-b)*log((b*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 - co
s(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)))/(b*f
*cos(f*x + e)^2 - b*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)) + 8*sqrt(b/cos(f*x + e))*cos(f*x + e)^2 + (cos(f*x + e)^2 - 1)*sqrt(b)*log((b*cos(f*x + e)^2 - 4*(c
os(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)))/(b*f*cos(f*x + e)^2 - b*f)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc ^{3}{\left (e + f x \right )}}{\sqrt {b \sec {\left (e + f x \right )}}}\, dx \end {gather*}

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

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Giac [A]
time = 4.31, size = 110, normalized size = 1.18 \begin {gather*} \frac {b^{2} {\left (\frac {2 \, \sqrt {b \cos \left (f x + e\right )} \cos \left (f x + e\right )}{{\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} b} + \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}}\right )}}{4 \, f \mathrm {sgn}\left (\cos \left (f x + e\right )\right )} \end {gather*}

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^2*(2*sqrt(b*cos(f*x + e))*cos(f*x + e)/((b^2*cos(f*x + e)^2 - b^2)*b) + arctan(sqrt(b*cos(f*x + e))/sqrt
(-b))/(sqrt(-b)*b^2) + arctan(sqrt(b*cos(f*x + e))/sqrt(b))/b^(5/2))/(f*sgn(cos(f*x + e)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\sin \left (e+f\,x\right )}^3\,\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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