3.2.1 \(\int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) [101]

Optimal. Leaf size=82 \[ -\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}-\frac {\csc (a+b x) F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{3 b d^2} \]

[Out]

-2/3*csc(b*x+a)/b/d/(d*tan(b*x+a))^(1/2)+1/3*csc(b*x+a)*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*Elliptic
F(cos(a+1/4*Pi+b*x),2^(1/2))*sin(2*b*x+2*a)^(1/2)*(d*tan(b*x+a))^(1/2)/b/d^2

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Rubi [A]
time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2677, 2681, 2653, 2720} \begin {gather*} -\frac {\sqrt {\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{3 b d^2}-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]/(d*Tan[a + b*x])^(3/2),x]

[Out]

(-2*Csc[a + b*x])/(3*b*d*Sqrt[d*Tan[a + b*x]]) - (Csc[a + b*x]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b
*x]]*Sqrt[d*Tan[a + b*x]])/(3*b*d^2)

Rule 2653

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2677

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*Sin[e + f
*x])^m*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n + 1))), x] - Dist[(n + 1)/(b^2*(m + n + 1)), Int[(a*Sin[e + f*x])
^m*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && LtQ[n, -1] && NeQ[m + n + 1, 0] && Integer
sQ[2*m, 2*n] &&  !(EqQ[n, -3/2] && EqQ[m, 1])

Rule 2681

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[e + f*x]
^n*((b*Tan[e + f*x])^n/(a*Sin[e + f*x])^n), Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\csc (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}-\frac {\int \csc (a+b x) \sqrt {d \tan (a+b x)} \, dx}{3 d^2}\\ &=-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}-\frac {\left (\sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{3 d^2 \sqrt {\sin (a+b x)}}\\ &=-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}-\frac {\left (\csc (a+b x) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx}{3 d^2}\\ &=-\frac {2 \csc (a+b x)}{3 b d \sqrt {d \tan (a+b x)}}-\frac {\csc (a+b x) F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {\sin (2 a+2 b x)} \sqrt {d \tan (a+b x)}}{3 b d^2}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.80, size = 110, normalized size = 1.34 \begin {gather*} \frac {2 \cos (2 (a+b x)) \sec (a+b x) \sqrt {\sec ^2(a+b x)} \left (\sqrt {\sec ^2(a+b x)}-\sqrt [4]{-1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\tan (a+b x)}\right )\right |-1\right ) \tan ^{\frac {3}{2}}(a+b x)\right )}{3 b (d \tan (a+b x))^{3/2} \left (-1+\tan ^2(a+b x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]/(d*Tan[a + b*x])^(3/2),x]

[Out]

(2*Cos[2*(a + b*x)]*Sec[a + b*x]*Sqrt[Sec[a + b*x]^2]*(Sqrt[Sec[a + b*x]^2] - (-1)^(1/4)*EllipticF[I*ArcSinh[(
-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Tan[a + b*x]^(3/2)))/(3*b*(d*Tan[a + b*x])^(3/2)*(-1 + Tan[a + b*x]^2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(301\) vs. \(2(97)=194\).
time = 0.33, size = 302, normalized size = 3.68

method	result	size
default	\(-\frac {\left (-1+\cos \left (b x +a \right )\right )^{2} \left (\EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sin \left (b x +a \right ) \cos \left (b x +a \right )+\sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sin \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+\cos \left (b x +a \right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{3 b \sin \left (b x +a \right )^{4} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \cos \left (b x +a \right )^{2}}\)	\(302\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)/(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/b*(-1+cos(b*x+a))^2*(EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((-1+cos(b*x+a))
/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*s
in(b*x+a)*cos(b*x+a)+((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x
+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*sin(b*x+a)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1
/2))+cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^2/sin(b*x+a)^4/(d*sin(b*x+a)/cos(b*x+a))^(3/2)/cos(b*x+a)^2*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)/(d*tan(b*x + a))^(3/2), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.10, size = 117, normalized size = 1.43 \begin {gather*} \frac {{\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {i \, d} {\rm ellipticF}\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ), -1\right ) + {\left (\cos \left (b x + a\right )^{2} - 1\right )} \sqrt {-i \, d} {\rm ellipticF}\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ), -1\right ) + 2 \, \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}} \cos \left (b x + a\right )}{3 \, {\left (b d^{2} \cos \left (b x + a\right )^{2} - b d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

1/3*((cos(b*x + a)^2 - 1)*sqrt(I*d)*ellipticF(cos(b*x + a) + I*sin(b*x + a), -1) + (cos(b*x + a)^2 - 1)*sqrt(-
I*d)*ellipticF(cos(b*x + a) - I*sin(b*x + a), -1) + 2*sqrt(d*sin(b*x + a)/cos(b*x + a))*cos(b*x + a))/(b*d^2*c
os(b*x + a)^2 - b*d^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\csc {\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*tan(b*x+a))**(3/2),x)

[Out]

Integral(csc(a + b*x)/(d*tan(a + b*x))**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)/(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)/(d*tan(b*x + a))^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\sin \left (a+b\,x\right )\,{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(a + b*x)*(d*tan(a + b*x))^(3/2)),x)

[Out]

int(1/(sin(a + b*x)*(d*tan(a + b*x))^(3/2)), x)

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